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 (1) 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 .

⇒ 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 (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 (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)

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 (1)

5000 molecules of H2 =V

5000 molecules of N2 =V

2.5 *108   ÷5000=5 x 10⁴ V

(2)

6* 1022     present in 2.24 dm3 of Cl

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 (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 (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 (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 (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 (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 