Stoichiometry

(a) Molecular weight of carbon dioxide CO₂ = 12 + (2 x 16) = 12 + 32 = 44

44 g of CO₂ contains 12 g of carbon

∴ 100 g of CO₂ contains $\frac{12}{44}$ x 100 = 27.27% of carbon

Hence, percentage by weight of C in CO₂ is 27.3%

(b) Molecular weight of sodium carbonate Na₂CO₃ = (2 x 23) + 12 + (3 x 16) = 46 + 12 + 48 = 106

106 g of sodium carbonate Na₂CO₃ contains 46 g of sodium

∴ 100 g of Na₂CO₃ contains $\frac{46}{106}$ x 100 = 43.39% ≈ 43.4% of sodium.

Hence, percentage by weight of sodium in sodium carbonate is 43.4%

(c) Molecular weight of Aluminium Nitride = AlN = 27 + 14 = 41

41 g of Aluminium Nitride contains 27 g of Aluminium

∴ 100 g of Aluminium Nitride contains $\frac{27}{41}$ x 100 = 65.85% of Aluminium

Hence, percentage by weight of Aluminium in Aluminium Nitride is 65.85%

Molecular weight of K₃Fe(CN)₆ = 3(39) + 56 + 6(12 + 14) = 117 + 56 + 156 = 329 g

329 g of K₃Fe(CN)₆ contains 56 g of iron

∴ 100 g of K₃Fe(CN)₆ contains $\frac{56}{329}$ x 100 = 17.02% of iron.

Hence, percentage by weight of iron in K₃Fe(CN)₆ is 17.02%

Molecular weight of calcium nitrate Ca[NO₃]₂ = 40 + 2[(14 + 3(16))] = 40 + 2[14 + 48] = 164 g

164 g of calcium nitrate contains 28 g of nitrogen

∴ 100 g of calcium nitrate contains $\frac{28}{164}$ x 100 = 17.07% of nitrogen

Percentage by weight of nitrogen in calcium nitrate is 17.07%

Molecular weight of ammonium sulphate [NH₄]₂SO₄ = 2[14 + 4(1)] + 32 + 4(16) = 36 + 32 + 64 = 132

132 g of ammonium sulphate contains 28 g of nitrogen

∴ 100 g of ammonium sulphate contains $\frac{28}{132}$ x 100 = 21.21% of nitrogen

Percentage by weight of nitrogen in ammonium sulphate is 21.21 %

Hence, ammonium sulphate has higher concentration of nitrogen than calcium nitrate.

90% of 10 kg = 9000 g

Molecular weight of aluminium oxide [Al₂O₃] = 2(27) + 3(16) = 54 + 48 = 102 g

102 g of aluminium oxide contains 54 g of aluminium

∴ 9000 g of aluminium oxide contains $\frac{54}{102}$ x 9000 = 4764 g = 4.764 kg of aluminium

∴ Percentage of pure Al in 10 kg. of aluminium oxide [Al₂O₃] = $\frac{4.764}{10}$ x 100 = 47.64%

Hence, percentage of pure aluminium in 10 kg. of aluminium oxide [Al₂O₃] of 90% purity is 47.64%

(i) Molecular weight of K₃PO₄ = 3(39) + 31 + 4(16) = 117 + 31 + 64 = 212 g

Percentage of K in K₃PO₄ = $\frac{117}{212}$ x 100 = 55.18%

Molecular weight of [KNO₃] = 39 + 14 + 3(16) = 39 + 14 + 48 = 101 g

Percentage of K in KNO₃ = $\frac{39}{101}$ x 100 = 38.61%

Hence, potassium phosphate (% of K = 55.18 %) is better fertilizer than potassium nitrate (% of K = 38.61 %).

