# ICSE Solutions for Chapter 5 Mole Concept and Stoichiometry Class 10 Selina Chemistry

### Exercise 5 A

Solution 1.

(a) Gay-Lussac’s law states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

(b) Avogadro’s law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Solution 2.

(a) 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.

(b) Nmeans 1 molecule of nitrogen and 2N means two atoms of nitrogen.
N2 can exist independently but 2N cannot exist independently.

Solution 3.

(a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
Now the volume of hydrogen gas = volume of helium gas
n molecules of hydrogen = n molecules of helium gas
nH2 = nHe
1 mol. of hydrogen has 2 atoms of hydrogen and 1 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 to violate Boyles law as the volume is increasing with the increase in pressure. Since the mass of gas is also increasing.

Solution 4.

From the equation, 2V of hydrogen reacts with 1V of oxygen
so 200cm3 of Hydrogen reacts with = 200/2= 100 cm3
Hence, the unreacted oxygen is 150 – 100 = 50cm3 of oxygen.

Solution 5:
From the equation, 1V of CHreacts with = 2V of O2
so, 80 cm3 CH4 reacts with = 80×2 = 160cm3 O2
Remaining O2 is 200-160 = 40cm3
From the equation 1V of methane gives 1V of CO2
So, 80 cm3 gives 80cm3 CO2 and H2O is negligible.

Solution 6:

From equation, 2V of C2H requires = 5 V of O2
So, for 400ml C2H2, O2 required = 400 × 5/2 = 1000 ml
Similarly, 2 V of C2H2 gives = 4 V of CO2
So, 400ml of C2H2  gives CO2 = 400 × 4/2 = 800ml

Solution 7:
Balanced chemical equation:

(i) At STP, 1 mole gas occupies 22.4 L.
As 1 mole H2S gas produces 2 moles HCl gas,
22.4 L H2S gas produces 22.4 × 2 = 44.8 L HCl gas.
Hence, 112 cmH2S gas will produce 112 × 2 = 224 cm3 HCl gas.

(ii) 1 mole H2S gas consumes 1 mole Cl2 gas.
This means 22.4 L H2S gas consumes 22.4 L Cl2 gas at STP.
Hence, 112 cm3 H2S gas consumes 112 cm3 Cl2 gas.
120 cm3 - 112 cm3 = 8 cm3 Cl2 gas remains unreacted.
Thus, the composition of the resulting mixture is 224 cm3 HCl gas + 8 cm3
Cl2 gas.

Solution 8:

Now from equation, 2V of ethane reacts with = 7 V of oxygen
So, 600cc of ethane reacts with = 600 × 7/2 = 2100cc
Hence, unused Ois = 2500 - 2100 = 400 cc
From 2V of ethane = 4 V of CO2 is produced
So, 600cc of ethane will produce = (4 × 600/2) = 1200cc CO2

Solution 9:

P1V1/T1 = P2V2/T2
V= P1V1T2/P2T1
= (380 × 33 × 273)/(549 ×760) = 8.25 litres

Solution 10:

From equation, 1V of CH4 gives = 2V HCI
so, 40 ml of methane gives 80 ml HCI
For 1V of methane = 2V of Cl2 required
So, for 40ml of methane = 40 × 2 = 80 ml of Cl2

Solution 11.
From equation, 5V of O2 required = 1V of propane
So, 100 cm3 of O2 will require = 20 cm3 of propane

Solution 12:

From equation, 1V of O2 reacts with = 2 V of NO
200cm3 oxygen will react with = 200 × 2 = 400 cm3 NO
Hence, remaining NO is 450 - 400 = 50 cm3
NO2 produced = 400cm3 because 1V oxygen gives 2 V NO2
Total mixture = 400 + 50 = 450 cm3

Solution 13.

2 V of CO requires = 1V of O2
so, 100 litres of CO requires = 50 litres of O2

Solution 14:
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.

Since 1 V hydrogen requires 1 V of oxygen and 4cm3 of H2 remained behind so the mixture had com ">16 cm3 hydrogen and 16 cm3 chlorine.
Therefore Resulting mixture is H2 = 4cm3, HCl = 32cm3

Solution 16.

From the equations, we can see that
1V CH4 requires oxygen = 2V O2
So, 10cm3 CH4 will require = 20 cm3 O2
Similarly 2V C2H2 requires = 5V O2
So, 10 cm3 C2H2 will require = 25 cm3 O2
Now, 20 V O2 will be present in 100 V air and 25V O2 will be present in 125 V air, so the volume of air required is 225 cm3

Solution 17:

C3H8 + 5O2 + 3CO2 + 4H2O
2C4H10 + 13O2 → 8CO2 + 10H2O
60 ml of propane (C3H8) gives 3 × 60 = 180 ml CO2
40 ml of butane (C4H10) gives = 8 × 40/2 = 160 ml of CO2
Total carbon dioxide produced = 340 ml
So, when 10 litres of the mixture is burnt = 34 litres of CO2 is produced.

Solution 18:

2C2H2(g)  + 5O2(g) → 4CO2(g) + 2H2O(g)
4V CO2 is collected with 2 V C2H2
So, 200m3 CO2 will be collected with = 100cm3 C2H2
Similarly, 4V of CO2 is produced by 5 V of O2
So, 200cm3 CO2 will be produced by = 250 ml of O2

Solution 19:

This experiment supports Gay lussac’s law of combining volumes.
Since the unchanged or remaining O2 is 58 cc so, used oxygen 106 - 58 = 48cc
According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.

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

Solution 20:

Solution 21:

100 cm3 of oxygen contains = Y molecules
50 cm3 nitrogen contains = 50Y/100 = Y/2

Exercise 5(B)

Solution 1:

a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C-12.
b) The value of Avogadro's number is 6.023 × 1023
c) The molar volume of a gas at STP is 22.4 dm3 at STP

Solution 2:

(a) The vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.

(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm3.

(c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.

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

(e) The number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 × 1023 atoms.

(f) The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.
(g) Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.

Solution 3:

(a) Applications of Avogadro’s Law :
1. It explains Gay-Lussac’s law.
2. It determines the atomicity of the gases.
3. It determines the molecular formula of a gas.
4. It determines the relation between molecular mass and vapour density.
5. It gives the relationship between gram molecular mass and gram molecular volume.

(b) 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 a simple ratio by a 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.
n molecules : n molecules : 2n molecules (By Avogadro's law)

Solution 4:

(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5×6 = 444
(b) KClO3 = (K)39 + (Cl)35.5 + (3O)48 = 122.5
(c) (Cu)63.5 + (S)32 + (4O)64 + (5H2O)5×18 = 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)3×35.5 = 119.5
(g) (2N)28 + (8H)8 + (2Cr)2×51.9 + (7O)7×16 = 252

Solution 5:

(a) No. of molecules in 73 g HCl = (6.023×1023 ×73)/36.5(mol. mass of HCl)
= 12.04×1023

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

(c) No. of molecules in 1.8 g H2O = (6.023×1023 ×1.8)/18
= 6.023×1022

(d) No. of moles in 10g of CaCO3 = 10/100(mol. Mass CaCO3)
= 0.1 mole

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

(f) No. of molecules in 3.2 g of SO2 = (6.023×1023×3.2)/64
= 3.023×1022

Solution 6:

The molecular mass of H2O is 18, CO2  is 44, NH3 is 17 and CO is 28
So, the weight of 1 mole of CO2  is more than the other three.

