Class 10 SELINA Solutions Chemistry Chapter 5  Mole Concept And Stoichiometry
Mole Concept And Stoichiometry Exercise Ex. 5A
Solution 1
(a)2CO + O_{2}_{}2CO_{2}
2 V 1 V 2 V
2 V of CO requires = 1V of O_{2}
so, 100 litres of CO requires = 50 litres of O_{2}
(b) 2H_{2 }+ O_{2}_{ }_{ }2H_{2}O
2 V 1V 2V
From the equation, 2V of hydrogen reacts with 1V of oxygen
so 200cm^{3} of Hydrogen reacts with = 200/2= 100 cm^{3}
Hence, the unreacted oxygen is 150  100 = 50cm^{3 }of oxygen.
Solution 2
This experiment supports Gay lussac's law of combining volumes.
Since the unchanged or remaining O_{2} is 58 cc so, used oxygen 106  58 = 48cc
According to Gay lussac's law, the volumes of gases reacting should be in a simple ratio.
CH_{4}+2O_{2}_{}CO_{2} + 2H_{2}O
1 V2 V
24 cc48 cc
i.e. methane and oxygen react in a 1:2 ratio.
Solution 3
2C_{2}H_{2}+ 5O_{2}_{}4CO_{2} + 2H_{2}O (l)
2 V5 V4 V
From equation, 2 V of C_{2}H_{2 }requires = 5 V of O_{2}
So, for 400ml C_{2}H_{2 }, O_{2} required = 400 _{}5/2 =1000 ml
Similarly, 2 V of C_{2}H_{2} gives = 4 V of CO_{2}
So, 400ml of C_{2}H_{2} gives CO_{2} = 400 _{}4/2 = 800ml
Solution 4
Balanced chemical equation:
_{}
(i)At STP, 1 mole gas occupies 22.4 L.
As 1 mole H_{2}S gas produces 2 moles HCl gas,
22.4 L H_{2}S gas produces 22.4 × 2 = 44.8 L HCl gas.
Hence, 112 cm^{3} H_{2}S gas will produce 112 × 2 = 224 cm^{3} HCl gas.
(ii) 1 mole H_{2}S gas consumes 1 mole Cl_{2} gas.
This means 22.4 L H_{2}S gas consumes 22.4 L Cl_{2} gas at STP.
Hence, 112 cm^{3} H_{2}S gas consumes 112 cm^{3} Cl_{2} gas.
120 cm^{3}  112 cm^{3 }= 8 cm^{3 }Cl_{2} gas remains unreacted.
Thus, the composition of the resulting mixture is 224 cm^{3}HCl gas + 8 cm^{3} Cl_{2} gas.
Solution 5
From the equation, 2V of ethane reacts with 7V oxygen.
So, 300 cc of ethane reacts with
Hence, unused O_{2} = 1250  1050 = 200 cc
From 2V of ethane, 4V of CO_{2} is produced.
So, 300 cc of ethane will produce
Solution 6
C_{2}H_{4}+3O_{2}2CO_{2} + 2H_{2}O
1V 3V
11litre 33 litre
Solution 7
CH_{4} + 2Cl_{2} _{}CH_{2}Cl_{2 }+2HCl
1 V2 V1 V2 V
From equation, 1V of CH_{4} gives = 2 V HCl
so, 40 ml of methane gives = 80 ml HCl
For 1V of methane= 2V of Cl_{2} required
So, for 40ml of methane = 40 _{}2 = 80 ml of Cl_{2}
Solution 8
C_{3}H_{8 }+ 5O_{2} _{}3CO_{2} + 4H_{2}O
1 V5 V3 V
From equation, 5 V of O_{2} required = 1V of propane
so, 100 cm^{3} of O_{2}will require = 20 cm^{3} ofpropane
Solution 9
2NO + O_{2}_{}2NO_{2}
2 V1 V2 V
From equation, 1V of O_{2} reacts with = 2 V of NO
200cm^{3} oxygen will react with = 200 _{}2 =400 cm^{3 }NO
Hence, remaining NO is 450  400 = 50 cm^{3}
NO_{2} produced = 400cm^{3} because 1V oxygen gives 2 V NO_{2}
Total mixture = 400 + 50 = 450 cm^{3}
Solution 10
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 11
4NH_{3} + 5O_{2}_{}4NO + 6H_{2}O
4 V5 V4 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 12
H_{2} + Cl_{2 }2HCl
1V1V2 V
Since 1 V hydrogen requires 1 V of oxygen and 4cm^{3} of H_{2} remained behind so the mixture had composition: 16 cm^{3} hydrogen and 16 cm^{3 }chlorine.
Therefore Resulting mixture is H_{2} =4cm^{3},HCl=32cm^{3}
Solution 13
CH_{4} + 2O_{2 }_{}CO_{2} + 2H_{2}O
1 V2 V1 V
2C_{2}H_{2} + 5O_{2} _{}4CO_{2} + 2H_{2}O
2 V5 V4 V
From the equations, we can see that
1V CH_{4}requires oxygen =2 V O_{2}
So, 10cm^{3} CH_{4} will require =20 cm^{3} O_{2}
Similarly 2 V C_{2}H_{2} requires = 5 V O_{2}
So, 10 cm^{3} C_{2}H_{2 }will require= 25 cm^{3} O_{2}
Now, 20 V O_{2} will be present in 100 V air and 25 V O_{2} will be present in 125 V air ,so the volume of air required is 225cm^{3 }
Solution 14
C_{3}H_{8} + 5O_{2}_{}3CO_{2} + 4H_{2}O
2C_{4}H_{10} + 13O_{2} _{}8CO_{2} + 10H_{2}O
60 ml of propane (C_{3}H_{8}) gives 3 _{}60 = 180 ml CO_{2}
40 ml of butane (C_{4}H_{10}) gives = 8 _{}40/2 = 160 ml of CO_{2}
Total carbon dioxide produced = 340 ml
So, when 10 litres of the mixture is burnt = 34 litres of CO_{2 }is produced.
Solution 15
2C_{2}H_{2}(g) + 5O_{2}(g)_{}4CO_{2}(g)+ 2H_{2}O(g)
4 V CO_{2} is collected with 2 V C_{2}H_{2}
So, 200cm^{3} CO_{2 }will be collected with = 100cm^{3} C_{2}H_{2}
Similarly, 4V of CO_{2} is produced by 5 V of O_{2}
So, 200cm^{3} CO^{2 }will be produced by = 250 ml of O_{2}
Solution 16
According to Avogadro's law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and least volume will contain least molecules.
So,
(a) 5 litres of hydrogen has greatest no. of molecules with the maximum volume.
(b) 1 litre of SO_{2} contains the least number of molecules since it has the smallest volume.
Solution 17
Gas 
Volume (in litres) 
Number of molecules 
Chlorine 
10 
x/2 
Nitrogen 
20 
x 
Ammonia 
20 
X 
Sulphur dioxide 
5 
x/4 
Solution 18
(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.
Mole Concept And Stoichiometry Exercise Ex. 5B
Solution 1
(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444
(b) KClO_{3} = (K)39 + (Cl)35.5 + (3O)48 = 122.5
(c) (Cu)63.5 + (S)32 + (4O)64 + (5H_{2}O)5 x 18 = 249.5
(d) (2N)28 + (8H)8 + (S)32 + (4O)64 = 132
(e) (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82
(f) (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5
(g) (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252
Solution 2
(a) No. of molecules in 73 g HCl = 6.023 x10^{23 }x 73/36.5(mol.
mass of HCl)
= 12.04 x 10^{23}
(b) Weight of 0.5 mole of O_{2} is = 32(mol. Mass of O_{2}) x 0.5=16 g
(c) No. of molecules in 1.8 g H_{2}O = 6.023 x 10^{23 }x 1.8/18
= 6.023 x 10^{22}
(d) No. of moles in 10g of CaCO_{3} = 10/100(mol. Mass CaCO_{3})
= 0.1 mole
(e) Weight of 0.2 mole H_{2} gas = 2(Mol. Mass) x 0.2 = 0.4 g
(f) No. of molecules in 3.2 g of SO_{2} = 6.023 x 10^{23}x 3.2/64
= 3.023 x 10^{22}
Solution 3
Molecular mass of H_{2}O is 18, CO_{2} is 44, NH_{3} is 17 and CO is 28
So, the weight of 1 mole of CO_{2} is more than the other three.
Solution 4
4g of NH_{3} having minimum molecular mass contain maximum molecules.