(ii) Molecular weight of NH₂CONH₂ = 14 + 2(1) + 12 + 16 + 14 + 2(1) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g

Percentage of N in NH₂CONH₂ = $\frac{28}{60}$ x 100 = 46.67%

Molecular weight of (NH₄)₃PO₄ = 3[14 + 4(1)] + 31 + 4(16) = 54 + 31 + 64 = 149 g

Percentage of N in (NH₄)₃PO₄ = $\frac{42}{149}$ x 100 = 28.19%

Hence, Urea is a better fertilizer as percentage of N in urea is 46.67% and in (NH₄)₃PO₄ is 28.19 %

Molecular weight of CaCO₃ = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g

100 g of CaCO₃ contains 12 g of carbon.

Given, purity is 55% therefore,

55% of 12 g = $\frac{55}{100}$ x 12 = 6.6%

Hence, percentage of carbon is 6.6%.

Molecular weight of hydrated copper sulphate CuSO₄.5H₂O = 63.5 + 32 + 4(16) + 5[2(1) + 16] = 63.5 + 32 + 64 + 5[18] = 63.5 + 32 + 64 + 90 = 249.5 g

249.5 g of CuSO₄.5H₂O contains 90 g of water of crystallization.

∴ Percentage of water of crystallization = $\frac{90}{249.5}$ x 100 = 36%

Hence, percentage of water of crystallization is 36%

Given,

CaSO₄.xH₂O contains 21% of water of crystallization, so % of CaSO₄ = 100 - 21 = 79%

Molecular weight of anhydrous calcium sulphate CaSO₄ = 40 + 32 + 4(16) = 40 + 32 + 64 = 136 g

i.e. 79% of CaSO₄.xH₂O = 136 g

∴ CaSO₄.xH₂O = $\frac{136}{79}$ x 100 = 172.15 g

21% of CaSO₄.xH₂O = $\frac{21}{100}$ x 172.15 = 36.15 g

Molecular weight of water H₂O = 2(1) + 16 = 18

Number of molecules (x) = $\frac{36.15}{18}$ = 2 molecules

∴ CaSO₄.xH₂O = CaSO₄.2H₂O

C : O : H = 2 : 1 : $\frac{3}{2}$ = 4 : 2 : 3

Simplest ratio of whole numbers = 4 : 2 : 3

Hence, empirical formula is C₄O₂H₃ or C₄H₃O₂

Empirical formula weight = (4 x 12) + (3 x 1) + (2 x 16) = 48 + 3 + 32 = 83

Vapour density (V.D.) = 83

Molecular weight = 2 x V.D. = 2 x 83 = 166

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{166}{83} = 2\end{aligned}$$

Molecular formula = n[E.F.] = 2[C₄H₃O₂] = C₈H₆O₄

∴ Molecular formula = C₈H₆O₄

4 g of metallic chloride contains 1.89 g of metal X

∴ 100 g of metallic chloride contains $\frac{1.89}{4}$ x 100 = 47.25 g of metal X

∴ % of Cl in the metallic chloride = 100 - 47.25 = 52.75%.

X : Cl = 1 : 2

Hence, empirical formula is XCl₂

Empirical formula is CH₂O

Empirical formula weight = 12 + 2(1) + 16 = 12 + 2 + 16 = 30

Vapour density (V.D.) = 30

Molecular weight = 2 x V.D. = 2 x 30 = 60

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{empirical formula weight}} \\[0.5em] = \frac{60}{30} = 2\end{aligned}$$

Molecular formula = n[E.F.] = 2[CH₂O] = C₂H₄O₂

Hence, Molecular formula is C₂H₄O₂

Original weight = 1.30 g

Al : P : O = 1 : 1 : 4

Hence, Simplest ratio of whole numbers = 1 : 1 : 4

Hence, empirical formula is AlPO₄

Empirical formula weight = 27 + 31+ 4(16) = 27 + 31 + 64 = 122

Molecular weight = 122

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{empirical formula weight}} \\[0.5em] = \frac{122}{122} = 1\end{aligned}$$

∴ Molecular formula = n[E.F.] = 1[AlPO₄] = AlPO₄

Hence, Molecular formula = AlPO₄

Given,

Vapour density_X = 13

Vapour density_Y = 39

Molecular weight_X = 2 x V.D. = 2 x 13 = 26

Molecular weight_Y = 2 x V.D. = 2 x 39 = 78

Let formula for X be C_aH_b

where a and b are simple whole numbers.