Solution 7:

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

Solution 8:

a) No. of particles in s1 mole = 6.023 × 1023
So, particles in 0.1 mole = 6.023 × 1023 × 0.1 = 6.023 × 1022
b) 1 mole of H2SO4  contains = 2 × 6.023 × 1023
So, 0.1 mole of H2SO4 contains = 2 × 6.023 × 1023 × 0.1
= 1.2 × 1023 atoms of hydrogen
c) 111g CaCl2 contains = 6.023 × 1023 molecules
So, 1000 g contains = 5.42 × 1024 molecules

Solution 9:

(a) 1 mole of aluminium has mass = 27 g
So, 0.2 mole of aluminium has mass = 0.2 × 27 = 5.4 g
(b) 0.1 mole of HCl has mass = 0.1 × 36.5 (mass of 1 mole)
= 3.65 g
(c) 0.2 mole of H2O has mass = 0.2×18 = 3.6 g
(d) 0.1 mole of CO2 has mass = 0.1×44 = 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 × 22.4/5.6
= 48g(molar mass)
(b) 1 mole of SO2 has volume = 22.4 litres
So, 2 moles will have = 22.4 × 2 = 44.8 litre

Solution 11:

(a) 1 mole of CO2 contains O2 = 32g
So, CO2 having 8 gm of O2 has no. of moles = 8/32 = 0.25 moles
(b) 16 g of methane has no. of moles = 1
So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles

Solution 12:

(a) 6.023×1023  atoms of oxygen has mass = 16 g
So, 1 atom has mass = 16/6.023 × 1023 = 2.656 × 10-23  g
(b) 1 atom of Hydrogen has mass = 1/6.023 × 1023  = 1.666 × 10-24
(c) 1 molecule of NH3 has mass = 17/6.023 × 1023 = 2.82 × 10-23  g
(d) 1 atom of silver has mass = 108/6.023 ×1023 = 1.701 × 10-22
(e) 1 molecule of O2  has mass = 32/6.023 × 1023  = 5.314 × 10-23  g
(f) 0.25 gram atom of calcium has mass = 0.25 × 40 = 10g

Solution 13:

(a) 0.1 mole of CaCO3 has mass = 100(molar mass) × 0.1 = 10 g
(b) 0.1 mole of Na2SO4.10H2O has mass = 322 × 0.1 = 32.2 g
(c) 0.1 mole of CaCl2 has mass = 111 × 0.1 = 11.1g
(d) 0.1 mole of Mg has mass = 24 × 0.1 = 2.4 g

Solution 14:

1molecule of Na2CO3.10H2O contains oxygen atoms = 13
So, 6.023 ×1023  molecules (1mole) has atoms = 13 × 6.023 × 1023
So, 0.1 mole will have atoms = 0.1 × 13 × 6.023 × 1023  = 7.8 × 1023

Solution 15:

3.2 g of S has number of atoms = (6.023 × 1023 × 3.2)/32
= 0.6023 × 1023
So, 0.6023 × 1023 atoms of Ca has mass = (40 × 0.6023 × 1023)/(6.023 × 1023)
= 4g

Solution 16:

(a) No. of atoms = 52 × 6.023 × 1023  = 3.131 × 1025
(b) 4 amu = 1 atom of He
so, 52 amu = 13 atoms of He
(c) 4 g of He has atoms = 6.023 × 1023
So, 52 g will have = (6.023 ×1023 × 52)/4 = 7.828 ×1024 atoms

Solution 17:

Molecular mass of Na2CO= 106 g
106 g has 2 x 6.023 x 1023  atoms of Na
So, 5.3g will have = (2 × 6.023 × 1023 × 5.3)/106 = (6.022 × 1022 )atoms
Number of atoms of C = (6.023 ×1023 ×5.3)/106 = 3.01 ×1022  atoms
And atoms of O = (3 ×6.023 ×1023 ×5.3)/106 = 9.03×1022 atoms

Solution 18:

(a) 60 g urea has mass of nitrogen(N2) = 28 g
So, 5000 g urea will have mass = 28 × 5000/60 = 2.33 kg
(b) 64 g has volume = 22.4 litre
So, 320 g will have volume = 22.4×320/64 = 112 litres

Solution 19:

(a) Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.
(b) Vapour density of Chlorine atom is 35.5.

Solution 20:

22400 cm3 of CO has mass = 28 g
So, 56 cm3 will have mass = 56 ×28/22400 = 0.07 g

Solution 21:

18 g of water has number of molecules = 6.023 ×1023
So, 0.09 g of water will have no. of molecules =  (6.023× 1023 ×0.09)/18 = 3.01 × 1021 molecules

Solution 22:

(a) No. of moles in 256 g S8 = 1 mole
So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles
(b) No. of molecules = 0.02 × 6.023 × 1023 = 1.2 × 1022 molecules
No. of atoms in 1 molecule of S = 8
So, no. of atoms in 1.2 × 1022 molecules = 1.2 × 1022 × 8
= 9.635 × 1022 molecules

Solution 23:

The atomic mass of phosphorus P = 30.97 g
Hence, the molar mass of P4 = 123.88 g
If phosphorus is considered as P4 molecules,
then 1 mole P4 ≡ 123.88 g
Therefore, 100 g of P4 = 0.807 g

Solution 24:

(a) 308 cm3 of chlorine weighs = 0.979g
So, 22400 cm3 will weigh = gram molecular mass
= 0.979 ×22400/308 = 71.2 g

(b) 2 g(molar mass) H2 at 1 atm has volume = 22.4 litres
So, 4g H2 at 1 atm will have volume = 44.8 litres
Now, at 1 atm(P1) 4g H2 has volume (V1) = 44.8 litres
So, at 4 atm(P2) the volume(V2) will be P1V1/P2 = (1 × 44.8)/4 = 11.2 litres

(c) Mass of oxygen in 22.4 litres = 32 g(molar mass)
So, mass of oxygen in 2.2 litres = (2.2 × 32)/22.4 = 3.14 g

Solution 25:

No. of atoms in 12 g C = 6.023 × 1023
So, no. of carbon atoms in 10-12 g = (10-12 × 6.023 × 1023)/12
= 5.019 × 1010 atoms

Solution 26:

Given:
P = 1140 mm Hg
Density = D = 2.4 g / L
T = 273°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 an unknown gas at a standard temperature of 273 K.
V1= 1 L, T1 = 546 K
V2 = ?, T= 273 K
V1/T1 = V2/ T2
V2 = (V1 × T2)/T1
= (1 L × 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.
P1 = 1140 mm Hg, V1 = 0.5 L
P2 = 760 mm Hg, V2 = ?
P1 × V1 = P2 × V2
V2 = (P1 × V1)/P2
= (1140 mm Hg × 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 = (342 × 40)/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 x 1023 atoms of C weigh = 12 g
So, 3 x 1025 atoms will weigh = (12 × 3 × 1025)/(6.023 × 1023) = 597.79

(c) 1 mole of sulphur weighs = 32g

(d) 7 g of silver
So, 7 grams of silver weighs least.