Solution 5
a) No. of particles in s1 mole = 6.023 x 10^{23}
So, particles in 0.1 mole = 6.023 x 10 ^{23} x 0.1 = 6.023 x 10^{22}
b)1 mole of H_{2}SO_{4} contains =2 x 6.023 x 10^{23}
So, 0.1 mole of H_{2}SO_{4} contains =2 x 6.023 x 10^{23} x0.1
= 1.2x10^{23 }atoms of hydrogen
c)111g CaCl_{2} contains = 6.023 x 10^{23 }molecules
So, 1000 g contains = 5.42 x 10^{24}molecules
Solution 6
(a) 1 mole of aluminium has mass = 27 g
So, 0.2 mole of aluminium has mass = 0.2 x 27 = 5.4 g
(b) 0.1 mole of HCl has mass = 0.1 x 36.5(mass of 1 mole)
= 3.65 g
(c) 0.2 mole of H_{2}O has mass = 0.2 x 18 = 3.6 g
(d) 0.1 mole of CO_{2} has mass = 0.1 x 44 = 4.4 g
Solution 7
(a) 5.6 litres of gas at STP has mass = 12 g
So, 22.4 litre (molar volume) has mass =12 x 22.4/5.6
= 48g(molar mass)
(b)1 mole of SO_{2} has volume = 22.4 litres
So, 2 moles will have = 22.4 x 2 = 44.8 litre
Solution 8
(a) 1 moleof CO_{2} contains O_{2} = 32g
So, CO_{2 }having 8 gm of O_{2 }has no. of moles = 8/32 = 0.25 moles
(b) 16 g of methane has no. of moles = 1
So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles
Solution 9
(a) 6.023 x 10 ^{23 }atoms of oxygen has mass = 16 g
So, 1 atom has mass = 16/6.023 x 10^{23 }= 2.656 x 10^{23 }g
(b) 1 atom of Hydrogen has mass = 1/6.023 x 10^{23 }= 1.666 x 10^{24}
(c) 1 molecule of NH_{3} has mass = 17/6.023 x10^{23 }= 2.82 x 10^{23 }g
(d) 1 atom of silver has mass = 12 × 10^{22} /6.023 x 10^{23 }= 12/60 = 1/5 = 0.2 g
(e) 1 molecule of O_{2} has mass = 32/6.023 x 10^{23 }= 5.314 x 10^{23 }g
(f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g
Solution 10
(a) 0.1 mole of CaCO_{3} has mass =100(molar mass) x 0.1=10 g
(b) 0.1 mole of Na_{2}SO_{4}.10H_{2}O has mass = 322 x 0.1 = 32.2 g
(c) 0.1 mole of CaCl_{2} has mass = 111 x 0.1 = 11.1g
(d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g
Solution 11
(a) 1molecule of Na_{2}CO_{3}.10H_{2}O contains oxygen atoms = 13
So, 6.023 x10^{23 }molecules (1mole) has atoms=13 x 6.023 x 10^{23}
So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 10^{23} =7.8x10^{23}
(b) Given Na = 4.6 gm
At. mass = 23
No. of gram atoms of Na= 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 12
3.2 g of S has number of atoms = 6.023 x10^{23} x 3.2 /32
= 0.6023 x 10^{23}
So, 0.6023 x 10^{23} atoms of Ca has mass=40 x0.6023x10^{23}/6.023
x10^{23}
= 4g
Solution 13
(a) No. of atoms = 52 x 6.023 x10^{23} = 3.131 x 10^{25 }
(b) 4 amu = 1 atom of He
so, 52 amu = 13 atoms of He
(c) 4 g of He has atoms =6.023 x10^{23}
So, 52 g will have = 6.023 x 10^{23} x 52/4 = 7.828 x10^{24} atoms
Solution 14
Molecular mass of Na_{2}CO_{3} = 106 g
106 g has 2 x 6.023 x10^{23} atoms of Na
So, 5.3g will have = 2 x 6.023 x10^{23}x 5.3/106=6.022 x10^{22 }atoms
Number of atoms of C = 6.023 x10^{23 }x 5.3/106 = 3.01 x 10^{22 }atoms
And atoms of O = 3 x 6.023 x 10^{23 }x 5.3/106= 9.03 x10^{22 }atoms
Solution 15
(a) 60 g urea has mass of nitrogen(N_{2}) = 28 g
So, 5000 g urea will have mass = 28 x 5000/60 = 2.33 kg
(b) 64 g has volume = 22.4 litre
So, 320 g will have volume = 22.4 x 320/64=112 litres
Solution 16
(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 17
22400 cm^{3} of CO has mass = 28 g
So, 56 cm^{3} will have mass = 56 x 28/22400 = 0.07 g
Solution 18
18 g of water has number of molecules = 6.023 x 10^{23}
So, 0.09 g of water will have no. of molecules = 6.023 x 10^{23} x 0.09/18 = 3.01 x 10^{21}molecules
Solution 19
(a) No. of moles in 256 g S_{8} = 1 mole
So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles
(b) No. of molecules = 0.02 x 6.023 x 10^{23} = 1.2 x 10^{22 }molecules
No. of atoms in 1 molecule of S= 8
So, no. of atoms in 1.2 x 10^{22 }molecules = 1.2 x 10^{22 }x 8
= 9.635x 10^{22 }molecules
Solution 20
Atomic mass of phosphorus P = 30.97 g
Hence, molar mass of P_{4} = 123.88 g
If phosphorus is considered as P_{4} molecules,
then 1 mole P_{4 }≡ 123.88 g
Therefore, 100 g of P_{4 }= 0.807 g
Solution 21
(a) 308 cm^{3 }of chlorine weighs = 0.979 g
So, 22400 cm^{3} will weigh = gram molecular mass
= 0.979 x 22400/308 =71.2 g
(b) 2 g(molar mass) H_{2 }at 1 atm has volume = 22.4 litres
So, 4 g H_{2 }at 1 atm will have volume = 44.8 litres
Now, at 1 atm(P_{1}) 4 g H_{2 }has volume (V_{1}) = 44.8 litres
So, at 4 atm(P_{2}) the volume(V_{2}) will be = _{}
(c) Mass of oxygen in 22.4 litres = 32 g(molar mass)
So, mass of oxygen in 2.2 litres = 2.2 x 32/22.4=3.14 g
Solution 22
No. of atoms in 12 g C = 6.023 x10^{23}
So, no. of carbon atoms in 10^{12} g = 10^{12} x 6.023 x10^{23}/12
= 5.019 x 10^{10 }atoms
Solution 23
Given:
P= 1140 mm Hg
Density = D = 2.4 g / L
T = 273 ^{0}C = 273+273 = 546 K
M = ?
We know that, at STP, the volume of one mole of any gas is 22.4 L
Hence we have to find out the volume of the unknown gas at STP.
First, apply Charle’s law.
We have to find out the volume of one litre of unknown gas at standard temperature 273 K.
V_{1}= 1 L T_{1} = 546 K
V_{2}=? T_{2} = 273 K
V_{1}/T_{1} = V_{2}/ T_{2}
V_{2} = (V_{1} x T_{2})/T_{1}
= (1 L x 273 K)/546 K
= 0.5 L
We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.
Apply Boyle’s law.
P _{1} = 1140 mm Hg V_{1} = 0.5 L
P_{2} = 760 mm Hg V_{2} = ?
P_{1} x V_{1 }= P_{2} x V_{2}
V_{2} = (P_{1} x V_{1})/P_{2}
= (1140 mm Hg x 0.5 L)/760 mm Hg
= 0.75 L
Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP
X moles = 0.75 L / 22.4 L
= 0.0335 moles
The original mass is 2.4 g
n = m / M
0.0335 moles = 2.4 g / M
M = 2.4 g / 0.0335 moles
M= 71.6 g / mole
Hence, the gram molecular mass of the unknown gas is 71.6 g
Solution 24
1000 g of sugar costs = Rs. 40
So, 342g(molar mass) of sugar will cost=342x40/1000=Rs. 13.68
Mole Concept And Stoichiometry Exercise Ex. 5C
Solution 1
(a) CH
(b) C_{2}H_{6}O
(c) CH
(d) CH_{2}O
Solution 2
Solution 3
(a) Molecular mass of Ca(H_{2}PO_{4})_{2 }= 234
So, % of P = 2 _{}31 _{}100/234 = 26.5%
(b) Molecular mass of Ca_{3}(PO_{4})_{2 }= 310
% of P = 2 _{}31 _{}100/310 = 20%
Solution 4
Molecular mass of KClO_{3 }= 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 5
Element%At. massAtomic ratioSimple ratio
Pb62.5207_{}1
N8.514_{}2
O29.016_{}6
So, Pb(NO_{3})_{2}is the empirical formula.
