Mole Concept And Stoichiometry ICSE Class-10 Concise Selina Solutions Chapter-5. We Provide Step by Step Answer of Exercise – 5 (A) , Numericals – 5 (A), Exercise – 5 (B), Numerical – 5(B), Exercise-5(C), Exercise – 5(D), Miscellaneous Exercise and Previous Year Questions of Exercise-5 Mole Concept And Stoichiometry ICSE Class-10. Visit official Website CISCE for detail information about ICSE Board Class-10.
Board | ICSE |
Publications | Selina Publishers PVT LTD |
Subject | Concise Chemistry |
Class | 10th |
writer | Dr SP Singh |
Chapter-5 | Mole Concept And Stoichiometry |
Topics | Solutions of Exe – 5 (A) , Num – 5 (A), Exe – 5 (B), Num – 5(B), Exe-5(C), Exe – 5(D), Miscellaneous Exe and Previous Year Questions |
Edition | 2021-2022 |
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Exercise – 5 (A) Mole Concept And Stoichiometry
Page-75
Question 1
State :
(a) Gay-Lussac’s Law of combining volumes.
(b) Avogadro’s law
Answer 1
(a) Gay-Lussac’s law states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.
(b) Avogadro’s law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
Question 2
(a) What do you mean by stoichiometry?
(b) Define atomicity of a gas. State the atomicity of Hydrogen, Phosphorus and Sulphur.
(c) Differentiate between N_{2} and 2N.
Answer 2
(a) stoichiometry measures quantitative relationships and is used to determine the amount of products/reactants that are produced/needed in a given reaction.
(b) The number of atoms in a molecule of an element is called its atomicity. Atomicity of Hydrogen is 2, phosphorus is 4 and sulphur is 8.
(c) N_{2 }means 1 molecule of nitrogen and 2N means two atoms of nitrogen.
N_{2} can exist independently but 2N cannot exist independently.
Question 3
Explain Why?
(a) “The number of atoms in a certain volume of hydrogen is twice the number of atoms in the same volume of helium at the same temperature and pressure.”
(b) “When stating the volume of a gas, the pressure and temperature should also be given.”
(c) Inflating a balloon seems to violate Boyle’s law.
Answer 3
(a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
Now volume of hydrogen gas =volume of helium gas
n molecules of hydrogen =n molecules of helium gas
nH_{2}=nHe
1 mol. of hydrogen has 2 atoms of hydrogen and I molecule of helium has 1 atom of helium
Therefore 2H=He
Therefore atoms in hydrogen is double the atoms of helium.
(b) For a given volume of gas under given temperature and pressure, a change in any one of the variable i.e., pressure or temperature changes the volume.
(c) Inflating a balloon seems violating Boyles law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.
Numerical Problems:- 5 (A) Mole Concept And Stoichiometry Selina Solutions
Question 4
(a) Calculate the volume of oxygen at S.T.P required for the complete combustion of 100 litres of carbon monoxide at the same temperature and pressure.
2CO + O_{2 }2CO_{2}
(b) 200 cm^{3} of hydrogen and 150 cm^{3} of oxygen are mixed and ignited, as per the following reaction,
2H_{2 }+ O_{2 }2H_{2}O
What volume of oxygen remains unreacted?
Answer 4
(a) 2CO + O_{2 }2CO_{2}
2 V 1 V 2 V
2 V of CO requires = 1V of O_{2}
so, 100 litres of CO requires = 50 litres of O_{2}
(b) 2H_{2 }+ O_{2 }_{ }_{ }2H_{2}O
2 V 1V 2V
From the equation, 2V of hydrogen reacts with 1V of oxygen
so 200cm^{3} of Hydrogen reacts with = 200/2= 100 cm^{3}
Hence, the unreacted oxygen is 150 – 100 = 50cm^{3 }of oxygen.
Question 5
24 cc Marsh gas (CH_{4}) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculation.
Answer 5
This experiment supports Gay lussac’s law of combining volumes.
Since the unchanged or remaining O_{2} is 58 cc so, used oxygen 106 – 58 = 48cc
According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.
CH_{4}+2O_{2 }CO_{2} + 2H_{2}O
1 V2 V
24 cc48 cc
i.e. methane and oxygen react in a 1:2 ratio.
Question 6
What volume of oxygen would be required to burn completely 400 ml of acetylene [C_{2}H_{2}]? Also calculate the volume of carbon dioxide formed.
2C_{2}H_{2} + 5O_{2 }4CO_{2} + 2H_{2}O (l)
Answer 6
2C_{2}H_{2} + 5O_{2 }4CO_{2} + 2H_{2}O (l)
2 V 5 V 4 V
From equation, 2 V of C_{2}H_{2} requires = 5 V of O_{2}
So, for 400ml C_{2}H_{2} , O_{2} required = 400 5/2 =1000 ml
Similarly, 2 V of C_{2}H_{2} gives = 4 V of CO_{2}
So, 400ml of C_{2}H_{2} gives CO_{2} = 400 4/2 = 800ml
Question 7
112 cm^{3} of H_{2}S(g) is mixed with 120 cm^{3} of Cl_{2}(g) at STP to produce HCl(g) and sulphur(s). Write a balanced equation for this reaction and calculate (i) the volume of gaseous product formed (ii) composition of the resulting mixture.
Answer 7
Balanced chemical equation:
(i) At STP, 1 mole gas occupies 22.4 L.
As 1 mole H_{2}S gas produces 2 moles HCl gas,
22.4 L H_{2}S gas produces 22.4 × 2 = 44.8 L HCl gas.
Hence, 112 cm^{3} H_{2}S gas will produce 112 × 2 = 224 cm^{3} HCl gas.
(ii) 1 mole H_{2}S gas consumes 1 mole Cl_{2} gas.
This means 22.4 L H_{2}S gas consumes 22.4 L Cl_{2} gas at STP.
Hence, 112 cm^{3} H_{2}S gas consumes 112 cm^{3} Cl_{2} gas.
120 cm^{3} – 112 cm^{3 }= 8 cm^{3 }Cl_{2} gas remains unreacted.
Thus, the composition of the resulting mixture is 224 cm^{3}HCl gas + 8 cm^{3} Cl_{2} gas.
Question 8
1250cc of oxygen was burnt with 300cc of ethane [C_{2}H_{6}]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed;
Answer 8
From the equation, 2V of ethane reacts with 7V oxygen.
So, 300 cc of ethane reacts with
Hence, unused O_{2} = 1250 – 1050 = 200 cc
From 2V of ethane, 4V of CO_{2} is produced.
So, 300 cc of ethane will produce
Question 9
What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C_{2}H_{4}] at 273^{o} C and 380 mm of Hg pressure?
C_{2}H_{4}+3O_{2 }2CO_{2} + 2H_{2}O
Answer 9
C_{2}H_{4}+3O_{2 }2CO_{2} + 2H_{2}O
1V 3V
11litre 33 litre
Question 10
Calculate the volume of HCl gas formed and chlorine gas required when 40 ml of methane reacts completely with chlorine at S.T.P.
CH_{4} + 2Cl_{2 }CH_{2}Cl_{2}+2HCl
Answer 10
CH_{4} + 2Cl_{2} CH_{2}Cl_{2 }+2HCl
1 V 2 V 1 V 2 V
From equation, 1V of CH_{4} gives = 2 V HCl
so, 40 ml of methane gives = 80 ml HCl
For 1V of methane = 2V of Cl_{2} required
So, for 40ml of methane = 40 2 = 80 ml of Cl_{2}
Question 11
What volume of propane is burnt for every 500 cm^{3} of air used in the reaction under the same conditions? (assuming oxygen is 1/5^{th} of air)
C_{3}H_{8 }+ 5O_{2 } 3CO_{2} + 4H_{2}O
Answer 11
C_{3}H_{8 }+ 5O_{2} 3CO_{2} + 4H_{2}O
1 V 5 V 3 V
From equation, 5 V of O_{2} required = 1V of propane
_{so, 100 cm3 of O2 will require = 20 cm3 of propane}
Question 12
450 cm^{3} of nitrogen monoxide and 200 cm^{3} of oxygen are mixed together and ignited. Caclulate the composition of resulting mixture.
2NO + O_{2} 2NO_{2}
Answer 12
2NO + O_{2 } 2NO_{2}
2 V 1 V 2 V
From equation, 1V of O_{2} reacts with = 2 V of NO
200cm^{3} oxygen will react with = 200 2 =400 cm^{3 }NO
Hence, remaining NO is 450 – 400 = 50 cm^{3}
NO_{2} produced = 400cm^{3} because 1V oxygen gives 2 V NO_{2}
Total mixture = 400 + 50 = 450 cm^{3}
Question 13
If 6 liters of hydrogen and 4 liters of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.
Answer 13
(i) 6 litres of hydrogen and 4 litres of chlorine when mixed, results in the formation of 8 litres of HCl gas.
(ii) When water is added to it, it results in the formation of hydrochloric acid. Chlorine acts as a limiting agent leving behind only 2 litres of hydrogen gas.
(iii) Therefore, the volume of the residual gas will be 2 litres.
Question 14
Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.
4NH_{3} + 5O_{2 } 4NO + 6H_{2}O
If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?
Solution 14
4NH_{3} + 5O_{2} 4NO + 6H_{2}O
4 V 5 V 4 V
9 litres of reactants gives 4 litres of NO
So, 27 litres of reactants will give = 27 4/9 = 12 litres of NO
Question 15
A mixture of hydrogen and chlorine occupying 36 cm^{3} was exploded. On shaking it with water, 4cm^{3 }of hydrogen was left behind. Find the composition of the mixture.
Answer 15
H_{2} + Cl_{2 }2HCl
1V 1V 2 V
Since 1 V hydrogen requires 1 V of oxygen and 4cm^{3} of H_{2} remained behind so the mixture had composition: 16 cm^{3} hydrogen and 16 cm^{3 }chlorine.
Therefore Resulting mixture is H_{2} =4cm^{3},HCl=32cm^{3}
Question 16
What volume of air (containing 20% O_{2} by volume) will be required to burn completely 10 cm^{3} each of methane and acetylene?
CH_{4} + 2O_{2 }CO_{2} + 2H_{2}O
2C_{2}H_{2} + 5O_{2 } 4CO_{2} + 2H_{2}O
Answer 16
CH_{4} + 2O_{2 }CO_{2} + 2H_{2}O
1 V 2 V 1 V
2C_{2}H_{2} + 5O_{2} 4CO_{2} + 2H_{2}O
2 V 5 V 4 V
From the equations, we can see that
1V CH_{4} requires oxygen = 2 V O_{2}
So, 10cm^{3} CH_{4} will require =20 cm^{3} O_{2}
Similarly 2 V C_{2}H_{2} requires = 5 V O_{2}
So, 10 cm^{3} C_{2}H_{2 }will require_{ }= 25 cm^{3} O_{2}
Now, 20 V O_{2} will be present in 100 V air and 25 V O_{2} will be present in 125 V air ,so the volume of air required is 225cm^{3}
Question 17
LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.
C_{3}H_{8} + 5O_{2} 3CO_{2} + 4H_{2}O
2C_{4}H_{10} + 13O_{2} 8CO_{2} + 10H_{2}O
Answer 17
C_{3}H_{8} + 5O_{2} 3CO_{2} + 4H_{2}O
2C_{4}H_{10} + 13O_{2} 8CO_{2} + 10H_{2}O
60 ml of propane (C_{3}H_{8}) gives 3 60 = 180 ml CO_{2}
40 ml of butane (C_{4}H_{10}) gives = 8 40/2 = 160 ml of CO_{2}
Total carbon dioxide produced = 340 ml
So, when 10 litres of the mixture is burnt = 34 litres of CO_{2 }is produced.
Question 18
200 cm^{3} of CO_{2} is collected at S.T.P when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at S.T.P. in original mixture.
2C_{2}H_{2}(g) + 5O_{2}(g) 4CO_{2}(g)+ 2H_{2}O(g)
Answer 18
2C_{2}H_{2}(g) + 5O_{2}(g) 4CO_{2}(g)+ 2H_{2}O(g)
4 V CO_{2} is collected with 2 V C_{2}H_{2}
So, 200cm^{3} CO_{2 }will be collected with = 100cm^{3} C_{2}H_{2}
Similarly, 4V of CO_{2} is produced by 5 V of O_{2}
So, 200cm^{3} CO^{2 }will be produced by = 250 ml of O_{2}
Page-76
Question 19
You have collected (a) 2 litres of CO_{2} (b) 3 litres of chlorine (c) 5 litres of hydrogen (d) 4 litres of nitrogen and (e) 1 litres of SO_{2}, under similar conditions of temperature and pressure. Which gas sample will have :
(a) the greatest number of molecules, and
(b) The least number of molecules?
