Goyal Brothers Class-10 Mole Concept and Stoichiometry Ch-5

Goyal Brothers Class-10 Mole Concept and Stoichiometry Ch-5. Step by Step Solutions of Practice Problems , Exercise and Objective Type Questions of Goyal Brothers Prakashan Chapter-5 Mole Concept and Stoichiometry for ICSE Class 10 Chemistry.

Practice Problems Numerical based on Volume to Volume relation, Volume to Weight relation and  Weight to Weight relation of mole concept and Stoichiometry for ICSE Class-10 Goyal Brother Prakashan Chemistry. Visit official Website CISCE  for detail information about ICSE Board Class-10 Chemistry .

Goyal Brothers Class-10 Mole Concept and Stoichiometry Ch-5


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Practice Problem 

Exercise 1

Exercise 2

Objective Type Questions


Practice Problems Mole Concept and Stoichiometry Ch-5 ICSE Class-10

(Numerical Problems Based on Gay Lussac’s Law )

Page-71 Practice Problems-1

Patrice Problem 1

Answer 

(1)

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Thus, gaseous composition after reaction is: 

Excess oxygen = 200 – 160 = 40 cm3

CO= 80 cm3

Water = negligible

(2)

2 C₂H₂ + 5 O₂ ======> 4 CO₂ + 2 H₂O

100 cm³ of acetylene is mixed with 300 cm³ of pure oxygen and ignited ,

According to the equation :

2 volumes of C₂H₂ + 5 volumes of O₂ ===> 4 volumes of CO₂ + 2 volumes of H₂O

Now it is given that C₂H₅ has 100 cm³ volume .

So 2 volumes = 100 cm³ .

⇒ 1 volume = 100 cm³/2

⇒ 1 volume = 50 cm³

300 cm³ is ignited .

Given 5 volumes of oxygen react .

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⇒ 5 × 50 cm³ react .

⇒ 250 cm³ react .

Excess oxygen can be calculated by subtracting the amount of reacted oxygen by total volume of oxygen ignited .

Unused Oxygen = 300 cm³ – 250 cm³

⇒ Unused oxygen = 50 cm³ .

Total volume of the resulting mixture :

4 volumes of CO₂ + 2 volumes of  H₂O

⇒ 4 × 50 cm³ + Nil

⇒ 200 cm³ + Nil

200  cm³


Page-72 Practice Problems-2

Patrice Problem 2

Answer 

(1)

Given react ion is :
N2 + O2 → 2NO
According to Gay -Lussac’s law in the above reaction 1 volume of nitrogen combines with 1 volume of oxygen to produce 2 volumes of nitric oxide.
i .e. N2 + O2 → 2NO
1 vol. 1 vol. 2 vol.
The volume of nitric oxide produced is = 1400cm3.
Let the volumes of nitrogen and oxygen gases be = x
Then, N2 + O→ 2NO
x        x    1400cm3
So, x + x = 1400
2x = 1400
x= 1400/ 2 = 700cm3
Hence the volumes of reacting gases i .e. nitrogen and oxygen is 700 cm3 each.

(2)

2C2H2(g) + 5O2(g) → 4CO2(g)+ 2H2O(g)

4 V CO2 is collected with 2 V C2H2

So, 200cm3 COwill be collected with = 100cm3 C2H2

Similarly, 4V of CO2 is produced by 5 V of O2

So, 200cm3 COwill be produced by = 250 ml of O2


Numerical Problems Based on Avogadro’s Law

(Class-10 Mole Concept and Stoichiometry Ch-5)

Page-79 Practice Problems-1

PP 1

Answer 

(1)

At STP,  gas occupies  volume.

So,  gas have 

Number of moles of dry Nitrogen at STP 

Number of molecules of dry Nitrogen 

 Avogardo Number 

So, Nitrogen have  or 

 (2)

According to Avogadro’s law,
If 112cc of dry CO contains P molecules,then 112 cc of dry Cleaners will also contain P molecules.

1dm3=1000 cc

so 336 dm3 =1000 x 336 cm3
Applying unitary method,
112cc => P molecules
336000cc => 336000 x P/ 112 = 3000P.


