SELINA Solutions for Class 10 Chemistry Chapter 5 - Mole Concept And Stoichiometry

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Chapter 5 - Mole Concept And Stoichiometry Exercise Ex. 5A

Solution 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.

Solution 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) N2 means 1 molecule of nitrogen and 2N means two atoms of nitrogen.

N2 can exist independently but 2N cannot exist independently.

Solution 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

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 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.

Solution 4

(a)2CO + O22CO2

     2 V    1 V     2 V

2 V of CO requires = 1V of O2

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


(b) 2H2 + O2   2H2O

       2 V         1V            2V

 

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

 

so 200cm3 of Hydrogen reacts with = 200/2= 100 cm3

 

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


Solution 5

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+2O2CO2 + 2H2O

1 V2 V

24 cc48 cc

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

Solution 6

2C2H2 + 5O2 4CO2 + 2H2O (l)

2 V 5 V 4 V

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

So, for 400ml C2H2 , O2 required = 400 5/2 =1000 ml

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

So, 400ml of C2H2 gives CO2 = 400 4/2 = 800ml

Solution 7

Balanced chemical equation:

(i)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) 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.

Solution 8

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

So, 300 cc of ethane reacts with

Hence, unused O2 = 1250 - 1050 = 200 cc

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

So, 300 cc of ethane will produce

Solution 9

  C2H4+3O22CO2 + 2H2O

  1V      3V

11litre   33 litre

 

Solution 10

CH4 + 2Cl2 CH2Cl2 +2HCl

1 V 2 V 1 V 2 V

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 2 = 80 ml of Cl2

Solution 11

C3H8 + 5O2 3CO2 + 4H2O

1 V 5 V 3 V

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

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

Solution 12

2NO + O2 2NO2

2 V 1 V 2 V

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

200cm3 oxygen will react with = 200 2 =400 cm3 NO

Hence, remaining NO is 450 - 400 = 50 cm3

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

Total mixture = 400 + 50 = 450 cm3

Solution 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.

Solution 14

4NH3 + 5O2 4NO + 6H2O

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

Solution 15

H2 + Cl2 2HCl

1V 1V 2 V

Since 1 V hydrogen requires 1 V 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

Solution 16

CH4 + 2O2 CO2 + 2H2O

1 V 2 V 1 V

2C2H2 + 5O2 4CO2 + 2H2O

2 V 5 V 4 V

From the equations, we can see that

1V CH4 requires oxygen = 2 V O2

So, 10cm3 CH4 will require =20 cm3 O2

Similarly 2 V C2H2 requires = 5 V 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

Solution 17

C3H8 + 5O2 3CO2 + 4H2O

2C4H10 + 13O2 8CO2 + 10H2O

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

40 ml of butane (C4H10) gives = 8 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.

Solution 18

2C2H2(g) + 5O2(g) 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

Solution 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 SO2 contains the least number of molecules since it has the smallest volume.

Solution 20

 

Gas

Volume (in litres)

Number of molecules

Chlorine 

10

x/2

Nitrogen

20

x

Ammonia

20

X

Sulphur dioxide

5

x/4

 

Solution 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. 

Chapter 5 - Mole Concept And Stoichiometry Exercise Ex. 5B

Solution 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 1023

c) The molar volume of a gas at STP is 22.4 dm3 at STP

Solution 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 dm3.

(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 moleculae of the substance is heavier than 1/12 of the mass of an atom of carbon-12.

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

(f) The 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.

Solution 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.

H2 + Cl2 ? 2HCl

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

n molecules n molecules 2n molecules (By Avogadros law)

Solution 4

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

(b) KClO3 = (K)39 + (Cl)35.5 + (3O)48 = 122.5

(c) (Cu)63.5 + (S)32 + (4O)64 + (5H2O)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

Solution 5

(a) No. of molecules in 73 g HCl = 6.023 x1023 x 73/36.5(mol.

mass of HCl)

= 12.04 x 1023

(b) Weight of 0.5 mole of O2 is = 32(mol. Mass of O2) x 0.5=16 g

(c) No. of molecules in 1.8 g H2O = 6.023 x 1023 x 1.8/18

= 6.023 x 1022

(d) No. of moles in 10g of CaCO3 = 10/100(mol. Mass CaCO3)

= 0.1 mole

(e) Weight of 0.2 mole H2 gas = 2(Mol. Mass) x 0.2 = 0.4 g

(f) No. of molecules in 3.2 g of SO2 = 6.023 x 1023 x 3.2/64

= 3.023 x 1022

Solution 6

Molecular mass of H2O is 18, CO2 is 44, NH3 is 17 and CO is 28

So, the weight of 1 mole of CO2 is more than the other three.

