## Mole Concept and Stoichiometry Solutions for Class 10 Science ICSE Board (Concise - Selina Publishers)

Free download of step by step solutions for class 10 Science (Chemistry) Chapter 5 - Mole Concept and Stoichiometry of ICSE Board (Concise - Selina Publishers). All exercise questions are solved & explained by an expert teacher and as per ICSE board guidelines.

## Access ICSE Selina Solutions for Grade 10 Chemistry Chapter 5. - Mole Concept and Stoichiometry

1. State:

(a) Gay-Lussac's Law of combining volumes.

Ans: 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

Ans: Avogadro's law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

2. (a) What do you mean by stoichiometry?

Ans: Stoichiometry determines the amount of products/reactants produced/needed in a given reaction by measuring quantitative relationships.

(b) Define atomicity of a gas. State the atomicity of Hydrogen, Phosphorus and Sulphur.

Ans: Atomicity refers to the no of atoms present in a molecule. Atomicity of hydrogen is 2, phosphorus is 4 and sulphur is 8.

(c) Differentiate between N2 and 2N.

Ans:

N2 | 2N |

It consists of one molecule of nitrogen. | It consists of two atoms of nitrogen. |

Exist independently. | Cannot exist independently. |

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."

Ans: This is due to Avogadro's 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

nH2 = 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 are double the atoms of helium.

(b) "When stating the volume of a gas, the pressure and temperature should also be given."

Ans: For a given volume of gas under given temperature and pressure, a change in any one of the variables i.e., pressure or temperature changes the volume.

(c) Inflating a balloon seems to violate Boyle's law.

Ans: Inflating a balloon seems violating Boyle's law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.

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 + O2 $\to ~$2CO2

Ans: 2CO + O2 $\to $ 2CO2

2V 1V 2V

2 V of CO requires = 1V of O2

so, 100 litres of CO requires = 50 litres of O2

(b) 200 cm3 of hydrogen and 150 cm3 of oxygen are mixed and ignited, as per the following reaction,

2H2 + O2 $\to $2H2O

What volume of oxygen remains unreacted?

Ans: 2H2 + O2 $\to $2H2O

2V 1V 2V

From the equation, 2V of hydrogen reacts with 1V of oxygen

so 200cm3 of Hydrogen reacts with = $\frac{200}{2}$= 100 cm3

Hence, the unreacted oxygen is 150 - 100 = 50cm3 of oxygen.

5. 24 cc Marsh gas (CH4) 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.

Ans: This experiment supports Gay lussac's law of combining volumes.

Since the unchanged or remaining O2 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.

CH4 + 2O2$\to$CO2 + 2H2O

1V 2V

24cc 48 cc

i.e. methane and oxygen react in a 1:2 ratio.

6. What volume of oxygen would be required to burn 400 ml of acetylene [C2H2]? Also calculate the volume of carbon dioxide formed.

2C2H2 + 5O2 $\to $4CO2 + 2H2O (l)

Ans: 2C2H2 + 5O2 $\to $ 4CO2 + 2H2O (l)

2V 5V 4V

From equation, 2 V of C2H2 requires = 5 V of O2

So, for 400ml C2H2, O2 required = 400 \[\times\] $\frac{~~5~~}{2}$ =1000 ml

Similarly, 2 V of C2H2 gives = 4 V of CO2

So, 400ml of C2H2 gives CO2 = 400 ×$\frac{~4}{2}$ = 800ml

7. 112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Write a balanced equation for this reaction

Ans: H2S + Cl_{2} $\to$ 2HCl + S

and calculate

(i) the volume of gaseous product formed

Ans: At STP, 1 mole gas occupies 22.4 L.

As 1 mole H2S gas produces 2 moles HCl gas,

22.4 L H2S gas produces 22.4 × 2 = 44.8 L HCl gas.

Hence, 112 cm3 H2S gas will produce 112 × 2 = 224 cm3 HCl gas.

(ii) composition of the resulting mixture.

Ans: 1 mole H2S gas consumes 1 mole Cl2 gas.

This means 22.4 L H2S gas consumes 22.4 L Cl2 gas at STP.

Hence, 112 cm3 H2S gas consumes 112 cm3 Cl2 gas.

120 cm3 - 112 cm3 = 8 cm3 Cl2 gas remains unreacted.

Thus, the composition of the resulting mixture is 224 cm3HCl gas + 8 cm3 Cl2 gas.

8. 1250cc of oxygen was burnt with 300cc of ethane [C2H6]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed:

2C2H6+7O_{2} $\to $ 4CO2+6H2O

Ans: From the equation, 2V of ethane reacts with 7V oxygen.

So, 300 cc of ethane reacts with $\frac{~~~300\times 7~~}{2}$=1050cc

Hence, unused O2 = 1250 - 1050 = 200 cc

From 2V of ethane, 4V of CO2 is produced.

So, 300 cc of ethane will produce $\frac{~~~300\times 4~~}{2}$=600cc of CO2

9. What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C2H4] at 273o C and 380 mm of Hg pressure?

C2H4+3O2$\to $2CO2 + 2H2O

Ans:

C2H4+3O2$\to $2CO2 + 2H2O

1V 3V

$\frac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}}{{{\text{T}}_{\text{1}}}}$= $\frac{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}}{{{\text{T}}_{\text{2}}}}$

V2= \[\frac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{2}}}}{{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{1}}}}\]=$\frac{~~380\times 33\times 273~~}{549\times 760}$=8.25 litres

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.

CH4 + 2Cl2$\to $CH2Cl2 + 2HCl

Ans: CH4 + 2Cl2$\to $ CH2Cl2 +2HCl

1V 2V V 2V

From equation, 1V of CH4 gives = 2 V HCl

so, 40 ml of methane gives = 80 ml HCl

For 1V of methane = 2V of Cl2 required

So, for 40ml of methane = 40 x 2 = 80 ml of Cl2

11. What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions? (assuming oxygen is 1/5th of air)

C3H8 + 5O2$\to $3CO2 + 4H2O

Ans: C3H8 + 5O2\[\rightarrow\] 3CO2 + 4H2O

1V 5V 3V

From equation, 5V of O2 required = 1V of propane

so, 100 cm3 of O2 will require = 20 cm3 of propane

12. 450 cm3 of nitrogen monoxide and 200 cm3 of oxygen are mixed together and ignited. Calculate the composition of the resulting mixture.

2NO + O2$\to $2NO2

Ans: 2NO + O2$\to $2NO2

2V V 2V

From equation, 1V of O2 reacts with = 2V of NO

200cm3 oxygen will react with = 200 $\times $2 =400 cm3 NO

Hence, remaining NO is 450 - 400 = 50 cm3

NO2 produced = 400cm3 because 1V oxygen gives 2V NO2

Total mixture = 400 + 50 = 450 cm3

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.

Ans: 6 litres of hydrogen and 4 litres of chlorine when mixed, results in the formation of 8 litres of HCl gas.

When water is added to it, it results in the formation of hydrochloric acid. Chlorine acts as a limiting agent leaving behind only 2 litres of hydrogen gas.

Therefore, the volume of the residual gas will be 2 litres.

14. Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.

4NH3 + 5O2$\to $4NO + 6H2O

If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?

Ans: 4NH3 + 5O2$\to $4NO + 6H2O

4V 5V 4V

9 litres of reactants gives 4 litres of NO

So, 27 litres of reactants will give = 27 \[\times \frac{~~4~~}{9}\] = 12 litres of NO

15. A mixture of hydrogen and chlorine occupying 36 cm3 exploded. On shaking it with water, 4cm3 of hydrogen was left behind. Find the composition of the mixture.

Ans: H2 + Cl2 $\to $2HCl

1V 1V 2V

Since 1V hydrogen requires 1V of oxygen and 4cm3 of H2 remained behind so the mixture had composition: 16 cm3 hydrogen and 16 cm3 chlorine.

Therefore Resulting mixture is H2 = 4cm3, HCl = 32cm3

16. What volume of air (containing 20% O2 by volume) will be required to burn completely 10 cm3 each of methane and acetylene?