∴ Molecular mass of X = 12a + b = 26

Hence, a = 2 and b = 2 (only values which hold true for the above equation)

So, Molecular formula of X = C₂H₂

Let formula for Y be C_cH_d

where c and d are simple whole numbers.

∴ Molecular mass of Y = 12c + d = 78

Hence, c = 6 and d = 6 (only values which hold true for the above equation)

So, Molecular formula of Y = C₆H₆

Simplest ratio of whole numbers = Zn : S : O : H = 1 : 1 : 11 : 14

Hence, empirical formula is ZnSO₁₁H₁₄

Molecular weight = 287

Empirical formula weight = 65 + 32 + 11(16) + 14(1) = 65 + 32 + 176 + 14 = 287

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{287}{287} = 1\end{aligned}$$

Molecular formula = n[E.F.] = 1[ZnSO₁₁H₁₄] = ZnSO₁₁H₁₄

Since all the hydrogen in the compound is in combination with oxygen as water of crystallization .

Therefore, 14 atoms of hydrogen and 7 atoms of oxygen = 7H₂O and hence, 4 atoms of oxygen remain.

Hence, molecular formula is ZnSO₄.7H₂O.

C : H = 1 : 2.5 = 1 : $\frac{5}{2}$ = 2 : 5

Simplest ratio of whole numbers = 2 : 5

Hence, empirical formula is C₂H₅

Empirical formula weight = 2(12) + 5(1) = 24 + 5 = 29

Vapour density (V.D.) = 29

Molecular weight = 2 x V.D. = 2 x 29 = 58

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{58}{29} = 2\end{aligned}$$

Molecular formula = n[E.F.] = 2[C₂H₅] = C₄H₁₀

Hence, Molecular formula = C₄H₁₀

Simplest ratio of whole numbers = O : H : C = 1 : 2 : 1

As number of carbon atoms = 12

Therefore, O : H : C = 12 : 24 : 12

Hence, empirical formula is C₁₂H₂₄O₁₂

Hence, Simplest ratio of whole numbers = Cu : S : O : H₂O = 1 : 1 : 4 : 5

Hence, empirical formula is CuSO₄.5H₂O

H = 0.70 - 0.60 = 0.1 g

Simplest ratio of whole numbers = C : H = 1 : 2

Hence, empirical formula is CH₂

Empirical formula weight = 12 + 2(1) = 14

Molecular weight = 70

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{70}{14} = 5\end{aligned}$$

Molecular formula = n[E.F.] = 5[CH₂] = C₅H₁₀

Hence, Molecular formula = C₅H₁₀

Simplest ratio of whole numbers = Al : K : S : O = 1 : 1 : 2 : 8

Hence, empirical formula is AlKS₂O₈ or AlK(SO₄)₂

Molecular Weight of Silver Nitrate (AgNO₃) = 108 + 14 + 48 = 170 g

Molecular Weight of Silver Chloride (AgCl) = 108 + 35.5 = 143.5 g

Chemical Equation:

$\underset{170 g}{\text{AgNO}_3} + \text{HCl} \longrightarrow \underset{143.5 g}{\text{AgCl}} + \text{HNO}_{3}$

170 g of AgNO₃ gives 143.5 g of AgCl

∴ 0.34 g of AgNO₃ will give $\frac{143.5}{170}$ x 0.34 = 0.287 g of AgCl

Hence, 0.287 g of silver chloride will be obtained

$$\begin{matrix} 2\text{KClO}_3 & \xrightarrow{\space \Delta \space} & 2\text{KCl} & + & 3\text{O}_2 \\ 2[39 + 35.5 & & & & 3 \times 22.4 \\ + 48] = 245 & & & & = 67.2 \\ \end{matrix}$$