Solution 7

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

Solution 8

a) No. of particles in s1 mole = 6.023 x 1023

So, particles in 0.1 mole = 6.023 x 10 23 x 0.1 = 6.023 x 1022

b)1 mole of H2SO4 contains =2 x 6.023 x 1023

So, 0.1 mole of H2SO4 contains =2 x 6.023 x 1023 x0.1

= 1.2x1023 atoms of hydrogen

c)111g CaCl2 contains = 6.023 x 1023 molecules

So, 1000 g contains = 5.42 x 1024 molecules

Solution 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 H2O has mass = 0.2 x 18 = 3.6 g

 

(d) 0.1 mole of CO2 has mass = 0.1 x 44 = 4.4 g

Solution 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 SO2 has volume = 22.4 litres

So, 2 moles will have = 22.4 x 2 = 44.8 litre

Solution 11

(a) 1 mole of CO2 contains O2 = 32g

So, CO2 having 8 gm of O2 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

Solution 12

(a) 6.023 x 10 23 atoms of oxygen has mass = 16 g

So, 1 atom has mass = 16/6.023 x 1023 = 2.656 x 10-23 g

(b) 1 atom of Hydrogen has mass = 1/6.023 x 1023 = 1.666 x 10-24

(c) 1 molecule of NH3 has mass = 17/6.023 x1023 = 2.82 x 10-23 g

(d) 1 atom of silver has mass = 108/6.023 x 1023 =1.701 x 10-22

(e) 1 molecule of O2 has mass = 32/6.023 x 1023 = 5.314 x 10-23 g

(f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g

Solution 13

(a) 0.1 mole of CaCO3 has mass =100(molar mass) x 0.1=10 g

(b) 0.1 mole of Na2SO4.10H2O has mass = 322 x 0.1 = 32.2 g

(c) 0.1 mole of CaCl2 has mass = 111 x 0.1 = 11.1g

(d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g

Solution 14

(a) 1molecule of Na2CO3.10H2O contains oxygen atoms = 13

 

So, 6.023 x1023 molecules (1mole) has atoms=13 x 6.023 x 1023

 

So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 1023 =7.8x1023

(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

 


                                                              

Solution 15

3.2 g of S has number of atoms = 6.023 x1023 x 3.2 /32

= 0.6023 x 1023

So, 0.6023 x 1023 atoms of Ca has mass=40 x0.6023x1023/6.023

x1023

= 4g

Solution 16

(a) No. of atoms = 52 x 6.023 x1023 = 3.131 x 1025

(b) 4 amu = 1 atom of He

so, 52 amu = 13 atoms of He

(c) 4 g of He has atoms = 6.023 x1023

So, 52 g will have = 6.023 x 1023 x 52/4 = 7.828 x1024 atoms

Solution 17

Molecular mass of Na2CO3 = 106 g

106 g has 2 x 6.023 x1023 atoms of Na

So, 5.3g will have = 2 x 6.023 x1023x 5.3/106=6.022 x1022 atoms

Number of atoms of C = 6.023 x1023 x 5.3/106 = 3.01 x 1022 atoms

And atoms of O = 3 x 6.023 x 1023 x 5.3/106= 9.03 x1022 atoms

Solution 18

(a) 60 g urea has mass of nitrogen(N2) = 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

Solution 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.

Solution 20

22400 cm3 of CO has mass = 28 g

So, 56 cm3 will have mass = 56 x 28/22400 = 0.07 g

Solution 21

18 g of water has number of molecules = 6.023 x 1023

So, 0.09 g of water will have no. of molecules = 6.023 x 1023 x 0.09/18 = 3.01 x 1021 molecules

Solution 22

(a) No. of moles in 256 g S8 = 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 1023 = 1.2 x 1022 molecules

No. of atoms in 1 molecule of S = 8

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

= 9.635x 1022 molecules

Solution 23

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

Solution 24

(a) 308 cm3 of chlorine weighs = 0.979 g

So, 22400 cm3 will weigh = gram molecular mass

= 0.979 x 22400/308 =71.2 g

(b) 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 =

(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

Solution 25

No. of atoms in 12 g C = 6.023 x1023

So, no. of carbon atoms in 10-12 g = 10-12 x 6.023 x1023/12

= 5.019 x 1010 atoms

Solution 26

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

V1/T1 = V2/ T2

V2 = (V1 x T2)/T1

      = (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  V1 = 0.5 L

P2 = 760 mm Hg  V2 = ?