Solution 6
In Fe_{2}O_{3} , Fe = 56 and O = 16
Molecular mass of Fe_{2}O_{3 }= 2 _{}56 + 3 _{}16 = 160 g
Iron present in 80% of Fe_{2}O_{3 }= _{}
So, mass of iron in 100 g of ore = 56 g
_{}mass of Fe in 10000 g of ore = 56 _{}10000/100
= 5.6 kg
Solution 7
For acetylene , molecular mass = 2 _{}V.D = 2 _{}13 = 26 g
The empirical mass = 12(C) + 1(H) = 13 g
n = _{}
Molecular formula of acetylene= 2 _{}Empirical formula =C_{2}H_{2}
Similarly, for benzene molecular mass= 2 _{}V.D = 2 _{}39 = 78
n = 78/13=6
So, the molecular formula = C_{6}H_{6}
Solution 8
Element 
% 
Atomic mass 
Atomic ratio 
Simple ratio 
H 
17.64 
1 
17.64/1 =17.64 
17.64/5.88 = 3 
N 
82.35 
14 
82.35/14 = 5.88 
5.88/5.88 = 1 
Solution 9
Element%at. massatomic ratiosimple ratio
C54.5412_{}2
H9.091_{}4
O36.3616_{}1
(a) So, its empirical formula= C_{2}H_{4}O
(b) empirical formula mass = 44
Since, vapour density = 44
So, molecular mass = 2 _{}V.D = 88
Or n = 2
so, molecular formula = (C_{2}H_{4}O)_{2} = C_{4}H_{8}O_{2}
Solution 10
Element%at. massatomic ratiosimple ratio
C26.5912_{}1
H2.221_{}1
O71.1916_{}2
(a) its empirical formula = CHO_{2}
(b) empirical formula mass = 45
Vapour density = 45
So, molecular mass = V.D _{}2 = 90
so, molecular formula = C_{2}H_{2}O_{4}
Solution 11
Element%at. massatomic ratiosimple ratio
Cl71.6535.5_{}1
H4.071_{}2
C24.2812_{}1
(a) its empirical formula = CH_{2}Cl
(b) empirical formula mass = 49.5
Since, molecular mass = 98.96
so, molecular formula = (CH_{2}Cl)_{2} = C_{2}H_{4}Cl_{2}
Solution 12
(a) The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1=1
(b) ElementGiven massAt. massGram atomRatio
C4.8120.412
H1112.55
So, the empirical formula = C_{2}H_{5}
(c) Empirical formula mass = 29
Molecular mass = V.D _{}2 = 29 _{}2 = 58
So, molecular formula = C_{4}H_{10}
Solution 13
Since, g atom of Si = given mass/mol. Mass
so, given mass = 0.2 _{}28 = 5.6 g
ElementmassAt. massGram atomRatio
Si5.6280.21
Cl21.335.5_{}3
Empirical formula = SiCl_{3}
Solution 14
% of carbon = 82.76%
% of hydrogen = 100  82.76 = 17.24%
Element 
% Weight 
Atomic Weight 
Relative No. of Moles 
Simplest Ratio 
C 
82.76 
12 
82.76/12 = 6.89 
6.89/6.89 = 1 x 2 = 2 
H 
17.24 
1 
17.24/1 = 17.24 
17.24/6.89 = 2.5 x 2 = 5 
Empirical formula = C_{2}H_{5}
Empirical formula weight = 2 x 12 + 1 x 5 = 24 + 5 = 29
Vapour Density = 29
Relative molecular mass = 29 x 2 = 58
N =
Molecular formula = n x empirical formula
= 2 x C_{2}H_{5}
= C_{4}H_{10}
Solution 15
(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 gramatoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg_{3 }N_{2}
Solution 16
Barium chloride = BaCl_{2}.x H_{2}O
Ba + 2Cl + x[H_{2} + O]
=137+ 235.5 + x [2+16]
=[208 + 18x] containswater = 14.8% water in BaCl_{2}.x H_{2}O
=[208 + 18 x] 14.8/100 = 18x
=[104 + 9x] 2148=18000x
=[104+9x] 37=250x
=3848 + 333x =2250x
1917x =3848
x = 2molecules of water
Solution 17
Molar mass of urea; CON_{2}H_{4 }= 60 g
So, % of Nitrogen = 28 _{}100/60 = 46.66%
Solution 18
Element%At. massAtomic ratioSimple ratio
C42.1123.51
H6.4816.482
O51.42163.21
The empirical formula is CH_{2}O
Since the compound has 12 atoms of carbon, so the formula is
C_{12} H_{24} O_{12.}
Solution 19
(a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, the molecular formula is A_{2}B_{4}.
(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D
Empirical formula weight = V.D/3
So, n = molecular mass/ Empirical formula weight = 6
Hence, the molecular formula is A_{6}B_{6}
(c)
Given:
Wt. of the compound: 10.47g
Wt. of metal A: 6.25g
Wt. of nonmetal B: 10.47 – 6.25 = 4.22g
Element 
mass 
At. Wt. 
Relative no. of atoms 
Simplest ratio 
A 
6.25g 
207 
6.25/207=0.03 
0.03/0.03=1 
B 
4.22g 
35.5 
4.26/35.5=0.12 
0.12/0.03=4 
Hence, the empirical formula is AB_{4}
Solution 20
Atomic ratio of N = 87.5/14 =6.25
Atomic ratio of H= 12.5/1 = 12.5
This gives us the simplest ratio as 1:2
So, the molecular formula is NH_{2}
Solution 21
Element%at. massatomic ratiosimple ratio
Zn22.65650.3481
H4.8814.8814
S11.15320.3481
O61.32163.8311
Empirical formula of the given compound =ZnSH_{14}O_{11}
Empiricala formula mass = 65.37+32+141+11+16=287.37
Molecular mass = 287
n= Molecular mass/Empirical formula mass = 287/287=1
Molecular formula = ZnSO_{11}H_{14}
=ZnSO_{4}.7H_{2}O
Mole Concept And Stoichiometry Exercise Ex. 5D
Solution 1
(a) 100 g of CaCO_{3 }produces = 164 g of Ca(NO_{3})_{2}
So, 15 g CaCO_{3 }will produce = 164 _{}15/100 = 24.6 g Ca(NO_{3})_{2}
(b) 1 V of CaCO_{3 }produces 1 V of CO_{2}
100 g of CaCO_{3 }has volume = 22.4 litres
So, 15 g will have volume = 22.4_{}15/100 = 3.36 litres CO_{2}
Solution 2
2NH_{3} + H_{2}SO_{4} _{}(NH_{4})_{2}SO_{4}
66 g
(a) 2NH_{3} + H_{2}SO_{4} (NH_{4})_{2}SO_{4}
34 g98 g132 g
For 132 g (NH_{4})_{2}SO_{4} = 34 g of NH_{3} is required
So, for 66 g (NH_{4})_{2}SO_{4} = 66 _{}32/132 = 17 g of NH_{3} is required
(b) 17g of NH_{3} requires volume = 22.4 litres
(c) Mass of acid required, for producing 132g (NH_{4})_{2}SO_{4} = 98g
So, Mass of acid required, for 66g (NH_{4})_{2}SO_{4 }= 66 _{}98/132 = 49g
Solution 3
(a) Molecular mass of Pb_{3}O_{4 }= 3 _{}207.2 + 4 _{}16 = 685 g
685 g of Pb_{3}O_{4 }gives = 834 g of PbCl_{2}
Hence, 6.85 g of Pb_{3}O_{4 }will give = 6.85 _{}834/685 = 8.34 g
(b) 685g of Pb_{3}O_{4 }gives = 71g of Cl_{2}
Hence, 6.85 g of Pb_{3}O_{4 }will give = 6.85 _{}71/685 = 0.71 g Cl_{2}
(c) 1 V Pb_{3}O_{4}produces 1 V Cl_{2}
685g of Pb_{3}O_{4}has volume = 22.4 litres = volume of Cl_{2} produced
So, 6.85 Pb_{3}O_{4} will produce = 6.85 _{}22.4/685 = 0.224 litres of Cl_{2}
Solution 4
Molecular mass of KNO_{3 }= 101 g
63 g of HNO_{3} is formed by = 101 g of KNO_{3}
So, 126000 g of HNO_{3} is formed by = 126000 _{}101/63 = 202 kg
Similarly,126 g of HNO_{3} is formed by 170 kg of NaNO_{3}
So, smaller mass of NaNO_{3 }is required.