Justify your answers.
Answer 19
According to Avogadro’s law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and least volume will contain least molecules. So,
(a) 5 litres of hydrogen has greatest no. of molecules with the maximum volume.
(b) 1 litre of SO_{2} contains the least number of molecules since it has the smallest volume.
Question 20
The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. The following table gives the volumes of gases collected and the number of molecules (x) in 20 litres of nitrogen. You are to complete the table giving the number of molecules in the other gases in terms of x.
Gas | Volume (in litres) | Number of molecules |
Chlorine
Nitrogen Ammonia Sulphur dioxide |
10
20 20 5 |
x |
Answer 20
Gas | Volume (in litres) | Number of molecules |
Chlorine | 10 | x/2 |
Nitrogen | 20 | x |
Ammonia | 20 | X |
Sulphur dioxide | 5 | x/4 |
Question 21
(i) If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc of B?
The gases A and B are under the same conditions of temperature and pressure.
(ii) Name the law on which the above problem is based.
Answer 21
(i) According to Avogadro’s law, under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.
As 150 cc of gas A contains X molecules, 150 cc of gas B also contains X molecules.
So, 75 cc of B will contain X/2 molecules.
(ii) The problem is based on Avogadro’s law.
Exercise – 5(B) Mole Concept And Stoichiometry Concise Selina ICSE Class-10
(Relative Atomic Mass ,Relative Molecular Mass and Mole Concept)
Page-83
Question 1
(a) The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.
(b) What is the value of Avogadro’s number?
(c) What is the value of molar volume of a gas at S.T.P?
Answer 1
(a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C-12.
(b)The value of Avogadro’s number is 6.023 10^{23}
(c) The molar volume of a gas at STP is 22.4 dm^{3}at STP
Question 2
Define or explain the terms
(a) Vapour density
(b) Molar volume
(c) Relative atomic mass
(d) Relative molecular mass
(e) Avogadro’s number
(f) Gram atom
(g) Mole
Answer 2
(a) The vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.
(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm^{3}.
(c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.
(d) The relative molecular mass of an compound is the number that represents how many times one molecular of the substance is heavier than 1/12 of the mass of an atom of carbon-12.
(e) Avogadro’s number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 x10^{23} atoms.
(f) Gram atom quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.
(g) Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.
Question 3
(a) What are the main applications of Avogadro’s Law?
(b) How dose Avogadro’s Law explain Gay-Lussac’s Law of combining volumes?
Answer 3
(a) Applications of Avogadro’s Law :
(1) It explains Gay-Lussac’s law.
(2) It determines atomicity of the gases.
(3) It determines the molecular formula of a gas.
(4) It determines the relation between molecular mass and vapour density.
(5) It gives the relationship between gram molecular mass and gram molecular volume.
(b) According to Avogadro’s law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.
Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another.This what Gay Lussac’s Law says.
H_{2} + Cl_{2} ? 2HCl
1V 1V 2V(By Gay-Lussacs law)
n molecules n molecules 2n molecules (By Avogadros law)
Numericals of – 5 (B) Exe
Question 4
Calculate the relative molecular masses of:
(a) Ammonium chloroplatinate, (NH_{4})_{2} PtCl_{6}
(b) Potassium chlorate
(c) CuSO_{4}. 5H_{2}O
(d) (NH_{4})_{2}SO_{4}
(e) CH_{3}COONa
(f) CHCl_{3}
(g) (NH_{4})_{2} Cr_{2}O_{7}
Answer 4
(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444
(b) KClO_{3} = (K)39 + (Cl)35.5 + (3O)48 = 122.5
(c) (Cu)63.5 + (S)32 + (4O)64 + (5H_{2}O)5 x 18 = 249.5
(d) (2N)28 + (8H)8 + (S)32 + (4O)64 = 132
(e) (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82
(f) (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5
(g) (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252
Page-84
Question 5
Find the
(a) number of molecules in 73 g of HCl,
(b) weight of 0.5 mole of O_{2},
(c) number of molecules in 1.8 g of H_{2}O
(d) number of moles in 10 g of CaCO_{3}
(e) Weight of 0.2 mole of H_{2} gas,
(f) Number of molecules in 3.2 g of SO_{2.}
Answer 5
(a) No. of molecules in 73 g HCl = 6.023 x10^{23 }x 73/36.5(mol.
mass of HCl)
= 12.04 x 10^{23}
(b) Weight of 0.5 mole of O_{2} is = 32(mol. Mass of O_{2}) x 0.5=16 g
(c) No. of molecules in 1.8 g H_{2}O = 6.023 x 10^{23 }x 1.8/18
= 6.023 x 10^{22}
(d) No. of moles in 10g of CaCO_{3} = 10/100(mol. Mass CaCO_{3})
= 0.1 mole
(e) Weight of 0.2 mole H_{2} gas = 2(Mol. Mass) x 0.2 = 0.4 g
(f) No. of molecules in 3.2 g of SO_{2} = 6.023 x 10^{23} x 3.2/64
= 3.023 x 10^{22}
Question 6
Which of the following would weigh most?
(a) 1 mole of H_{2}O
(b) 1 mole of CO_{2}
(c) 1 mole of NH_{3}
(d) 1 mole of CO
Answer 6
Molecular mass of H_{2}O is 18, CO_{2} is 44, NH_{3} is 17 and CO is 28
So, the weight of 1 mole of CO_{2} is more than the other three.
Question 7
Which of the following contains maximum number of molecules?
(a) 4 g of O_{2}
(b) 4 g of NH_{3}
(c) 4 g of CO_{2}
(d) 4 g of SO_{2}
Answer 7
4g of NH_{3} having minimum molecular mass contain maximum molecules.
Question 8
Calculate the number of
(a) Particles in 0.1 mole of any substance.
(b) Hydrogen atoms in 0.1 mole of H_{2}SO_{4}.
(c) Molecules in one Kg of calcium chloride.
Answer 8
(a) No. of particles in s1 mole = 6.023 x 10^{23}
So, particles in 0.1 mole = 6.023 x 10 ^{23} x 0.1 = 6.023 x 10^{22}
(b)1 mole of H_{2}SO_{4} contains =2 x 6.023 x 10^{23}
So, 0.1 mole of H_{2}SO_{4} contains =2 x 6.023 x 10^{23} x0.1
= 1.2×10^{23 }atoms of hydrogen
(c) 111g CaCl_{2} contains = 6.023 x 10^{23 }molecules
So, 1000 g contains = 5.42 x 10^{24 }molecules
Question 9
How many grams of
(a) Al are present in 0.2 mole of it?
(b) HCl are present in 0.1 mole of it?
(c) H_{2}O are present in 0.2 mole of it?
(d) CO_{2} is present in 0.1 mole of it?
Answer 9
(a) 1 mole of aluminium has mass = 27 g
So, 0.2 mole of aluminium has mass = 0.2 x 27 = 5.4 g
(b) 0.1 mole of HCl has mass = 0.1 x 36.5(mass of 1 mole)
= 3.65 g
(c) 0.2 mole of H_{2}O has mass = 0.2 x 18 = 3.6 g
(d) 0.1 mole of CO_{2} has mass = 0.1 x 44 = 4.4 g
Question 10
(a) The mass of 5.6 litres of a certain gas at S.T.P. is 12 g. What is the relative molecular mass or molar mass of the gas? (2017)
(b) Calculate the volume occupied at S.T.P. by 2 moles of SO_{2}.
Answer 10
(a) 5.6 litres of gas at STP has mass = 12 g
So, 22.4 litre (molar volume) has mass =12 x 22.4/5.6
= 48g(molar mass)
(b)1 mole of SO_{2} has volume = 22.4 litres
So, 2 moles will have = 22.4 x 2 = 44.8 litre
Question 11
Calculate the number of moles of
(a) CO_{2} which contain 8.00 g of O_{2}
(b) Methane in 0.80 g of methane.
Answer 11
(a) 1 mole of CO_{2} contains O_{2} = 32g
So, CO_{2 }having 8 gm of O_{2 }has no. of moles = 8/32 = 0.25 moles
(b) 16 g of methane has no. of moles = 1
So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles
Question 12
Calculate the actual mass of
(a) An atom of oxygen
(b) an atom of hydrogen
(c) a molecule of NH_{3}
(d) the atom of silver
(e) the molecule of oxygen
(f) 0.25 gram atom of calcium
Answer 12
(a) 6.023 x 10 ^{23 }atoms of oxygen has mass = 16 g
So, 1 atom has mass = 16/6.023 x 10^{23 }= 2.656 x 10^{-23 }g
(b) 1 atom of Hydrogen has mass = 1/6.023 x 10^{23 }= 1.666 x 10^{-24}
(c) 1 molecule of NH_{3} has mass = 17/6.023 x10^{23 }= 2.82 x 10^{-23 }g
(d) 1 atom of silver has mass = 108/6.023 x 10^{23 }=1.701 x 10^{-22}
(e) 1 molecule of O_{2} has mass = 32/6.023 x 10^{23 }= 5.314 x 10^{-23 }g
(f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g
Question 13
Calculate the mass of 0.1 mole of each of the following
(a) CaCO_{3}
(b) Na_{2}SO_{4}.10H_{2}O
(c) CaCl_{2}
(d) Mg
(Ca = 40, Na=23, Mg =24, S=32, C = 12, Cl = 35.5, O=16, H=1)
Answer 13
(a) 0.1 mole of CaCO_{3} has mass =100(molar mass) x 0.1=10 g
(b) 0.1 mole of Na_{2}SO_{4}.10H_{2}O has mass = 322 x 0.1 = 32.2 g
(c) 0.1 mole of CaCl_{2} has mass = 111 x 0.1 = 11.1g
(d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g
Question 14
Calculate the number of
(a) oxygen atoms in 0.10 mole of Na_{2}CO_{3}.10H_{2}O. (2017)
(b) gram atoms in 4.6 gram of sodium (2019)
(c) Mole in 12 g of oxygen
Answer 14
(a) 1 molecule of Na_{2}CO_{3}.10H_{2}O contains oxygen atoms = 13
So, 6.023 x10^{23 }molecules (1mole) has atoms=13 x 6.023 x 10^{23}
So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 10^{23} =7.8×10^{23}
(b) Given Na = 4.6 gm
At. mass = 23
No. of gram atoms of Na= Mass of Na
At. mass of Na
= 4.6
23
= 0.2 gms
(c) 32 g of oxygen gas = 1 mole
1 gram of oxygen gas = 1/32 mole
Given that 12 g of oxygen gas
No: of moles = given mass / molar mass
= 12 / 32 = 0.375 mole
Question 15
What mass of Ca will contain the same number of atoms as are present in 3.2 g of S? (2015)
Answer 15
3.2 g of S has number of atoms = 6.023 x10^{23} x 3.2 /32
= 0.6023 x 10^{23}
So, 0.6023 x 10^{23} atoms of Ca has mass=40 x0.6023×10^{23}/6.023
x10^{23}
= 4g
Question 16
Calculate the number of atoms in each of the following:
(a) 52 moles of He
(b) 52 amu of He
(c) 52 g of He
Answer 16
(a) No. of atoms = 52 x 6.023 x10^{23} = 3.131 x 10^{25}
(b) 4 amu = 1 atom of He
so, 52 amu = 13 atoms of He
(c) 4 g of He has atoms = 6.023 x10^{23}
So, 52 g will have = 6.023 x 10^{23} x 52/4 = 7.828 x10^{24} atoms
Question 17
Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.
Answer 17
Molecular mass of Na_{2}CO_{3} = 106 g
106 g has 2 x 6.023 x10^{23} atoms of Na
So, 5.3g will have = 2 x 6.023 x10^{23}x 5.3/106=6.022 x10^{22 }atoms
Number of atoms of C = 6.023 x10^{23 }x 5.3/106 = 3.01 x 10^{22 }atoms
And atoms of O = 3 x 6.023 x 10^{23 }x 5.3/106= 9.03 x10^{22 }atoms
Question 18
(a) Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH_{2})_{2}] [O = 16; N = 14; C = 12 ; H = 1 ]
(b) Calculate the volume occupied by 320 g of sulphur dioxide at S.T.P. [S = 32; O = 16]
Answer 18
(a) 60 g urea has mass of nitrogen(N_{2}) = 28 g
So, 5000 g urea will have mass = 28 x 5000/60 = 2.33 kg
(b) 64 g has volume = 22.4 litre
So, 320 g will have volume = 22.4 x 320/64=112 litres
Question 19
(a) What do you understand by the statement that ‘vapour density of carbon dioxide is 22’?