245 g of KClO₃ gives 67.2 lit of O₂

∴ 20 g of KClO₃ will give $\frac{67.2}{245}$ x 20 = 5.486 lit of O₂

Hence, 5.486 lit of oxygen will be obtained

$$\begin{matrix} 3\text{Cu} & + & 8\text{HNO}_3 & \longrightarrow & 3\text{Cu[NO}_3]_2 & + & 4\text{H}_2\text{O} & + & 2\text{NO} \\ 3(64) & & 8[1 + 14 + 48] \\ = 192 \text{ g} & & = 8 \times 63 = 504 \text{ g} \\ \end{matrix}$$

504 g of nitric acid reacts with 192 g of copper

∴ 63 g of nitric acid will react with $\frac{192}{504}$ x 63 = 24 g of copper

Hence, 24 g of copper is needed to react with 63 g of nitric acid

(ii)

$3\text{Cu} + \underset{504 \text{ g}}{8\text{HNO}_3} \longrightarrow 3\text{Cu[NO}_3]_2 + 4\text{H}_2\text{O} + \underset{2(22.4) \text {lit.}}{2\text{NO}}$

504 g of nitric acid liberates (2 x 22.4) lit of Nitric Oxide

∴ 63 g of nitric acid will liberate $\frac{2 \times 22.4}{504}$ x 63 = 5.6 lits of Nitric Oxide

Hence, 5.6 lits. of nitric oxide is collected.

$\underset{65 + 32 = 97 \text{ g}}{\text{ZnS}} + \text{O}_2 \longrightarrow \underset{1 \text{ mole}}{\text{SO}_2}$

97 g of ZnS gives 1 mole of SO₂

∴ 776 g of ZnS will give $\frac{1}{97}$ x 776 = 8 moles of SO₂

Hence, 8 moles of sulphur dioxide is liberated by 776 g of ZnS

(ii)

$\underset{65 + 32 = 97 \text{ g}}{\text{ZnS}} + \text{O}_2 \longrightarrow \underset{1 \text{ mole}}{\text{SO}_2}$

97 g of ZnS liberates 1 mole of SO₂

As 1 mole occupies 22.4 lit volume at s.t.p.

∴ 22.4 lit of SO₂ is produced by 97 g of ZnS.

$$\begin{matrix} 2\text{NH}_3 & + & \text{H}_2\text{SO}_4 & \longrightarrow & [\text{NH}_4]_2\text{SO}_4 \\ 2(22.4 \text{ lit.}) & & & & 2[14 + (4 \times 1)] \\ & & & & + 32 + (4 \times 16) \\ & & & & = 132 \text{g} \end{matrix}$$

132 g of ammonium sulphate is produced by 2 x 22.4 lit of Ammonia

∴ 59 g of ammonium sulphate is produced by $\frac{2 \times 22.4}{132}$ x 59 = 20.02 lit. of Ammonia

Hence, 20.02 lit. of ammonia [at s.t.p.] is used to form 59 g of ammonium sulphate

$$\begin{matrix} 2\text{Pb[NO}_3]_2 & \xrightarrow{\space \Delta \space} & 2\text{PbO} & + & 4\text{NO}_2 & + & \text{O}_2 \\ 2[207 + 2(14 + 48)] & & & & (4 \times 22.4) & & 22.4 \text{ lit} \\ = 2[207 + 124] & & & & = 89.6 \text{ lit} & & \\ = 662 \text{ g} & & & & & & \end{matrix}$$

662 g of lead nitrate gives 89.6 lit. of NO₂

∴ 8.5 g of lead nitrate will give $\frac{89.6}{662}$ x 8.5 = 1.15 lit. of NO₂

Hence, NO₂ produced is 1.15 lit

662 g of lead nitrate gives 22.4 lit. of O₂

∴ 8.5 g of lead nitrate will give $\frac{22.4}{662}$ x 8.5 = 0.287 lit of O₂

Hence, O₂ produced is 0.287 lit

∴ Total volume produced = 1.15 + 0.287 = 1.437 lits.