Solution 29:

40 g of NaOH contains 6.023 × 1023 molecules
So, 4 g of NaOH contains = (6.02 × 1023 × 4)/40
= 6.02 ×1022  molecules

Solution 30:

The number of molecules in 18 g of ammonia = 6.02 ×1023
So, no. of molecules in 4.25 g of ammonia = (6.02 ×1023 ×4.25)/18
= 1.5 × 1023

Solution 31:

(a) One mole of chlorine contains 6.023 ×1023 atoms of chlorine.
(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.
(c) The relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon-12.
(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

### Exercise 5 C

Solution 1:

Information conveyed by H2O
1. That H2O contains 2 volumes of hydrogen and 1 volume of oxygen.
2. That ratio by weight of hydrogen and oxygen is 1:8.
3. That molecular weight of H2O is 18g.

Solution 2:

The empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.
The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

Solution 3.
(a) CH
(b) CH2
(c) CH
(d) CH2O

Solution 4:

Relative mol. mass of CuSO4.5H2O = 63.5 + 32 + (16 × 4) + 5(1×2 + 16)
= 249.5 g
249.5 g of CuSO4.5H2O contains water of crystallization = 90 g
So, 100 g will contain = (90 ×100)/249.5 = 36.07 g
So, % of H2O = 36.07 × 100 = 36.07%

Solution 5:

(a) Molecular mass of Ca(H2PO4)2 = 234
So, % of P = (2 × 31 × 100)/234 = 26.5%

(b) Molecular mass of Ca3(PO4)2 = 310
% of P = (2 × 31 × 100)/310 = 20%

Solution 6:

Molecular mass of KClO3 = 122.5 g
% of K = 39 /122.5 = 31.8%
% of Cl = 35.5/122.5 = 28.98%
% of O = 3 × 16/122.5 = 39.18%

Solution 7:

 Element % At. mass Atomic ratio Simple ratio Pb 62.5 207 62.5/207 0.3019 N 8.5 14 8.5/14 0.6071 O 29.0 16 29.0/16 1.81
So, Pb(NO3)2 is the empirical formula.

Solution 8:

In Fe2O3, Fe = 56 and 0 = 16
Molecular mass of Fe2O3= 2 × 56 + 3 × 16 = 160 g
Iron present in 80% of Fe2O3 = 112/(160 × 80) = 56g
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 mass/Empirical formula weight  = 26/13 = 2
Molecular formula of acetylene = 2 × Empirical formula = C2H2
Similarly, for benzene molecular mass = 2 × V.D = 2 × 39 = 78
n = 78/13 = 6
So, the molecular formula = C6H6

Solution 10:

 Element % At. mass Atomic ratio Simple ratio H 17.71 1 17.7/1 = 17.7 17.7/5.87 = 3 N 82.3 14 82.3/14 = 5.87 5.87/5.87 = 1
So, the empirical formula = NH3

Solution 11:

 Element % At. mass Atomic ratio Simple ratio C 54.54 12 54.54/12 4.55 H 9.09 1 9.09/1 9.09 O 36.36 16 36.36/16 2.27
(a) So, its empirical formula C2H4O
(b) empirical formula mass = 44
Since, vapour density = 44
So, molecular mass = 2 × V.D = 88
Or, n = 2
So, molecular formula = (C2H4O)2 = C4H8O2

Solution 12:

 Element % At. mass Atomic ratio Simple ratio C 26.59 12 26.59/12 2.21 H 2.22 1 2.22/1 2.22 O 71.19 16 71.19/16 4.44
(a) its empirical formula = CHO2
(b) empirical formula mass = 45
Vapour density = 45
So, molecular mass = V.D × 2 = 90
So, molecular formula = C2H2O4

Solution 13:

 Element % At. mass Atomic ratio Simple ratio Cl 71.65 35.5 71.65/35.5 2.01 H 4.07 1 4.07/1 4.07 C 24.28 12 24.28/12 2.02
(a) its empirical formula = CH2Cl
(b) empirical formula mass = 49.5
Since, molecular mass = 98.96
so, molecular formula = (CH2Cl)2 = C2H4Cl2

Solution 14:

(a) The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1 = 1
(b)
 Element Given mass At. mass Gram atom Ratio C 4.8 12 0.4 12 H 1 1 1 5
So, the empirical formula = C2H5
(c) Empirical formula mass = 29
Molecular mass = V.D × 2 = 29 × 2 = 58
So, molecular formula = C4H10

Solution 15:

Since, g atom of Si = given mass/mol. Mass
So, given mass = 0.2 × 28 = 5.6 g
 Element mass At. mass Gram atom Ratio Si 5.6 28 0.2 1 Cl 21.3 35.5 21.3/35.5 0.6
Empirical formula = SiCl3

Solution 16:

 Element mass At. mass Gram atom Ratio C 92.3 12 92.3/12 7.7 H 7.7 1 7.7/1 7.7
So, empirical formula is CH
Empirical formula mass = 13
Since molecular mass = 78
So, n = 6
∴ molecular formula is C6H6

Solution 17:

(a) G atoms of magnesium = 18/24 = 0.75 or g- atom of Mg
(b) G atoms of nitrogen = 7/14 = 0.5 or 1/2 g- atoms of N
(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg3N2

Solution 18:

Barium chloride = BaCl2.x H2O
Ba + 2Cl + x[H2 + O]
= 137+ 235.5 + x[2+16]
= [208 + 18x] contains water = 14.8% water in BaCl2.xH2O
= [208 + 18x] 14.8/100 = 18x
= [104 + 9x] 2148 = 18000x
= [104+9x] 37 = 250x
= 3848 + 333x  = 2250x
1917x = 3848
x = 2 molecules of water

Solution 19:

Molar mass of urea; CON2H4 = 60 g
So, % of Nitrogen = 28 × 100/60 = 46.66%

Solution 20:

 Element % At. mass Atomic ratio Simple ratio C 42.1 12 3.5 1 H 6.48 1 6.48 2 O 51.42 16 3.2 1
The empirical formula is CH2O
Since the compound has 12 atoms of carbon, so the formula is
C12H24O12.