(b) Atomic mass of Chlorine is 35.5.What is its vapour density?
Answer 19
(a) Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.
(b) Vapour density of Chlorine atom is 35.5.
Question 20
What is the mass of 56 cm^{3} of carbon monoxide at STP?
(C=12 ,O=16)
Answer 20
22400 cm^{3} of CO has mass = 28 g
So, 56 cm^{3} will have mass = 56 x 28/22400 = 0.07 g
Question 21
Determine the number of molecules in a drop of water which weighs 0.09g.
Answer 21
18 g of water has number of molecules = 6.023 x 10^{23}
So, 0.09 g of water will have no. of molecules = 6.023 x 10^{23} x 0.09/18 = 3.01 x 10^{21 }molecules
Question 22
The molecular formula for elemental sulphur is S_{8}.In sample of 5.12 g of sulphur
(a) How many moles of sulphur are present?
(b) How many molecules and atoms are present?
Answer 22
(a) No. of moles in 256 g S_{8} = 1 mole
So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles
(b) No. of molecules = 0.02 x 6.023 x 10^{23} = 1.2 x 10^{22 }molecules
No. of atoms in 1 molecule of S_{ }= 8
So, no. of atoms in 1.2 x 10^{22 }molecules = 1.2 x 10^{22 }x 8
= 9.635x 10^{22 }molecules
Question 23
If phosphorus is considered to contain P_{4} molecules, then calculate the number of moles in 100g of phosphorus?
Answer 23
Atomic mass of phosphorus P = 30.97 g
Hence, molar mass of P_{4} = 123.88 g
If phosphorus is considered as P_{4} molecules,
then 1 mole P_{4 }≡ 123.88 g
Therefore, 100 g of P_{4 }= 0.807 g
Question 24
Calculate:
(a) The gram molecular mass of chlorine if 308cm^{3} of it at STP weighs 0.979 g
(b) The volume of 4g of H_{2} at 4 atmospheres.
(c) The mass of oxygen in 2.2 litres of CO_{2} at STP.
Answer 24
(a) 308 cm^{3 }of chlorine weighs = 0.979 g
So, 22400 cm^{3} will weigh = gram molecular mass
= 0.979 x 22400/308 =71.2 g
(b) 2 g(molar mass) H_{2 }at 1 atm has volume = 22.4 litres
So, 4 g H_{2 }at 1 atm will have volume = 44.8 litres
Now, at 1 atm(P_{1}) 4 g H_{2 }has volume (V_{1}) = 44.8 litres
So, at 4 atm(P_{2}) the volume(V_{2}) will be =
(c) Mass of oxygen in 22.4 litres = 32 g(molar mass)
So, mass of oxygen in 2.2 litres = 2.2 x 32/22.4=3.14 g
Question 25
A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10^{-12} g, calculate the number of carbon atoms in the signature.
Answer 25
No. of atoms in 12 g C = 6.023 x10^{23}
So, no. of carbon atoms in 10^{-12} g = 10^{-12} x 6.023 x10^{23}/12
= 5.019 x 10^{10 }atoms
Question 26
An unknown gas shows a density of 3 g per litre at 273^{0}C and 1140 mm Hg pressure. What is the gram molecular mass of this gas?
Answer 26
Given:
P= 1140 mm Hg
Density = D = 2.4 g / L
T = 273 ^{0}C = 273+273 = 546 K
M = ?
We know that, at STP, the volume of one mole of any gas is 22.4 L
Hence we have to find out the volume of the unknown gas at STP.
First, apply Charle’s law.
We have to find out the volume of one litre of unknown gas at standard temperature 273 K.
V_{1}= 1 L T_{1} = 546 K
V_{2}=? T_{2} = 273 K
V_{1}/T_{1} = V_{2}/ T_{2}
V_{2} = (V_{1} x T_{2})/T_{1}
= (1 L x 273 K)/546 K
= 0.5 L
We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.
Apply Boyle’s law.
P _{1} = 1140 mm Hg V_{1} = 0.5 L
P_{2} = 760 mm Hg V_{2} = ?
P_{1} x V_{1 }= P_{2} x V_{2}
V_{2} = (P_{1} x V_{1})/P_{2}
= (1140 mm Hg x 0.5 L)/760 mm Hg
= 0.75 L
Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP
X moles = 0.75 L / 22.4 L
= 0.0335 moles
The original mass is 2.4 g
n = m / M
0.0335 moles = 2.4 g / M
M = 2.4 g / 0.0335 moles
M= 71.6 g / mole
Hence, the gram molecular mass of the unknown gas is 71.6 g
Question 27
Cost of Sugar (C_{12}H_{22} O_{11}) is Rs 40 per kg; calculate its cost per mole.
Answer 27
1000 g of sugar costs = Rs. 40
So, 342g(molar mass) of sugar will cost=342×40/1000 = Rs. 13.68
Question 28
Which of the following weighs the least?
(a) 2 g atom of N
(b) 3 x10^{25} atoms of carbon
(c)1mole of sulphur
(d) 7 g of silver
Answer 28
(a) Weight of 1 g atom N = 14 g
So, weight of 2 g atom of N = 28 g
(b) 6.023 x10^{23 }atoms of C weigh = 12 g
So, 3 x10^{25} atoms will weigh =
(c) 1 mole of sulphur weighs = 32 g
(d) 7 g of silver
So, 7 grams of silver weighs least.
Question 29
Four grams of caustic soda contains:
(a) 6.02 x 10^{23} atoms of it
(b) 4 g atom of sodium
(c) 6.02 x10^{22} molecules
(d) 4 moles of NaOH
Answer 29
Option C is correct.
40 g of NaOH contains 6.023 x 10^{23 }molecules
So, 4 g of NaOH contains = 6.02 x10^{23} x 4/40
= 6.02 x10^{22} molecules
Page-85
Question 30
The number of molecules in 4.25 g of ammonia is:
(a) 1.0 x 10^{23}
(b) 1.5 x 10^{23 }
(c)^{ }2.0 x 10^{23 }
(d)^{ }3.5x 10^{23}
Answer 30
The number of molecules in 18 g of ammonia= 6.02 x10^{23}
So, no. of molecules in 4.25 g of ammonia = 6.02 x10^{23 }x 4.25/18
= 1.5 x 10^{23}
Question 31
Correct the statements, if required
(a) One mole of chlorine contains 6.023 x 10^{10} atoms of chlorine.
(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with two volumes of oxygen will give two volumes of water vapour.
(c) Relative atomic mass of an element is the number of times one molecule of an element is heavier than 1/12 the mass of an atom of carbon-12.
(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of atoms.
Answer 31
(a) One mole of chlorine contains 6.023 x 10^{23} atoms of chlorine.
(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.
(c) Relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon-12.
(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.
Exercise – 5 (C) Selina Solutions of ICSE Class-10 Chemistry
(Percentage Composition, Empirical and Molecular Formula)
Page-90
Question 1
Give three kinds of information conveyed by the formula H_{2}O.
Answer 1
Information conveyed by H_{2}O
(1)That H_{2}O contains 2 volumes of hydrogen and 1 volume of oxygen.
(2)That ratio by weight of hydrogen and oxygen is 1:8.
(3)That molecular weight of H_{2}O is 18g.
Question 2
Explain the terms empirical formula and molecular formula.
Answer 2
The empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.
The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.
Question 3
Give the empirical formula of:
(a) C_{6}H_{6}
(b) C_{6}H_{12}O_{6}
(c) C_{2}H_{2}
(d) CH_{3}COOH
Answer 3
(a) CH
(b) CH_{2}O
(c) CH
(d) CH_{2}O
Question 4
Find the percentage of water of crystallisation in CuSO_{4}.5H_{2}O. (At. Mass Cu = 64, H = 1, O = 16, S = 32) (2017)
Answer 4
Question 5
Calculate the percentage of phosphorus in
(a) Calcium hydrogen phosphate Ca(H_{2}PO_{4})_{2}
(b) Calcium phosphate Ca_{3}(PO_{4})_{2}
Answer 5
(a) Molecular mass of Ca(H_{2}PO_{4})_{2 }= 234
So, % of P = 2 31 100/234 = 26.5%
(b) Molecular mass of Ca_{3}(PO_{4})_{2 }= 310
% of P = 2 31 100/310 = 20%
Question 6
Calculate the percent composition of Potassium chlorate KClO_{3}.
Answer 6
Molecular mass of KClO_{3 }= 122.5 g
And , % of K = 39 /122.5 = 31.8%
So % of Cl = 35.5/122.5 = 28.98%
Hence % of O = 3 16/122.5 = 39.18%
Question 7
Find the empirical formula of the compounds with the following percentage composition:
Pb = 62.5%, N = 8.5%, O = 29.0%
Answer 7
Element % At. mass Atomic ratio Simple ratio
Pb 62.5 207 1
N 8.5 14 2
O 29.0 16 6
So, Pb(NO_{3})_{2 }is the empirical formula.
Question 8
Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.
Answer 8
In Fe_{2}O_{3} , Fe = 56 and O = 16
Molecular mass of Fe_{2}O_{3 }= 2 56 + 3 16 = 160 g
Iron present in 80% of Fe_{2}O_{3 }=
So, mass of iron in 100 g of ore = 56 g
mass of Fe in 10000 g of ore = 56 10000/100
= 5.6 kg
Question 9
If the empirical formula of two compounds is CH and their Vapour densities are 13 to 39 respectively, find their molecular formula. (2015)
Answer 9
For acetylene , molecular mass = 2 V.D = 2 13 = 26 g
The empirical mass = 12(C) + 1(H) = 13 g
n =
Molecular formula of acetylene= 2 Empirical formula =C_{2}H_{2}
Similarly, for benzene molecular mass= 2 V.D = 2 39 = 78
n = 78/13=6
So, the molecular formula = C_{6}H_{6}
Question 10
Find the empirical formula of a compound containing 17.7% hydrogen and 82.3% nitrogen. (2018)
Answer 10
Element % At. mass Atomic ratio Simple ratio
H 17.7 1
N 82.3 14
So, the empirical formula = NH_{3}
Question 11
On analysis, a substance was found to contain
C = 54.54%, H = 9.09%, O = 36.36%
The vapour density of the substance is 44,calculate;
(a) its empirical formula, and
(b) its molecular formula
Answer 11
Element % at. mass atomic ratio simple ratio
C 54.54 12 2
H 9.09 1 4
O 36.36 16 1
(a) So, its empirical formula = C_{2}H_{4}O
(b) empirical formula mass = 44
Since, vapour density = 44
So, molecular mass = 2 V.D = 88
Or n = 2
so, molecular formula = (C_{2}H_{4}O)_{2} = C_{4}H_{8}O_{2}
Question 12
An organic compound ,whose vapour density is 45, has the following percentage composition
H=2.22%, O = 71.19% and remaining carbon. Calculate ,
(a) its empirical formula, and
(b) its molecular formula
Answer 12
Element % at. mass atomic ratio simple ratio
C 26.59 12 1
H 2.22 1 1
O 71.19 16 2
(a) its empirical formula = CHO_{2}
(b) empirical formula mass = 45
Vapour density = 45
So, molecular mass = V.D 2 = 90
so, molecular formula = C_{2}H_{2}O_{4}
Question 13
An organic compound contains H = 4.07%, Cl = 71.65% chlorine and remaining carbon. Its molar mass = 98.96. Find,
(a) Empirical formula, and
(b) Molecular formula
Answer 13
Element % at. mass atomic ratio simple ratio
Cl71.6535.5 1
H4.071 2
C24.2812 1
(a) its empirical formula = CH_{2}Cl
(b) empirical formula mass = 49.5
Since, molecular mass = 98.96
so, molecular formula = (CH_{2}Cl)_{2} = C_{2}H_{4}Cl_{2}
Question 14
A hydrocarbon contains 4.8g of carbon per gram of hydrogen. Calculate
(a) the g atom of each
(b) find the empirical formula
(c) Find molecular formula, if its vapour density is 29.
Answer 14
(a) The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1=1
(b) Element Given mass At. mass Gram atom Ratio
C 4.8 12 0.4 1 2
H 1 1 1 2.5 5
So, the empirical formula = C_{2}H_{5}
(c) Empirical formula mass = 29
Molecular mass = V.D 2 = 29 2 = 58
So, molecular formula = C_{4}H_{10}
Question 15
0.2 g atom of silicon Combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.
