Class 10 SELINA Solutions Chemistry Chapter 5  Mole Concept And Stoichiometry
Mole Concept And Stoichiometry Exercise Ex. 5A
Solution 1
(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 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 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 4
(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 5
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 6
2C_{2}H_{2} + 5O_{2 }_{}4CO_{2} + 2H_{2}O (l)
2 V 5 V 4 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 7
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 8
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 9
C_{2}H_{4}+3O_{2}2CO_{2} + 2H_{2}O
1V 3V
11litre 33 litre
Solution 10
CH_{4} + 2Cl_{2} _{}CH_{2}Cl_{2 }+2HCl
1 V 2 V 1 V 2 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 11
C_{3}H_{8 }+ 5O_{2} _{}3CO_{2} + 4H_{2}O
1 V 5 V 3 V
From equation, 5 V of O_{2} required = 1V of propane
so, 100 cm^{3} of O_{2 }will require = 20 cm^{3} of propane
Solution 12
2NO + O_{2} _{}2NO_{2}
2 V 1 V 2 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 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
4NH_{3} + 5O_{2} _{}4NO + 6H_{2}O
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
H_{2} + Cl_{2 }2HCl
1V 1V 2 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 16
CH_{4} + 2O_{2 }_{}CO_{2} + 2H_{2}O
1 V 2 V 1 V
2C_{2}H_{2} + 5O_{2} _{}4CO_{2} + 2H_{2}O
2 V 5 V 4 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 17
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 18
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 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 SO_{2} 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.
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 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 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 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 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
1V 1V 2V(By GayLussacs 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) 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 5
(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 6
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 7
4g of NH_{3} having minimum molecular mass contain maximum molecules.
Solution 8
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 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 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 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 SO_{2} has volume = 22.4 litres
So, 2 moles will have = 22.4 x 2 = 44.8 litre
Solution 11
(a) 1 mole of 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 12
(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 = 108/6.023 x 10^{23 }=1.701 x 10^{22}
(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 13
(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 14
(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
= 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 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 16
(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 17
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 18
(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 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 cm^{3} of CO has mass = 28 g
So, 56 cm^{3} will have mass = 56 x 28/22400 = 0.07 g
Solution 21
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 22
(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 23
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 24
(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 25
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 26
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 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 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 29
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 30
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 31
(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.
Mole Concept And Stoichiometry Exercise Ex. 5C
Solution 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 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) CH_{2}O (c) CH (d) CH_{2}O
Solution 4
Solution 5
(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 6
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 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(NO_{3})_{2 }is the empirical formula.
Solution 8
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 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 =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 10
Element % At. mass Atomic ratio Simple ratio
H 17.7 1 _{}_{}
N 82.3 14 _{}_{}
So, the empirical formula = NH_{3}
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 = 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 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 = 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 13
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 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 = 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 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
Cl21.335.5_{}3
Empirical formula = SiCl_{3}
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 = 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 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 gramatoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg_{3 }N_{2}
Solution 18
Barium chloride = BaCl_{2}.x H_{2}O
Ba + 2Cl + x[H_{2} + O]
=137+ 235.5 + x [2+16]
=[208 + 18x] contains water = 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 19
Molar mass of urea; CON_{2}H_{4 }= 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 CH_{2}O
Since the compound has 12 atoms of carbon, so the formula is
C_{12} H_{24} O_{12.}
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 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 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 NH_{2}
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 =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) 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 2
(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 3
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 4
(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 5
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 6
CaCO_{3} + 2HCl_{}CaCl_{2} + H_{2}O + CO_{2}
100g73g22.4L
(a)V_{1} =2 litresV_{2} =?
T_{1} = (273+27)=300KT_{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 =100/22.4 2274/300 =.125g
(b)22.4 litres are CO_{2} are prepared from acid =73g
_{}litres are prepared from = 73/22.4 2273/300=5.9g
Solution 7
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 8
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 9
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 10
Solution 11
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 12
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 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 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 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 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 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)
_{}
(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 = 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 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 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 = XO_{2}
Solution 10
(a) V.D = _{}
(b) Molecular mass = 17(V.D) x 2= 34g
Solution 11
(a) CO_{2} + 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 + O_{2} _{}2CO_{2}
2 V 1 V 2 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 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 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 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 =SiCl_{3}
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 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 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 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 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 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}=760mm P_{2} =740mm
V_{1}=1400cc V_{2} =?
T_{1} =273 K, T_{2} = 27 +73 = 300K
(d) P_{1}= 750mm P_{2}=760mm
V_{1}= 44lit. V_{2}=?
T_{1}= 298K T_{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 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 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 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 25
2H_{2}S + 3O_{2} _{}2H_{2}O + 2SO_{2}
2 V 3 V 2 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 26
From equation, 2NH_{3} + 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 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 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 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 30
Na_{2}CO_{3}.10H_{2}O _{}Na_{2}CO_{3} + 10H_{2}O
286 g 106 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 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 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 33
4N_{2}O + CH_{4} _{}CO_{2} + 2H_{2}O + 4N_{2}
4 V 1 V 1 V 2 V 4 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 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 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 36
CH_{4} + 2O_{2} _{}CO_{2} + 2H_{2}O
1 V 2 V 1 V 2 V
From equation:
22.4 litres of methane requires oxygen = 44.8 litres O_{2}
2H_{2} + O_{2} _{}2H_{2}O
2 V 1 V 2 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 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 _{}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 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 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 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}
(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 = PbCl_{4}
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.
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 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 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 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 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 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 V 1 V 2 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 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 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}
ii. 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.
i. Molecules of nitrogen gas in a cylinder = 24 x 10^{24}
Avogadro's number = 6 x 10^{23 }
1. Mass of nitrogen in a cylinder =
=1120g
2. 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}
Solution 48
a.
i. 10 litres of LPG contains
Propane
Butane
18+16=34 L
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
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 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 
2 
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. 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
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 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 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 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