Solution 5
CaCO_{3} + 2HCl ⟶ CaCl_{2} + H_{2}O + CO_{2}
100g 73g 2 L
(a)V_{1} = 2 litres
V_{2}= ?
T_{1} = (273 + 27) = 300K
T_{2} = 273K
V_{1}/T_{1 }= V_{2}/T_{2}
V_{2} = V_{1}T_{2}/T_{1} =
Now at STP 22.4 litres of CO_{2} are produced using CaCO_{3} = 100g
So, litres are produced by = (2 × 273 × 100) / (300 × 22.4) = 8.125 g
(b)22.4 litres are CO_{2} are prepared from acid =73 g
litres are prepared from = (2 × 273 × 73) / (300 × 22.4) =5.9 g
Solution 6
2H_{2}O_{}2H_{2} + O_{2}
2 V2 V1 V
2 moles of H_{2}O gives = 1 mole of O_{2}
So, 1 mole of H_{2}O will give = 0.5 moles of O_{2}
so, mass of O_{2} = no. of moles x molecular mass
= 0.5 _{}32 = 16 g of O_{2}
and 1 mole of O_{2} occupies volume =22.4 litre
so, 0.5 moles will occupy = 22.4 _{}0.5 = 11.2 litres at S.T.P.
Solution 7
2Na_{2}O_{2} + 2H_{2}O_{}4NaOH + O_{2}
2 V4 V1 V
(a) Mol. Mass of Na_{2}O_{2} = 2 _{}23 + 2 _{}16 = 78 g
Mass of 2Na_{2}O_{2}= 156 g
156 g Na_{2}O_{2 }gives = 160 g of NaOH (4 _{}40 g)
So, 1.56 Na_{2}O_{2 }will give = 160 _{}1.56/156 = 1.6 g
(b) 156 g Na_{2}O_{2 }gives = 22.4 litres of oxygen
So, 1.56 g will give = 22.4 _{}1.56/156 = 0.224 litres
= 224 cm^{3}
(c)156 g Na_{2}O_{2 }gives = 32 g O_{2}
So, 1.56 g Na_{2}O_{2 }will give = 32 _{}1.56/156
= 32/100 = 0.32 g
Solution 8
2NH_{4}Cl + Ca(OH)_{2}_{}CaCl_{2}+2H_{2}O + 2NH_{3}
2 V1 V1 V2 V
Mol. Mass of 2NH_{4}Cl = 2[14 + (1 _{}4) + 35.5] = 2[53.5] = 107 g
(a) 107 g NH_{4}Cl gives = 34 g NH_{3}
So, 21.4 g NH_{4}Cl will give = 21.4 _{}34/107 = 6.8 g NH_{3}
(b) The volume of 17 g NH_{3} is 22.4 litre
So, volume of 6.8 g will be = 6.8 _{}22.4/17 = 8.96 litre
Solution 9
Solution 10
MnO_{2} + 4HCl _{}MnCl_{2} + 2H_{2}O +Cl_{2}
1 V4 V1 V1 V
(a) 1 mole of MnO_{2} weighs = 87 g (mol. Mass)
So, 0.02 mole will weigh = 87 _{}0.02 = 1.74 g MnO_{2}
(b) 1 mole MnO_{2} gives = 1 mole of MnCl_{2}
So, 0.02 mole MnO_{2}will give =0.02 mole of MnCl_{2}
(c) 1 mole MnCl_{2} weighs = 126 g(mol mass)
So, 0.02 mole MnCl_{2} will weigh = 126 _{}0.02 g = 2.52 g
(d) 0.02 mole MnO_{2}will form =0.02 mole of Cl_{2}
(e) 1 mole of Cl_{2 }weighs = 35.5 g
So, 0.02 mole will weigh = 71 _{}0.02 = 1.42 g of Cl_{2}
(f) 1 mole of chlorine gas has volume = 22.4 litres
So, 0.02 mole will have volume = 22.4 _{}0.02 = 0.448 litre
(g) 1 mole MnO_{2}requires HCl = 4 mole
So, 0.02 mole MnO_{2} will require =4 _{}0.02 = 0.08 mole
(h) For 1 mole MnO_{2 }, acid required = 4 mole of HCl
So, for 0.02 mole, acid required = 4 _{}0.02 =0.08 mole
Mass of HCl = 0.08 x 36.5 = 2.92 g
Solution 11
N_{2} + 3H_{2} _{}2NH_{3}
28g6g34g
28g of nitrogen requires hydrogen = 6g
2000g of nitrogen requires hydrogen = 6/28 _{}2000=3000/7g
So mass of hydrogen left unreacted =10003000/7=571.4g of H_{2}
(b)28g of nitrogen forms NH_{3} = 34g
2000g of N_{2} forms NH_{3}
= 34/28 _{}2000
=2428.6g
Mole Concept And Stoichiometry Exercise Misc. Ex.
Solution A 1
(a) Weight of 1 g atom N = 14 g
So, weight of 2 g atom of N = 28 g
(b) 6.023 x10^{23 }atoms of C weigh = 12 g
So, 3 x10^{25} atoms will weigh = _{}
(c) 1 mole of sulphur weighs = 32 g
(d) 7 g of silver
So, 7 grams of silver weighs least.
Solution A 2
Option C is correct.
40 g of NaOH contains 6.023 x 10^{23 }molecules
So, 4 g of NaOH contains = 6.02 x10^{23} x 4/40
= 6.02 x10^{22} molecules
Solution A 3
The number of molecules in 18 g of ammonia= 6.02 x10^{23}
So, no. of molecules in 4.25 g of ammonia = 6.02 x10^{23 }x 4.25/18
= 1.5 x 10^{23}
Solution A 4
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
Solution A 5
Correct option: (c) molecular mass
Solution:
The relative molecular mass of a gas or vapour is twice its vapour density.
2 × Rel. V. D.= Rel. molecular mass of a gas or vapour
Solution A 6
Correct option: (a) C_{3}H_{6}O_{3}
Solution:
Molecular formula = n × Empirical formula
Where, n = an integer
In the given case, if n = 3, then
C_{3}H_{6}O_{3} = 3 × CH_{2}O
Solution A 7
Correct option: (b) 29
Solution:
2 × Rel. V. D.= Rel. molecular mass of a gas or vapour
2 × Rel. V. D.= 58
Rel. V. D. = 58 / 2
Rel. V. D. = 29
Solution A 8
Correct option: (d) P_{3}Q_{6}
Solution:
Molecular formula = n × Empirical formula
Where, n = an integer
Also,
Molecular formula mass = n × Empirical formula mass
30 = n × 10
n = 3
Molecular formula = n × Empirical formula
Molecular formula = 3 × PQ_{2}
Molecular formula = P_{3}Q_{6}
Solution A 9
Correct option: (c) 1 : 1
Solution:
Molecular mass of hydrogen is 1 g = 1 mole of hydrogen = 6.023 × 10^{23} molecules of hydrogen
So, 2 g of hydrogen = 2 moles of hydrogen = 2 × 6.023 × 10^{23} molecules of hydrogen
Molecular mass of oxygen is 16 g = 1 mole of oxygen = 6.023 × 10^{23} molecules of oxygen
So, 32 g of oxygen = 2 moles of oxygen = 2 × 6.023 × 10^{23} molecules of oxygen
Therefore,
Ratio of molecules in 2 g of hydrogen to molecules in 32 g of oxygen is 1:1.
Solution B 1
(a) Moles:1 mole + 2 mole _{}1 mole + 2 mole
(b) Grams: 42g + 36g_{}74g+ 4 g
(c) Molecules = 6.02 _{}10^{23 }+ 12.046 _{}10^{23 }_{}6.02 _{}10^{23}+ 12.046 _{}10^{23}
Solution B 2
(a) One mole of chlorine contains 6.023 x 10^{23} atoms of chlorine.
(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.
(c) Relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon12.
(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.
Solution B 3
According to Avogadros law:
Equal volumes of all gases, under similar conditions of temperature and pressure ,contain equal number of molecules.