P1 x V1 = P2 x V2

V2 = (P1 x V1)/P2

      = (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

Solution 27

1000 g of sugar costs = Rs. 40

So, 342g(molar mass) of sugar will cost=342x40/1000=Rs. 13.68

Solution 28

(a) Weight of 1 g atom N = 14 g

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

(b) 6.023 x1023 atoms of C weigh = 12 g

So, 3 x1025 atoms will weigh =

(c) 1 mole of sulphur weighs = 32 g

(d) 7 g of silver

So, 7 grams of silver weighs least.

Solution 29

Option C is correct.

40 g of NaOH contains 6.023 x 1023 molecules

So, 4 g of NaOH contains = 6.02 x1023 x 4/40

= 6.02 x1022 molecules

Solution 30

The number of molecules in 18 g of ammonia= 6.02 x1023

So, no. of molecules in 4.25 g of ammonia = 6.02 x1023 x 4.25/18

= 1.5 x 1023

Solution 31

(a) One mole of chlorine contains 6.023 x 1023 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.

Chapter 5 - Mole Concept And Stoichiometry Exercise Ex. 5C

Solution 1

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)That molecular weight of H2O is 18g.

Solution 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.

Solution 3

(a) CH (b) CH2O (c) CH (d) CH2O

Solution 4

Solution 5

(a) Molecular mass of Ca(H2PO4)2 = 234

So, % of P = 2 31 100/234 = 26.5%

(b) Molecular mass of Ca3(PO4)2 = 310

% of P = 2 31 100/310 = 20%

Solution 6

Molecular mass of KClO3 = 122.5 g

% of K = 39 /122.5 = 31.8%

% of Cl = 35.5/122.5 = 28.98%

% of O = 3 16/122.5 = 39.18%

Solution 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(NO3)2 is the empirical formula.

Solution 8

In Fe2O3 , Fe = 56 and O = 16

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

Iron present in 80% of Fe2O3 =

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

Solution 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 =C2H2

 

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

n = 78/13=6

So, the molecular formula = C6H6

Solution 10

Element % At. mass Atomic ratio Simple ratio

H 17.7 1

N 82.3 14

So, the empirical formula = NH3

Solution 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 = C2H4O

(b) empirical formula mass = 44

Since, vapour density = 44

So, molecular mass = 2 V.D = 88

Or n = 2

so, molecular formula = (C2H4O)2 = C4H8O2

Solution 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 = CHO2

(b) empirical formula mass = 45

Vapour density = 45

So, molecular mass = V.D 2 = 90

so, molecular formula = C2H2O4

Solution 13

Element%at. massatomic ratiosimple ratio

Cl71.6535.51

H4.0712

C24.28121

 

(a) its empirical formula = CH2Cl

(b) empirical formula mass = 49.5

Since, molecular mass = 98.96

so, molecular formula = (CH2Cl)2 = C2H4Cl2

Solution 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 = C2H5

(c) Empirical formula mass = 29

Molecular mass = V.D 2 = 29 2 = 58

So, molecular formula = C4H10

Solution 15

Since, g atom of Si = given mass/mol. Mass

so, given mass = 0.2 28 = 5.6 g

ElementmassAt. massGram atomRatio

Si5.6280.21

Cl21.335.53

 

Empirical formula = SiCl3

Solution 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 = C2H5

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 C2H5

 = C4H10

Solution 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 Mg3 N2

Solution 18

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] 14.8/100 = 18x

=[104 + 9x] 2148=18000x

=[104+9x] 37=250x

=3848 + 333x =2250x

1917x =3848

x = 2molecules of water

Solution 19

Molar mass of urea; CON2H4 = 60 g

So, % of Nitrogen = 28 100/60 = 46.66%

Solution 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 CH2O

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

C12 H24 O12.

Solution 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 A2B4.