CH4 + 2O2 $\to $CO2 + 2H2O

2C2H2 + 5O2$\to $4CO2 + 2H2O

Ans: CH4 + 2O2 $\to $CO2 + 2H2O

1V 2V 1V

2C2H2 + 5O2$\to $4CO2 + 2H2O

2V 5V 4V

From the equations, we can see that

1V CH4 requires oxygen = 2V O2

So, 10cm3 CH4 will require =20 cm3 O2

Similarly 2V C2H2 requires = 5V O2

So, 10 cm3 C2H2 will require= 25 cm3 O2

Now, 20 V O2 will be present in 100 V air and 25 V O2 will be present in 125 V air ,so the volume of air required is 225cm3

17. LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to the atmosphere.

C3H8 + 5O2$\to $3CO2 + 4H2O

2C4H10 + 13O2$\to$8CO2 + 10H2O

Ans: C3H8 + 5O2$\to $3CO2 + 4H2O

2C4H10 + 13O2$\to $8CO2 + 10H2O

60 ml of propane (C3H8) gives 3 $\times $60 = 180 ml CO2

40 ml of butane (C4H10) gives = 8 $\times $ \[\frac{40}{2}\] = 160 ml of CO2

Total carbon dioxide produced = 340 ml

So, when 10 litres of the mixture is burnt = 34 litres of CO2 is produced.

18. 200 cm3 of CO2 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 the original mixture.

2C2H2(g) + 5O2(g) $\to $4CO2(g)+ 2H2O(g)

Ans: 2C2H2(g) + 5O2(g) $\to $4CO2(g)+ 2H2O(g)

4 V CO2 is collected with 2 V C2H2

So, 200cm3 CO2 will be collected with = 100cm3 C2H2

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

So, 200cm3 CO2 will be produced by = 250 ml of O2

19. You have collected (a) 2 litres of CO2 (b) 3 litres of chlorine (c) 5 litres of hydrogen (d) 4 litres of nitrogen and (e) 1 litres of SO2, 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.

Ans: 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 less volume will contain less molecules.

(a) 5 litres of hydrogen has the greatest no. of molecules with the maximum volume.

(b) 1 litre of SO2 contains the least number of molecules since it has the smallest volume.

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 |

Ans:

Gas | Volume (in litres) | Number of molecules |

Chlorine | 10 | \[\frac{x}{2}\] |

Nitrogen | 20 | x |

Ammonia | 20 | x |

Sulphur dioxide | 5 | \[\frac{x}{4}\] |

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.

Ans: 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) Name the law on which the above problem is based.

Ans: The problem is based on Avogadro's law.

### Exercise – 5B

1. (a) The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.

Ans: This statement states that one atom of chlorine is 35.5 times heavier than 1/12 times of the mass of an atom C-12.

(b) What is the value of Avogadro's number?

Ans: 6.023 $\times $1023

(c) What is the value of molar volume of a gas at S.T.P?

Ans: The molar volume of a gas at STP is 22.4 dm3 at STP.

2. Define or explain the terms

(a) Vapour density

Ans: The ratio between the masses of equal quantities of gas and hydrogen at normal temperature and pressure is known as the vapour density.

(b) Molar volume

Ans: Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm3.

(c) Relative atomic mass

Ans: 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) Relative molecular mass

Ans: The relative molecular mass of a compound is the number that represents how many times one molecule of the substance is heavier than 1/12 of the mass of an atom of carbon-12.

(e) Avogadro's number

Ans: The number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 x1023 atoms.

(f) Gram atom

Ans: The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.

(g) Mole

Ans: Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.

3. (a) What are the main applications of Avogadro's Law?

Ans: (a) Applications of Avogadro's Law:

(1) It explains Gay-Lussac's law.

(2) It determines the 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) How dose Avogadro's Law explain Gay-Lussac's Law of combining volumes?

Ans: 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.

H2 + Cl2 $\to $ 2HCl

1V 1V 2V(By Gay-Lussacs law)

4. Calculate the relative molecular masses of:

(a) Ammonium chloroplatinate, (NH4)2 PtCl6

Ans: (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444

(b) Potassium chlorate

Ans: KClO3

= (K)39 + (Cl)35.5 + (3O)48 = 122.5

(c) CuSO4. 5H2O

Ans: (Cu)63.5 + (S)32 + (4O)64 + (5H2O)5 x 18 = 249.5

(d) (NH4)2SO4

Ans: (2N)28 + (8H)8 + (S)32 + (4O)64 = 132

(e) CH3COONa

Ans: (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82

(f) CHCl3

Ans: (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5

(g) (NH4)2 Cr2O7

Ans: (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252

5. Find the

(a) number of molecules in 73 g of HCl,

Ans: No. of molecules in 73 g HCl

= 6.023 $\times $1023$\times $ 73/36.5(mol. mass of HCl)

= 12.04 $\times $ 1023

(b) weight of 0.5 mole of O2,

Ans: Weight of 0.5 mole of O2 is

= 32(mol. Mass of O2) $\times $ 0.5

=16 g

(c) number of molecules in 1.8 g of H2O

Ans: No. of molecules in 1.8 g H2O

= 6.023 $\times $ 1023 $\times \frac{~~~18~~}{18}$

= 6.023 $\times $ 1022

(d) number of moles in 10 g of CaCO3

Ans: No. of moles in 10g of CaCO3

= $\frac{~~~10~~}{100}$(mol. Mass CaCO3)

= 0.1 mole

(e) Weight of 0.2 mole of H2 gas,

Ans: Weight of 0.2 mole H2 gas

= 2(Mol. Mass) $\times $ 0.2

= 0.4 g

(f) Number of molecules in 3.2 g of SO2.

Ans: No. of molecules in 3.2 g of SO2

= 6.023 × 1023 × \[\frac{3.2}{64}\]

= 3.023 × 1022

6. Which of the following would weigh most?

(a) 1 mole of H2O

(b) 1 mole of CO2

(c) 1 mole of NH3

(d) 1 mole of CO

Ans: (b) 1 mole of CO2

Weight of H2O= 2+16=18

Weight of CO2= 12+32=44

Weight of NH3= 14+3=17

Weight of CO= 12+16=28

7. Which of the following contains the maximum number of molecules?

(a) 4 g of O2

(b) 4 g of NH3

(c) 4 g of CO2

(d) 4 g of SO2

Ans: (b) 4 g of NH3

4g of NH3 having minimum molecular mass contains maximum molecules.

8. Calculate the number of

(a) Particles in 0.1 mole of any substance.

Ans: No. of particles in s1 mole = 6.023 $\times $ 1023

So, particles in 0.1 mole = 6.023 $\times $ 1023 $\times $ 0.1

= 6.023 $\times $ 1022

(b) Hydrogen atoms in 0.1 mole of H2SO4.

Ans: 1 mole of H2SO4 contains =2 $\times $ 6.023 $\times $1023

So, 0.1 mole of H2SO4 contains =2 $\times $6.023 $\times $ 1023 $\times $ 0.1

= 1.2$\times $1023 atoms of hydrogen

(c) Molecules in one Kg of calcium chloride.

Ans: 111g CaCl2 contains = 6.023 $\times $ 1023 molecules

So, 1000 g contains = 5.42 $\times $ 1024 molecules

9. How many grams of

(a) Al are present in 0.2 mole of it?

Ans: 1 mole of aluminium has mass = 27 g

So, 0.2 mole of aluminium has mass = 0.2 $\times $ 27 = 5.4 g

(b) HCl are present in 0.1 mole of it?

Ans: 0.1 mole of HCl has mass

=0.1 $\times $ 36.5(mass of 1 mole)

= 3.65 g

(c) H2O are present in 0.2 mole of it?

Ans: 0.2 mole of H2O has mass = 0.2 $\times $ 18 = 3.6 g

(d) CO2 is present in 0.1 mole of it?

Ans: 0.1 mole of CO2 has mass = 0.1 $\times $ 44 = 4.4 g

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?

Ans: 5.6 litres of gas at STP has mass = 12 g

So, 22.4 litre (molar volume) has mass =12 $\times $ \[\frac{22.4}{5.6}\]

= 48 g(molar mass)

(b) Calculate the volume occupied at S.T.P. by 2 moles of SO2.

Ans: 1 mole of SO2 has volume = 22.4 litres

So, 2 moles will have = 22.4 $\times $ 2 = 44.8 litre

11. Calculate the number of moles of

(a) CO2 which contain 8.00 g of O2

Ans: 1 mole of CO2 contains O2 = 32g

So, CO2 having 8 gm of O2 has no. of moles = $\frac{~~8~~}{32}$

= 0.25 moles

(b) Methane in 0.80 g of methane.