So, 1 mole of each gas contains = 6.02 _{}10^{23} molecules
Mol. Mass of H_{2} (2),O_{2}(32) ,CO_{2}(44),SO_{2}(64),Cl_{2}(71)
(1)Now 2 g of hydrogen contains molecules =6.02 _{}10^{23}
So, 8g of hydrogen contains molecules = 8/2 _{}6.02 _{}10^{23}
=4 _{}6.02 _{}10^{23} = 4M molecules
(2)32g of oxygen contains molecules = 8/32 _{}6.02 _{}10^{23}=M/4
(3)44g of carbon dioxide contains molecules = 8/44 6.02 10^{23}=2M/11
(4)64g of sulphur dioxide contains molecules =6.02 _{}10^{23}
So, 8g of sulphur dioxide molecules = 8/64 _{}6.02 _{}10^{23}= M/8
(5)71 g of chlorine contains molecules =6.02 _{}10^{23}
So, 8g of chlorine molecules = 8/72 _{}6.02 _{}10^{23} = 8M/71
Since 8M/71<M/8<2M/11<M/4<4M
Thus Cl_{2}<SO_{2}<CO_{2}<O_{2}<H_{2}
(i)Least number of molecules in Cl_{2}
(ii)Most number of molecules in H_{2}
Solution C 1
(a) The vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.
(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm^{3}.
(c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon12.
(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 carbon12.
(e) The number of atoms present in 12g (gram atomic mass) of C12 isotope, i.e. 6.023 x10^{23} 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 carbon12.
Solution C 2
(a) GayLussac'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 C 3
(a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
Now volume of hydrogen gas =volume of helium gas
n molecules of hydrogen =n molecules of helium gas
nH_{2}=nHe
1 mol. of hydrogen has 2 atoms of hydrogen and I molecule of helium has 1 atom of helium
Therefore 2H=He
Therefore atoms in hydrogen is double the atoms of helium.
(b) For a given volume of gas under given temperature and pressure, a change in any one of the variable i.e., pressure or temperature changes the volume.
(c) Inflating a balloon seems violating Boyles law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.
Solution C 4
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 D 1
Information conveyed by H_{2}O
(1) That H_{2}O contains 2 volumes of hydrogen and 1 volume of oxygen.
(2) That ratio by weight of hydrogen and oxygen is 1:8.
(3) That molecular weight of H_{2}O is 18g.
Solution D 2
(a) stoichiometry measures quantitative relationships and is used to determine the amount of products/reactants that are produced/needed in a given reaction.
(b) The number of atoms in a molecule of an element is called its atomicity. Atomicity of Hydrogen is 2, phosphorus is 4 and sulphur is 8.
(c) N_{2 }means 1 molecule of nitrogen and 2N means two atoms of nitrogen.
N_{2} can exist independently but 2N cannot exist independently.
Solution D 3
(a) Applications of Avogadro's Law :
(1) It explains GayLussac's law.
(2) It determines atomicity of the gases.
(3) It determines the molecular formula of a gas.
(4) It determines the relation between molecular mass and vapour density.
(5) It gives the relationship between gram molecular mass and gram molecular volume.
(b) According to Avogadro's law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.
Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another.This what Gay Lussac's Law says.
H_{2}+Cl_{2}?2HCl
1V1V2V(By GayLussacs law)
n moleculesnmolecules2n molecules(By Avogadros law)
Solution D 4
a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C12.
b)The value of avogadro's number is 6.023 _{}10^{23}
c) The molar volume of a gas at STP is 22.4 dm^{3} at STP
Solution E 1
From equation: 2H_{2}+ O_{2}_{}2H_{2}O
1 mole of Oxygen gives = 2 moles of steam
so, 0.5 mole oxygen will give = 2 _{}0.5 = 1mole of steam
Solution E 2
3Cu + 8HNO_{3} _{}3Cu (NO_{3})_{2}+ 4H_{2}O + 2NO
1 V8 V3 V2 V
Mol. Mass of 8HNO_{3} = 8 _{}63 = 504 g
(a) For 504 g HNO_{3}, Cu required is = 192 g
So, for 63g HNO_{3}Cu required = 192 _{}63/504 = 24g
(b) 504 g of HNO_{3} gives = 2 _{}22.4 litre volume of NO
So, 63g of HNO_{3} gives =2 _{}22.4 _{}63/504 =5.6 litre of NO
Solution E 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) 22400cm^{3} volume have mass =28 g of CO(molar mass)
So, 56cm^{3} volume will have mass =28 _{}56/22400= 0.07 g
Solution E 4
% of N in NaNO_{3}= _{}
% of N in (NH_{4})_{2}SO_{4} = _{}
% of N in CO(NH_{2})_{2} = _{}
So, highest percentage of N is in urea.
Solution E 5
2H_{2}O_{}2H_{2}+O_{2}
2 V2 V1 V
(a) From equation, 2 V of water gives 2 V of H_{2} and 1 V of O_{2}
where 2 V = 2500 cm^{3}
so, volume of O_{2} liberated = 2V/V = 1250 cm^{3}
(b)
P_{1} = 1 Atm.
P_{2} = 2.5 Atm.
V_{1} = 2500
V_{2} = ?
P_{1} V_{1} = P_{2} V_{2}
1 x 2500 = 2.5 x V_{2}
V_{2} = 2500 x 10/25
V_{2} = 1000 cm^{3}
(c)
(If volume of H_{2} = 1000 cm^{3}, temp. is T_{1}, V_{2} = 2500 and New temp. T_{2}
V_{1} / T_{1} = V_{2} / T_{2}
1000/T_{1} = 2500/T_{2}
or
T_{2}/T_{1}= 2500/1000
= 2.5
T_{2} = 2.5 T_{1}
It must be 2.5 times of original temperature.
Solution E 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 E 7
% of hydrogen = 20%
% of carbon = 100  20 = 80%

% Weight 
Atomic Weight 
Relative No. of Moles 
Simplest Ratio 
C 
80 
12 
80/12 = 6.667 
6.667/6.667 = 1 
H 
20 
1 
20/1 = 20 
20/6.667 = 2.99 ≈ 3 
Empirical formula = CH_{3}
Empirical formula weight = 1 x 12 + 1 x 3 = 12 + 3 = 15
Vapour Density = 15
Relative molecular mass = 15 x 2 = 30
N =
Molecular formula = n x empirical formula
= 2 x CH_{3}
= C_{2}H_{6}
Solution E 8
22400cm^{3} CO_{2 }has mass = 44g
so, 224 cm^{3} CO_{2 }will have mass= 0.44 g
Now since CO_{2 }is being formed and X is a hydrocarbon so it contains C and H.
In 0.44g CO_{2}, mass of carbon=0.440.32=0.12g=0.01g atom
So, mass of Hydrogen in X = 0.1450.12 = 0.025g
= 0.025g atom
Now the ratio of C:H is C=1: H=2.5 or C=2 : H=5
i.e. the formula of hydrocarbon is C_{2}H_{5}
(a) C and H
(b) Copper (II) oxide was used for reduction of the hydrocarbon.
(c)
(i) no. of moles of CO_{2}= 0.44/44 = 0.01 moles
(ii) mass of C = 0.12 g
(iii) mass of H = 0.025 g
(iv) The empirical formula of X = C_{2}H_{5}
Solution E 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 = 24/88100 = 27.3%
% of oxygen in the compound=64/88 100 =72.7%
Element%At. MassAtomic ratioSimplest ratio
X27.31227.3/12=2.271
O72.71672.2/16=4.542
So simplest formula = XO_{2}
Solution E 10
(a) V.D = _{}
(b) Molecular mass = 17(V.D) x 2= 34g
Solution E 11
(a) CO_{2}+C_{}2CO
1 V1 V2 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 + O_{2}_{}2CO_{2}
2 V1 V2 V
(i) 2 V CO requires oxygen = 1 V
so, 24 cm^{3} CO will require = 24/2 =12 cm^{3}
(ii) 2 x 22400 cm^{3} CO gives = 2 x 22400 cm^{3 }CO_{2}
so, 24cm^{3} CO will give = 24 cm^{3 }CO_{2}
Solution E 12
Molecular weight of =
=328g
Molecular weight of CaO =2(40+16)
=112g
a. 328g of Ca(NO_{3})_{2 }liberates 4 moles of NO_{2 }
328g of Ca(NO_{3})_{2 }liberates L of NO_{2 } 82g will liberate
=22.4dm^{3 }of NO_{2 }
b. 328 g of calcium nitrate gives 112g of CaO
82 g will give
=28 g of CaO
Solution E 13
2C_{8}H_{18} + 25O_{2}_{}16CO_{2} + 18H_{2}O
2 V25 V16 V18 V
(i) 2 moles of octane gives = 16 moles of CO_{2}
so, 1 mole octane will give = 8 moles of CO_{2}
(ii) 1 mole CO_{2 }occupies volume = 22.4 litre
so, 8 moles will occupy volume = 8 _{}22.4 = 179.2 litre
(iii) 1 mole CO_{2} has mass = 44 g
so, 16 moles will have mass = 44 _{}16 = 704 g
(iv) Empirical formula is C_{4}H_{9}.