Solution 21:

(a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, the molecular formula is A2B4.
(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D
Empirical formula weight = V.D/3
So, n = molecular mass/ Empirical formula weight = 6
Hence, the molecular formula is A6B6

Solution 22:

Atomic ratio of N = 87.5/14 = 6.25
Atomic ratio of H = 12.5/1 = 12.5
This gives us the simplest ratio as 1:2
So, the molecular formula is NH2

Solution 23:

 Element % At. mass Atomic ratio Simple ratio Zn 22.65 65 0.348 1 H 4.88 1 4.88 14 S 11.15 32 0.348 1 O 61.32 16 3.83 11
Empirical formula of the given compound = ZnSH14O11
Empirical formula mass = 65.37+ 32 + 141 + 11 + 16 = 287.37
Molecular mass = 287
n = Molecular mass/Empirical formula mass = 287/287=1
Molecular formula = ZnSO11H14
= ZnSO4.7H2O

### Exercise 5 D

Solution 1:

(a) Moles: 1 mole + 2 mole → 1 mole + 2 mole
(b) Grams: 42g + 36g → 745 +4g
(c) Molecules = 6.02 × 1023 + 12.046 × 1023 → 6.02 × 1023 + 12.046 × 1023

Solution 2:

(a)100 g of CaCO3 produces = 164 g of Ca(NO3)2
So, 15 g CaCO3 will produce = (164 × 15/100) = 24.6 g Ca(NO3)2
(b) 1 V of CaCO3 produces 1 V of CO2
100 g of CaCO3 has volume = 22.4 litres
So, 15 g will have volume = (22.4 × 15/100) = 3.36 litres CO2

Solution 3:

2NH3 + H2SO4 → (NH4)2SO4
66 g
(a) 2NH3 + H2SO4 → (NH4)2SO4
For 132 g (NH4)2SO4 = 34 g of NH3 is required
So, for 66 g (NH4)2SO4 = (66 × 32/132) = 17 g of NH3 is required
(b) 17g of NH3 requires volume = 22.4 litres
(c) Mass of acid required, for producing 132g (NH4)2SO4 = 98g
So, Mass of acid required, for 66g (NH4)2SO4 = (66 × 98/132) = 49g

Solution 4:

(a) Molecular mass of Pb3O4 = (3 × 207.2 + 4×16) = 685 g
685 g of Pb3O4 gives = 834 g of PbCl2
Hence, 6.85 g of Pb3O4 will give = (6.85 × 834)/685 = 8.34 g

(b) 685g of Pb3O4 gives = 71g of Cl2
Hence, 6.85 g of Pb3O4 will give = (6.85 × 71)/685 = 0.71 g Cl2

(c) 1 V Pb3Oproduces 1 V Cl2
685g of  Pb3O4 has volume = 22.4 litres = volume of Cl2 produced
So, 6.85 Pb3O4 will produce = (6.85 × 22.4/685) = 0.224 litres of Cl2

Solution 5:

The molecular mass of KNO3 = 101 g
63 g of HNO3 is formed by = 101 g of KNO3
So, 126000 g of HNO3 is formed by = (126000 × 101)/63 = 202 kg
Similarly, 126 g of HNOis formed by 170 kg of NaNO3
So, a smaller mass of NaNO3 is required.

Solution 6:

CaCO3 + 2HCI → CaCl2 + H2O + CO2

(a) V1 = 2 litres V= ?
T1 = (273 + 27) = 300K T2 = 273K
V1/T= V2/T2
V2= V1T2/T= [2 × 273/300]L
Now at STP 22.4 litres of CO2 are produced using CaCO3 = 100g
So, [2 × 273/300]  litres are produced by = 100/22.4  2274/300 = 125g
(b) 22.4 litres are CO2 are prepared from acid = 73g
[2 × 273/300] litres are prepared from = (73/22.4) × (2×273/300) = 5.9g

Solution 7:

2 moles of H2O gives = 1 mole of O2
So, 1 mole of H2O will give = 0.5 moles of O2
So, mass of O2 = no. of moles x molecular mass
= 0.5 × 32 = 16 g of O2
and 1 mole of O2 occupies volume = 22.4 litre
So, 0.5 moles will occupy = 22.4 × 0.5 = 11.2 litres at S.T.P.

Solution 8:

(a) Mol. Mass of Na2O2 = (2 × 23 + 2 × 16) = 78 g
Mass of 2Na2O2 = 156 g
156 g Na2O2  gives = 160 g of NaOH (4 × 40 g)
So, 1.56 Na2O2 will give = (160 × 1.56/156) = 1.6 g

(b) 156 g Na2O2 gives = 22.4 litres of oxygen
So, 1.56 g will give = (22.4 × 1.56/156) = 0.224 litres = 224 cm3

(c) 156 g Na2O2 gives = 32 g O2
So, 1.56 g Na2Owill give = (32 × 1.56/156)
= 32/100 = 0.32 g

Solution 9:

Mol. Mass of 2NH4Cl = 2(14 + (1 × 4) + 35.5] = 2[53.5] = 107 g
(a) 107 g NH4Cl gives = 34 g NH3
So, 21.4 g NH4Cl will give = (21.4 × 34/107) = 6.8 g NH3
(b) The volume of 17 g NH3 is 22.4 litre
So, volume of 6.8 g will be = 6.8 × 22.4/17 = 8.96 litre

Solution 10:

144g 3 × 22.4/volume
Now, since 144 g of Al4C3 gives = 3 × 22.4 litre of CH4
So, 14.4 g of Al4C3 willgive = (3 × 22.4 × 14.4)/144 = 6.72 litres CH4

Solution 11:

(a) 1 mole of MnO2 weighs = 87 g (mol. Mass)
So, 0.02 mole will weigh = 87 × 0.02 = 1.74 g MnO2

(b) 1 mole MnO2 gives = 1 mole of MnCl2
So, 0.02 mole MnO2 will give = 0.02 mole of MnCl2

(c) 1 mole MnCl2 weighs = 126 (g/mol mass)
So, 0.02 mole MnCl2 will weigh = 126 × 0.02 g = 2.52 g

(d) 0.02 mole MnO2 will form = 0.02 mole of Cl2

(e) 1 mole of Cl2  weighs = 35.5 g
So, 0.02 mole will weigh = 71 × 0.02 = 1.42 g of Cl2

(f) 1 mole of chlorine gas has volume = 22.4 litres
So, 0.02 mole will have volume = (22.4 × 0.02) = 0.448 litre

(g) 1 mole MnO2 requires HCl = 4 mole
So, 0.02 mole MnO2 will require = 4 × 0.02 = 0.08 mole

(h) For 1 mole MnO2, acid required = 4 mole of HCI
So, for 0.02 mole, acid required = 4 × 0.02 =0.08 mole
Mass of HCI = 0.08 × 36.5 = 2.92 g

Solution 12:

28g of nitrogen requires hydrogen = 6g
2000g of nitrogen requires hydrogen = 6/28 × 2000 = 3000/7g
So mass of hydrogen left unreacted = 1000 - 3000/7 = 571.4g of H2