Answer 15
Since, g atom of Si = given mass/mol. Mass
so, given mass = 0.2 28 = 5.6 g
Element mass At. mass Gram atom Ratio
Si5.6280.21
Cl21.335.5 3
Empirical formula = SiCl_{3}
Question 16
A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula. (2016)
Answer 16
% of carbon = 82.76%
% of hydrogen = 100 – 82.76 = 17.24%
Element | % Weight | Atomic Weight | Relative No. of Moles | Simplest Ratio |
C | 82.76 | 12 | 82.76/12 = 6.89 | 6.89/6.89 = 1 x 2 = 2 |
H | 17.24 | 1 | 17.24/1 = 17.24 | 17.24/6.89 = 2.5 x 2 = 5 |
Empirical formula = C_{2}H_{5}
Empirical formula weight = 2 x 12 + 1 x 5 = 24 + 5 = 29
Vapour Density = 29
Relative molecular mass = 29 x 2 = 58
N =
Molecular formula = n x empirical formula
= 2 x C_{2}H_{5}
= C_{4}H_{10}
Question 17
In a compound of magnesium (Mg = 24) and nitrogen (N = 14), 18 g of magnesium combines with 7g of nitrogen. Deduce the simplest formula by answering the following questions.
(a) How many gram- atoms of magnesium are equal to 18g?
Page-91
(b) How many gram- atoms of nitrogen are equal to 7g of nitrogen?
(c) Calculate simple ratio of gram- atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.
Answer 17
(a) G atoms of magnesium = 18/24 = 0.75 or g- atom of Mg
(b) G atoms of nitrogen = 7/14 = 0.5 or 1/2 g- atoms of N
(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg_{3 }N_{2}
Question 18
Barium chloride crystals contain 14.8% water of crystallization. Find the number of molecules of water of crystallization per molecule.
Answer 18
Barium chloride = BaCl_{2}.x H_{2}O
Ba + 2Cl + x[H_{2} + O]
=137+ 235.5 + x [2+16]
=[208 + 18x] contains water = 14.8% water in BaCl_{2}.x H_{2}O
=[208 + 18 x] 14.8/100 = 18x
=[104 + 9x] 2148=18000x
=[104+9x] 37=250x
=3848 + 333x =2250x
1917x =3848
x = 2 molecules of water
Question 19
Urea is very important nitrogenous fertilizer. Its formula is CON_{2}H_{4}.Calculate the percentage of nitrogen in urea. (C=12,O=16 ,N=14 and H=1).
Answer 19
Molar mass of urea; CON_{2}H_{4 }= 60 g
So, % of Nitrogen = 28 100/60 = 46.66%
Question 20
Determine the formula of the organic compound if its molecule contains 12 atoms of carbon. The percentage compositions of hydrogen and oxygen are 6.48 and 51.42 respectively.
Answer 20
Element % At. mass Atomic ratio Simple ratio
C 42.1 12 3.5 1
H 6.48 1 6.48 2
O 51.42 16 3.2 1
The empirical formula is CH_{2}O
Since the compound has 12 atoms of carbon, so the formula is
C_{12} H_{24} O_{12.}
Question 21
(a) A compound with empirical formula AB_{2}, has the vapour density equal to its empirical formula weight. Find its molecular formula.
(b) A compound with empirical formula AB has vapour density 3 times its empirical formula weight. Find the molecular formula.
(c)10.47 g of a compound contained 6.25 g of metal A and rest non-metal B. Calculate the empirical formula of the compound [At. wt of A = 207, B = 35.5]
Answer 21
(a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, the molecular formula is A_{2}B_{4}.
(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D
Empirical formula weight = V.D/3
So, n = molecular mass/ Empirical formula weight = 6
Hence, the molecular formula is A_{6}B_{6}
(c)
Given:
Wt. of the compound: 10.47g
Wt. of metal A: 6.25g
Wt. of non-metal B: 10.47 – 6.25 = 4.22g
Element | mass | At. Wt. |
Relative no. of atoms |
Simplest ratio |
A | 6.25g | 207 | 6.25/207=0.03 | 0.03/0.03=1 |
B | 4.22g | 35.5 | 4.26/35.5=0.12 | 0.12/0.03=4 |
Hence, the empirical formula is AB_{4}
Question 22
A hydride of nitrogen contains 87.5% per cent by mass of nitrogen. Determine the empirical formula of this compound.
Answer 22
Atomic ratio of N = 87.5/14 =6.25
Atomic ratio of H= 12.5/1 = 12.5
This gives us the simplest ratio as 1:2
So, the molecular formula is NH_{2}
Question 23
A compound has O=61.32%, S= 11.15%, H=4.88% and Zn=22.65%.The relative molecular mass of the compound is 287 amu. Find the molecular formula of the compound, assuming that all the hydrogen is present as water of crystallization.
Answer 23
Element % at. mass atomic ratio simple ratio
Zn 22.65 65 0.348 1
H 4.88 1 4.88 14
S 11.15 32 0.348 1
O 61.32 16 3.83 11
Empirical formula of the given compound =ZnSH_{14}O_{11}
Empiricala formula mass = 65.37+32+141+11+16=287.37
Molecular mass = 287
n = Molecular mass/Empirical formula mass = 287/287=1
Molecular formula = ZnSO_{11}H_{14}
=ZnSO_{4}.7H_{2}O
Exercise – 5(D) Mole Concept And Stoichiometry
Page-94
Question 1
Complete the following blanks in the equation as indicated.
CaH_{2} (s) + 2H_{2}O (aq) Ca(OH)_{2} (s) + 2H_{2} (g)
(a) Moles: 1 mole + ——- ——– + ————–
(b) Grams: 42g + ——- ——– + —————-
(c) Molecules: 6.02 x 10^{23} + ——- ——– + ———–
Answer 1
(a) Moles:1 mole + 2 mole 1 mole + 2 mole
(b) Grams: 42g + 36g 74g + 4 g
(c) Molecules = 6.02 10^{23 }+ 12.046 10^{23 }6.02 10^{23}+ 12.046 10^{23}
Question 2
The reaction between 15 g of marble and nitric acid is given by the following equation:
CaCO_{3} + 2HNO_{3 } Ca(NO_{3})_{2}+ H_{2}O + CO_{2}
Calculate: (a) the mass of anhydrous calcium nitrate formed, (b) the volume of carbon dioxide evolved at S.T.P.
Answer 2
(a) 100 g of CaCO_{3 }produces = 164 g of Ca(NO_{3})_{2}
So, 15 g CaCO_{3 }will produce = 164 15/100 = 24.6 g Ca(NO_{3})_{2}
(b) 1 V of CaCO_{3 }produces 1 V of CO_{2}
100 g of CaCO_{3 }has volume = 22.4 litres
So, 15 g will have volume = 22.4 15/100 = 3.36 litres CO_{2}
Question 3
66g of ammonium sulphate is produced by the action of ammonia on sulphuric acid.
Write a balanced equation and calculate:
(a) Mass of ammonia required.
(b) The volume of the gas used at the S.T.P.
(c) The mass of acid required.
Answer 3
2NH_{3} + H_{2}SO_{4} (NH_{4})_{2}SO_{4}
66 g
(a) 2NH_{3} + H_{2}SO_{4} (NH_{4})_{2}SO_{4}
34 g98 g132 g
For 132 g (NH_{4})_{2}SO_{4} = 34 g of NH_{3} is required
So, for 66 g (NH_{4})_{2}SO_{4} = 66 32/132 = 17 g of NH_{3} is required
(b) 17g of NH_{3} requires volume = 22.4 litres
(c) Mass of acid required, for producing 132g (NH_{4})_{2}SO_{4} = 98g
So, Mass of acid required, for 66g (NH_{4})_{2}SO_{4} = 66 98/132 = 49g
Question 4
The reaction between red lead and hydrochloric acid is given below:
Pb_{3}O_{4 }+ 8HCl 3PbCl_{2 }+ 4H_{2}O + Cl_{2}
Calculate:
(a) the mass of lead chloride formed by the action of the 6.85 g of red lead,
(b) the mass of the chlorine and
(c) the volume of the chlorine evolved at S.T.P.
Answer 4
(a) Molecular mass of Pb_{3}O_{4 }= 3 207.2 + 4 16 = 685 g
685 g of Pb_{3}O_{4 }gives = 834 g of PbCl_{2}
Hence, 6.85 g of Pb_{3}O_{4 }will give = 6.85 834/685 = 8.34 g
(b) 685g of Pb_{3}O_{4 }gives = 71g of Cl_{2}
Hence, 6.85 g of Pb_{3}O_{4 }will give = 6.85 71/685 = 0.71 g Cl_{2}
(c) 1 V Pb_{3}O_{4 }produces 1 V Cl_{2}
685g of Pb_{3}O_{4}has volume = 22.4 litres = volume of Cl_{2} produced
So, 6.85 Pb_{3}O_{4} will produce = 6.85 22.4/685 = 0.224 litres of Cl_{2}
Question 5
Find the mass of KNO_{3} required to produce 126 kg of nitric acid. Find whether a larger or smaller mass of NaNO_{3} is required for the same purpose.
KNO_{3}+ H_{2}SO_{4} KHSO_{4} + HNO_{3}
NaNO_{3 }+ H_{2}SO_{4} NaHSO_{4} + HNO_{3}
Answer 5
Molecular mass of KNO_{3 }= 101 g
63 g of HNO_{3} is formed by = 101 g of KNO_{3}
So, 126000 g of HNO_{3} is formed by = 126000 101/63 = 202 kg
Similarly,126 g of HNO_{3} is formed by 170 kg of NaNO_{3}
So, smaller mass of NaNO_{3 }is required.
Question 6
Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide was collected at 27^{o}C and normal pressure.
Calculate :
(a) The mass of salt required.
(b) The mass of the acid required to prepare the 2 litres of CO_{2} at 27 C and normal pressure.
CaCO_{3} + 2HCl CaCl_{2} + H_{2}O + CO_{2}
Answer 6
CaCO_{3} + 2HCl CaCl_{2} + H_{2}O + CO_{2}
100g73g22.4L
(a) V_{1} =2 litresV_{2} =?
T_{1} = (273+27)=300KT_{2}=273K
V_{1}/T_{1}=V_{2}/T_{2}
V_{2}=V_{1}T_{2}/T_{1}=
Now at STP 22.4 litres of CO_{2} are produced using CaCO_{3} =100g
So, litres are produced by =100/22.4 2274/300 =.125g
(b) 22.4 litres are CO_{2} are prepared from acid =73g
litres are prepared from = 73/22.4 2273/300=5.9g
Question 7
Calculate the mass and volume of oxygen at S.T.P., which will be evolved on electrolysis of 1 mole (18g) of water
Answer 7
2H_{2}O 2H_{2} + O_{2}
2 V2 V1 V
2 moles of H_{2}O gives = 1 mole of O_{2}
So, 1 mole of H_{2}O will give = 0.5 moles of O_{2}
so, mass of O_{2} = no. of moles x molecular mass
= 0.5 32 = 16 g of O_{2}
and 1 mole of O_{2} occupies volume =22.4 litre
so, 0.5 moles will occupy = 22.4 0.5 = 11.2 litres at S.T.P.
Question 8
1.56 g of sodium peroxide reacts with water according to the following equation:
2Na_{2}O_{2} + 2H_{2}O 4NaOH + O_{2}
Calculate:
(a) mass of sodium hydroxide formed,
(b) Volume of oxygen liberated at S.T.P.
(c) Mass of oxygen liberated.
Answer 8
2Na_{2}O_{2} + 2H_{2}O 4NaOH + O_{2}
2 V4 V1 V
(a) Mol. Mass of Na_{2}O_{2} = 2 23 + 2 16 = 78 g
Mass of 2Na_{2}O_{2}= 156 g
156 g Na_{2}O_{2 }gives = 160 g of NaOH (4 40 g)
So, 1.56 Na_{2}O_{2 }will give = 160 1.56/156 = 1.6 g
(b) 156 g Na_{2}O_{2 }gives = 22.4 litres of oxygen
So, 1.56 g will give = 22.4 1.56/156 = 0.224 litres
= 224 cm^{3}
(c)156 g Na_{2}O_{2 }gives = 32 g O_{2}
So, 1.56 g Na_{2}O_{2 }will give = 32 1.56/156
= 32/100 = 0.32 g
Question 9
(a) Calculate the mass of ammonia that can be obtained from 21.4 g of NH_{4}Cl by the reaction:
2NH_{4}Cl + Ca(OH)_{2 }CaCl_{2} +2H_{2}O + 2NH_{3}
(b) What will be the volume of ammonia when measured at S.T.P?
The molar volume of a gas = 22.4 litres at STP.