Solution E 14
The relative atomic mass of Cl = (35 _{}3 + 1 _{}37)/4=35.5 amu
Solution E 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. MassAt. RatioSimplest ratio
Si20.822820.82/28=0.741
Cl79.18 35.579.18/35.5=2.233
So the empirical formula of the given compound =SiCl_{3}
Solution E 16
% compositionAtomic ratioSimple ratio
P=38.27%38.27/31 =1.231
H= 2.47%2.47/1 = 2.472
O= 59.26%59.26/16 = 3.703
So, empirical formula is PH_{2}O_{3} or H_{2}PO_{3}
Empirical formula mass = 31+ 2 _{}1 + 3 _{}16 = 81
The molecular formula is = H_{4}P_{2}O_{6}, because n = 162/81=2
Solution E 17
a) V_{1} = 10 litres V_{2}=?
T_{1}= 27+ 273 = 300KT_{2}=273K
P_{1}=700 mmP_{2} = 760 mm
Using the gas equation
b)
Solution E 18
(a) Molecular mass of CO_{2} = 12+ 2x16 = 44 g
So, vapour density (V.D) = mol. Mass/2 = 44/2 = 22
V.D = _{}
_{}
So, mass of CO_{2} = 22 kg
(b) According to Avogadros law ,equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
So, number of molecules of carbon dioxide in the cylinder =number of molecules of hydrogen in the cylinder=X
Solution E 19
(a) The volume occupied by 1 mole of chlorine = 22.4 litre
(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.
(c) V_{1}/V_{2} = T_{1}/T_{2}
22.4/V_{2} =273/546
V_{2} = 44.8 litres
(d) Mass of 1 mole Cl_{2} gas =35.5 x 2 =71 g
Solution E 20
(a) Total molar mass of hydrated CaSO_{4}.xH_{2}O = 136+18x
Since 21% is water of crystallization, so
_{}
So, x = 2 i.e. water of crystallization is 2.
(b) For 18 g water, vol. of hydrogen needed = 22.4 litre
So, for 1.8 g, vol. of H_{2} needed= 1.8 x 22.4/18 = 2.24 litre
Now 2 vols. of water = 1 vol. of oxygen
1 vol. of water =1/2 vol. of O_{2} =22.4/2=11.2 lit.
18 g of water = 11.2 lit. of O_{2}
1.8 g of water = 11.2/18 18/10=1.12 lit.
(c) 32g of dry oxygen at STP = 22400cc
2g will occupy = 224002/32=1400cc
P_{1}=760mmP_{2} =740mm
V_{1}=1400ccV_{2} =?
T_{1} =273 K, T_{2} = 27 +73 = 300K
(d) P_{1}= 750mmP_{2}=760mm
V_{1}= 44lit.V_{2}=?
T_{1}= 298KT_{2}=273K
(e) Since 143.5g of AgCl is produced from =58.5 g of NaCl
so, 1.435 g of AgCl is formed by =0.585 g of NaCl
% of NaCl =0.585 x100 = 58.5%
Solution E 21
a.
For CO_{2 }12+32
i. Molecular mass of sulphuric acid = 2(2+32+64)
= 196
196 g of suphuric acid oxidized 12g of Carbon
49 g of suphuric acid will
=3 g
ii. 196 g of sulphuric acid gives 2(22.4)
=44.8L
49 g og sulphuric acid will give
=11.2 L of SO_{2 }
b.
i.
Element 
% Weight 
Atomic Weight 
Atomic Ratio 
Simplest Ratio 
C 
14.4 
12 
14.4/12 = 1.2 
1.2/1.2=1 
H 
1.2 
1 
1.2/1 =1.2 
1.2/1.2=1 
Cl 
84.5 
35.5 
84.5/35.5=2.3 
2.3/1.2=1.9=2 
Empirical formula = CHCl_{2}
ii.
Empirical formula = CHCl_{2 }
Empirical formula weight = 1 x 12 + 1 x 1+(2 x35.5)
= 12 + 1+70
= 83
Relative molecular mass = 168
N = 2.02≈2
Molecular formula = n x empirical formula
= 2 x CHCl_{2}
= C_{2}H_{2}Cl_{4}
Solution E 22
a. Relative molecular mass of [Mg (NO_{3}) 6H_{2}O]
=24+14+(3 x16)+(6 x18)=194
Since, 194g of [Mg (NO_{3}) 6H_{2}O] contains 144g of oxygen
100g of [Mg (NO_{3}) 6H_{2}O] contains of oxygen = 74.22%
b. Relative molecular mass of Na_{2}B_{4}O_{7}.10H_{2}O
(23 × 2) + (4 × 11) + (7 × 16) + 10(18) = 382
Since 382g of Na_{2}B_{4}O_{7}.10H_{2}O contains 44g of boron
100g Na_{2}B_{4}O_{7}.10H_{2}O of contains of boron
=11.5%
c. Relative molecular mass of Ca(H_{2}PO_{4})_{2}
= 40 + 2(2 + 31 + 64) = 234
Since,234g of Ca(H_{2}PO_{4})_{2} contains 62g of phosphorus
100g of Ca(H_{2}PO_{4})_{2} contains
=26.5%
Solution E 23
The given equation is,
2NaOH + CuSO_{4} → Na_{2}SO_{4} + Cu(OH)_{2}
Molecular weight of NaOH, sodium hydroxide = 23 + 16 + 1 = 40
Molecular weight of Cu(OH)_{2} ,
Copper hydroxide = 64 + 16 + 1 + 16 + 1 = 98
Now, 40 g of NaOH is used to precipitate 98 g of Cu(OH)_{2}
Hence, 200 g of NaOH will be used to precipitate (98/40)×200 g of Cu(OH)_{2} = 490 g of Cu(OH)_{2}
So, 490 g of copper hydroxide would be prepared using 200 g of sodium hydroxide.
Solution E 24
Solid ammonium dichromate decomposes as:
(a)Molecular mass of ammonium dichromate
= 2(14+4)+104+112
= 252 g
Number of moles=
=
=0.25moles
(b)
252 g of ammonium dichromate gives 22.4 dm^{3} of N_{2}
63 g of ammonium dichromate gives
=5.6 L
= 0.25 moles
(c)
252 g of ammonium dichromate gives 22.4 dm^{3} of N_{2}
63 g of ammonium dichromate gives
= 5.6 L
(d)
Number of moles=
=
=0.25moles
0.25 moles of ammonium dichromate gives
0.25 moles of N_{2}=7 g
1 mole of H_{2}O =18 g
Therefore, total loss of mass=7+18
=25 g
(e)
252 g of ammonium dichromate gives 152 g of CrO_{3}
63 g of ammonium dichromate gives
=38 g
Solution E 25
2H_{2}S + 3O_{2}_{}2H_{2}O + 2SO_{2}
2 V3 V2 V
128 g of SO_{2}gives = 2 _{}22.4 litres volume
So, 12.8 g of SO_{2} gives = 2 _{}22.4 _{}12.8/128
= 4.48 litre volume
Or one can say 4.48 litres of hydrogen sulphide.