Ans: 16 g of methane has no. of moles = 1

So, 0.80 g of methane has no. of moles = $\frac{~0.8}{16}$

= 0.05 moles

12. Calculate the actual mass of

(a) An atom of oxygen

Ans: 6.023 $\times $ 10 23 atoms of oxygen has mass = 16 g

So, 1 atom has mass = $\frac{~~~~16~~}{6.023}\times $ 1023

= 2.656 $\times $ 10-23 g

(b) an atom of hydrogen

Ans: 1 atom of Hydrogen has mass = $\frac{~~~~1.6~~}{6.023}\times $ 1023 = 1.666 $\times $ 10-24

(c) a molecule of NH3

Ans: 1 molecule of NH3 has mass = $\frac{~~~~17~~}{6.023}\times $ 1023 = 2.82 $\times $ 10-23 g

(d) the atom of silver

Ans: 1 atom of silver has mass = $\frac{~~~~108~~}{6.023}\times $ 1023 =1.701 $\times $ 10-22

(e) the molecule of oxygen

Ans: 1 molecule of O2 has mass = $\frac{~~~~32~~}{6.023}\times $ 1023 = 5.314 $\times $ 10-23 g

(f) 0.25 gram atom of calcium

Ans: 0.25 gram atom of calcium has mass = 0.25 $\times $ 40 = 10g

13. Calculate the mass of 0.1 mole of each of the following

(Ca = 40, Na=23, Mg =24, S=32, C = 12, Cl = 35.5, O=16, H=1)

(a) CaCO3

Ans: 0.1 mole of CaCO3 has mass =100(molar mass) $\times $ 0.1 = 10 g

(b) Na2SO4.10H2O

Ans: 0.1 mole of Na2SO4.10H2O has mass = 322 $\times $ 0.1 = 32.2 g

(c) CaCl2

Ans: 0.1 mole of CaCl2 has mass = 111 $\times $ 0.1 = 11.1g

(d) Mg

Ans: 0.1 mole of Mg has mass = 24 $\times $ 0.1 = 2.4 g

14. Calculate the number of

(a) oxygen atoms in 0.10 mole of Na2CO3.10H2O.

Ans: 1molecule of Na2CO3.10H2O contains oxygen atoms = 13

So, 6.023 $\times $ 1023 molecules (1mole) has atoms = 13 $\times $ 6.023 $\times $ 1023

So, 0.1 mole will have atoms = 0.1 $\times $ 13 $\times $ 6.023 $\times $ 1023

=7.8 $\times $ 1023

(b) gram atoms in 4.6 gram of sodium

Ans: Given Na = 4.6 gm

Atomic mass= 23

No. of gram atoms of Na = $\frac{\text{ }\!\!~\!\!\text{ Mass}\,\text{of}\,\text{Na}}{\text{Atomic}\,\text{mass}\,\text{of}\,\text{Na}}$

= $\frac{4.6}{23}$ = 0.2

(c) moles in 12 g of oxygen gas

Ans: 32 g of oxygen gas = 1 mole

1 gram of oxygen gas = $\frac{~~1~~}{32}$ mole

Given that 12 g of oxygen gas

No: of moles = $\frac{given\,mas}{molar\,mass}$

= $\frac{~~~12~~~}{32}$ = 0.375 mole

15. What mass of Ca will contain the same number of atoms as are present in 3.2 g of S?

Ans: 3.2 g of S has number of atoms = 6.023 $\times $1023$\times $$\frac{\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }3.2\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}{32}$

= 0.6023 $\times $ 1023

So, 0.6023 $\times $ 1023 atoms of Ca has mass=40 $\times $$\frac{\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }0.6023\times {{10}^{23}}}{6.023\times {{10}^{23}}}$

= 4g

16. Calculate the number of atoms in each of the following:

(a) 52 moles of He

Ans:No. of atoms = 52 $\times $6.023 $\times $1023 = 3.131 $\times $ 1025

(b) 52 amu of He

Ans: 4 amu = 1 atom of He

so, 52 amu = 13 atoms of He

(c) 52 g of He

Ans: 4 g of He has atoms = 6.023 $\times $1023

So, 52 g will have = 6.023 $\times $ 1023 $\times$$\frac{~~~52~~~}{4}$= 7.828 $\times $1024 atoms

17. Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.

Ans: Molecular mass of Na2CO3 = 106 g

106 g has 2 $\times $ 6.023 $\times $1023 atoms of Na

So, 5.3g will have = 2 $\times $ 6.023 $\times $1023 $\times \frac{~~~5.3~~~}{106}$= 6.022 $\times $1022 atoms

Number of atoms of C = 6.023 $\times $1023 $\times \frac{~~~5.3~~~}{106}$ = 3.01 $\times $ 1022 atoms

And atoms of O = 3 $\times $ 6.023 $\times $1023 $\times \frac{~~~5.3~~~}{106}$ = 9.03 $\times $ 1022 atoms

18. (a) Calculate the mass of nitrogen supplied to soil by 5 kg of urea (CO(NH2)2) (O = 16; N = 14; C = 12 ; H = 1 )

Ans: 60 g urea has mass of nitrogen(N2) = 28 g

So, 5000 g urea will have mass = 28 $\times \frac{5000}{60}$ = 2.33 kg

(b) Calculate the volume occupied by 320 g of sulphur dioxide at S.T.P. (S = 32; O = 16)

Ans: 64 g has volume = 22.4 litre

So, 320 g will have volume = 22.4 $\times $$\frac{320}{64}$ = 112 litres

19. (a) What do you understand by the statement that 'vapour density of carbon dioxide is 22'?

Ans: Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.

(b) Atomic mass of Chlorine is 35.5.What is its vapour density?

Ans: Vapour density of Chlorine atom is 35.5.

20. What is the mass of 56 cm3 of carbon monoxide at STP?

(C=12, O=16)

Ans: 22400 cm3 of CO has mass = 28 g

So, 56 cm3 will have mass = 56 $\times \frac{28}{22400}$ = 0.07 g

21. Determine the number of molecules in a drop of water which weighs 0.09g.

Ans: 18 g of water has number of molecules = 6.023 $\times $ 1023

So, 0.09 g of water will have no. of molecules = 6.023 $\times $ 1023 $\times$ $\frac{0.09}{18}$ = 3.01 $\times $ 1021 molecules

22. The molecular formula for elemental sulphur is S8.In sample of 5.12 g of sulphur

(a) How many moles of sulphur are present?

Ans: No. of moles in 256 g S8 = 1 mole

So, no. of moles in 5.12 g = $\frac{~~5.12~~~}{256}$= 0.02 moles

(b) How many molecules and atoms are present?

Ans: No. of molecules = 0.02 $\times $ 6.023 $\times $ 1023 = 1.2 $\times $ 1022 molecules

No. of atoms in 1 molecule of S = 8

So, no. of atoms in 1.2 $\times $ 1022 molecules = 1.2 x$\times $ 1022 $\times$ 8

= 9.635$\times $ 1022 molecules

23. If phosphorus is considered to contain P4 molecules, then calculate the number of moles in 100g of phosphorus?

Ans: Atomic mass of phosphorus P = 30.97 g

Hence, molar mass of P4 = 123.88 g

If phosphorus is considered as P4 molecules,

then 1 mole P4 ≡ 123.88 g

Therefore, 100 g of P4 = 0.807 g

24. Calculate:

(a) The gram molecular mass of chlorine if 308cm3 of it at STP weighs 0.979 g

Ans: 308 cm3 of chlorine weighs = 0.979 g

So, 22400 cm3 will weigh = gram molecular mass

= 0.979 $\times \frac{~~~22400~~}{308}$ =71.2 g

(b) The volume of 4g of H2 at 4 atmospheres.

Ans: 2 g(molar mass) H2 at 1 atm has volume = 22.4 litres

So, 4 g H2 at 1 atm will have volume = 44.8 litres

Now, at 1 atm(P1) 4 g H2 has volume (V1) = 44.8 litres

So, at 4 atm(P2) the volume(V2) will be = $\frac{{{P}_{1}}{{V}_{1}}}{{{P}_{2}}}$ = $\frac{1\times 44.8}{4}$ = 11.2 litres

(c) The mass of oxygen in 2.2 litres of CO2 at STP.

Ans: Mass of oxygen in 22.4 litres = 32 g(molar mass)

So, mass of oxygen in 2.2 litres = 2.2 $\times \frac{~~~32~~}{22.4~}$=3.14 g

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.

Ans: No. of atoms in 12 g C = 6.023 $\times $ 1023

So, no. of carbon atoms in 10-12 g = 10-12 $\times $$\frac{6.023\times{{10}^{23}}}{12}$

= 5.019 $\times $ 1010 atoms

26. An unknown gas shows a density of 3 g per litre at 2730C and 1140 mm Hg pressure. What is the gram molecular mass of this gas?

Ans: Given:

P = 1140 mm Hg

Density = D = 2.4 g / L

T = 273 0C = 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.

V1= 1 L

T1 = 546 K

V2=?