(b) 28g of nitrogen forms NH3 = 34g
2000g of N2 forms NH3
= 34/28 × 2000
= 2428.6g

### Miscellaneous Exercise

Solution 1:

From equation: 2H2 + O2 → 2H2O
1 mole of Oxygen gives = 2 moles of steam
So, 0.5 mole oxygen will give = 2 × 0.5 = 1mole of steam

Solution 2:

Mol. Mass of 8HNO3 = 8 × 63 = 504 g
(a) For 504 g HNO3, Cu required is = 192 g
So, for 63g HNO3 Cu required = (192 × 63/504) = 24g

(b) 504 g of HNO3 gives = 2 × 22.4 litre volume of NO
So, 63g of HNO3 gives = (2 x 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 moles

(b) Volume of 71 g of Cl2 at STP = 22.4 litres
Volume of 7.1 g chlorine = (22.4 × 7.1)/71 = 2.24 litre
(c) 22400cm3 volume have mass = 28 g of CO(molar mass)
So, 56cm3 volume will have mass = (28 × 56)/22400 = 0.07 g

Solution 4:

% of Nin NaNO = (14/85 × 100) = 16.47%
% of Nin (NH4)2SO= (14/132 × 100) = 21.21%
% of N in CO(NH2)2 = (14/60 × 100) = 46.66%
So, highest percentage of N is in urea.

Solution 5:

(a) From equation, 2V of water gives 2V of H2 and 1 V of O2
where 2 V = 2500 cm3
so, volume of O2 liberated = 2V/V = 1250 cm3

(b) P1V1/T1  =  P2V2/T2
P1V1/T1  =  (7P1 × V2)/(2 × T1)
V2 = (2500×2)/7
V2 = 5000/7 cm3

(c) V1/V2  = T1/T2
5000/(7 × 2500)  = (T1/T2)
T2  = 3.5 T1
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:

(a) 80% C and 20% H
So, atomic ratio of C and H are: C = 80/12 = 6.66; H = 20/1 = 20
Simple ratio of C:H = 1: 3
So, empirical formula is CH3

(b) Empirical formula mass = 12 + (3 × 1) = 15 g
Vapour density = 15
So, the molecular mass = 15(V.D) × 2 = 30 g
Hence, n = 2 so the molecular formula is C2H3

Solution 8:

22400 cm3 CO2 has mass = 44g
So, 224 cm3 CO2 will have mass = 0.44g
Now since CO2 is being formed and X is a hydrocarbon so it contains C and H.
In 0.44g CO2, mass of carbon = 0.44 - 0.32 = 0.12g = 0.012 g atom
So, mass of Hydrogen in X = 0.145 - 0.12 = 0.025g = 0.025g atom
Now the ratio of C : H is C = 1:H = 2.5 or C = 2 : H = 5
i.e. the formula of hydrocarbon is C2H5
(a) C and H
(b) Copper (11) oxide was used for reduction of the hydrocarbon.
(c) (i) no. of moles of CO2 = 0.44/44 = 0.01 moles
(ii) mass of C = 0.12g
(iii) mass of H = 0.025 g
(iv) The empirical formula of X = C2H5

Solution 9:

Mass of X in the given compound = 24g
Mass of oxygen in the given compound = 64g
So total mass of the compound  = 24 + 64 = 88g
% of X in the compound = 24/88 100 = 27.3%
% of oxygen in the compound = 64/88 100 = 72.7%
 Element % At. Mass Atomic ratio Simplest ratio X 27.3 12 27.3/12 2.27 O 72.7 16 72.2/16 4.54
So simplest formula = XO2

Solution 10:

(a) V.D = mass of gas at STP/mass of equal volume of H2 = 85/5 = 17
(b) Molecular mass = 17(V.D) × 2 = 34g

Solution 11:

(a)
12 g of C gives = 44.8 litre volume of CO
So, 3 g of C will give = 11.2 litre of CO

(b)
(i) 2V CO requires oxygen = 1 V
so, 24 cm3 CO will require = 24/2 = 12 cm3
(ii) 2 x 22400 cm3 CO gives = 2 × 22400 cm3 CO2
So, 24cm3 CO will give = 24 cm3 CO2

Solution 12:

(a) 56 g of CaO is obtained with NO2 = 2 × 22.4 litre of NO2
So, 5.6g of CaO is obtained with NO2 = (2 × 22.4 × 5.6/56) = 4.48 litre

(b) 56 g of CaO is obtained by = 164 g Ca(NO3)2
So, 5.6 g Cao is obtained by = 5.6 × 56/164 g Ca(NO3)2
= 16.4 g of Ca(NO3)2 is heated.

Solution 13:

(a) Number of molecules in 100cm3 of oxygen = Y
According to Avogadro's law, Equal volumes of all gases under similar conditions of temperature and pressure contain an equal number of molecules. Therefore, the number of molecules in 100 cm3 of nitrogen under the same conditions of temperature and pressure = Y
So, the number of molecules in 50 cm3 of nitrogen under the same conditions of temperature and pressure = Y/100 50 = Y/2

(b) (i) The empirical formula is the formula which tells about the simplest ratio of combining the capacity of elements present in a compound.
(ii) The empirical formula is CH3
(iii) The empirical formula mass for CH2O = 30
V.D = 30
Molecular formula mass = V.D 2 = 60
Hence, n = mol. Formula mass/empirical formula mass = 2
So, molecular formula = (CH2O)2 = C2H4O2

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 0.741 Cl 79.18 35.5 79.18/35.5 2.233
So the empirical formula of the given compound = SiCl3

Solution 16:

 Element % composition Atomic ratio Simple ratio P 38.27 38.27/31 1.23 H 2.47 2.47/1 2.47 O 59.26 59.26/16 3.703

So, empirical formula is PH2O3 or H2PO3
Empirical formula mass = (31 + 2×1 + 3×16) = 81
The molecular formula is = H4P2O6, because n = 162/81 = 2

Solution 17:

V1 = 10 litres V2 = ?
T1 = 27+ 273 = 300K T2 = 273K
P1 =700 mm P2 = 760 mm
Using the gas equation,
P1V1/T1 = P2V2/T2
V2 = P1V1T2/T1P2 = (700 × 10 × 273)/(300 × 760)
Molecular weight A = 60
So, weight of 22.4 litres of A at STP = 60 g
Weight of = (700 × 10 × 273)/(300 × 760)  litres of A at STP
= 60/22.4 × (700 × 10 × 273/300 × 760)  g or 22.45g

Solution 18:

(a) Molecular mass of CO2 = 12 + 2×16 = 44 g
So, vapour density (V.D) = mol. Mass/2  = 44/2 = 22
V.D = mass of certain amount of CO2/mass of equal volume of hydrogen = m/1
22 = m/1
So, mass of CO2 = 22 kg

(b) According to Avogadros law ,equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
So, number of molecules of carbon dioxide in the cylinder = number of molecules of hydrogen in the cylinder = X

Solution 19:

(a) The volume occupied by 1 mole of chlorine = 22.4 litre

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

(c) V1/V2 = T1/T2
22.4/V2 = 273/546
V2 = 44.8 litres

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

Solution 20:

(a) Total molar mass of hydrated CaSO4.xH2O = 136 + 18x
Since 21% is water of crystallization, so
18x/(136 + 18x) = 21/100
So, x = 2 i.e. water of crystallization is 2.