Page-79  Practice Problems-2  

pp2

Answer 

(1)

5000 molecules of H2 =V

By Avogadro’s law

5000 molecules of N2 =V

2.5 *108   ÷5000=5 x 10⁴ V

(2)

6* 1022     present in 2.24 dm3 of Cl

By Avogadro’s law

3 *1019 molecule  =  3 *1019 (2.24/6* 1022 )          

 1.12*10-3  dm3


Numerical Problems Based on Molecular Weight

(Class-10 Mole Concept and Stoichiometry Ch-5)

Page-80  Practice Problems-1  

Molecular 1

Answer 

 (i)

 Al+3 (O+ H)

=27+3(16+1)

=27+51

=78 amu

(ii)

= 2 K+ 2 Cr+ 7 O

= 2x 39 + 2 x 52 + 7 x 16

= 78 +  104 + 112

=294 amu

(iii)

The molar mass of ammonium nitrate= sum of the molar mass of all its constituents in the correct proportion.

Molecular weight of nitrogen = 14.

Molecular weight of hydrogen= 1.

The molecular weight of oxygen= 16

The molecular weight of ammonium nitrate= 2 * weight of nitrogen atom + 4 * weight of hydrogen atom + 3 * weight of an oxygen atom

= 28 + 4. + 48

= 80 amu


Page-80  Practice Problems-2 

Molecular 2

Answer 

(i)

MM of Cu + S +4 O + 5 (2H + O)

=64+32+ 4* 16  + 5(2*1 +16)

=64+32+64+90

=250amu

(ii)

MM of 2 Na + S +4 O + 10 (2H + O)

= 2 *23 +32 +  4* 16  + 10 (2*1 +16)

= 46+32 +64+ 180

= 322 amu


Numerical Problems Based on Avogadro’s Number

(Goyal Brothers Class-10 Mole Concept and Stoichiometry for ICSE Chemistry Ch-5)

Page-80  Practice Problems-1 

pp1

Answer 

 (1)

Number of molecules in 12.8g of sulphur dioxide gas.
Molecular mass of SO2 = 64a.m.u .
So, 64g = 1 mole
12.8g = 12.8/ 64 = 0.2 mole
Now 1 mole of SO2 contains = 6 x 1023molecules
0 .2 mole of SO2 contains = 0 .2 x 6 x 1023
= 1.2 x 1023molecules

(2)

71g of chlorine has 6×10^23 molecules

7.1g has 0.1×6×10^23 molecules

=6×10^22


Page-81  Practice Problems-2 

pp2

Answer 

 (1)

We know
One mole contains 6×10^23 molecules
Weight of one mole of nitrogen gas= 28gm
Weight of 2.8×10^24 molecules
= (2.8×10^24/6×10^23) * 28
= 28×28/6
= 130.67 gm

 (2)

gram molecular mass of Coso4 is =159

6*x 1023molecules  weight = 159 g

1 molecule weight is = 159/ 6 *x 1023      

2*x 1023molecules weight is = (159 * 2*x 1023)/ (6 *x 1023 )

=5.33 gm


Page-81  Practice Problems-3 

pp2

Answer 

(1)

O2 is a diatomic molecule.

The mass of an oxygen atom = 16 amu.

∴ Mass of O2 molecule = 2 × 16 = 32 amu.

Mass of one molecule of oxygen

⇒ 32/ Avogadro constant ( 6.0 × 1023)

⇒ 5.33 × 10-23g.

(2)

O3  is a triatomic molecule.

The mass of an oxygen atom = 16 amu.

∴ Mass of O3 molecule = 3 × 16 = 48 amu.

Mass of one molecule of oxygen

⇒ 48/ Avogadro constant ( 6.0 × 1023)

⇒ 8 × 10-23g.


Numerical Based on Gram Molecular Volume Mole Concept and Vapour Density

(Goyal Brothers Class-10 Mole Concept and Stoichiometry for ICSE Chemistry Ch-5)

 Page -82 Practice Problem -1

pp1

Answer 

(1)

Mass of one mole of nitrogen dioxide = 14 + 16 + 16 = 46g
No. of moles in 0.23g on NO2 = 0.005
volume occupied by 1 mole = 22.4 L
volume occupied by 0.005 mole = 22.4*0.005

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