T2 = 273 K

\[\frac{{{\text{V}}_{\text{1 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}}{{{\text{T}}_{\text{1}}}}\]= $\frac{{{\text{V}}_{\text{2 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}}{{{\text{T}}_{\text{2}}}}$

V2 = $\frac{{{V}_{1}}\times {{T}_{2}}}{{{T}_{1}}}$

= $\frac{1l\times 273}{546}$ = 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

V1 = 0.5 L

P2 = 760 mm Hg

V2 = ?

P1 $\times $ V1 = P2 $\times $ V2

V2 = \[\frac{{{P}_{1}}\times {{V}_{1}}}{{{P}_{2}}}\]

=$\frac{1140\,mm\,hg\,\times \,0.5~L}{760mm\,hg}$ = 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 = \[\frac{0.75}{22.4}\]

= 0.0335 moles

The original mass is 2.4 g

n = $\frac{~~m~}{M}$ = 0.0335 moles

= 2.4 g / M

M = $\frac{~~~2.4~~~}{0.0335}$

M = 71.6 g / mole

Hence, the gram molecular mass of the unknown gas is 71.6 g

27. Cost of Sugar (C12H22 O11) is Rs 40 per kg; calculate its cost per mole.

Ans: 1000 g of sugar costs = Rs. 40

So, 342g(molar mass) of sugar will cost= $\frac{~~342\times 40~~~}{1000}$ =Rs. 13.68

28. Which of the following weighs the least?

(a) 2 g atom of N

(b) 3 x1025 atoms of carbon

(c) 1mole of sulphur

(d)7 g of silver

Ans: d) 7 g of silver

a. Weight of 1 g atom N = 14 g

So, weight of 2 g atom of N = 28 g

b. 6.023 $\times $1023 atoms of C weigh = 12 g

So, 3 $\times $1025 atoms will weigh = $\frac{~~~12\times 3\times {{10}^{25}}}{6.022~\times ~{{10}^{23}}}$=597.7g

c. 1 mole of sulphur weighs = 32 g

d. 7 g of silver

So, 7 grams of silver weighs the least.

29. Four grams of caustic soda contains:

(a) 6.02 x 1023 atoms of it

(b) 4 g atom of sodium

(c) 6.02 x1022 molecules

(d) 4 moles of NaOH

Ans: (c) 6.02 $\times $1022 molecule

40 g of NaOH contains 6.023 $\times $1023 molecules

So, 4 g of NaOH contains = $\frac{~~~6.02\times {{10}^{23}}\times 4~~~~}{40}$6.02 $\times $1023 $\times $ \[\frac{4}{40}\]

= 6.02 $\times $1022 molecules

30. The number of molecules in 4.25 g of ammonia is:

(a) 1.0 $\times $ 1023

(b) 1.5 $\times $ 1023

(c) 2.0 $\times $ 1023

(d) 3.5 $\times $ 1023

Ans: (b) 1.5 $\times $ 1023

The number of molecules in 18 g of ammonia = 6.02 $\times $1023

So, no. of molecules in 4.25 g of ammonia = $\frac{~~~6.02\times {{10}^{23}}\times ~4.25~~~}{18}$

= 1.5 $\times $ 1023

31. Correct the statements, if required

(a) One mole of chlorine contains 6.023 $\times $ 1010 atoms of chlorine.

Ans: One mole of chlorine contains 6.023 $\times $ 1010 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.

Ans: 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.

Ans: 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.

Ans: Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of atoms.

### Exercise – 5C

1. Give three kinds of information conveyed by the formula H2O.

Ans: Information conveyed by H2O

(1)That H2O contains 2 volumes of hydrogen and 1 volume of oxygen.

(2)That ratio by weight of hydrogen and oxygen is 1:8.

(3)The molecular weight of H2O is 18g.

2. Explain the terms empirical formula and molecular formula.

Ans: 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.

3. Give the empirical formula of:

(a) C6H6

Ans: CH

(b) C6H12O6

Ans: CH2O

(c) C2H2

Ans: CH

(d) CH3COOH

Ans: CH2O

4. Find the percentage of water of crystallisation in CuSO4.5H2O. (At. Mass Cu = 64, H = 1, O = 16, S = 32)

Ans: Relative molecular mass of CuSO4. 5H2O

=64+32+4$\times $16+5(2+16)

=160+90=250

250g of CuSO4.5H2O contains 90g of water of crystallisation

=$\frac{90}{250}\times $100= 36%

5. Calculate the percentage of phosphorus in

(a) Calcium hydrogen phosphate Ca(H2PO4)2

Ans: Molecular mass of Ca(H2PO4)2 = 234

So, % of P = 2 $\times $ 31 $\times $ \[\frac{100}{234}\] = 26.5%

(b) Calcium phosphate Ca3(PO4)2

Ans: Molecular mass of Ca3(PO4)2 = 310

% of P = 2 $\times $31 $\times $ \[\frac{100}{310}\] = 20%

6. Calculate the percent composition of Potassium chlorate KClO3.

Ans: Molecular mass of KClO3 = 122.5 g

% of K = \[\frac{39}{122.5}\] = 31.8%

% of Cl = \[\frac{35.5}{122.5}\] = 28.98%

% of O = 3 $\times $ \[\frac{16}{122.5}\] = 39.18%

7. Find the empirical formula of the compounds with the following percentage composition:

Pb = 62.5%, N = 8.5%, O = 29.0%

Ans:

Element | % | At. mass | Atomic ratio | Simple ratio |

Pb | 62.5% | 207 | $\frac{~~~62.5~~~}{207}$=0.3019 | $\frac{~~0.3019~}{0.3019}$=1 |

N | 8.5% | 15 | $\frac{~~~~8.5~~~}{15}$= 0.6071 | $\frac{~~~0.6071~~}{0.3019}$=2 |

O | 29.0 | 16 | $\frac{~~~~~29~~~}{16}$= 1.81 | $\frac{~~1.81~~}{0.3019}$=6 |

So, Pb(NO3)2 is the empirical formula.

8. Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.

Ans: In Fe2O3, Fe = 56 and O = 16

Molecular mass of Fe2O3 = 2 $\times $56 + 3 $\times $16 = 160 g

Iron present in 80% of Fe2O3 = $\frac{~~112~~}{160}\times $80 = 56g

So, mass of iron in 100 g of ore = 56 g

mass of Fe in 10000 g of ore = 56 $\times $ \[\frac{10000}{100}\]

= 5.6 kg

9. If the empirical formula of two compounds is CH and their Vapour densities are 13 to 39 respectively, find their molecular formula.

Ans: For acetylene, molecular mass = 2 $\times $V.D = 2 $\times $13 = 26 g

The empirical mass = 12(C) + 1(H) = 13 g

n =$\frac{\text{Molecular formula mass}}{\text{Empirical formula weight}}$ = $\frac{26}{13}$ = 2

Molecular formula of acetylene= 2 $\times $Empirical formula =C2H2

Similarly, for benzene molecular mass= 2 $\times $V.D = 2 $\times $ 39 = 78

n = $\frac{78}{13}$ = 6

So, the molecular formula = C6H6

10. Find the empirical formula of a compound containing 17.7% hydrogen and 82.3% nitrogen.

Ans: Element % At. mass Atomic ratio Simple ratio

Element | % | Atomic mass | Atomic ratio | Simple ratio |

H | 17.7 | 1 | $\frac{~~~17.7~~}{1}$= 17.7 | $\frac{~~17.7~~}{5.87}$=3 |

N | 82.3 | 14 | $\frac{~~82.3~~}{14}$= 5.87 | $\frac{~~~5.87~~}{5.87}$=1 |

So, the empirical formula = NH3

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

Ans:

Element | % | Atomic mass | Atomic ratio | Simple ratio |

C | 54.54 | 2 | $\frac{~~54.54~~}{12}$=4.55 | $\frac{~~4.55~~}{2.27}$=2 |

H | 9.09 | 4 | $\frac{~~~9.09~~}{1}$=9.09 | $\frac{~~~~9.09~~}{2.27}$= 4 |

O | 36.36 | 1 | $\frac{~~36.36~~}{16}$= 2.27 | $\frac{~~~2.27~~}{2.27}$=1 |

(a) So, its empirical formula = C2H4O

(b) empirical formula mass = 44

Since, vapour density = 44

So, molecular mass = 2 $\times $V.D = 88

or n = 2

so, molecular formula = (C2H4O)2 = C4H8O2

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

Ans:

Elements | % | at. mass | atomic ratio | simple ratio |

C | 26.59 | 12 | $\frac{~~~26.59~~}{12}$=2.21 | $\frac{~~2.21~~}{2.21}$= 1 |

H | 2.22 | 1 | $\frac{~~~~2.22~~~}{1}$=2.22 | $\frac{~~~2.22~~}{2.21}$= 1 |

O | 71.19 | 16 | $\frac{~~~71.19~~}{16}$= 4.44 | $\frac{~~~4.44~~}{2}$= 2 |

(a) its empirical formula = CHO2

(b) empirical formula mass = 45

Vapour density = 45

So, molecular mass = V.D $\times $2 = 90

so, molecular formula = C2H2O4

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

Ans:

Element | % | Atomic mass | Atomic ratio | Simple ratio |

Cl | 71.65 | 35.5 | $\frac{~~71.65~~}{35.5}$=2.01 | $\frac{~~~2.01~~}{2.02}$=1 |

H | 4.07 | 1 | $\frac{~~~4.07~~}{1}$=4.07 | $\frac{~~~4.07~~}{2.01}$=2 |

C | 24.28 | 12 | $\frac{~~~24.28~~~}{12}$=2.02 | $\frac{~~~24.28~~}{12}$=1 |

(a) its empirical formula = CH2Cl

(b) empirical formula mass = 49.5

Since, molecular mass = 98.96

so, molecular formula = (CH2Cl)2 = C2H4Cl2

14. A hydrocarbon contains 4.8g of carbon per gram of hydrogen. Calculate

(a) the g atom of each

Ans: The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1=1

(b) find the empirical formula

Ans:

Element | Given mass | At. mass | Atomic ratio | Simplest ratio |

C | 4.8 | 12 | $\frac{~~~4.8~~}{12}$=0.4 | $\frac{~~~0.4~~}{0.4~}$=1 |

H | 1 | 1 | $\frac{~~1~~}{1}$=1 | $\frac{~~1~~}{0.4}$=$\frac{~~5~~}{2}$ |

So, the empirical formula = C2H5

(c) Find molecular formula, if its vapour density is 29.

Ans: Empirical formula mass = 29

Molecular mass = V.D $\times $2 = 29 $\times $ 2 = 58

So, molecular formula = C4H10

15. 0.2 g atom of silicon Combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.

Ans: 0.2g atom i.e 0.2 mole silicon

Moles of chlorine = $\frac{given\,mass}{molar\,mass}$ = $\frac{21.3}{35.5}$ = 0.6mol

Ratio of Si : Cl = $\frac{0.2}{0.2}$ : $\frac{0.6}{0.2}$ = 1:3

Empirical formula = SiCl3

16. A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.

Ans: % 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 | $\frac{~~~82.76~~}{12}$= 6.89 | $\frac{~~~6.89~~~}{6.89}$= 1 $\times $ 2 = 2 |

H | 17.24 | 1 | $\frac{~~~17.24~~~}{17.24}$=1 | $\frac{~~~17.24~~~}{6.89}$= 2.5 $\times $ 2 = 5. |

Empirical formula = C2H5

Empirical formula weight = 2 $\times $ 12 + 1 $\times $ 5 = 24 + 5 = 29

Vapour Density = 29

Relative molecular mass = 29 $\times $ 2 = 58

N = $\frac{\text{Relative molecular mass}}{\text{Empirical}\,\text{weight}}$ = $\frac{58}{29}$ = 2

Molecular Formula = n x empirical formula

= 2 $\times $ C2H5

= C4H10

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?

Ans: G atoms of magnesium = $\frac{18}{24}$ = 0.75 or g- atom of Mg

(b) How many gram- atoms of nitrogen are equal to 7g of nitrogen?

Ans: G atoms of nitrogen = $\frac{7}{14}$ = 0.5 or 1/2 g- atoms of N

(c) Calculate simple ratio of gram- atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.

Ans: Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3

So, the formula is Mg3 N2

18. Barium chloride crystals contain 14.8% water of crystallization. Find the number of molecules of water of crystallization per molecule.

Ans: Barium chloride = BaCl2. x H2O

Ba + 2Cl + x(H2 + O)

=137+ 235.5 + x(2+16)

= (208 + 18x) contains water = 14.8% water in BaCl2.x H2O

= (208 + 18 x) $\times \frac{14.8}{100}$ = 18x

= (104 + 9x) 2148=18000x

= (104+9x) 37=250x

= 3848 + 333x =2250x

1917x =3848

x = 2 molecules of water

19. Urea is a very important nitrogenous fertilizer. Its formula is CON2H4.Calculate the percentage of nitrogen in urea. (C=12, O=16, N=14 and H=1).

Ans: Molar mass of urea; CON2H4 = 60 g

So, % of Nitrogen = 28 $\times \frac{100}{60}$ = 46.66%

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.

Ans:

Element | % | Atomic mass | Atomic ratio | Simple ratio |

C | 42.1 | 12 | $\frac{~~~42.1~~}{12}$=3.5 | $\frac{~~3.5~~}{3.2}$=1 |

H | 6.48 | 1 | $\frac{~~~~6.48~~}{1}$=6.48 | $\frac{~~~6.48~~}{3.2}$=2 |

O | 51.42 | 16 | $\frac{~~~51.42~~}{16}$=3.2 | $\frac{~~3.2~~}{3.2}$=1 |

The empirical formula is CH2O

Since the compound has 12 atoms of carbon, so the formula is

C12 H24 O12.

21. (a) A compound with empirical formula AB2, has the vapour density equal to its empirical formula weight. Find its molecular formula.

Ans: 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 A2B4.

(b) A compound with empirical formula AB has vapour density 3 times its empirical formula weight. Find the molecular formula.

Ans: Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D

Empirical formula weight = $\frac{\text{V}.\text{D}}{3}$

So, n = $\frac{\text{Molecular mass}}{\text{Empirical formula}\,\text{weight}}$ = 6

Hence, the molecular formula is A6B6

(c)10.47 g of a compound contains 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)

Ans: 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 | $\frac{~~~6.25~~}{207}$= 0.03 | $\frac{~~~0.03~~}{0.03}$=1 |

B | 4.22g | 35.5 | $\frac{~~~4.26~~}{35.5}$=0.12 | $\frac{~~~0.12~~}{0.03}$=4 |

Empirical formula AB4

22. A hydride of nitrogen contains 87.5% percent by mass of nitrogen. Determine the empirical formula of this compound.

Ans: Atomic ratio of N = $\frac{87.5}{14}$ = 6.25

Atomic ratio of H = $\frac{12.5}{1}$ = 12.5

This gives us the simplest ratio as 1:2

So, the molecular formula is NH2

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.

Ans:

Element | % | Atomic mass | Atomic ratio | Simple ratio |

Zn | 22.65 | 65 | $\frac{~~~~22.65~~}{65}$=0.348 | $\frac{~~0.348~~}{0.348}$=1 |

H | 4.88 | 1 | $\frac{~~4.88~~~}{1}$=4.88 | $\frac{~~4.88~~}{0.348}$=14 |

S | 11.15 | 32 | $\frac{~~~11.15~~}{32}$=0.348 | $\frac{~~~0.348~~}{0.348}$=1 |

O | 61.32 | 16 | $\frac{~~61.32~~}{16}$=3.83 | $\frac{~~~~3.83~~}{0.348}$=11 |

Empirical formula of the given compound =ZnSH14O11

Empirical formula mass = 65.37+32+141+11+16=287.37

Molecular mass = 287

\[ n = \frac{\text{Molecular mass}}{\text{Empirical Formula}} = \frac{287}{287} = 1\]

Molecular formula = ZnSO11H14

=ZnSO4.7H2O

### Exercise – 5D

1. Complete the following blanks in the equation as indicated.

CaH2 (s) + 2H2O (aq) $\to $Ca(OH)2 (s) + 2H2 (g)

(a) Moles: 1 mol + ------- $\to $ -------- + --------------

Ans: (a) Moles:1 mole + 2 mole \[\rightarrow\]1 mole + 2 mole

(b) Grams: 42g + ------- $\to $ -------- + ----------------

Ans: Grams: 42g + 36g \[\rightarrow\] 74g + 4 g

(c) Molecules: 6.02 x 1023 + ------- $\to $-------- + -----------

Ans: Molecules = 6.02 \[\times\] 1023 + 12.046 \[\times\] 1023 \[\rightarrow\] 6.02 \[\times\] 1023+ 12.046 \[\times\] 1023

2. The reaction between 15 g of marble and nitric acid is given by the following equation:

CaCO3 + 2HNO3$\to $Ca(NO3)2+ H2O + CO2

Calculate:

(a) the mass of anhydrous calcium nitrate formed

Ans: 100 g of CaCO3 produces = 164 g of Ca(NO3)2

So, 15 g CaCO3 will produce = 164 \[\times\] 15/100 = 24.6 g Ca(NO3)2

(b) the volume of carbon dioxide evolved at S.T.P.