(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 A6B6

(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 AB4

 

Solution 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 NH2

Solution 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 =ZnSH14O11

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 = ZnSO11H14

=ZnSO4.7H2O

Chapter 5 - Mole Concept And Stoichiometry Exercise Ex. 5D

Solution 1

(a) Moles:1 mole + 2 mole 1 mole + 2 mole

(b) Grams: 42g + 36g 74g + 4 g

(c) Molecules = 6.02 1023 + 12.046 1023 6.02 1023+ 12.046 1023

Solution 2

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

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

(b) 1 V of CaCO3 produces 1 V of CO2

100 g of CaCO3 has volume = 22.4 litres

So, 15 g will have volume = 22.415/100 = 3.36 litres CO2

Solution 3

2NH3 + H2SO4 (NH4)2SO4

66 g

 

(a) 2NH3 + H2SO4 (NH4)2SO4

34 g98 g132 g

 

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) 17g of NH3 requires volume = 22.4 litres

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

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

Solution 4

(a) 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) 685g of Pb3O4 gives = 71g of Cl2

Hence, 6.85 g of Pb3O4 will give = 6.85 71/685 = 0.71 g Cl2

 

(c) 1 V Pb3O4produces 1 V Cl2

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

So, 6.85 Pb3O4 will produce = 6.85 22.4/685 = 0.224 litres of Cl2

Solution 5

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 101/63 = 202 kg

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

So, smaller mass of NaNO3 is required.

Solution 6

CaCO3 + 2HClCaCl2 + H2O + CO2

100g73g22.4L

(a)V1 =2 litresV2 =?

T1 = (273+27)=300KT2=273K

V1/T1=V2/T2

V2=V1T2/T1=

Now at STP 22.4 litres of CO2 are produced using CaCO3 =100g

So, litres are produced by =100/22.4 2274/300 =.125g

(b)22.4 litres are CO2 are prepared from acid =73g

litres are prepared from = 73/22.4 2273/300=5.9g

Solution 7

2H2O2H2 + O2

2 V2 V1 V

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 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.

Solution 8

2Na2O2 + 2H2O4NaOH + O2

2 V4 V1 V

(a) Mol. Mass of Na2O2 = 2 23 + 2 16 = 78 g

Mass of 2Na2O2= 156 g

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

So, 1.56 Na2O2 will give = 160 1.56/156 = 1.6 g

(b) 156 g Na2O2 gives = 22.4 litres of oxygen

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

= 224 cm3

(c)156 g Na2O2 gives = 32 g O2

So, 1.56 g Na2O2 will give = 32 1.56/156

= 32/100 = 0.32 g

Solution 9

2NH4Cl + Ca(OH)2CaCl2+2H2O + 2NH3

2 V1 V1 V2 V

Mol. Mass of 2NH4Cl = 2[14 + (1 4) + 35.5] = 2[53.5] = 107 g

(a) 107 g NH4Cl gives = 34 g NH3

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

(b) The volume of 17 g NH3 is 22.4 litre

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

Solution 10

  

Solution 11

MnO2 + 4HCl MnCl2 + 2H2O +Cl2

1 V4 V1 V1 V

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

So, 0.02 mole will weigh = 87 0.02 = 1.74 g MnO2

(b) 1 mole MnO2 gives = 1 mole of MnCl2

So, 0.02 mole MnO2will give =0.02 mole of MnCl2

 

(c) 1 mole MnCl2 weighs = 126 g(mol mass)

So, 0.02 mole MnCl2 will weigh = 126 0.02 g = 2.52 g

(d) 0.02 mole MnO2will form =0.02 mole of Cl2

(e) 1 mole of Cl2 weighs = 35.5 g

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

(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 MnO2requires HCl = 4 mole

So, 0.02 mole MnO2 will require =4 0.02 = 0.08 mole

(h) For 1 mole MnO2 , 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

Solution 12

N2 + 3H2 2NH3

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 H2

(b)28g of nitrogen forms NH3 = 34g

2000g of N2 forms NH3

= 34/28 2000

=2428.6g

Chapter 5 - Mole Concept And Stoichiometry Exercise Misc. Ex.