Answer 9
2NH_{4}Cl + Ca(OH)_{2 }CaCl_{2}+2H_{2}O + 2NH_{3}
2 V1 V1 V2 V
Mol. Mass of 2NH_{4}Cl = 2[14 + (1 4) + 35.5] = 2[53.5] = 107 g
(a) 107 g NH_{4}Cl gives = 34 g NH_{3}
So, 21.4 g NH_{4}Cl will give = 21.4 34/107 = 6.8 g NH_{3}
(b) The volume of 17 g NH_{3} is 22.4 litre
So, volume of 6.8 g will be = 6.8 22.4/17 = 8.96 litre
Question 10
Aluminium carbide reacts with water according to the following equation.
Al_{4}C_{3} + 12H_{2}O → 4Al(OH)_{3} + 3CH_{4}
(a) What mass of aluminium hydroxide is formed from 12g of aluminium carbide?
(b) What volume of methane s.t.p. is obtained from 12g of aluminium carbide? (2018)
Answer 10
Question 11
MnO_{2} + 4HCl MnCl_{2} + 2H_{2}O +Cl_{2}
0.02 moles of pure MnO_{2}is heated strongly with conc. HCl. Calculate:
(a) mass of MnO_{2 }used
(b) moles of salt formed,
(c) mass of salt formed,
(d) moles of chlorine gas formed,
(e) mass of chlorine gas formed,
(f) volume of chlorine gas formed at S.T.P.,
(g) moles of acid required,
(h) Mass of acid required.
Answer 11
MnO_{2} + 4HCl MnCl_{2} + 2H_{2}O +Cl_{2}
1 V4 V1 V1 V
(a) 1 mole of MnO_{2} weighs = 87 g (mol. Mass)
So, 0.02 mole will weigh = 87 0.02 = 1.74 g MnO_{2}
(b) 1 mole MnO_{2} gives = 1 mole of MnCl_{2}
So, 0.02 mole MnO_{2}will give =0.02 mole of MnCl_{2}
(c) 1 mole MnCl_{2} weighs = 126 g(mol mass)
So, 0.02 mole MnCl_{2} will weigh = 126 0.02 g = 2.52 g
(d) 0.02 mole MnO_{2}will form =0.02 mole of Cl_{2}
(e) 1 mole of Cl_{2 }weighs = 35.5 g
So, 0.02 mole will weigh = 71 0.02 = 1.42 g of Cl_{2}
(f) 1 mole of chlorine gas has volume = 22.4 litres
So, 0.02 mole will have volume = 22.4 0.02 = 0.448 litre
(g) 1 mole MnO_{2}requires HCl = 4 mole
So, 0.02 mole MnO_{2} will require =4 0.02 = 0.08 mole
(h) For 1 mole MnO_{2 }, acid required = 4 mole of HCl
So, for 0.02 mole, acid required = 4 0.02 =0.08 mole
Mass of HCl = 0.08 x 36.5 = 2.92 g
Question 12
Nitrogen and hydrogen react to form ammonia.
N_{2} (g) + 3H_{2} (g) 2NH_{3} (g)
If 1000g H_{2} react with 2000g of N_{2}:
(a) Will any of the two reactants remain unreacted? If yes, which one and what will be its mass?
(b) Calculate the mass of ammonia(NH_{3}) that will be formed?
Answer 12
N_{2} + 3H_{2} 2NH_{3}
28g6g34g
28g of nitrogen requires hydrogen = 6g
2000g of nitrogen requires hydrogen = 6/28 2000=3000/7g
So mass of hydrogen left unreacted =1000-3000/7=571.4g of H_{2}
(b) 28g of nitrogen forms NH_{3} = 34g
2000g of N_{2} forms NH_{3}
= 34/28 2000
=2428.6g
Misc.- Exercise of Mole Concept And Stoichiometry
Page-95
Question 1
From the equation for burning of hydrogen and oxygen
2H_{2} + O_{2} 2H_{2}O (Steam)
Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.
Solution 1
From equation: 2H_{2} + O_{2} 2H_{2}O
1 mole of Oxygen gives = 2 moles of steam
so, 0.5 mole oxygen will give = 2 0.5 = 1mole of steam
Question 2
From the equation
3Cu + 8HNO_{3} 3Cu (NO_{3})_{2}+ 4H_{2}O + 2NO
(At.mass Cu=64, H=1, N=14,O=16)
Calculate:
(a) Mass of copper needed to react with 63g of HNO_{3}
(b) Volume of nitric oxide at S.T.P. that can be collected.
Answer 2
3Cu + 8HNO_{3} 3Cu (NO_{3})_{2}+ 4H_{2}O + 2NO
1 V8 V3 V2 V
Mol. Mass of 8HNO_{3} = 8 63 = 504 g
(a) For 504 g HNO_{3}, Cu required is = 192 g
So, for 63g HNO_{3}Cu required = 192 63/504 = 24g
(b) 504 g of HNO_{3} gives = 2 22.4 litre volume of NO
So, 63g of HNO_{3} gives =2 22.4 63/504 =5.6 litre of NO
Question 3
(a) Calculate the number of moles in 7g of nitrogen.
(b) What is the volume at S.T.P. of 7.1 g of chlorine?
(c) What is the mass of 56 cm^{3} of carbon monoxide at S.T.P?
Answer 3
(a) 28g of nitrogen = 1mole
So, 7g of nitrogen = 1/28 7= 0.25 moles
(b) Volume of 71 g of Cl2 at STP =22.4 litres
Volume of 7.1 g chlorine =22.4 7.1/71=2.24 litre
(c) 22400cm^{3} volume have mass =28 g of CO(molar mass)
So, 56cm^{3} volume will have mass =28 56/22400= 0.07 g
Question 4
Some of the fertilizers are sodium nitrate NaNO_{3}, ammonium sulphate (NH_{4})_{2}SO_{4} and urea CO(NH_{2})_{2}. Which of these contains the highest percentage of nitrogen?
Answer 4
% of N in NaNO_{3}=
% of N in (NH_{4})_{2}SO_{4} =
% of N in CO(NH_{2})_{2} =
So, highest percentage of N is in urea.
Question 5
Water decomposes to O_{2} and H_{2} under suitable conditions as represented by the equation below:
2H_{2}O 2H_{2}+O_{2}
(a) If 2500 cm^{3} of H_{2} is produced, what volume of O_{2} is liberated at the same time and under the same conditions of temperature and pressure?
(b) The 2500 cm^{3} of H_{2} is subjected to 2 times increase in pressure (temp. remaining constant). What volume of H_{2} will now occupy?
(c) Taking the value of H_{2} calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm^{3} pressure remaining constant.
Answer 5
2H_{2}O 2H_{2}+O_{2}
2 V2 V1 V
(a) From equation, 2 V of water gives 2 V of H_{2} and 1 V of O_{2}
where 2 V = 2500 cm^{3}
so, volume of O_{2} liberated = 2V/V = 1250 cm^{3}
(b)
(c)
i.e. temperature should be increased by 3.5 times.
Question 6
Urea [CO(NH_{2})_{2}] is an important nitrogeneous fertilizer. Urea is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?
Answer 6
Molecular mass of urea=12 + 16+2(14+2) =60g
60g of urea contains nitrogen =28g
So, in 50g of urea, nitrogen present =23.33 g
50 kg of urea contains nitrogen=23.33kg
Question 7
Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.
Answer 7
% of hydrogen = 20%
% of carbon = 100 – 20 = 80%
% Weight | Atomic Weight | Relative No. of Moles | Simplest Ratio | |
C | 80 | 12 | 80/12 = 6.667 | 6.667/6.667 = 1 |
H | 20 | 1 | 20/1 = 20 | 20/6.667 = 2.99 ≈ 3 |
Empirical formula = CH_{3}
Empirical formula weight = 1 x 12 + 1 x 3 = 12 + 3 = 15
Vapour Density = 15
Relative molecular mass = 15 x 2 = 30
N =
Molecular formula = n x empirical formula
= 2 x CH_{3}
= C_{2}H_{6}
Question 8
The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon X is purified by fractional distillation.
0.145 g of X was heated with dry copper (II) oxide and 224 cm^{3} of carbon dioxide was collected at S.T.P.
(a) Which elements does X contain?
(b) What was the purpose of copper (II) oxide?
(c) Calculate the empirical formula of X by the following steps:
(i) Calculate the number of moles of carbon dioxide gas.
(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.
(iii) Calculate the mass of hydrogen in sample X.
(iv) Deduce the ratio of atoms of each element in X (empirical formula).
Answer 8
22400cm^{3} CO_{2 }has mass = 44g
so, 224 cm^{3} CO_{2 }will have mass= 0.44 g
Now since CO_{2 }is being formed and X is a hydrocarbon so it contains C and H.
In 0.44g CO_{2}, mass of carbon=0.44-0.32=0.12g=0.01g atom
So, mass of Hydrogen in X = 0.145-0.12 = 0.025g
= 0.025g atom
Now the ratio of C:H is C=1: H=2.5 or C=2 : H=5
i.e. the formula of hydrocarbon is C_{2}H_{5}
(a) C and H
(b) Copper (II) oxide was used for reduction of the hydrocarbon.
(c)
(i) no. of moles of CO_{2}= 0.44/44 = 0.01 moles
(ii) mass of C = 0.12 g
(iii) mass of H = 0.025 g
(iv) The empirical formula of X = C_{2}H_{5}
Question 9
A compound is formed by 24g of X and 64g of oxygen. If atomic mass of X=12 and O=16, calculate the simplest formula of compound.
Answer 9
Mass of X in the given compound =24g
Mass of oxygen in the given compound =64g
So total mass of the compound =24+64=88g
% of X in the compound = 24/88 100 = 27.3%
% of oxygen in the compound=64/88 100 =72.7%
Element % At. Mass Atomic ratio Simplest ratio
X 27.3 12 27.3/12=2.27 1
O 72.7 16 72.2/16=4.54 2
So simplest formula = XO_{2}
Question 10
A gas cylinder filled with hydrogen holds 5g of the gas. The same cylinder holds 85 g of gas X under same temperature and pressure. Calculate :
(a) Vapour density of gas X.
(b) Molecular weight of gas X.
Answer 10
(a) V.D =
(b) Molecular mass = 17(V.D) x 2= 34g
Question 11
(a) When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation :
CO_{2} + C 2CO
What volume of carbon monoxide at S.T.P. can be obtained from 3 g of carbon?
(b) 60 cm^{3} of oxygen was added to 24 cm^{3} of carbon monoxide and mixture ignited. Calculate:
(i) volume of oxygen used up and
(ii) Volume of carbon dioxide formed.
Answer 11
(a) CO_{2} + C 2CO
1 V 1 V 2 V
12 g of C gives = 44.8 litre volume of CO
So, 3 g of C will give = 11.2 litre of CO
Page-96
(b) 2CO + O_{2 } 2CO_{2}
2 V 1 V 2 V
(i) 2 V CO requires oxygen = 1 V
so, 24 cm^{3} CO will require = 24/2 =12 cm^{3}
(ii) 2 x 22400 cm^{3} CO gives = 2 x 22400 cm^{3 }CO_{2}
so, 24cm^{3} CO will give = 24 cm^{3 }CO_{2}
Question 12
How much calcium oxide is obtained by heating 82 g of calcium nitrate? Also find the volume of NO_{2} evolved: (2016)
Answer 12
Molecular weight of ==
=328g
Molecular weight of CaO =2(40+16)
=112g
(a) 328g of Ca(NO_{3})_{2}liberates 4 moles of NO_{2}
328g of Ca(NO_{3})_{2 }liberates L of NO_{2 } 82g will liberate
=22.4dm^{3 }of NO_{2}
(b) 328 g of calcium nitrate gives 112g ofCaO
82 g will give
=28 g of CaO
Question 13
The equation for the burning of octane is:
2C_{8}H_{18} + 25O_{2 }16CO_{2} + 18H_{2}O
(i) How many moles of carbon dioxide are produced when one mole of octane burns?
(ii) What volume at S.T.P. is occupied by the number of moles determined in (i)?
(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane?
(iv) What is the empirical formula of octane?
Answer 13
2C_{8}H_{18} + 25O_{2 }16CO_{2} + 18H_{2}O
25 V 16 V 18 V
(i) 2 moles of octane gives = 16 moles of CO_{2}
so, 1 mole octane will give = 8 moles of CO_{2}
(ii) 1 mole CO_{2 }occupies volume = 22.4 litre
so, 8 moles will occupy volume = 8 22.4 = 179.2 litre
(iii) 1 mole CO_{2} has mass = 44 g
so, 16 moles will have mass = 44 16 = 704 g
(iv) Empirical formula is C_{4}H_{9}.
Question 14
Ordinary chlorine gas has two isotopes ^{35}_{17}Cl and ^{37}_{17}Cl in the ratio of 3:1. Calculate the relative atomic mass of chlorine.