(b) For 18 g water, vol. of hydrogen needed = 22.4 litre
So, for 1.8 g, vol. of H2 needed = (1.8 × 22.4/18) = 2.24 litre
Now 2 vols. of water = 1 vol. of oxygen
1 vol. of water = 1/2 vol. of O2 = 22.4/2 = 11.2 lit.
18 g of water = 11.2 lit. of O2
1.8 g of water = 11.2/18 18/10 = 1.12 lit.

(c) 32g of dry oxygen at STP =  22400cc
2g will occupy = 224002/32 = 1400cc
P1 = 760mm P2 = 740mm
V1 = 1400cc V= ?
T1 = 273K, T2 = 27 + 73 = 300K
P1V1/T1 = P2V2/T2
V2 = P1V1T2/T1V2 = (760 × 1400 × 300)/(273 × 740) = 1580cc
1580/1000 = 1.581

(d) P1 = 750mm P2 = 760mm
V1 = 44lit. V2 = ?
T1 = 298K T2 = 273K
P1V1/T1 = P2V2/T2
V2 = P1V1T2/T1P2  = (750 × 44 × 273)/(298 × 760) = 39.78 lit.
22.4 lit. of CO2 at STP has mass = 44g
39.78 lit. of CO2 at STP has masss =  (44 × 39.78)/22.4 = 78.14 g

(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 × 100 = 58.5%

Solution 21:

P1V1/T1 = P2V2/T2
P1 × 22.4/273 = 2P2V2/546
V2 = 22.4 litre

Solution 22:

(a) The molecular mass of (Mg(NO3)2.6H2O = 256.4 g
% of Oxygen = 12 × 16/256 = 75%

(b) The molecular mass of boron in Na2B4O7.10H2O = 382 g
% of B = 4 × 11/382 = 11.5%

Solution 23:

(V×760)/273 = (360 × 380)/360
V = (360 × 380 × 273)/(760 × 360) = 136.5cm3
136.5cm3 of the gas weigh = 0.546
22400 cm3 of the gas weight = (0.546 × 22400)/136.5  = 89.6 a.m.u
Relative molecular mass = 89.6 a.m.u

Solution 24:

(a) 252 g of solid ammonium dichromate decomposes to give 152 g of solid chromium oxide, so the loss in mass in terms of solid formed = 100 g
Now, if 63 g ammonium dichromate is decomposed, the loss in mass would be = (100 × 63)/252 = 25 g

(b) If 252 g of ammonium dichromate produces Cr2O3 = 152 g
So, 63 g ammonium dichromate will produce = (63 × 152/252) = 38 g

Solution 25 :

128 g of SO2 gives = 2 × 22.4 litres volume
So, 12.8 g of SO2 gives = (2 × 22.4 × 12.8)/128 = 4.48 litre volume
Or one can say 4.48 litres of hydrogen sulphide.
2 x 22.4 litre H2S requires oxygen = 3 × 22.4 litre
So, 4.48 litres H2S will require = 6.72 litres of oxygen

Solution 26 :

From equation, 2NH3 + 2O2 → 2NO + 3H2O
When 60 g NO is formed, the mass of steam produced = 54g
So, 1.5 g NO is formed, the mass of steam produced =  (54 × 1.5/60) = 1.35 g

Solution 27:

In 1 hectare of soil, N2 removed = 20 kg
So, in 10 hectare N2 removed = 200 kg
The molecular mass of Ca(NO3)2 = 164
Now, 28 g N2 present in fertilizer = 164 g Ca(NO3)2
So, 200000 g of N2 is present in = (164 × 200000)/28
= 1171.42 kg

Solution 28:

(a) 1 mole of phosphorus atom = 31 g of phosphorus
31 g of P = 1 mole of P
6.2g of P = (6.2 × 1)/31 = 0.2 mole of P
(b) 31 g P reacts with HNO3 = 315 g
So, 6.2 g P will react with HNO3 = (315 × 6.2/31) = 63 g

(c) Moles of steam formed from 31g phosphorus = 18g/18g = 1mol
Moles of steam formed from 6.2 g phosphorus = 1mol/3196.2 = 0.2 mol
Volume of steam produced at STP = (0.2 × 22.4l)/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˚C = 4.48 × 2 = 8.96litre

Solution 29:

(a) 1 mole of gas occupies volume = 22.4 litre
(b) 112cm3 of gaseous fluoride has mass = 0.63 g
So, 22400cm3 will have mass = 0.63 × 22400/112 = 126 g
The molecular mass = At mass P + At mass of F
126 = 31 + At. Mass of F
So,  At. Mass of F = 95 g
But, at. mass of F = 19 so 95/19 = 5
Hence, there are 5 atoms of F so the molecular formula = PF5

Solution 30:

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

Solution 31:

(a) The molecular mass of Ca(H2PO4)2 = 234
The % of P = (2 × 31/234) = 26.49%
(b) Simple ratio of M = 34.5/56 = 0.616 = 1
Simple ratio of CI = 65.5/35.5 = 1.845 = 3
Empirical formula = MCI3
Empirical formula mass = 162.5, Molecular mass = 2 ×V.D. = 325
So, n = 2
So, molecular formula = M2Cl6

Solution 32:

V1/V2 = n1/n2
So, no. of moles of Cl = x/2 (since V is directly proportional to n)
No. of moles of NH3 = x
No. of moles of SO2 = x/4
This is because of Avogadro's law which states Equal volumes of all gases, under similar conditions of temperature and pressure, contain an equal number of molecules.
So, 20 litre nitrogen contains x molecules
So, 10 litre of chlorine will contain = x × 10/20 = x/2 mols.
And, 20 litre of ammonia will also contain = x molecules
And, 5 litre of sulphur dioxide will contain = x × 5/20 = x/4 mols.

Solution 33:
2 × 22400 litre steam is produced by N2O = 4 × 22400 cm3
So, 150 cm3 steam will be produced by = (4 × 22400 ×150)/(2 ×22400)
=  300 cm3 N2O

Solution 34:

(a) Volume of O2 = V
Since O2 and N2 have same no. of molecules = x
so, the volume of N2 = V

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

(c) 32 g oxygen is contained in = 44 g of CO2
So, 8 g oxygen is contained in = (44 × 8)/32 = 11 g
(d) Avogadro’s law is used in the above questions.