Solution 1

From equation: 2H2 + O2 2H2O

1 mole of Oxygen gives = 2 moles of steam

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

Solution 2

3Cu + 8HNO3 3Cu (NO3)2+ 4H2O + 2NO

1 V8 V3 V2 V

 

Mol. Mass of 8HNO3 = 8 63 = 504 g

 

(a) For 504 g HNO3, Cu required is = 192 g

So, for 63g HNO3Cu required = 192 63/504 = 24g

 

(b) 504 g of HNO3 gives = 2 22.4 litre volume of NO

So, 63g of HNO3 gives =2 22.4 63/504 =5.6 litre of NO

Solution 3

(a) 28g of nitrogen = 1mole

So, 7g of nitrogen = 1/28 7= 0.25 moless

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

So, 56cm3 volume will have mass =28 56/22400= 0.07 g

Solution 4

% of N in NaNO3=

% of N in (NH4)2SO4 =

% of N in CO(NH2)2 =

So, highest percentage of N is in urea.

Solution 5

2H2O2H2+O2

2 V2 V1 V

 

(a) 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 = 2V/V = 1250 cm3

 

(b)

 

(c)

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

Solution 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

Solution 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 = CH3

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 CH3

 = C2H6

Solution 8

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) C and H

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

(c)

(i) no. of moles of CO2= 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 = C2H5

Solution 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 = XO2

Solution 10

(a) V.D =

(b) Molecular mass = 17(V.D) x 2= 34g

Solution 11

(a) CO2 + 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

(b) 2CO + O2 2CO2

2 V 1 V 2 V

(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 x 22400 cm3 CO2

so, 24cm3 CO will give = 24 cm3 CO2

Solution 12

Molecular weight of   =

 =328g

 Molecular weight of CaO =2(40+16)

 =112g

 

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

328g of Ca(NO3)2 liberates  L of NO2   82g will liberate 

 =22.4dm3 of NO2

 

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

82 g will give 

 =28 g of CaO

Solution 13

2C8H18 + 25O216CO2 + 18H2O

2 V25 V16 V18 V

(i) 2 moles of octane gives = 16 moles of CO2

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

(ii) 1 mole CO2 occupies volume = 22.4 litre

so, 8 moles will occupy volume = 8 22.4 = 179.2 litre

(iii) 1 mole CO2 has mass = 44 g

so, 16 moles will have mass = 44 16 = 704 g

(iv) Empirical formula is C4H9.

Solution 14

The relative atomic mass of Cl = (35 3 + 1 37)/4=35.5 amu

Solution 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 =SiCl3

Solution 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 PH2O3 or H2PO3

Empirical formula mass = 31+ 2 1 + 3 16 = 81

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

Solution 17

a) V1 = 10 litres V2=?

T1= 27+ 273 = 300KT2=273K

P1=700 mmP2 = 760 mm

Using the gas equation

b)

Solution 18

(a) Molecular mass of CO2 = 12+ 2x16 = 44 g

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

V.D =

So, mass of CO2 = 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

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) V1/V2 = T1/T2

22.4/V2 =273/546

V2 = 44.8 litres

(d) Mass of 1 mole Cl2 gas =35.5 x 2 =71 g

Solution 20

(a) Total molar mass of hydrated CaSO4.xH2O = 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 H2 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 O2 =22.4/2=11.2 lit.

18 g of water = 11.2 lit. of O2

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

P1=760mm P2 =740mm

V1=1400cc V2 =?

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

(d) P1= 750mm P2=760mm

V1= 44lit. V2=?

T1= 298K T2=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%

Solution 21

a. 

 For CO2 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 SO2

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 = CHCl2

 

 ii. 

Empirical formula = CHCl2

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 CHCl2

   = C2H2Cl4

Solution 22

a. Relative molecular mass of [Mg (NO3) 6H2O]

 =24+14+(3 x16)+(6 x18)=194

   Since, 194g of [Mg (NO3) 6H2O] contains 144g of oxygen

  100g of [Mg (NO3) 6H2O] contains   of oxygen = 74.22%  

b. Relative molecular mass of Na2B4O7.10H2O

 (23 × 2) + (4 × 11) + (7 × 16) + 10(18) = 382

 Since 382g of Na2B4O7.10H2O contains 44g of boron

 100g Na2B4O7.10H2O of contains  of boron

   =11.5%

c. Relative molecular mass of Ca(H2PO4)2

  = 40 + 2(2 + 31 + 64) = 234

 Since,234g of Ca(H2PO4)2 contains 62g of phosphorus 

  100g of Ca(H2PO4)2 contains

 =26.5%

Solution 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 dm3 of N2

 63 g of ammonium dichromate gives 

  =5.6 L

 