Ans: 1 V of CaCO3 produces 1 V of CO2

100 g of CaCO3 has volume = 22.4 litres

So, 15 g will have volume = 22.4 \[\times\] 15/100 = 3.36 litres CO2

3. 66g of ammonium sulphate is produced by the action of ammonia on sulphuric acid.

Write a balanced equation and calculate:

2NH3 + H2SO4\[\rightarrow\](NH4)2SO4

(a) Mass of ammonia required.

Ans: 2NH3 + H2SO4 → (NH4)2SO4

34g 98g 132g

For 132 g (NH4)2SO4 = 34 g of NH3 is required

So, for 66 g (NH4)2SO4 = 66 × 32/132 = 17 g of NH3 is required

(b) The volume of the gas used at the S.T.P.

Ans: 17g of NH3 requires volume = 22.4 litres

(c) The mass of acid required.

Ans: Mass of acid required, for producing 132g (NH4)2SO4 = 98g

So, Mass of acid required, for 66g (NH4)2SO4 = 66 × 98/132 = 49g

4. The reaction between red lead and hydrochloric acid is given below:

Pb3O4 + 8HCl $\to $ 3PbCl2 + 4H2O + Cl2

Calculate

(a) the mass of lead chloride formed by the action of the 6.85 g of red lead,

Ans: Molecular mass of Pb3O4 = 3 × 207.2 + 4 × 16 = 685 g

685 g of Pb3O4 gives = 834 g of PbCl2

Hence, 6.85 g of Pb3O4 will give = 6.85 × 834/685 = 8.34 g

(b) the mass of the chlorine

Ans: 685g of Pb3O4 gives = 71g of Cl2

Hence, 6.85 g of Pb3O4 will give = 6.85 × \[\frac{71}{685}\] = 0.71 g Cl2

(c) the volume of the chlorine evolved at S.T.P.

Ans: 1 V Pb3O4 produces 1 V Cl2

685g of Pb3O4has volume = 22.4 litres = volume of Cl2 produced

So, 6.85 Pb3O4 will produce = 6.85 × \[\frac{22.4}{685}\] = 0.224 litres of Cl2

5. Find the mass of KNO3 required to produce 126 kg of nitric acid. Find whether a larger or smaller mass of NaNO3 is required for the same purpose.

KNO3+ H2So4 $\to$ KHSO4 + HNO3

NaNO3 + H2SO4 $\to$ NaHSO4 + HNO3

Ans: Molecular mass of KNO3 = 101 g

63 g of HNO3 is formed by = 101 g of KNO3

So, 126000 g of HNO3 is formed by = 126000 × \[\frac{101}{63}\] = 202 kg

Similarly,126 g of HNO3 is formed by 170 kg of NaNO3

So, a smaller mass of NaNO3 is required.

6. Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide was collected at 27oC and normal pressure.

CaCO3 + 2HCl →CaCl2 + H2O + CO2

Calculate:

(a) The mass of salt required.

Ans: CaCO3 + 2HCl → CaCl2 + H2O + CO2

100g 73g 22.4L

According to given reaction

2L of O2 = $\frac{~~~2~~}{22.4}$ of CaCO3

Mass of salt required= 100$\times \frac{~~~2~~}{22.4~}$ = 8.93g

(b) The mass of the acid required to prepare the 2 litres of CO2 at 27 C and normal pressure.

CaCO3 + 2HCl $\to $ CaCl2 + H2O + CO2

Ans: 2L CO2 given by = 2$\times \frac{~~2~~}{22.4~}$mole HCl

Mass of HCl = 36.5$\times \frac{~~~4~~}{22.4}$ = 6.52g

7. Calculate the mass and volume of oxygen at S.T.P., which will be evolved on electrolysis of 1 mole (18g) of water

Ans: 2H2O $\to $ 2H2 + O2

2 moles of H2O gives = 1 mole of O2

So, 1 mole of H2O will give = 0.5 moles of O2

so, mass of O2 = no. of moles x molecular mass

= 0.5 $\times $32 = 16 g of O2

and 1 mole of O2 occupies volume =22.4 litre

so, 0.5 moles will occupy = 22.4 × 0.5 = 11.2 litres at S.T.P.

8. 1.56 g of sodium peroxide reacts with water according to the following equation:

2Na2O2 + 2H2O$\to $4NaOH + O2

Calculate:

(a) mass of sodium hydroxide formed,

Ans: Mol. Mass of Na2O2 = 2 $\times $23 + 2 $\times $16 = 78 g

Mass of 2Na2O2= 156 g

156 g Na2O2 gives = 160 g of NaOH (4 $\times $40 g)

So, 1.56 Na2O2 will give = 160 $\times $ \[\frac{1.56}{156}\] = 1.6 g

(b) Volume of oxygen liberated at S.T.P.

Ans: 156 g Na2O2 gives = 22.4 litres of oxygen

So, 1.56 g will give = 22.4 $\times $1.56/156 = 0.224 litres

= 224 cm3

(c) Mass of oxygen liberated.

Ans: 156 g Na2O2 gives = 32 g O2

So, 1.56 g Na2O2 will give = 32 $\times $1.56/156

= 32/100 = 0.32 g

9. (a) Calculate the mass of ammonia that can be obtained from 21.4 g of NH4Cl by the reaction:

2NH4Cl + Ca(OH)2 $\to$ CaCl2 +2H2O + 2NH3

Ans: 2NH4Cl + Ca(OH)2 $\to$ CaCl2+2H2O + 2NH3

Mol. Mass of 2NH4Cl = 2(14 + (1 $\times $ 4) + 35.5)= 2(53.5) = 107 g

107 g NH4Cl gives = 34 g NH3

So, 21.4 g NH4Cl will give = 21.4 $\times $34/107 = 6.8 g NH3

(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.

Ans: The volume of 17 g NH3 is 22.4 litre

So, volume of 6.8 g will be = 6.8 $\times $22.4/17 = 8.96 litre

10. Aluminium carbide reacts with water according to the following equation.

Al4C3 + 12H2O → 4Al(OH)3 + 3CH4

(a) What mass of aluminium hydroxide is formed from 12g of aluminium carbide?

Ans: Al4C3 + 12H2O → 4Al(OH)3 + 3CH4

1V 3V

144g 3$\times $22.4 Vol

Since 144g of Al4C3 gives 312g of Al(OH)3

So, 12g of Al4C3 will give = $\frac{312\times 12}{144}$ = 26g Al(OH)3

(b) What volume of methane s.t.p. is obtained from 12g of aluminium carbide?

Ans: 144g of Al4C3 gives 3$\times $22.4 litre of CH4

So 12g of Al4C3 will give = $\frac{3\times 22.4\times 12}{144}$ = 5.6 litre CH4

11. MnO2 + 4HCl $\to $MnCl2 + 2H2O +Cl2

0.02 moles of pure MnO2is heated strongly with conc. HCl. Calculate:

(a) mass of MnO2 used

Ans: 1 mole of MnO2 weighs = 87 g (mol. Mass)

So, 0.02 mole will weigh = 87 $\times $ 0.02 = 1.74 g MnO2

(b) moles of salt formed,

Ans: 1 mole MnO2 gives = 1 mole of MnCl2

So, 0.02 mole MnO2will give = 0.02 mole of MnCl2

(c) mass of salt formed,

Ans: 1 mole MnCl2 weighs = 126 g(mol mass)

So, 0.02 mole MnCl2 will weigh = 126 $\times $ 0.02 g = 2.52 g

(d) moles of chlorine gas formed,

Ans: 0.02 mole MnO2will form =0.02 mol of Cl2

(e) mass of chlorine gas formed,

Ans: 1 mole of Cl2 weighs = 35.5 g

So, 0.02 mole will weigh = 71 $\times $ 0.02 = 1.42 g of Cl2

(f) volume of chlorine gas formed at S.T.P.,

Ans: 1 mole of chlorine gas has volume = 22.4 litres

So, 0.02 mole will have volume = 22.4 $\times $ 0.02 = 0.448 litre

(g) moles of acid required,

Ans: 1 mole MnO2requires HCl = 4 mole

So, 0.02 mole MnO2 will require = 4 $\times $ 0.02 = 0.08 mole

(h) Mass of acid required.

Ans: For 1 mole MnO2 , acid required = 4 mole of HCl

So, for 0.02 mole, acid required = 4 $\times $ 0.02 = 0.08 mole

Mass of HCl = 0.08 $\times $ 36.5 = 2.92 g

12. Nitrogen and hydrogen react to form ammonia.

N2 (g) + 3H2 (g) $\to $ 2NH3 (g)

If 1000g H2 react with 2000g of N2:

(a) Will any of the two reactants remain unreacted? If yes, which one and what will be its mass?