Hence, total volume of NO₂ and O₂ produced on heating 8.5 of lead nitrate is 1.437 lits.

$$\begin{matrix} \text{C} & + & \text {O}_2 & \xrightarrow{\Delta} & \text{CO}_2 \\ 12 \text{ g} & & & & 22.4 \text{ lit} \end{matrix}$$

12 g of C gives 22.4 lit of CO₂

∴ 24 g of C will give $\frac{22.4}{12}$ x 24 = 44.8 lit. of CO₂

Hence, 24 g of C will give 44.8 of CO₂

$$\begin{matrix} 2\text{KClO}_3 & \xrightarrow{\space \Delta \space} & & 2\text{KCl} & + & 3\text{O}_2 \\ 2[39 + 35.5 & & & & & (3 \times 22.4) \\ + 48] = 245 \text{ g} & & & & & = 67.2 \text{ lit} \\ \end{matrix}$$

67.2 lit of O₂ is produced by 245 g of KClO₃

∴ 44.8 lit of O₂ is produced by $\frac{245}{67.2}$ x 44.8 = 163.33 g

Hence, amount of KClO₃ is 163.33 g.

(a) Molecular Weight of Ammonia (NH₃) = 14 + 3(1) = 17 g

Molecular Weight of Ammonium Sulphate ((NH₄)₂SO₄) = 2[14 + 4(1)] + 32 + 4(16) = 132 g

$\underset{2(17) = 34 \text{ g}}{2\text{NH}_3} + \text{H}_2\text{SO}_4 \longrightarrow \underset{132 \text{ g}}{(\text{NH}_4)_2\text{SO}_4}$

132 g of ammonium sulphate is given by 34 g of ammonia

∴ 78 g of ammonium sulphate is given by $\frac{34}{132}$ x 78 = 20.09 g of ammonia

Hence, 20.09 g of ammonia gas is required.

(b) Molecular Weight of Ammonium Chloride (NH₄Cl) = 14 + 4(1) + 35.5 = 53.5 g

$\underset{2(53.5) = 107 \text{ g}}{2\text{NH}_4\text{Cl}} + \text{Ca}(\text{OH})_2 \longrightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} + \underset{2(17) = 34 \text{ g}}{2\text{NH}_3}$

107 g of ammonium chloride gives 34 g of ammonia

∴ 32.6 g of ammonium chloride will give $\frac{34}{107}$ x 32.6 = 10.36 g of ammonia

Hence, 10.36 g of ammonia gas is obtained.

Given,

300 g. of Na₂CO₃ is of 80% purity

∴ Mass of pure Na₂CO₃ present in it = $\frac{80}{100}$ x 300 = 240 g

We have,

$$\begin{matrix} \text{Na}_2\text{CO}_3 & + & \text{H}_2\text{SO}_4 & \longrightarrow & \text{Na}_2\text{SO}_4 & + & \text{H}_2\text{O} & + & \text{CO}_2 \\ (2 \times 23) + 12 & & & & (2 \times 23) + 32 \\ + (3 \times 16 ) & & & & + (4 \times 16) \\ = 106 \text{ g} & & & & = 142 \text{ g} \end{matrix}$$

106 g of Na₂CO₃ gives 142 g of Na₂SO₄

∴ 240 g of Na₂CO₃ gives $\frac{142}{106}$ x 240 = 321.51 g

Hence, mass of pure salt formed is 321.51 g

$$\begin{matrix} 2\text{KNO}_3 & \longrightarrow & 2\text{KNO}_2 & + & \text{O}_2 \\ 2[39 + 14 & & & & 2 \times 16 \\ + (16 \times 3)] & & & & = 32 \text{ g} \\ = 202 \text{ g} & & & & \end{matrix}$$

$$\begin{matrix} \text{S} & + & \text{O}_2 & \longrightarrow & \text{SO}_2 \\ 32 \text {g} \\ \end{matrix}$$

32 g of sulphur require 202 g of potassium nitrate to be heated

∴ 16 g of sulphur will require $\frac{202}{32}$ x 16 = 101 g of potassium nitrate.