 

 = 0.25 moles  

(c)

 252 g of ammonium dichromate gives 22.4 dm3 of N2

 63 g of ammonium dichromate gives 

 = 5.6 L

(d)

 Number of moles=

 =

 =0.25moles

 

 0.25 moles of ammonium dichromate gives

 0.25 moles of N2=7 g

 1 mole of H2O =18 g

Therefore, total loss of mass=7+18

 =25 g

(e)

 252 g of ammonium dichromate gives 152 g of CrO3

 63 g of ammonium dichromate gives 

  =38 g

Solution 25

2H2S + 3O2 2H2O + 2SO2

2 V 3 V 2 V

128 g of SO2 gives = 2 22.4 litres volume

So, 12.8 g of SO2 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 H2S requires oxygen = 3 22.4 litre

So, 4.48 litres H2S will require = 6.72 litre of oxygen

Solution 26

From equation, 2NH3 + 2 O2 2NO + 3H2O

When 60 g NO is formed, mass of steam produced = 54 g

So, 1.5 g NO is formed, mass of steam produced = 54 1.5/60

=1.35 g

Solution 27

In 1 hectare of soil, N2 removed = 20 kg

So, in 10 hectare N2 removed = 200 kg

The molecular mass of Ca(NO3)2 =164

Now, 28 g N2 present in fertilizer = 164 g Ca(NO3)2

So, 200000 g of N2 is present in = 164 200000/28

= 1171.42 kg

Solution 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 HNO3 = 315 g

so, 6.2 g P will react with HNO3 = 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 2730C=4.48 2=8.96litre

Solution 29

112cm3 of gaseous fluoride has mass = 0.63 g

so, 22400cm3 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 = PF5

Solution 30

Na2CO3.10H2O Na2CO3 + 10H2O

286 g 106 g

So, for 57.2 g Na2CO3.10H2O = 106 57.2/286 = 21.2 g Na2CO3

Solution 31

Simple ratio of M = 34.5/56 = 0.616 = 1

Simple ratio of Cl = 65.5/35.5 = 1.845 = 3

Empirical formula = MCl3

Empirical formula mass = 162.5, Molecular mass = 2 V.D = 325

So, n = 2

So, molecular formula = M2Cl6

Solution 32

(i) Element%atomic massatomic ratiosimple ratio

C4.8121

Br95.2803

 

So, empirical formula is CBr3

 

(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 = C2Br6

Solution 33

4N2O + CH4 CO2 + 2H2O + 4N2

4 V 1 V 1 V 2 V 4 V

2 x 22400 litre steam is produced by N2O = 4 x 22400 cm3

So, 150 cm3 steam will be produced by= 4 22400 150/2 x 22400

= 300 cm3 N2O

Solution 34

(a) Volume of O2 = V

Since O2 and N2 have same no. of molecules = x

so, the volume of N2 = V

(b) 3x molecules means 3V volume of CO

(c) 32 g oxygen is contained in = 44 g of CO2

So, 8 g oxygen is contained in = 44 x 8/32 = 11 g

(d) Avogadro's law is used in the above questions.

Solution 35

simple ratio of Na = 42.1/23 = 1.83 = 3

simple ratio of P = 18.9/31 = 0.609 = 1

simple ratio of O = 39/16 = 2.43 = 4

So, the empirical formula is Na3PO4

Solution 36

CH4 + 2O2 CO2 + 2H2O

1 V 2 V 1 V 2 V

From equation:

22.4 litres of methane requires oxygen = 44.8 litres O2

2H2 + O2 2H2O

2 V 1 V 2 V

From equation,

44.8 litres hydrogen requires oxygen = 22.4 litres O2

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

Solution 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 1023 molecules

Mol. Mass of H2 (2),O2(32) ,CO2(44),SO2(64),Cl2(71)