Answer 14
The relative atomic mass of Cl = (35 3 + 1 37)/4=35.5 amu
Question 15
Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of the compound.
Answer 15
Mass of silicon in the given compound =5.6g
Mass of the chlorine in the given compound=21.3g
Total mass of the compound=5.6g+21.3g=26.9g
% of silicon in the compound = 56/26.9 100 = 20.82%
% of chlorine in the compound = 21.2/26.9 100 = 79.18%
Element % At. Mass At. Ratio Simplest ratio
Si 20.82 28 20.82/28=0.74 1
Cl 79.18 35.5 79.18/35.5=2.23 3
So the empirical formula of the given compound =SiCl_{3}
Question 16
An acid of phosphorus has the following percentage composition; Phosphorus = 38.27%; hydrogen = 2.47 %; oxygen = 59.26 %. Find the empirical formula of the acid and its molecular formula, given that its relative molecular mass is 162.
Answer 16
% composition Atomic ratio Simple ratio
P = 38.27% 38.27/31 =1.23 1
H = 2.47% 2.47/1 = 2.47 2
O = 59.26% 59.26/16 = 3.70 3
So, empirical formula is PH_{2}O_{3} or H_{2}PO_{3}
Empirical formula mass = 31+ 2 1 + 3 16 = 81
The molecular formula is = H_{4}P_{2}O_{6}, because n = 162/81=2
Question 17
Calculate the mass of substance ‘A’ which in gaseous form occupies 10 litres at 27^{0}C and 700 mm pressure. The molecular mass of ‘A’ is 60.
Answer 17
V_{1} = 10 litres V_{2}=?
T_{1}= 27+ 273 = 300K T_{2}=273K
P_{1}=700 mm P_{2} = 760 mm
Using the gas equation
Question 18
A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure.
(a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure?
(b) If the number of molecules of hydrogen in the cylinder is X, calculate the number of carbon dioxide molecules in the cylinder. State the law that helped you to arrive at the above result
Answer 18
(a) Molecular mass of CO_{2} = 12+ 2×16 = 44 g
So, vapour density (V.D) = mol. Mass/2 = 44/2 = 22
V.D =
So, mass of CO_{2} = 22 kg
(b) According to Avogadros law ,equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
So, number of molecules of carbon dioxide in the cylinder =number of molecules of hydrogen in the cylinder=X
Question 19
Following questions refer to one mole of chlorine gas.
(a) What is the volume occupied by this gas at S.T.P.?
(b) What will happen to volume of gas, if pressure is doubled?
(c) What volume will it occupy at 273^{0}C?
(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?
Solution 19
(a) The volume occupied by 1 mole of chlorine = 22.4 litre
(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.
(c) V_{1}/V_{2} = T_{1}/T_{2}
22.4/V_{2} =273/546
V_{2} = 44.8 litres
(d) Mass of 1 mole Cl_{2} gas =35.5 x 2 =71 g
Question 20
(a) A hydrate of calcium sulphate CaSO_{4}.xH_{2}O contains 21% water of crystallisation. Find the value of x.
(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.
(c) How much volume will be occupied by 2g of dry oxygen at 27^{0}C and 740 mm pressure?
(d) What would be the mass of CO_{2} occupying a volume of 44 litres at 25^{0} C and 750 mm pressure?
(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.
AgNO_{3} (aq) + NaCl (aq) AgCl (s) + NaNO_{3}
Calculate the percentage of NaCl in the mixture.
Answer 20
(a) Total molar mass of hydrated CaSO_{4}.xH_{2}O = 136+18x
Since 21% is water of crystallization, so
So, x = 2 i.e. water of crystallization is 2.
(b) For 18 g water, vol. of hydrogen needed = 22.4 litre
So, for 1.8 g, vol. of H_{2} needed= 1.8 x 22.4/18 = 2.24 litre
Now 2 vols. of water = 1 vol. of oxygen
1 vol. of water =1/2 vol. of O_{2} =22.4/2=11.2 lit.
18 g of water = 11.2 lit. of O_{2}
1.8 g of water = 11.2/18 18/10=1.12 lit.
(c) 32g of dry oxygen at STP = 22400cc
2g will occupy = 224002/32=1400cc
P_{1}=760mm P_{2} =740mm
V_{1}=1400cc V_{2} =?
T_{1} =273 K, T_{2} = 27 +73 = 300K
(d) P_{1}= 750mm P_{2}=760mm
V_{1}= 44lit. V_{2}=?
T_{1}= 298K T_{2}=273K
(e) Since 143.5g of AgCl is produced from =58.5 g of NaCl
so, 1.435 g of AgCl is formed by =0.585 g of NaCl
% of NaCl =0.585 x100 = 58.5%
Question 21
(a) From the equation :
Calculate :
(i) The mass of carbon oxidized by 49 g of sulphuric acid.
(ii) The volume of sulphur dioxide measured at STP, liberated at the same time.
(b)
(i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)
(ii) The relative molecular mass of this compound is 168, so what is its molecular formula?
Answer 21
(a)
For CO_{2 }12+32
(i) Molecular mass of sulphuric acid = 2(2+32+64)
= 196
196 g of suphuric acid oxidized 12g of Carbon
49 g of suphuric acid will
=3 g
(ii) 196 g of sulphuric acid gives 2(22.4)
=44.8L
49 g og sulphuric acid will give
=11.2 L of SO_{2}
_{(b) }
_{(i) }
Element | % Weight | Atomic Weight | Atomic Ratio |
Simplest Ratio |
C | 14.4 | 12 | 14.4/12 = 1.2 | 1.2/1.2=1 |
H | 1.2 | 1 | 1.2/1 =1.2 | 1.2/1.2=1 |
Cl | 84.5 | 35.5 | 84.5/35.5=2.3 | 2.3/1.2=1.9=2 |
Empirical formula = CHCl_{2}
_{(ii) }
Empirical formula = CHCl_{2}
Empirical formula weight = 1 x 12 + 1 x 1+(2 x35.5)
= 12 + 1+70
= 83
Relative molecular mass = 168
N = 2.02≈2
Molecular formula = n x empirical formula
= 2 x CHCl_{2}
= C_{2}H_{2}Cl_{4}
Question 22
Find the percentage of
(a) oxygen in magnesium nitrate crystals [Mg (NO_{3}) 6H_{2}O].
(b) boron in Na_{2}B_{4}O_{7}.10H_{2}O. [H = 1, B = 11, O =16, Na = 23].
(c) phosphorus in the fertilizer superphosphate Ca(H_{2}PO_{4})_{2}
Answer 22
(a) Relative molecular mass of [Mg (NO_{3}) 6H_{2}O]
=24+14+(3 x16)+(6 x18)=194
Since, 194g of [Mg (NO_{3}) 6H_{2}O] contains 144g of oxygen
100g of [Mg (NO_{3}) 6H_{2}O] contains of oxygen = 74.22%
(b) Relative molecular mass of Na_{2}B_{4}O_{7}.10H_{2}O
(23 × 2) + (4 × 11) + (7 × 16) + 10(18) = 382
Since 382g of Na_{2}B_{4}O_{7}.10H_{2}O contains 44g of boron
100g Na_{2}B_{4}O_{7}.10H_{2}O of contains of boron
=11.5%
(c) Relative molecular mass of Ca(H_{2}PO_{4})_{2}
= 40 + 2(2 + 31 + 64) = 234
Since,234g of Ca(H_{2}PO_{4})_{2} contains 62g of phosphorus
100g of Ca(H_{2}PO_{4})_{2} contains
=26.5%
Question 23
A gas occupied 360 cm^{3} at 87^{0}C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find its relative molecular mass.
Answer 23
Question 24
Solid ammonium dichromate decomposes as:
If 63 g of ammonium dichromate decomposes. Calculate
(a) the quantity in moles of (NH_{4})_{2}Cr_{2}O_{7}
(b) the quantity in moles of nitrogen formed
(c) the volume of N_{2} evolved at STP.
Page-97
(d) what will be the loss of mass ?
(e) calculate the mass of chromium (III) oxide formed at the same time.
Answer 24
Solid ammonium dichromate decomposes as:
(a)Molecular mass of ammonium dichromate
= 2(14+4)+104+112
= 252 g
Number of moles=
=
=0.25moles
(b)
252 g of ammonium dichromate gives 22.4 dm^{3} of N_{2}
63 g of ammonium dichromate gives
=5.6 L
= 0.25 moles
(c)
252 g of ammonium dichromate gives 22.4 dm^{3} of N_{2}
63 g of ammonium dichromate gives
= 5.6 L
(d)
Number of moles=
=
=0.25 moles
0.25 moles of ammonium dichromate gives
0.25 moles of N_{2}=7 g
1 mole of H_{2}O =18 g
Therefore, total loss of mass=7+18
=25 g
(e)
252 g of ammonium dichromate gives 152 g of CrO_{3}
63 g of ammonium dichromate gives
=38 g
Question 25
Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under :
2H_{2}S + 3O_{2} 2H_{2}O + 2SO_{2}
Calculate the volume of hydrogen sulphide at S.T.P. Also, calculate the volume of oxygen required at S.T.P. which will complete the combustion of hydrogen sulphide determined in (litres).
Answer 25
2H_{2}S + 3O_{2 } 2H_{2}O + 2SO_{2}
2 V 3 V 2 V
128 g of SO_{2} gives = 2 22.4 litres volume
So, 12.8 g of SO_{2} gives = 2 22.4 12.8/128
= 4.48 litre volume
Or one can say 4.48 litres of hydrogen sulphide.
2 22.4 litre H_{2}S requires oxygen = 3 22.4 litre
So, 4.48 litres H_{2}S will require = 6.72 litre of oxygen
Question 26
Ammonia burns in oxygen and the combustion, in the presence of a catalyst. May be represented by 2NH_{3} + 2O_{2} _{?} 2NO + 3H_{2}O
[H= 1, N= 14, O=16]
What mass of steam is produced when 1.5 g of nitrogen monoxide is formed?
Answer 26
From equation, 2NH_{3} + 2 O_{2} 2NO + 3H_{2}O
When 60 g NO is formed, mass of steam produced = 54 g
So, 1.5 g NO is formed, mass of steam produced = 541.5/60
=1.35 g
Question 27
If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass of the fertilizer, calcium nitrate Ca(NO_{3})_{2} would be required to replace the nitrogen in a 10 hectare field ?
Answer 27
In 1 hectare of soil, N_{2} removed = 20 kg
So, in 10 hectare N_{2} removed = 200 kg
The molecular mass of Ca(NO_{3})_{2} =164
Now, 28 g N_{2 }present in fertilizer = 164 g Ca(NO_{3})_{2}
So, 200000 g of N_{2} is present in = 164 200000/28
= 1171.42 kg
Question 28
Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:
P+ 5HNO_{3} H_{3}PO_{4}+ 5NO_{2} + H_{2}O
If 6.2g of phosphorus was used in the reaction calculate:
(a) Number of moles of phosphorus taken and mass of phosphoric acid formed.
(b) mass of nitric acid will be consumed at the same time?
(c) The volume of steam produced at the same time if measured at 760 mm Hg pressure and 273^{0}C?
Answer 28
(a) 1 mole of phosphorus atom = 31 g of phosphorus
31 g of P =1 mole of P
6.2g of P = =0.2 mole of P
(b) 31 g P reacts with HNO_{3 }= 315 g
so, 6.2 g P will react with HNO_{3} = 315 6.2/31 = 63 g
(c)
Moles of steam formed from 31g phosphorus = 18g/18g = 1mol
Moles of steam formed from 6.2 g phosphorus = 1mol/31g6.2=0.2 mol
Volume of steam produced at STP =0.2 22.4 l/MOL=4.48 litre
Since the pressure (760mm) remains constant , but the temperature (273+273)=546 is double, the volume of the steam also gets doubled
So, Volume of steam produced at 760mm Hg and 273^{0}C=4.48 2=8.96litre
Question 29
112 cm^{3} of a gaseous fluoride of phosphorus has a mass of 0.63 g. Calculate the relative molecular mass of the fluoride. If the molecule of the fluoride contains only one atom of phosphorus, then determine the formula of the phosphorus fluoride. [ F=19, P=31].
Answer 29
112cm^{3 }of gaseous fluoride has mass = 0.63 g
so, 22400cm^{3} will have mass = 0.63 22400/112
= 126 g
The molecular mass = At mass P + At. mass of F
126= 31 + At. Mass of F
So, At. Mass of F = 95 g
But, at. mass of F = 19 so 95/19 = 5
Hence, there are 5 atoms of F so the molecular formula = PF_{5}
Question 30
Washing soda has formula Na_{2}CO_{3}.10H_{2}O. What mass of anhydrous sodium carbonate is left when all the water of crystallization is expelled by heating 57.2 g of washing soda?