2 _{}22.4 litre H_{2}S requires oxygen = 3 _{}22.4 litre
So, 4.48 litres H_{2}S will require = 6.72 litre of oxygen
Solution E 26
From equation, 2NH_{3} + 2½ 2 O_{2}_{}2NO + 3H_{2}O
When 60 g NO is formed, mass of steam produced = 54 g
So, 1.5 g NO is formed, mass of steam produced = 54 _{}1.5/60
=1.35 g
Solution E 27
In 1 hectare of soil, N_{2} removed = 20 kg
So, in 10 hectare N_{2} removed = 200 kg
The molecular mass of Ca(NO_{3})_{2} =164
Now, 28 g N_{2 }present in fertilizer = 164 g Ca(NO_{3})_{2}
So, 200000 g of N_{2} is present in = 164 _{}200000/28
= 1171.42 kg
Solution E 28
(a) 1 mole of phosphorus atom = 31 g of phosphorus
31 g of P =1 mole of P
6.2g of P = _{}=0.2 mole of P
(b) 31 g P reacts with HNO_{3 }= 315 g
so, 6.2 g P will react with HNO_{3} = 315 _{}6.2/31 = 63 g
(c)
Moles of steam formed from 31g phosphorus = 18g/18g = 1mol
Moles of steam formed from 6.2 g phosphorus = 1mol/31g6.2=0.2 mol
Volume of steam produced at STP =0.2 _{}22.4 l/MOL=4.48 litre
Since the pressure (760mm) remains constant , but the temperature (273+273)=546 is double, the volume of the steam also gets doubled
So,Volume of steam produced at 760mm Hg and 273^{0}C=4.48 _{}2=8.96litre
Solution E 29
112cm^{3 }of gaseous fluoride has mass = 0.63 g
so, 22400cm^{3} will have mass = 0.63 _{}22400/112
= 126 g
The molecular mass = At mass P + At. mass of F
126= 31 + At. Mass of F
So, At. Mass of F = 95 g
But, at. mass of F = 19 so 95/19 = 5
Hence, there are 5 atoms of F so the molecular formula = PF_{5}
Solution E 30
Na_{2}CO_{3}.10H_{2}O _{}Na_{2}CO_{3} + 10H_{2}O
286 g106 g
So, for 57.2 g Na_{2}CO_{3}.10H_{2}O = 106 _{}57.2/286 = 21.2 g Na_{2}CO_{3}
Solution E 31
Simple ratio of M = 34.5/56 = 0.616 = 1
Simple ratio of Cl = 65.5/35.5 = 1.845 = 3
Empirical formula = MCl_{3}
Empirical formula mass = 162.5, Molecular mass = 2 _{}V.D = 325
So, n = 2
So, molecular formula = M_{2}Cl_{6}
Solution E 32
(i) Element%atomic massatomic ratiosimple ratio
C4.812_{}1
Br95.280_{}3
So, empirical formula is CBr_{3}
(ii) Empirical formula mass = 12 + 3 _{}80 = 252 g
molecular formula mass = 2 _{}252(V.D) = 504 g
n= 504/252 = 2
so, molecular formula = C_{2}Br_{6}
Solution E 33
4N_{2}O + CH_{4} _{}CO_{2} + 2H_{2}O + 4N_{2}
4 V1 V1 V2 V4 V
2 x 22400 litre steam is produced by N_{2}O = 4 x 22400 cm^{3}
So, 150 cm^{3} steam will be produced by= 4 _{}22400 _{}150/2 x 22400
= 300 cm^{3}N_{2}O
Solution E 34
(a) Volume of O_{2} = V
Since O_{2 }and N_{2 }have same no. of molecules = x
so, the volume of N_{2 }= V
(b) 3x molecules means 3V volume of CO
(c) 32 g oxygen is contained in = 44 g of CO_{2}
So, 8 g oxygen is contained in = 44 x 8/32 = 11 g
(d) Avogadro's law is used in the above questions.
Solution E 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 Na_{3}PO_{4}
Solution E 36
CH_{4} + 2O_{2} _{}CO_{2} + 2H_{2}O
1 V2 V1 V2 V
From equation:
22.4 litres of methane requires oxygen = 44.8 litres O_{2}
2H_{2} + O_{2} _{}2H_{2}O
2 V1 V2 V
From equation,
44.8 litres hydrogen requires oxygen = 22.4 litres O_{2}
So, 11.2 litres will require = 22.4 x 11.2/44.8 = 5.6 litres
Total volume = 44.8 + 5.6 = 50.4 litres
Solution E 37
Gram atoms of Pb = 6.21/207=0.03 = 1
Gram atoms of Cl = 4.26/35.5 = 0.12 = 4
So, the empirical formula = PbCl_{4}
Solution E 38
Na_{2}SO_{4}+BaCl_{2}_{}BaSO_{4}+2NaCl
Molecular mass of BaSO_{4 }= 233 g
Now, 233 g of BaSO_{4 }is produced by Na_{2}SO_{4 }= 142 g
So, 6.99 g BaSO_{4 }will be produced by = 6.99 _{}142/233 = 4.26
The percentage of Na_{2}SO_{4 }in original mixture = 4.26 _{}100/10
= 42.6%
Solution E 39
(a) 1 litre of oxygen has mass = 1.32 g
So, 24 litres (molar vol. at room temp.) will have mass = 1.32 x 24
= 31.6 or 32 g
(b) 2KMnO_{4}_{}K_{2}MnO_{4} + MnO_{2} + O_{2}
316 g of KMnO_{4} gives oxygen = 24 litres
So, 15.8 g of KMnO_{4 }will give = 24 _{}316/15.8 = 1.2 litres
Solution E 40
(a)
(i) The no. of moles of SO_{2} = 3.2/64 = 0.05 moles
(ii) In 1 mole of SO_{2}, no. of molecules present = 6.02 _{}10^{23 }
So, in 0.05 moles, no. of molecules = 6.02_{}10^{23 }_{}0.05
= 3.0 _{}10^{22 }
(iii) The volume occupied by 64 g of SO_{2} = 22.4 dm^{3}
3.2 g of SO_{2} will be occupied by volume= 22.4 _{}3.2/64 =1.12 dm^{3}
Solution E 41
(i) D contains the maximum number of molecules because volume is directly proportional to the number of molecules.
(ii) The volume will become double because volume is directly proportional to the no. of molecules at constant temperature and pressure.
V_{1}/V_{2} = n_{1}/n_{2 }
V_{1}/V_{2} = n_{1}/2n_{1}
So, V_{2 }= 2V_{1}
(iii) Gay lussac's law of combining volume is being observed.
(iv) The volume of D = 5.6 _{}4 = 22.4 dm^{3}, so the number of molecules = 6 x 10^{23} because according to mole concept 22.4 litre volume at STP has = 6 x 10 ^{23} molecules
(v) No. of moles of D = 1 because volume is 22.4 litre
so, mass of N_{2}O = 1 _{}44 = 44 g
Solution E 42
(a) NaCl+NH_{3}+ CO_{2 }+ H_{2}O_{}NaHCO_{3}+NH_{4}Cl
2NaHCO_{3} _{}Na_{2}CO_{3}+H_{2}O + CO_{2}
From equation:
106 g of Na_{2}CO_{3 }is produced by = 168 g of NaHCO_{3}
So, 21.2 g of Na_{2}CO_{3 }will be produced by = 168 _{}21.2/106
= 33.6 g of NaHCO_{3}
(b) For 84 g of NaHCO_{3}, requiredvolume of CO_{2} = 22.4 litre
So, for 33.6 g of NaHCO_{3}, required volume of CO_{2} = 22.4 x 33.6/84
= 8.96 litre
Solution E 43
(a) NH_{4}NO_{3}_{}N_{2}O+2H_{2}O
1mole1mole2mole
1 V1 V2 V
44.8 litres of water produced by = 22.4 litres of NH_{4}NO_{3}
So, 8.96 litres will be produced by = 22.4 x 8.96/44.8
= 4.48 litres of NH_{4}NO_{3}
So, 4.48 litres of N_{2}O is produced.
(i) 44.8 litre H_{2}O is produced by = 80 g of NH_{4}NO_{3}
So, 8.96 litre H_{2}O will be produced by = 80 x 8.96/44.8
= 16g NH_{4}NO_{3}
(iii) % of O in NH_{4}NO_{3} = 3x16/80 = 60%
Solution E 44
Molecular mass of =(240) g
Molecular mass of = 2 × 22.4 = 44.8dm^{3}
Molecular mass of = (192)g
240 g of CuO requires 44.8 dm^{3 }of NH_{3}
120g of CuO will require
=22.4dm^{3 }
Solution E 45
(a) The molecular mass of ethylene(C_{2}H_{4}) is 28 g
No. of moles = 1.4/28 = 0.05 moles
No. of molecules = 6.023 x10^{23} x 0.05 = 3 x 10^{22 }molecules
Volume = 22.4 x 0.05 = 1.12 litres
(b) Molecular mass = 2 X V.D
S0, V.D = 28/2 = 14
Solution E 46
(a) Molecular mass of Na_{3}AlF_{6 }= 210
So, Percentage of Na = 3x23x100/210 = 32.85%
(b) 2CO + O_{2}_{}2CO_{2}
2 V1 V2 V
1 mole of O_{2} has volume = 22400 ml
Volume of oxygen used by 2 x 22400 ml CO = 22400 ml
So, Vol. of O_{2} used by 560 ml CO =22400 x 560/(2 x 22400)
= 280 ml
So, Volume of CO_{2} formed is 560 ml.