Ans: N2 + 3H2 → 2NH3

28g 6g 34g

28g of nitrogen requires hydrogen = 6g

2000g of nitrogen requires hydrogen = 6/28 $\times $ 2000 = 3000/7g

So mass of hydrogen left unreacted = 1000 - \[\frac{3000}{7}\] = 571.4g of H2

(b) Calculate the mass of ammonia(NH3) that will be formed?

Ans: 28g of nitrogen forms NH3 = 34g

2000g of N2 forms NH3

= \[\frac{34}{28}\] $\times $ 2000

= 2428.6g

### Miscellaneous Exercise

1. From the equation for burning of hydrogen and oxygen

2H2 + O2 $\to$ 2H2O (Steam)

Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.

Ans: As per the given equation

1 mole of Oxygen gives = 2 moles of steam

so, 0.5 mole oxygen will give = 2 $\times $0.5 = 1mole of steam

2. From the equation

3Cu + 8HNO3 → 3Cu (NO3)2+ 4H2O + 2NO

(At. mass Cu=64, H=1, N=14, O=16)

Calculate:

(a) Mass of copper needed to react with 63g of HNO3

Ans: Mol. Mass of 8HNO3 = 8 $\times $ 63 = 504 g

For 504 g HNO3, Cu required is = 192 g

So, for 63g HNO3Cu required = 192 $\times $ \[\frac{63}{504}\] = 24g

(b) Volume of nitric oxide at S.T.P. that can be collected.

Ans:504 g of HNO3 gives = 2 $\times $ 22.4 litre volume of NO

So, 63g of HNO3 gives =2 $\times $ 22.4 $\times $ \[\frac{63}{504}\] = 5.6 litre of NO

3. (a) Calculate the number of moles in 7g of nitrogen.

Ans: 28g of nitrogen = 1mole

So, 7g of nitrogen = \[\frac{1}{28}\] $\times $ 7 = 0.25 moles

(b) What is the volume at S.T.P. of 7.1 g of chlorine?

Ans: Volume of 71 g of Cl2 at STP = 22.4 litres

Volume of 7.1 g chlorine = 22.4 $\times $ \[\frac{7.1}{71}\] = 2.24 litre

(c) What is the mass of 56 cm3 of carbon monoxide at S.T.P?

Ans: 22400cm3 volume have mass = 28 g of CO(molar mass)

So, 56cm3 volume will have mass = 28 $\times $ \[\frac{56}{22400}\] = 0.07 g

4. Some of the fertilizers are sodium nitrate NaNO3, ammonium sulphate (NH4)2SO4 and urea CO(NH2)2. Which of these contains the highest percentage of nitrogen?

Ans:

% of N in NaNO3 = $\frac{~~~14~~~}{85}\times $ 100= 16.47%

% of N in (NH4)2SO4 = $\frac{~~~~~14~~~}{132}\times $100= 21.21%

% of N in CO(NH2)2 =$\frac{~~~~14~~~}{60}$ $\times $ 100= 46.66%

So, the highest percentage of N is in urea.

5. Water decomposes to O2 and H2 under suitable conditions as represented by the equation below:

2H2O$\to $2H2+O2

(a) If 2500 cm3 of H2 is produced, what volume of O2 is liberated at the same time and under the same conditions of temperature and pressure?

Ans: 2H2O $\to $ 2H2 + O2

2V 2V 1V

From equation, 2 V of water gives 2 V of H2 and 1 V of O2

where 2 V = 2500 cm3

so, volume of O2 liberated = \[\frac{\text{2V}}{\text{V}}\] = 1250 cm3

(b) The 2500 cm3 of H2 is subjected to 2 times increase in pressure (temp. remaining constant). What volume of H2 will now occupy?

Ans: $\frac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}}$=$\frac{{{\text{P}}_{2}}{{\text{V}}_{2}}}{{{\text{T}}_{2}}}$

$\frac{{{\text{P}}_{1}}{{\text{V}}_{1}}}{{{\text{T}}_{1}}}$ = $\frac{\text{7}{{\text{P}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{\text{V}}_{\text{2}}}}{\text{2 }\!\!\times\!\!\text{ }{{\text{T}}_{\text{1}}}}$

V2= $\frac{~~~2500\times 2~~~}{7}$

V2= $\frac{~~5000~~}{7}$cm3

(c) Taking the value of H2 calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 pressure remaining constant.

Ans:$\frac{~~{{V}_{1~~}}}{{{V}_{2}}}$ = $\frac{~~~~{{T}_{1~~~}}}{{{T}_{2}}}$

$\frac{~~~5000~~~}{7\times 2500}$= $\frac{~~~~{{\mathbf{T}}_{1~~~}}}{{{\mathbf{T}}_{2}}}$

T2= 3.5 T1

i.e. temperature should be increased by 3.5 times.

6. Urea (CO(NH2)2) is an important nitrogenous fertilizer. Urea is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?

Ans: 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

7. Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.

Ans:

% of hydrogen = 20%

% of carbon = 100 - 20 = 80%

% Weight | Atomic Weight | Relative No. of Moles | Simplest Ratio | |

C | 80 | 12 | $\frac{~~~80~~}{12}$= 6.667 | $\frac{~~~6.667}{6.667}$= 1 |

H | 20 | 1 | $\frac{~~~20~~}{1}$= 20 | $\frac{~~~20~~}{6.667}$= 2.99 ≈ 3 |

Empirical formula = CH3

Empirical formula weight = 1 $\times $ 12 + 1 $\times $ 3 = 12 + 3 = 15

Vapour Density = 15

Relative molecular mass = 15 $\times $ 2 = 30

N = $\frac{\text{Relative molecular mass}}{\text{Empirical weight}}$= $\frac{30}{15}$ = 2

Molecular formula = n x empirical formula

= 2 $\times $ CH3

= C2H6

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 cm3 of carbon dioxide was collected at S.T.P.

Ans: 22400cm3 CO2 has mass = 44g

so, 224 cm3 CO2 will have mass= 0.44 g

Now since CO2 is being formed and X is a hydrocarbon so it contains C and H.

In 0.44g CO2, 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 C2H5

(a) Which elements does X contain?

Ans: C and H

(b) What was the purpose of copper (II) oxide?

Ans: Copper (II) oxide was used for reduction of the hydrocarbon.

(c ) Calculate the empirical formula of X by the following steps:

(i) Calculate the number of moles of carbon dioxide gas.

Ans: no. of moles of CO2= 0.44/44 = 0.01 moles

(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.

Ans: mass of C = 0.12 g

(iii) Calculate the mass of hydrogen in sample X.

(iii) mass of H = 0.025 g

(iv) Deduce the ratio of atoms of each element in X (empirical formula).

(iv) The empirical formula of X = C2H5

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.

Ans: 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 | % | Atomic mass | Atomic ratio | Simplest ratio |

X | 27.3 | 12 | $\frac{~~~27.3~~}{12}$=2.27 | $\frac{~~~~2.27~~}{2.27}$=1 |

O | 72.7 | 16 | $\frac{~~~72.2~~}{16}$=4.54 | $\frac{~~~~4.54~~}{2.27}$=2 |

So simplest formula = XO2

10. A gas cylinder filled with hydrogen holds 5g of the gas. The same cylinder holds 85 g of gas X under the same temperature and pressure. Calculate:

(a) Vapour density of gas X.

Ans:

V.D = mass of gas at STP/mass of equal volume of H2 \[ = \frac{85}{5}\] = 17

(b) Molecular weight of gas X.

Ans: Molecular mass = 17(V.D) $\times $ 2= 34g

11. (a) When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation :

CO2 + C $\to $2CO

What volume of carbon monoxide at S.T.P. can be obtained from 3 g of carbon?

Ans: CO2 + C \[\rightarrow\] 2CO

1V 1V 2V

12 g of C gives = 44.8 litre volume of CO

So, 3 g of C will give = 11.2 litre of CO

(b) 60 cm3 of oxygen was added to 24 cm3 of carbon monoxide and mixture ignited. Calculate:

(i) volume of oxygen used up and

(ii) Volume of carbon dioxide formed.