Answer 30
Na_{2}CO_{3}.10H_{2}O Na_{2}CO_{3} + 10H_{2}O
286 g 106 g
So, for 57.2 g Na_{2}CO_{3}.10H_{2}O = 106 57.2/286 = 21.2 g Na_{2}CO_{3}
Question 31
A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride (M = 56).
Answer 31
Simple ratio of M = 34.5/56 = 0.616 = 1
Simple ratio of Cl = 65.5/35.5 = 1.845 = 3
Empirical formula = MCl_{3}
Empirical formula mass = 162.5, Molecular mass = 2V.D = 325
So, n = 2
So, molecular formula = M_{2}Cl_{6}
Question 32
A compound X consists of 4.8% carbon and 95.2% bromine by mass.
(i) Determine the empirical formula of this compound working correct to one decimal place (C = 12 ; Br = 80)
(ii) If the vapour density of the compound is 252, what is the molecular formula of the compound?
Answer 32
(i) Element % atomic mass atomic ratio simple ratio
C4.812 1
Br95.280 3
So, empirical formula is CBr_{3}
(ii) Empirical formula mass = 12 + 3 80 = 252 g
molecular formula mass = 2 252(V.D) = 504 g
n= 504/252 = 2
so, molecular formula = C_{2}Br_{6}
Question 33
The reaction: 4N_{2}O + CH_{4 } CO_{2} + 2H_{2}O + 4N_{2} takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of di nitrogen oxide (N_{2}O) required to give 150 cm^{3} of steam.
Answer 33
4N_{2}O + CH_{4} CO_{2} + 2H_{2}O + 4N_{2}
4 V 1 V 1 V 2 V 4 V
2 x 22400 litre steam is produced by N_{2}O = 4 x 22400 cm^{3}
So, 150 cm^{3} steam will be produced by= 4 22400 150/2 x 22400
= 300 cm^{3 }N_{2}O
Question 34
Samples of the gases O_{2}, N_{2}, CO_{2} and CO under the same conditions of temperature and pressure contain the same number of molecules x. The molecules of oxygen occupy V litres and have a mass of 8 g under the same of temperature and pressure
What is the volume occupied by:
(a) x molecules of N_{2}
(b) 3x molecules of CO
(c) What is the mass of CO_{2} in grams?
(d) In answering the above questions, which law have you used?
Solution 34
(a) Volume of O_{2} = V
Since O_{2 }and N_{2 }have same no. of molecules = x
so, the volume of N_{2 }= V
(b) 3x molecules means 3V volume of CO
(c) 32 g oxygen is contained in = 44 g of CO_{2}
So, 8 g oxygen is contained in = 44 x 8/32 = 11 g
(d) Avogadro’s law is used in the above questions.
Question 35
The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound?
Answer 35
simple ratio of Na = 42.1/23 = 1.83 = 3
and simple ratio of P = 18.9/31 = 0.609 = 1
Hence simple ratio of O = 39/16 = 2.43 = 4
So, the empirical formula is Na_{3}PO_{4}
Question 36
What volume of oxygen is required to burn completely a mixture of 22.4 dm^{3} of methane and 11.2 dm^{3} of hydrogen into carbon dioxide and steam?
CH_{4} + 2O_{2} CO_{2} + 2H_{2}O
2H_{2} + O_{2} 2H_{2}O
Answer 36
CH_{4} + 2O_{2} CO_{2} + 2H_{2}O
1 V 2 V 1 V 2 V
From equation:
22.4 litres of methane requires oxygen = 44.8 litres O_{2}
2H_{2} + O_{2} 2H_{2}O
2 V 1 V 2 V
From equation,
44.8 litres hydrogen requires oxygen = 22.4 litres O_{2}
So, 11.2 litres will require = 22.4 x 11.2/44.8 = 5.6 litres
Total volume = 44.8 + 5.6 = 50.4 litres
Question 37
The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular masses. Given 8 g of each gas at S.T.P., which gas will contain the least number of molecules and which gas the most?
Answer 37
According to Avogadros law:
Equal volumes of all gases, under similar conditions of temperature and pressure ,contain equal number of molecules.
So, 1 mole of each gas contains = 6.02 10^{23} molecules
Mol. Mass of H_{2} (2),O_{2}(32) ,CO_{2}(44),SO_{2}(64),Cl_{2}(71)
(1) Now 2 g of hydrogen contains molecules =6.02 10^{23}
So, 8g of hydrogen contains molecules = 8/2 6.02 10^{23}
=4 6.02 10^{23} = 4 M molecules
(2) 32g of oxygen contains molecules = 8/32 6.02 10^{23}=M/4
(3) 44g of carbon dioxide contains molecules = 8/44 6.02 10^{23}=2M/11
(4) 64g of sulphur dioxide contains molecules =6.02 10^{23}
So, 8g of sulphur dioxide molecules = 8/64 6.02 10^{23}= M/8
(5) 71 g of chlorine contains molecules =6.02 10^{23}
So, 8g of chlorine molecules = 8/72 6.02 10^{23} = 8M/71
Since 8M/71<M/8<2M/11<M/4<4M
Thus Cl_{2}<SO_{2}<CO_{2}<O_{2}<H_{2}
(i) Least number of molecules in Cl_{2}
(ii) Most number of molecules in H_{2}
Question 38
10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:
Na_{2}SO_{4} + BaCl_{2} BaSO_{4} + 2NaCl
Calculate the percentage of sodium sulphate in the original mixture.
Answer 38
Na_{2}SO_{4} + BaCl_{2} BaSO_{4} + 2NaCl
Molecular mass of BaSO_{4 }= 233 g
Now, 233 g of BaSO_{4 }is produced by Na_{2}SO_{4} = 142 g
So, 6.99 g BaSO_{4 }will be produced by = 6.99 142/233 = 4.26
The percentage of Na_{2}SO_{4} in original mixture = 4.26 100/10
= 42.6%
Question 39
When heated, potassium permanganate decomposes according to the following equation :
(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.
Page-98
(b) Given that the molecular mass of potassium permanganate is 158. What volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres)
Answer 39
(a) 1 litre of oxygen has mass = 1.32 g
So, 24 litres (molar vol. at room temp.) will have mass = 1.32 x 24
= 31.6 or 32 g
(b) 2KMnO_{4} K_{2}MnO_{4} + MnO_{2} + O_{2}
316 g of KMnO_{4} gives oxygen = 24 litres
So, 15.8 g of KMnO_{4 }will give = 24 316/15.8 = 1.2 litres
Question 40
(a) A flask contains 3.2 g of sulphur dioxide. Calculate the following:
(i) The moles of sulphur dioxide present in the flask.
(ii) The number of molecules of sulphur dioxide present in the flask.
(iii) The volume occupied by 3.2 g of sulphur dioxide at S.T.P.
(S= 32, O= 16)
(b) An Experiment showed that in a lead chloride solution, 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of this chlorine? (Pb = 207; Cl = 35.5)
Answer 40
(a)
(i) The no. of moles of SO_{2} = 3.2/64 = 0.05 moles
(ii) In 1 mole of SO_{2}, no. of molecules present = 6.02 10^{23}
So, in 0.05 moles, no. of molecules = 6.02 10^{23 }0.05
= 3.0 10^{22}
(iii) The volume occupied by 64 g of SO_{2} = 22.4 dm^{3}
3.2 g of SO_{2} will be occupied by volume = 22.4 3.2/64 =1.12 dm^{3}
(b) Gram atoms of Pb = 6.21/207=0.03 = 1
Gram atoms of Cl = 4.26/35.5 = 0.12 = 4
So, the empirical formula = PbCl_{4}
Question 41
The volume of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure .
(i) Which sample of gas contains the maximum number of molecules?
(ii) If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?
(iii) If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?
(iv) If the volume of A is actually 5.6 dm^{3} at S.T.P., calculate the number of molecules in the actual Volume of D at S.T.P. (Avogadro’s number is 6 10^{23}).
(v) Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N_{2}O)
Answer 41
(i) D contains the maximum number of molecules because volume is directly proportional to the number of molecules.
(ii) The volume will become double because volume is directly proportional to the no. of molecules at constant temperature and pressure.
V_{1}/V_{2} = n_{1}/n_{2}
V_{1}/V_{2} = n_{1}/2n_{1}
So, V_{2 }= 2V_{1}
(iii) Gay lussac’s law of combining volume is being observed.
(iv) The volume of D = 5.6 4 = 22.4 dm^{3}, so the number of molecules = 6 x 10^{23} because according to mole concept 22.4 litre volume at STP has = 6 x 10 ^{23} molecules
(v) No. of moles of D = 1 because volume is 22.4 litre
so, mass of N_{2}O = 1 44 = 44 g
Question 42
The equation given below relate the manufacture of sodium carbonate (molecular weight of Na_{2}CO_{3} = 106).
- NaCl+NH_{3}+ CO_{2}+ H_{2}O NaHCO_{3}+NH_{4}Cl
- 2NaHCO_{3 }Na_{2}CO_{3}+H_{2}O + CO_{2}
Equations (1) and (2) are based on the production of 21.2 g of sodium carbonate.
(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g of sodium carbonate?
(b) To produce the mass of sodium hydrogen carbonate calculate in (a), what volume of carbon dioxide, measured at S.T.P. would be required?
Answer 42
(a) NaCl+NH_{3}+ CO_{2 }+ H_{2}O NaHCO_{3}+NH_{4}Cl
2NaHCO_{3} Na_{2}CO_{3}+H_{2}O + CO_{2}
From equation:
106 g of Na_{2}CO_{3 }is produced by = 168 g of NaHCO_{3}
So, 21.2 g of Na_{2}CO_{3 }will be produced by = 168 21.2/106
= 33.6 g of NaHCO_{3}
(b) For 84 g of NaHCO_{3}, requiredvolume of CO_{2} = 22.4 litre
So, for 33.6 g of NaHCO_{3}, required volume of CO_{2} = 22.4 x 33.6/84
= 8.96 litre
Question 43
A sample of ammonium nitrate when heated yields 8.96 litres of steam (measure at S.T.P.)
NH_{4}NO_{3 }N_{2}O + 2H_{2}O
(i) What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam?
(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam? (Relative molecular mass of ammonium nitrate is 80)
(iii) Determine the percentage of oxygen in ammonium nitrate (O = 16).
Answer 43
(a) NH_{4}NO_{3 }N_{2}O+2H_{2}O
1mole 1mole 2 mole
1 V 1 V 2 V
44.8 litres of water produced by = 22.4 litres of NH_{4}NO_{3}
So, 8.96 litres will be produced by = 22.4 x 8.96/44.8
= 4.48 litres of NH_{4}NO_{3}
So, 4.48 litres of N_{2}O is produced.
(i) 44.8 litre H_{2}O is produced by = 80 g of NH_{4}NO_{3}
So, 8.96 litre H_{2}O will be produced by = 80 x 8.96/44.8
= 16g NH_{4}NO_{3}
(iii) % of O in NH_{4}NO_{3} = 3×16/80 = 60%
Question 44
Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120g of copper oxide? The equation for the reaction is :
Answer 44
Molecular mass of =(240) g
Molecular mass of = 2 × 22.4 = 44.8dm^{3}
Molecular mass of = (192)g
240 g of CuO requires 44.8 dm^{3 }of NH_{3}
120g of CuO will require
=22.4dm^{3}
Question 45
(a) Calculate the number of moles and the number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?
(b) What is the vapour density of ethylene?
Answer 45
(a) The molecular mass of ethylene(C_{2}H_{4}) is 28 g
No. of moles = 1.4/28 = 0.05 moles
No. of molecules = 6.023 x10^{23} x 0.05 = 3 x 10^{22 }molecules
Volume = 22.4 x 0.05 = 1.12 litres
(b) Molecular mass = 2 X V.D
S0, V.D = 28/2 = 14
Question 46
(a) Calculate the percentage of sodium in sodium aluminium fluoride (Na_{3}AlF_{6}) correct to the nearest whole number.
(F = 19; Na =23; Al = 27)
(b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:
2CO + O_{2 } 2CO_{2}
Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.
Answer 46
(a) Molecular mass of Na_{3}AlF_{6 }= 210
So, Percentage of Na = 3x23x100/210 = 32.85%
(b) 2CO + O_{2} 2CO_{2}
2 V 1 V 2 V
1 mole of O_{2} has volume = 22400 ml
Volume of oxygen used by 2 x 22400 ml CO = 22400 ml
So, Vol. of O_{2} used by 560 ml CO =22400 x 560/(2 x 22400)
= 280 ml
So, Volume of CO_{2} formed is 560 ml.