Solution E 47
(a)
i. The combustion reaction
According to GayLussac's law,
2 volume of acetylene requires 5 volume of oxygen to burn it
1 volume of acetylene requires 2.5 volume of oxygen to burn it
200cm^{3 }requires 2.5×200=500 cm^{3 }of oxygen
2 volume of acetylene on combustion gives 4CO_{2 }
1 volume of acetylene on combustion gives 2CO_{2 }
200cc of acetylene on combustion will give 200×2=400cc of CO_{2}
(b) Hydrogen = 12.5%
∴ Nitrogen= 10012.5= 87.5%
Element 
% Weight 
Atomic Weight 
Atomic Ratio 
Simplest Ratio 
N 
87.5 
14 
87.5/14=6.25 
6.25/6.25=1 
H 
12.5 
1 
12.5/1=12.5 
12.5/6.25=2 
The Empirical formula of the compound is NH_{2}
_{ }Empirical formula weight =14+2=16
Relative molecular mass =37
N = 2.3≈2
Molecular formula = n x empirical formula = 2 x NH_{2}
=N_{2}H_{4 }
(c)
Molecules of nitrogen gas in a cylinder = 24 x 10^{24}
Avogadro's number = 6 x 10^{23 }
(i) Mass of nitrogen in a cylinder =
=1120g
(ii) Volume of nitrogen at stp
Volume of 28 g of N_{2} = 22.4dm^{3 }
Volume of 1120g of N_{2 }= dm^{3}
=896 dm^{3}
(d) NaCl + AgNO_{3} ⟶ AgCl + NaNO_{3}
As, 143 g of AgCl is obtained from = 58 g of NaCl
So, 14.3 g of AgCl will be obtained from = 58 × 14.3 / 143 = 5.8 g of NaCl
Weight of commercial NaOH = 30 g
Percentage of NaCl in NaOH = 5.8 × 100 / 30 = 19.33 %
(e) At STP, 100 cm^{3} of gas weighs = 0.5 g
So, at STP 22400 cm^{3} of gas will weigh = 0.5 x 22400 / 100 = 112 g
Solution E 48
(a)
(i) C_{3}H_{8(g)} + 5O_{2(g)} ⟶ 3CO_{2(g)} + 4H_{2}O_{(g)}
2C_{4}H_{10(g)} + 13O_{2(g)} + 8CO_{2(g)} + 10H_{2}O_{(g)}
60 ml of propane (C_{3}H_{8}) gives 3 × 60 = 180 ml CO_{2}
40 ml of butane (C_{4}H_{10}) gives = 8 × 40 / 2 = 160 ml of CO_{2}
Total carbon dioxide produced = 340 ml
So, when 10 litres of the mixture is burnt = 34 litres of CO2 is produced.
(ii) Molecular mass of NH_{4}(NO_{3}) =80
H=1, N=14, O=16
% of Nitrogen
As 80 g of NH4(NO3) contains 28 g of nitrogen
100 g of of NH_{4}(NO_{3}) will contain
= 35%
% of Oxygen
As,80 g of NH4(NO3) contains 48 g of oxygen
100 g of of NH_{4}(NO_{3}) will contain
= 60%
(b)
(i) Equation for reaction of calcium carbonate with dilute hydrochloric acid:
(ii) Relative molecular mass of calcium carbonate=100
Mass of 4.5 moles of calcium carbonate
= No. of moles× Relative molecular mass
= 4.5×100
= 450g
(iii)
As, 100g of calcium carbonate gives 22.4dm^{3} of CO_{2}
450 g of calcium carbonate will give
=100.8 L
(iv) Molecular mass of calcium carbonate =100
Relative molecular mass of calcium chloride =111
As 100 g of calcium carbonate gives 111g of calcium chloride
450 g of calcium carbonate will give
=499.5 g
(v) Molecular mass of HCl=36.5
Molecular mass of calcium carbonate =100
As 100 g of calcium carbonate gives (2×36.5)= 73g of HCl
450 g of calcium carbonate will give
=328.5g
Number of moles of HCl=
=
= 9 moles
Solution E 49
(a)
(i) Atomic mass: S = 32 and O = 16
Molecular mass of SO_{2}=32+(2×16)
=64g
As 64 g of SO_{2} = 22.4dm^{3}
Then, 320 g of SO_{2} =
=112 L
(ii) GayLussac's law GayLussac's Law states "When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure."
(iii) C_{3}H_{8} + 5O_{2}→3CO_{2} + 4H_{2}O
Molar mass of propane = 44
44 g of propane requires 5 × 22.4 litres of oxygen at STP.
8.8 g of propane requires = 22.4 litres
(b)
(i)
Element 
Relative atomic mass 
%Compound 
Atomic ratio 
Simplest ratio 
H 
1 
2.13 
2.13/1=2.13 
2 
C 
12 
12.67 
12.67/12=1.055 
1 
Br 
80 
85.11 
85.11/80=1 
1 
Empirical formula = CH_{2}Br
n(Empirical formula mass of CH_{2}Br) = Molecular mass (2 × VD)
n(12 + 2 + 80) = 94 × 2
n = 2
Molecular formula = Empirical formula × 2
= (CH_{2}Br) × 2
= C_{2}H_{4}Br_{2}
(ii)
1. 10^{22} atoms of sulphur
6.022 × 10^{23} atoms of sulphur will have mass = 32 g
10^{22} atoms of sulphur will have mass =
= 0.533 g
2. 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 E 50
(a)
(i) Number of moles of phosphorus taken =
= 0.3 mol
(ii) 1 mole of phosphorus gives 98 gm of phosphoric acid.
So, 0.3 mole of phosphorus gives (0.3 × 98) gm of phosphoric acid
= 29.4 gm of phosphoric acid
(iii) 1 mole of phosphorus gives 112 L of NOgas at STP.
So, 0.3 mole of phosphorus gives (112 × 0.3) L of
NOgas at STP.
= 33.6 L of NOgas at STP
(b)
(i) According to the equation
3 volumes of hydrogen produce 2 volumes of ammonia
67.2 litres of hydrogen produce = 44.8 L
3 volumes of hydrogen combine with 1 volume of ammonia.
67.2 litres of hydrogen combine with =22.4L Nitrogen left = 44.8  22.4 = 22.4 litres
(ii) 5.6 dm^{3} of gas weighs 12 g
1 dm^{3} of gas weighs = (12/56) gm
22.4 dm^{3} of gas weighs = (12/56 × 22.2) gm = 48g
Therefore, the relative molecular mass of gas = 48 gm.
(iii) Molar mass of Mg (NO_{3})_{2}.6H_{2}O
= 24 × (14 × 2) + (16 × 12) + (1 × 12) = 256 g
Mass percent of magnesium =
Solution E 51
(a)
(i)
2 vols. of butane requires O_{2} = 13 vols
90 dm^{3} of butane will require O_{2} = × 90
= 585 dm^{3}
(ii) Molecular mass = 2 × Vapour density
So, molecular mass of gas = 2 × 8 = 16 g
As we know, molecular mass or molar mass occupies 22.4 litres.
That is,
16 g of gas occupies volume = 22.4 litres
So, 24 g of gas will occupy volume
=
(iii) According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
So, molecules of nitrogen gas present in the same vessel = X
(b)
(i)
3 vols. of oxygen require KClO_{3} = 2 vols.
So, 1 vol. of oxygen will require KClO_{3} =
So, 6.72 litres of oxygen will require KClO_{3}
=
22.4 litres of KClO_{3} has mass = 122.5 g
So, 4.48 litres of KClO_{3} will have mass
=
(ii) 22.4 litres of oxygen = 1 mole
So, 6.72 litres of oxygen =
No. of molecules present in 1 mole of O_{2}
= 6.023 × 10^{23}
^{ }So, no. of molecules present in 0.3 mole of O_{2}
= 6.023 × 10^{23} × 0.3
= 1.806 × 10^{23}
(iii) Volume occupied by 1 mole of CO_{2} at STP = 22.4 litres
So, volume occupied by 0.01 mole of CO_{2} at STP = 22.4 × 0.01= 0.224 litres
Solution E 52
(a)
(i)
2 moles of C_{2}H_{2}=4moles of CO_{2}
x dm^{3} of C_{2}H_{2 } =8.4 dm^{3} of CO_{2}
x=
=4.2 dm^{3} of C_{2}H_{2 }
(ii) Empirical formula= X_{2}Y
Atomic weight (X)= 10
Atomic weight (Y)= 5
Empirical formula weight = (2 × 10) + 5
=25
= 2
So, molecular formula = X_{2}Y×2
= X_{4}Y_{2 }
(b)
(i) A cylinder contains 68 g of ammonia gas at STP.
Molecular weight of ammonia = 17 g/mole
68 g of ammonia gas at STP =?
1 mole = 22.4 dm^{3}
∴ 4 mole = 22.4 × 4 = 89.6 dm^{3}
(ii) 4 moles of ammonia gas is present in the cylinder.
(iii) 1 mole = 6.023 × 10^{23} molecules
4 moles = 24.092 × 10^{23} molecules