(1)Now 2 g of hydrogen contains molecules =6.02 1023

So, 8g of hydrogen contains molecules = 8/2 6.02 1023

=4 6.02 1023 = 4M molecules

(2)32g of oxygen contains molecules = 8/32 6.02 1023=M/4

(3)44g of carbon dioxide contains molecules = 8/44 6.02 1023=2M/11

(4)64g of sulphur dioxide contains molecules =6.02 1023

So, 8g of sulphur dioxide molecules = 8/64 6.02 1023= M/8

(5)71 g of chlorine contains molecules =6.02 1023

So, 8g of chlorine molecules = 8/72 6.02 1023 = 8M/71

Since 8M/71<M/8<2M/11<M/4<4M

Thus Cl2<SO2<CO2<O2<H2

(i)Least number of molecules in Cl2

(ii)Most number of molecules in H2

Solution 38

Na2SO4 + BaCl2 BaSO4 + 2NaCl

Molecular mass of BaSO4 = 233 g

Now, 233 g of BaSO4 is produced by Na2SO4 = 142 g

So, 6.99 g BaSO4 will be produced by = 6.99 142/233 = 4.26

The percentage of Na2SO4 in original mixture = 4.26 100/10

= 42.6%

Solution 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) 2KMnO4 K2MnO4 + MnO2 + O2

316 g of KMnO4 gives oxygen = 24 litres

So, 15.8 g of KMnO4 will give = 24 316/15.8 = 1.2 litres

Solution 40

(a)

(i) The no. of moles of SO2 = 3.2/64 = 0.05 moles

(ii) In 1 mole of SO2, no. of molecules present = 6.02 1023

So, in 0.05 moles, no. of molecules = 6.02 1023 0.05

= 3.0 1022

(iii) The volume occupied by 64 g of SO2 = 22.4 dm3

3.2 g of SO2 will be occupied by volume = 22.4 3.2/64 =1.12 dm3

(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 = PbCl4

Solution 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.

V1/V2 = n1/n2

V1/V2 = n1/2n1

So, V2 = 2V1

(iii) Gay lussac's law of combining volume is being observed.

(iv) The volume of D = 5.6 4 = 22.4 dm3, so the number of molecules = 6 x 1023 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 N2O = 1 44 = 44 g

Solution 42

(a) NaCl+NH3+ CO2 + H2ONaHCO3+NH4Cl

2NaHCO3 Na2CO3+H2O + CO2

From equation:

106 g of Na2CO3 is produced by = 168 g of NaHCO3

So, 21.2 g of Na2CO3 will be produced by = 168 21.2/106

= 33.6 g of NaHCO3

(b) For 84 g of NaHCO3, requiredvolume of CO2 = 22.4 litre

So, for 33.6 g of NaHCO3, required volume of CO2 = 22.4 x 33.6/84

= 8.96 litre

Solution 43

(a) NH4NO3N2O+2H2O

1mole1mole2mole

1 V1 V2 V

 

44.8 litres of water produced by = 22.4 litres of NH4NO3

So, 8.96 litres will be produced by = 22.4 x 8.96/44.8

= 4.48 litres of NH4NO3

So, 4.48 litres of N2O is produced.

 

(i) 44.8 litre H2O is produced by = 80 g of NH4NO3

So, 8.96 litre H2O will be produced by = 80 x 8.96/44.8

= 16g NH4NO3

 

(iii) % of O in NH4NO3 = 3x16/80 = 60%

Solution 44

Molecular mass of   =(240) g

Molecular mass of   = 2 × 22.4 = 44.8dm3

Molecular mass of   = (192)g

240 g of CuO requires 44.8 dm3 of NH3

 120g of CuO will require   

 =22.4dm3

Solution 45

(a) The molecular mass of ethylene(C2H4) is 28 g

No. of moles = 1.4/28 = 0.05 moles

No. of molecules = 6.023 x1023 x 0.05 = 3 x 1022 molecules

Volume = 22.4 x 0.05 = 1.12 litres

(b) Molecular mass = 2 X V.D

S0, V.D = 28/2 = 14

Solution 46

(a) Molecular mass of Na3AlF6 = 210

So, Percentage of Na = 3x23x100/210 = 32.85%

(b) 2CO + O2 2CO2

2 V 1 V 2 V

1 mole of O2 has volume = 22400 ml

Volume of oxygen used by 2 x 22400 ml CO = 22400 ml

So, Vol. of O2 used by 560 ml CO =22400 x 560/(2 x 22400)

= 280 ml

So, Volume of CO2 formed is 560 ml.