Solution 35:

(a) 444 g is the molecular formula of (NH4)2 PtCl6
% of Pt = (195/444) × 100 = 43.91% or 44%
(b) simple ratio of Na = 42.1/23 = 1.83 = 3
simple ratio of P = 18.9/31 = 0.609 = 1
simple ratio of O = 39/16 = 2.43 = 4
So, the empirical formula is Na3PO4

Solution 36:

From equation:
22.4 litres of methane requires oxygen = 44.8 litres O2
From equation,
44.8 litres hydrogen requires oxygen = 22.4 litres O2
So, 11.2 litres will require = (22.4 × 11.2)/44.8 = 5.6 litres
Total volume = 44.8 + 5.6 = 50.4 litres

Solution 37:

Equal volumes of all gases, under similar conditions of temperature and pressure, contain an equal number of molecules.
So, 1 mole of each gas contains = 6.02 ×1023 molecules
Mol. Mass of H2 (2), O2(32), CO2(44), SO2(64),Cl2(71)

(1) Now 2 g of hydrogen contains molecules = 6.02 ×1023
So, 8g of hydrogen contains molecules = (8/2 ×6.02 ×1023) = 4 × 6.02 ×1023 = 4M molecules

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

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

(4) 64g of sulphur dioxide contains molecules = 6.02 ×1023
So, 8g of sulphur dioxide molecules = 8/64 × 6.02 ×1023  = M/8

(5) 71 g of chlorine contains molecules = 6.02 ×1023
So, 8g of chlorine molecules = 8/72 × 6.02 × 1023  = 8M/71
Since 8M/71<M/8<2M/11<M/4<4M
Thus Cl2<SO2<CO2<O2<H2
(i) Least number of molecules in Cl2
(ii) Most number of molecules in H2

Solution 38:

Na2SO4 + BaCl2 → BaSO4 + 2NaCI
Molecular mass of BaSO4 = 233 g
Now, 233 g of BaSO4 is produced by Na2SO4 = 142 g
So, 6.99 g BaSO4 will be produced by = (6.99 × 142/233) = 4.26
The percentage of Na2SO4 in original mixture = (4.26 × 100/10) = 42.6%

Solution 39:

(a) 1 litre of oxygen has mass = 1.32 g
So, 24 litres (molar vol. at room temp.) will have mass = 1.32 ×24 = 31.6 or 32g

(b) 2KMnO4 → K2MnO4 + MnO2 + O2
316 g of KMnO4 gives oxygen = 24 litres
So, 15.8 g of KMnO4 will give = (24 × 316/15.8) = 1.2 litres

Solution 40:

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

(ii) In 1 mole of SO2, no. of molecules present = 6.02 ×1023
So, in 0.05 moles, no. of molecules = 6.02 × 1023 × 0.05 = 3.0 ×1022

(iii) The volume occupied by 64 g of SO2 = 22.4 dm3
3.2 g of SO2 will be occupied by volume = (22.4 × 3.2/64) = 1.12 dm3

(b) Gram atoms of Pb = 6.21/207 = 0.03 = 1
Gram atoms of Cl = 4.26/35.5 = 0.12 = 4
So, the empirical formula = PbCl4

Solution 41:

(i) D contains the maximum number of molecules because the volume is directly proportional to the number of molecules.

(ii) The volume will become double because the volume is directly proportional to the no. of molecules at constant temperature and pressure.
V1/V2 = n1/n2
V1/V2 = n1/2n1
So, V2 = 2V1

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

(iv) The volume of D = 5.6 4 = 22.4 dm3, so the number of molecules = 6 ×1023 because according to mole concept 22.4 litre volume at STP has = 6 ×1023 molecules

(v) No. of moles of  D = 1 because volume is 22.4 litre
so, mass of N2O = 1 44 = 44 g

Solution 42:

(a) NaCl + NH3 + CO2 + H2O →  NaHCO3 + NH4CI
2NaHCO3 → Na2CO+ H2O + CO2
From equation:
106 g of Na2CO3 is produced by = 168g of NaHCO3
So, 21.2 g of Na2CO3 will be produced by = (168 × 21.2)/106
= 33.6 g of NaHCO3

(b) For 84 g of NaHCO3, required volume of CO2 = 22.4 litre
So, for 33.6 g of NaHCO3, required volume of CO2 = 22.4 × 33.6/84
=  8.96 litre

Solution 43:

(a)

44.8 litres of water produced by = 22.4 litres of NH4NO3
So, 8.96 litres will be produced by = (22.4 × 8.96)/44.8
= 4.48 litres of NH4NO3
So, 4.48 litres of N2O is produced.

(i) 44.8 litre H2O is produced by = 80 g of NH4NO3
So, 8.96 litres H2O will be produced by = (80 × 8.96/44.8)
= 16g NH4NO3

(iii) % of O in NH4NO3 = (3 × 16)/80 = 60%

Solution 44:

(a)
 Element % At. Mass Atomic ratio Simplest ratio K 47.9 39 1.22 2 Be 5.5 9 0.6 1 F 46.6 19 2.45 4
So, empirical formula is K2BeF4

(b)
3 × 80 g of CuO reacts with = 2 × 22.4 litre of NH3
so, 120 g of CuO will react with = (2 × 22.4 × 120)/(80 × 3) = 22.4 litres

Solution 45:

(a) The molecular mass of ethylene(C2H4) is 28 g
No. of moles = 1.4/28 = 0.05 moles
No. of molecules = (6.023 × 103 × 0.05)= 3 × 1022 molecules
Volume = 22.4 x 0.05 = 1.12 litres

(b) Molecular mass = 2 × V.D
S0, V.D = 28/2 = 14

Solution 46:

(a) Molecular mass of Na3AlF6 = 210
So, Percentage of Na = (3 × 23 × 100)/210 = 32.85%

(b)
1 mole of O2 has volume = 22400 ml
Volume of oxygen used by 2 x 22400 ml CO = 22400 ml
So, Vol. of O2 used by 560 ml CO = (22400 × 560)/(2 × 22400) = 280 ml
So, Volume of CO2 formed is 560 ml.