Previous Year Question of Mole Concept And Stoichiometry
Page-98
Question 47 (2009)
(a) A gas cylinder of capacity of 20 dm^{3}is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2 g, hence the relative molecular mass of the gas is:
(i) 5
(ii) 10
(iii).15
(iv) 20
(b)
(i) Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If 200 cm^{3} of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.
Page-99
(ii) A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find molecular formula of the compound if its relative molecular mass is 37. [N = 14, H = 1].
(c)
(i) A gas cylinder contains 24 x 10^{24} molecules of nitrogen gas. If Avogadro’s number is 6 x 10^{23} and the relative atomic mass of nitrogen is 14, calculate :
1 Mass of nitrogen gas in the cylinder
2. Volume of nitrogen at STP in dm^{3}.
(ii) Commercial sodium hydroxide weighing 30g has some sodium chloride in it. The Mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is
[Relative molecular mass of NaCl = 58; AgCl = 143]
(iii) A certain gas ‘X’ occupies a volume of 100 cm^{3} at S.T.P. and weighs 0.5 g. Find its relative molecular mass
Answer 47 (2009)
(a) Mass of gas X =10g
Mass of hydrogen gas= 2
Relative vapour density
= =5
Relative molecular mass of the gas= 2×relative vapour
density = 2×5
=10
(b)
(i) The combustion reaction
According to Gay-Lussac’s law,
2 volume of acetylene requires 5 volume of oxygen to burn it
1 volume of acetylene requires 2.5 volume of oxygen to burn it
200cm^{3 }requires 2.5×200=500 cm^{3 }of oxygen
2 volume of acetylene on combustion gives 4CO_{2 }
1 volume of acetylene on combustion gives 2CO_{2 }
200 cc of acetylene on combustion will give 200×2=400cc of CO_{2}
(iii) Hydrogen = 12.5%
∴ Nitrogen= 100-12.5= 87.5%
Element | Weight | Atomic Weight | Atomic Ratio | Simplest Ratio |
N | 87.5 | 14 | 87.5/14=6.25 | 6.25/6.25=1 |
H | 12.5 | 1 | 12.5/1=12.5 | 12.5/6.25=2 |
The Empirical formula of the compound is NH_{2}
_{ }Empirical formula weight =14+2=16
Relative molecular mass =37
N = 2.3≈2
Molecular formula = n x empirical formula = 2 x NH_{2}
=N_{2}H_{4}
(C)
(i) Molecules of nitrogen gas in a cylinder = 24 x 10^{24}
Avogadro’s number = 6 x 10^{23}
- Mass of nitrogen in a cylinder =
=1120g
- Volume of nitrogen at stp
and Volume of 28 g of N_{2} = 22.4dm^{3 }
So Volume of 1120g of N_{2}
hence =896 dm^{3}
Question 48 (2010)
(a)
(i) LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as :
(ii) Calculate the percentage of nitrogen and oxygen in ammonium nitrate. [Relative molecular mass of ammonium nitrate is 80, H = 1, N = 14, O = 16].
(b) 4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.
(i) Write the equation for the reaction.
(ii) What is the mass of 4.5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)
(iii). What is the volume of carbon dioxide liberated at STP?
(iv) What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111).
(v) How many moles of HCl are used in this reaction?
Answer 48 (2010)
(a)
(i) 10 litres of LPG contains
Propane
Butane
18+16=34 L
(ii) Molecular mass of NH_{4}(NO_{3}) =80
H=1, N=14, O=16
% of Nitrogen
As 80 g of NH4(NO3) contains 28 g of nitrogen
100 g of of NH_{4}(NO_{3}) will contain
= 35%
% of Oxygen
As,80 g of NH4 (NO3) contains 48 g of oxygen
100 g of of NH_{4}(NO_{3}) will contain
= 60%
(b)
(i) Equation for reaction of calcium carbonate with dilute hydrochloric acid:
(ii) Relative molecular mass of calcium carbonate=100
Mass of 4.5 moles of calcium carbonate
= No. of moles x Relative molecular mass
and = 4.5 x 100
therefore = 450g
(iii).
As, 100g of calcium carbonate gives 22.4dm^{3} of CO_{2}
450 g of calcium carbonate will give
=100.8 L
(iv). Molecular mass of calcium carbonate =100
Relative molecular mass of calcium chloride =111
As 100 g of calcium carbonate gives 111g of calcium chloride
450 g of calcium carbonate will give
=499.5 g
(v)
Molecular mass of HCl=36.5
Molecular mass of calcium carbonate =100
As 100 g of calcium carbonate gives (2×36.5)= 73g of HCl
450 g of calcium carbonate will give
=328.5g
Number of moles of HCl=
=
= 9 moles
Question 49 (2011)
(a)
(i) Calculate the volume of 320 g of SO_{2} at STP. (Atomic mass: S = 32 and O = 16).
(ii) State Gay-Lussac’s Law of combining volumes.
(iii). Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C_{3}H_{8}). (Atomic mass: C = 12, O = 16, H = I, Molar Volume = 22.4 dm^{3} at stp.)
(b)
(i) An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula. [Atomic mass : C = 12, H = 1, Br = 80]
(ii) Calculate the mass of:
1. 10^{22}atoms of sulphur.
2. 0.1 mole of carbon dioxide.
[Atomic mass : S = 32, C = 12 and O = 16 and
Avogadro’s Number = 6 x 10^{23}]
Answer 49 (2011)
(a)
(i) Atomic mass: S = 32 and O = 16
Molecular mass of SO_{2}=32+(2×16)
=64g
As 64 g of SO_{2} = 22.4dm^{3}
Then, 320 g of SO_{2} =
=112 L
(ii) Gay-Lussac’s law Gay-Lussac’s Law states “When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure.”
(iii). C_{3}H_{8} + 5O_{2}→3CO_{2} + 4H_{2}O
Molar mass of propane = 44
44 g of propane requires 5 × 22.4 litres of oxygen at STP.
8.8 g of propane requires = 22.4 litres
(b)
(i)
Element | Relative atomic mass | %Compound | Atomic ratio | Simplest ratio |
H | 1 | 2.13 | 2.13/1=2.13 | 2 |
C | 12 | 12.67 | 12.67/12=1.055 | 2 |
Br | 80 | 85.11 | 85.11/80=1 | 1 |
Empirical formula = CH_{2}Br
n(Empirical formula mass of CH_{2}Br) = Molecular mass (2 × VD)
n(12 + 2 + 80) = 94 × 2
n = 2
Molecular formula = Empirical formula × 2
= (CH_{2}Br) × 2
= C_{2}H_{4}Br_{2}
(ii) 10^{22} atoms of sulphur
6.022 × 10^{23} atoms of sulphur will have mass = 32 g
10^{22} atoms of sulphur will have mass =
= 0.533 g
(iii). 0.1 mole of carbon dioxide
1 mole of carbon dioxide will have mass = 44 g
0.1 mole of carbon dioxide will have mass = 4.4 g
Question 50 (2012)
(a) Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:
If 9.3 g of phosphorus was used in the reaction, calculate :
(i) Number of moles of phosphorus taken.
(ii) The mass of phosphoric acid formed.
(iii) The volume of nitrogen dioxide produced at STP.
(b)
(i) 67.2 litres of hydrogen combines with 44.8 litres of nitrogen to form ammonia under specific conditions as:
Calculate the volume of ammonia produced. What is the other substance, if any, that remains in the resultant mixture?
(ii) The mass of 5.6 dm3 of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.
(iii). Find the total percentage of Magnesium in magnesium nitrate crystals, Mg(NO_{3})_{2}-6H_{2}O.
[Mg = 24, N = 14, 0 = 16 and H = 1]
Answer 50
(a)
(i) Number of moles of phosphorus taken =
= 0.3 mol
(ii) 1 mole of phosphorus gives 98 gm of phosphoric acid.
So, 0.3 mole of phosphorus gives (0.3 × 98) gm of phosphoric acid
= 29.4 gm of phosphoric acid
(iii) 1 mole of phosphorus gives 112 L of NOgas at STP.
So, 0.3 mole of phosphorus gives (112 × 0.3) L of
NOgas at STP.
= 33.6 L of NOgas at STP
(b)
(i) According to the equation
3 volumes of hydrogen produce 2 volumes of ammonia
67.2 litres of hydrogen produce = 44.8 L
3 volumes of hydrogen combine with 1 volume of ammonia.
67.2 litres of hydrogen combine with =22.4L Nitrogen left = 44.8 – 22.4 = 22.4 litres
(ii) 5.6 dm^{3} of gas weighs 12 g
1 dm^{3} of gas weighs = (12/56) gm
22.4 dm^{3} of gas weighs = (12/56 × 22.2) gm = 48g
Therefore, the relative molecular mass of gas = 48 gm.
(iii). Molar mass of Mg (NO_{3})_{2}.6H_{2}O
= 24 × (14 × 2) + (16 × 12) + (1 × 12) = 256 g
Mass percent of magnesium =
Question 51 (2013)
(a)
(i) What volume of oxygen is required to burn completely 90 dm^{3} of butane under similar conditions of temperature and pressure?
Page-100
(ii) The vapour density of a gas is 8. What would be the volume occupied by 24.0 g of the gas at STP?
(iii). A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same conditions of temperature and pressure?
(b) O_{2}is evolved by heating KClO_{3} using MnO_{2} as a catalyst.
(i) Calculate the mass of KClO_{3} required to produce 6.72 litre of O_{2} at STP.
[atomic masses of K = 39, Cl = 35.5, 0 = 16].
(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.
(iii). Calculate the volume occupied by 0.01 mole of CO_{2} at STP.
Answer 51 (2013)
(a)
(i)
2 vols. of butane requires O_{2} = 13 vols
90 dm^{3} of butane will require O_{2} = × 90
= 585 dm^{3}
(ii) Molecular mass = 2 × Vapour density
So, molecular mass of gas = 2 × 8 = 16 g
As we know, molecular mass or molar mass occupies 22.4 litres.
That is,
16 g of gas occupies volume = 22.4 litres
So, 24 g of gas will occupy volume
=
(iii). According to Avogadro’s law, equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
So, molecules of nitrogen gas present in the same vessel = X
(b)
(i)
3 vols. of oxygen require KClO_{3} = 2 vols.
So, 1 vol. of oxygen will require KClO_{3} =
So, 6.72 litres of oxygen will require KClO_{3}
=
22.4 litres of KClO_{3} has mass = 122.5 g
So, 4.48 litres of KClO_{3} will have mass
=
22.4 litres of oxygen = 1 mole
So, 6.72 litres of oxygen =
No. of molecules present in 1 mole of O_{2}
= 6.023 × 10^{23}
^{ }So, no. of molecules present in 0.3 mole of O_{2}
= 6.023 × 10^{23} × 0.3
= 1.806 × 10^{23}
(iii). Volume occupied by 1 mole of CO_{2} at STP = 22.4 litres
So, volume occupied by 0.01 mole of CO_{2} at STP = 22.4 × 0.01= 0.224 litres
Question 52 (2014)
(a)
(i) Oxygen oxidises ethyne to carbon dioxide and water as shown by the equation:
What volume of ethyne gas at s.t.p. is require to produce 8.4 dm^{3} of carbon dioxide at STP [H = 1, C = 12, 0 = 16]
(ii) A compound made up of two elements X and Y has an empirical formula X_{2}Y. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density (V.D.) 25, find its molecular formula.
(b) A cylinder contains 68 g of Ammonia gas at STP
(i) What is the volume occupied by this gas?
(ii) How many moles of ammonia are present in the cylinder?
(iii). How many molecules of ammonia are present in the cylinder?
Answer 52 (2014)
(a)
(i)
2 moles of the C_{2}H_{2}=4 moles of CO_{2}
x dm^{3} of Ethine =8.4 dm^{3} of CO_{2}
x=
=4.2 dm^{3} of Etheline
(ii) Empirical formula= X_{2}Y
Atomic weight (X)= 10
Atomic weight (Y)= 5
Empirical formula weight = (2 × 10) + 5
=25
= 2
So, the molecular formula = X_{2}Y×2
= X_{4}Y_{2 }
_{(b) }
(i) A cylinder contains 68 g of ammonia gas at STP.
Molecular weight of ammonia = 17 g/mole
68 g of ammonia gas at STP =?
1 mole = 22.4 dm^{3}
∴ 4 mole = 22.4 × 4 = 89.6 dm^{3}
(ii) 4 moles of ammonia gas is present in the cylinder.
(iii). 1 mole = 6.023 × 10^{23} molecules
4 moles = 24.092 × 10^{23} molecules
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