Ans: 2CO + O2 $\to$ 2CO2

2V 1V 2V

(i) 2 V CO requires oxygen = 1 V

so, 24 cm3 CO will require = 24/2 =12 cm3

(ii) 2 x 22400 cm3 CO gives = 2 $\times $ 22400 cm3 CO2

so, 24cm3 CO will give = 24 cm3 CO2

12. How much calcium oxide is obtained by heating 82 g of calcium nitrate? Also find the volume of NO2 evolved:

2Ca(NO3)2 $\to$ 2CaO+4NO2+O2

Ans: Molecular weight of 2Ca(NO3)2= 2[40+2(14+48)]

=328g

Molecular weight of CaO =2(40+16)

=112g

a. 328g of Ca(NO3)2 liberates 4 moles of NO2

Ans: 328g of Ca(NO3)2 liberates 4$\times $22.4L of NO2 , 82g will liberate

$\frac{~~~4\times 22.4~\times 82~~~}{328}$

=22.4dm3 of NO2

b. 328 g of calcium nitrate gives 112g of CaO

82 g will give $\frac{~~~112\times 82~~~}{328}$

=28 g of CaO

13.The equation for the burning of octane is:

2C8H18 + 25O2 $\to$ 16CO2 + 18H2O

2V 25V 16V 18V

(i) How many moles of carbon dioxide are produced when one mole of octane burns?

Ans: 2 moles of octane gives = 16 moles of CO2

so, 1 mole octane will give = 8 moles of CO2

(ii) What volume at S.T.P. is occupied by the number of moles determined in (i)?

Ans: 1 mole CO2 occupies volume = 22.4 litre

so, 8 moles will occupy volume = 8 $\times $22.4 = 179.2 litre

(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?

Ans: 1 mole CO2 has mass = 44 g

so, 16 moles will have mass = 44 $\times $16 = 704 g

(iv) What is the empirical formula of octane?

Ans: Empirical formula is C4H9.

14. Ordinary chlorine gas has two isotopes 3517Cl and 3717Cl in the ratio of 3:1. Calculate the relative atomic mass of chlorine.

Ans: The relative atomic mass of Cl \[=\frac{\text{(35}\times \text{3+1}\times \text{37)}}{4}=35.5\] amu

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.

Ans: 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 \[\times\] 100 = 20.82%

% of chlorine in the compound = 21.2/26.9 \[\times\] 100 = 79.18%

Element | % | Atomic mass | Atomic ratio | Simplest ratio |

Si | 20.82 | 28 | $\frac{~~~20.82~~}{28}$=0.74 | $\frac{~~~0.74~~}{0.74}$=1 |

Cl | 79.18 | 35 | $\frac{~~~79.18~~}{35.5~}$=2.23 | $\frac{~~2.23~~~}{0.74~~}$=3 |

So the empirical formula of the given compound =SiCl3

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.

Ans:

Element | % | Atomic mass | Atomic ratio | Simplest ratio |

P | 38.72 | 31 | $\frac{~~~38.72~~}{31}$=1.23 | $\frac{~~~1.23~~}{1.23}$=1 |

H | 2.47 | 1 | $\frac{~~~2.47~~}{1}$=2.47 | $\frac{~~2.47~~~}{1.23}$=2 |

O | 59.26 | 16 | $\frac{~~~59.26~~}{16}$=3.70 | $\frac{3.70~~}{1.23~~}$=3 |

So, empirical formula is PH2O3 or H2PO3

Empirical formula mass = 31+ 2 $\times $1 + 3 $\times $16 = 81

The molecular formula is = H4P2O6, because n = 162/81=2

17. a) Calculate the mass of substance 'A' which in gaseous form occupies 10 litres at 27 oC and 700 mm pressure. The molecular mass of 'A' is 60.

Ans: V1 = 10 litres V2=?

T1= 27+ 273 = 300KT2=273K

P1=700 mmP2 = 760 mm

Using the gas equation

\[ \frac{P_{1}V_{1}}{T_{1}} \] = \[ \frac{P_{2}V_{2}}{T_{2}}\]

\[V_{2}\] = \[ \frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}} \] = \[ \frac{700 \times 10 \times 273}{300 \times 760}\]

Molecular weight A= 60

So weight of 22.4 litres of A at STP=60g

Weight of A at STP=$\frac{~~~~700\times 10\times 273~~~}{300\times 760}\times \frac{~~~60~}{22.4}$= 22.45g

b) A gas occupied 360 cm3 at 870C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find its relative molecular mass.

Ans:

V$\times \frac{~~~760~}{273}$= $\frac{~~~360\times 380~~~}{360}$

V=$\frac{~~~360\times ~380~\times 273~~~}{760\times }$=136.5cm3

136.5cm3 of gas weigh= 0.546

22400cm3 of gas weight= $\frac{~~0.546~~}{136.5}\times $22400= 89.6 amu

Relative molecular mass= 89.6 amu

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?

Ans: Molecular mass of CO2 = 12+ 2$\times $16 = 44 g

So, vapour density (V.D) = mol. Mass/2 = 44/2 = 22

V.D = $\frac{\text{mass of certain amount of C}{{\text{O}}_{\text{2}}}}{\text{mass of equal volume of hydrogen}}$ = $\frac{\text{m}}{1}$

22 = $\frac{\text{m}}{1}$

So, mass of CO2 = 22 kg

(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

Ans: According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain equal numbers of molecules.

So, number of molecules of carbon dioxide in the cylinder =number of molecules of hydrogen in the cylinder = X

19. Following questions refer to one mole of chlorine gas.

(a) What is the volume occupied by this gas at S.T.P.?

Ans: The volume occupied by 1 mole of chlorine = 22.4 litre

(b) What will happen to the volume of gas, if pressure is doubled?

Ans: Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.

(c) What volume will it occupy at 273oC?

Ans: \[\frac{{{\text{V}}_{\text{1}}}}{{{\text{V}}_{2}}}=\frac{{{\text{T}}_{\text{1}}}}{{{\text{T}}_{2}}}\]

\[\frac{\text{22}\text{.4}}{{{\text{V}}_{2}}}=\frac{273}{546}\]

V2 = 44.8 litres

(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?

Ans: Mass of 1 mole Cl2 gas =35.5 $\times $ 2 =71 g

20. (a) A hydrate of calcium sulphate CaSO4.xH2O contains 21% water of crystallisation. Find the value of x.

Ans: Total molar mass of hydrated CaSO4.xH2O = 136+18x

Since 21% is water of crystallization, so

$\frac{~~~18x~~~~}{136+18x}$= $\frac{~~~21~~}{100}$

So, x = 2 i.e. water of crystallization is 2.

(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.

Ans: For 18 g water, vol. of hydrogen needed = 22.4 litre

So, for 1.8 g, vol. of H2 needed = 1.8 $\times $ \[\frac{\text{22}\text{.4}}{18}\] = 2.24 litre

Now 2 vols. of water = 1 vol. of oxygen

1 vol. of water =1/2 vol. of O2 = \[\frac{\text{22}\text{.4}}{2}\] = 11.2 lit.

18 g of water = 11.2 lit. of O2

1.8 g of water = \[\frac{11.2}{18}=\frac{18}{10}\] = 1.12 lit.

(c) How much volume will be occupied by 2g of dry oxygen at 270C and 740 mm pressure?

Ans: 32g of dry oxygen at STP = 22400cc

2g will occupy = \[\frac{\text{224002}}{32}\] = 1400cc

P1 = 760 mm P2 = 740mm

V1 = 1400cc V2 =?

T1 = 273 K, T2 = 27 +73 = 300K

$\frac{{{\text{P}}_{1}}{{\text{V}}_{1}}}{{{\text{T}}_{1}}}$= $\frac{{{\text{P}}_{2}}{{\text{V}}_{2}}}{{{\text{T}}_{2}}}$

${{\text{V}}_{2}}=\frac{{{\text{P}}_{1}}{{\text{V}}_{1}}{{\text{T}}_{2}}}{{{\text{T}}_{1}}{{\text{P}}_{2}}}$ = $\frac{760\times 1400\times 300}{273\times 740}$ = 1580cc

$\frac{1580}{1000}$ = 1.58 litres

(d) What would be the mass of CO2 occupying a volume of 44 litres at 250 C and 750 mm pressure?

Ans: P1= 750 mm P2=760 mm

V1= 44lit. V2=?

T1= 298K T2=273K

$\frac{{{\text{P}}_{1}}{{\text{V}}_{1}}}{{{\text{T}}_{1}}}$= $\frac{{{\text{P}}_{2}}{{\text{V}}_{2}}}{{{\text{T}}_{2}}}$

${{\text{V}}_{\text{2}}}\text{=}\frac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{2}}}}{{{\text{T}}_{\text{1}}}{{\text{P}}_{\text{2}}}}$ = $\frac{750\times 44\times 273}{298\times 760}$ = 39.78litre

22.4 litre of CO2 at STP has mass = 44g

39.78 litre of CO2 at STP has mass = $\frac{\text{44 }\!\!\times\!\!\text{ 39}\text{.78}}{\text{22}\text{.4}}$ = 78.14g

(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.

AgNO3 (aq) + NaCl (aq) $\to $AgCl (s) + NaNO3

Calculate the percentage of NaCl in the mixture.