Solution 47

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

  200cm3 requires 2.5×200=500 cm3 of oxygen

  2 volume of acetylene on combustion gives 4CO2 

   1 volume of acetylene on combustion gives 2CO2 

  200cc of acetylene on combustion will give  200×2=400cc of CO2

 ii. 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 NH2

 Empirical formula weight =14+2=16

 Relative molecular mass =37

 N =  2.3≈2

 Molecular formula = n x empirical formula  = 2 x NH2

  =N2H4

c.  

 i. Molecules of nitrogen gas in a cylinder = 24 x 1024

 Avogadro's number = 6 x 1023

1. Mass of nitrogen in a cylinder  =

 =1120g

2. Volume of nitrogen at stp

 

Volume of 28 g of N2  = 22.4dm3 

Volume of 1120g of N2 =   dm3

   =896 dm3

Solution 48

a. 

 i. 10 litres of LPG contains 

 Propane

 Butane 

 

   

  18+16=34 L

 ii. Molecular mass of NH4(NO3) =80

 H=1, N=14, O=16

 % of Nitrogen

 As 80 g of NH4(NO3) contains 28 g of nitrogen

   100 g of of NH4(NO3) will contain   

 = 35%

 % of Oxygen

 As,80 g of NH4(NO3) contains 48 g of oxygen

   100 g of of NH4(NO3) 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× Relative molecular mass 

 = 4.5×100

 = 450g

 iii. 

As, 100g of calcium carbonate gives 22.4dm3 of CO2

  450 g of calcium carbonate will give

 =100.8 L

 iii. 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 

 iv. 

 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

Solution 49

a. 

 i. Atomic mass: S = 32 and O = 16

 Molecular mass of SO2=32+(2×16)

  =64g

 As 64 g of SO2 = 22.4dm3

 Then, 320 g of SO2 =  

 =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. C3H8 + 5O23CO2 + 4H2O

 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 = CH2Br 

n(Empirical formula mass of CH2Br) = Molecular mass (2 × VD)

n(12 + 2 + 80) = 94 × 2

  n = 2

 Molecular formula = Empirical formula × 2

 = (CH2Br) × 2

  = C2H4Br2

 ii. 1022 atoms of sulphur

 6.022 × 1023 atoms of sulphur will have mass = 32 g

 1022 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

Solution 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 NO gas at STP.

  So, 0.3 mole of phosphorus gives (112 × 0.3) L of

 NO gas at STP.

 = 33.6 L of NO gas 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 dm3 of gas weighs 12 g

 1 dm3 of gas weighs = (12/56) gm

 22.4 dm3 of gas weighs = (12/56 × 22.2) gm = 48g

 Therefore, the relative molecular mass of gas = 48 gm.

 iii. Molar mass of Mg (NO3)2.6H2O

 = 24 × (14 × 2) + (16 × 12) + (1 × 12) = 256 g

 Mass percent of magnesium =   

Solution 51

a.  

 i. 

  2 vols. of butane requires O2 = 13 vols

 90 dm3 of butane will require O2 =  × 90

  = 585 dm3

 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 KClO3 = 2 vols. 

 So, 1 vol. of oxygen will require KClO3 =

 So, 6.72 litres of oxygen will require KClO3

 =

 22.4 litres of KClO3 has mass = 122.5 g

 So, 4.48 litres of KClO3 will have mass

 = 

 ii. 22.4 litres of oxygen = 1 mole

 So, 6.72 litres of oxygen =

 No. of molecules present in 1 mole of O2

 = 6.023 × 1023

 So, no. of molecules present in 0.3 mole of O2

 = 6.023 × 1023 × 0.3

  = 1.806 × 1023

 iii. Volume occupied by 1 mole of CO2 at STP = 22.4  litres 

So, volume occupied by 0.01 mole of CO2 at STP = 22.4 × 0.01= 0.224 litres

Solution 52

a.  

 i. 

 2 moles of C2H2=4moles of CO2

 x dm3 of C2H2  =8.4 dm3 of CO2

 x=

 =4.2 dm3 of C2H2  

 ii. Empirical formula= X2Y

  Atomic weight (X)= 10

  Atomic weight (Y)= 5 

  Empirical formula weight = (2 × 10) + 5

 =25

   

  = 2

 So, molecular formula = X2Y×2

 = X4Y2    

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 dm3

  4 mole = 22.4 × 4 = 89.6 dm3 

 ii. 4 moles of ammonia gas is present in the cylinder.

 iii. 1 mole = 6.023 × 1023 molecules

  4 moles = 24.092 × 1023 molecules