Solution 47:

(a) Mass of gas X = 10g
Mass of hydrogen gas = 2
Relative vapour density
= Mass of volume of gas X under similar conditions/ Mass of volume of hydrogen gas under similar conditions = 10/2 = 5
Relative molecular mass of the gas = 2 × relative vapour
density = 2 × 5 = 10

(b) 2C2H(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

(i) The combustion reaction
According to Gay-Lussac's law,
2 volume of acetylene requires 5 volume of oxygen to burn it
∴ 1 volume of acetylene requires 2.5 volume of oxygen to burn it
∴ 200cm3 requires 2.5 ×200 = 500 cm3 of oxygen
2 volume of acetylene on combustion gives 4CO2
∴ 1 volume of acetylene on combustion gives 2CO2
∴ 200cc of acetylene on combustion will give 200 × 2 = 400cc of CO2

(ii) Hydrogen = 12.5%
∴ Nitrogen = 100 - 12.5 = 87.5
The Empirical formula of the compound is NH2
Empirical formula weight = 14 + 2 = 16
Relative molecular mass = 37
N = Relative molecular mass/Empirical Weight = 37/16 = 2.3≈2
Molecular formula = n ×empirical formula = 2 × NH= N2H4

(i) Molecules of nitrogen gas in a cylinder = 24 ×1024
1. Mass of nitrogen in a cylinder = (24 × 1024 × 28)/(6 ×1023) = 1120g
2. Volume of nitrogen at stp
Volume of 28 g of N2 = 22.4dm3
Volume of 1120g of N2 = (1120 × 22.4)/28 dm= 896 dm3

Solution 48:

a. (i) 10 litres of LPG contains
Propane = (60/100) × 10 = 6lites
Butane =  (40/100) × 10 = 4litres
CO2 = 18 + 16 = 34l

(ii) Molecular mass of NH4(NO3) = 80
H = 1, N = 14, O = 16
% of Nitrogen
As 80 g of NH4(NO3) contains 28 g of nitrogen
∴ 100 g of of NH4(NO3) will contain (28 × 100)/80 = 35%
% of Oxygen
As, 80 g of NH4(NO3) contains 48 g of oxygen
∴ 100 g of of NH4(NO3) will contain (100 × 48)/80 = 60%

b. (i) Equation for reaction of calcium carbonate with dilute hydrochloric acid:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
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) CaCO3 + 2HCl → CaCl2 + H2O + CO2
As, 100g of calcium carbonate gives 22.4dm3 of CO2
∴ 450 g of calcium carbonate will give 100 = 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  (450 × 111/100) = 499.5 g

(iv) The molecular mass of HCI = 36.5
The molecular mass of calcium carbonate = 100
As 100 g of calcium carbonate gives (2 ×36.5) = 73g of HCI
∴ 450 g of calcium carbonate will give (450 × 73/100) = 328.5g
Number of moles of HCl = Weight of HCl/Molecular weight of HCI = 328.5/36.5 = 9 moles

Solution 49:

a. (i) Atomic mass: S = 32 and O = 16
Molecular mass of SO2 = 32 + (2 ×16) = 64g
As  64 g of SO2 = 22.4dm3
Then, 320 g of SO2 = (320 × 22.4)/64
= 112 L

(ii) Gay-Lussac's law Gay-Lussac's Law states "When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure."

(iii) C3H8 + 5O2 → 3CO2 + 4H2O
Molar mass of propane = 44
44 g of propane requires 5 × 22.4 litres of oxygen at STP.
8.8 g of propane requires (5 × 22.4 × 8.8)/44 = 22.4 litres

b. (i)

Empirical formula = CH2Br
n(Empirical formula mass of CH2Br) = Molecular mass (2 × VD)
n(12 + 2 + 80) = 94 ×2
n = 2
Molecular formula = Empirical formula ×2
= (CH2Br) × 2
= C2H4Br2

(ii) 1022 atoms of sulphur
6.022 × 1023 atoms of sulphur will have mass = 32 g
1022 atoms of sulphur will have mass = (32 ×1022)/(6.022 ×1023)= 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. P + 5HNO3(conc.) → H3PO4+ H2O + 5NO2
(i) Number of moles of phosphorus taken = 9.3/31 = 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 NO2 gas at STP.
So, 0.3 mole of phosphorus gives (112 × 0.3) L of
NO2 gas at STP.
= 33.6 L of NO2 gas at STP

b. (i) According to the equation
N2(g) + 3H2(g) → 2NH3(g)
3 volumes of hydrogen produce 2 volumes of ammonia
67.2 litres of hydrogen produce (2 × 67.2)/3 = 44.8 L
3 volumes of hydrogen combine with 1 volume of ammonia.
67.2 litres of hydrogen combine with (1 × 67.2)/3 = 22.4L
Nitrogen left = 44.8 - 22.4 = 22.4 litres

(ii) 5.6 dm3 of gas weighs 12 g
1 dm3of gas weighs = (12/56) gm
22.4 dm3 of gas weighs = (12/56 × 22.2) gm = 48g
Therefore, the relative molecular mass of gas = 48 gm.

(iii) Molar mass of Mg (NO3)2.6H2O
= 24 × (14 × 2) + (16 × 12) + (1 × 12) = 256 g
24 × 100 = 9.37%
Mass percent of magnesium = (24 × 100)/256 = 9.37%

Solution 51:

a.

(i) 2 vols. of butane requires O2 = 13 vols
90 dm3 of butane will require O2 = 13/2 × 90 = 585 dm3

(ii) Molecular mass = 2 × Vapour density
So, molecular mass of gas = 2 × 8 = 16 g
As we know, molecular mass or molar mass occupies 22.4 litres.
That is,
16 g of gas occupies volume = 22.4 litres
So, 24 g of gas will occupy a volume
22.4 × 24 = 33.6 litres

(iii) According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
So, molecules of nitrogen gas present in the same vessel = X

b.

(i) 3 vols. of oxygen require KClO3 = 2 vols.
So, 1 vol. of oxygen will require KClO3 = 2/3 vols
So, 6.72 litres of oxygen will require KCIO3
So, 1 vol. of oxygen will require KClO3 = 2/3vols
So, 6.72 litres of oxygen will require KClO3
2/3 x 6.72 = 4.48 litres
22.4 litres of KClO3 has mass = 122.5 g
So, 4.48 litres of KÄŒIO3 will have mass
= (122.5/22.4 x 4.48) = 24.5g

(ii) 22.4 litres of oxygen = 1 mole
So, 6.72 litres of oxygen = 6.72/22.4 = 0.3 moles
No. of molecules present in 1 mole of O2 = 6.023 ×1023
So, no. of molecules present in 0.3 mole of O2
= 6.023 × 1023 × 0.3
= 1.806 × 1023

(iii) Volume occupied by 1 mole of CO2 at STP = 22.4 litres
So, volume occupied by 0.01 mole of CO2 at STP = 22.4 × 0.01= 0.224 litres

Solution 52:
a. (i) 2C2H2 + 5O2 → 4CO2 + 2H2O
2 moles of C2H2 = 4moles of CO2
x dm3 of C2H2 = 8.4 dm3 of CO2
x = (2 × 8.4)/4
= 4.2 dm3 of C2H2

(ii) Empirical formula= X2Y
Atomic weight (X) = 10
Atomic weight (Y) = 5
Empirical formula weight = (2 × 10) + 5 = 25
n = (Molecular weight/Emperical formula weight)
= (2 × V.D)/Emperical formula weight
= (2 × 25)/25 = 2
So, molecular formula = X2Y2

b. (i) A  cylinder contains  68g of ammonia gas at STP.
Molecular weight of ammonia = 17g/mole
68g of ammonia gas at STP =?
1 mole = 22.4 dm3
∴ 4 mole = 22.4 × 4 = 89.6 dm3

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

(iii) 1 mole = 6.023 ×1023 molecules
4 moles = 24.092 ×1023  molecules