Hence, amount of potassium nitrate which must be heated is 101 g

$$\begin{matrix} \text{Mg} & + & \text{H}_2\text{SO}_4 & \longrightarrow & \text{MgSO}_4 & + & \text{H}_2 \\ 24 \text{g} \\ \end{matrix}$$

22400 cc of H₂ is produced from 24 g of pure magnesium

∴ 298.6 cc. of H₂ is produced from $\frac{24}{22400}$ x 298.6 = 0.3199 g of pure magnesium

∴ Percentage Purity = 0.3199 x 100 = 31.99 %

Hence, the % purity of the sample is 31.99 %

[By Lussac's law]

$$\begin{matrix} \text{N}_2 & + & 3\text{H}_2 & \longrightarrow & 2 \text{NH}_3 \\ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$$

To calculate the volume of ammonia gas formed.

$$\begin{matrix} \text{H}_2 & : & \text{NH}_3 \\ 3 & : & 2 \\ 67.2 & : & x \end{matrix}$$

$\therefore \frac{2}{3} \times 67.2 = 44.8 \text{ lit}$

Hence, volume of NH₃ formed is 44.8 lit.

We know,

$$\begin{matrix}\text{H}_2 & : & \text{N}_2 \\ 3 & : & 1 \\ 67.2 & : & x \end{matrix}$$

$\therefore \frac{1}{3} \times 67.2 = 22.4 \text{ lit}$

∴ nitrogen left = 44.8 - 22.4 = 22.4 lit.

Hence, 22.4 lit of nitrogen remains in the resultant mixture.

[By Lussac's law]

$$\begin{matrix} 2\text{C}_4\text{H}_{10} & + & 13\text{O}_2 & \longrightarrow & 8\text{CO}_2 & + & 10\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} & : & 10\text{ vol.} \end{matrix}$$

To calculate the volume of oxygen

$$\begin{matrix}\text{C}_4\text{H}_{10} & : & \text{O}_2 \\ 2 & : & 13 \\ 90 & : & x \end{matrix}$$

∴

$\frac{13}{2} \times 90 = 585 \text{ dm}^3$

Hence, volume of oxygen is required is 585 dm³

[By Lussac's law]

$$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 &\longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} & \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} & : & 2\text{ vol.} \end{matrix}$$

To calculate the volume of ethyne gas

$$\begin{matrix}\text{CO}_2 & : & \text{C}_2\text{H}_2 \\ 4 & : & 2 \\ 8.4 & : & x \end{matrix}$$

∴

$\frac{2}{4} \times 8.4 = 4.2 \text{ dm}^3$

Hence, volume of ethyne gas required = 4.2 dm³.

[By Lussac's law]

$$\begin{matrix} \text{H}_2 & + & \text{Cl}_2 &\longrightarrow & 2\text{HCl} \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$$

To calculate the amount of hydrogen used:

$$\begin{matrix}\text{Cl}_2 & : & \text{H}_2 \\ 1 \text{ vol.} & : & 1 \text{ vol.} \\ 4 \text{ lit.} & : & \text{x} \end{matrix}$$

∴

$x = \frac{1}{1} \times 4 = 4 \text{ lit.}$

Remaining hydrogen = 6 - 4 = 2 lit.

To calculate the amount of HCl :

$$\begin{matrix}\text{Cl}_2 & : & \text{HCl} \\ 1 \text{ vol.} & : & 2 \text{ vol.} \\ 4 \text{ lit.} & : & \text{x} \end{matrix}$$

∴

$x = \frac{2}{1} \times 4 = 8 \text{ lit.}$

Hence, after reaction 8 lit of HCl is formed which dissolves in water and 2 lit of hydrogen is left.

∴ 2 lits. of hydrogen is left.

[By Lussac's law]

$$\begin{matrix} 4\text{NH}_3 & + & 5\text{O}_2 &\longrightarrow & 4\text{NO} & + & 6\text{H}_2\text{O} \\ 4 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \end{matrix}$$

To calculate the volume of oxygen required :

$$\begin{matrix}\text{NH}_3 & : & \text{O}_2 \\ 4 \text{ vol.} & : & 5 \text{ vol.} \\ 100 \text{ cm.}^3 & : & \text{x} \end{matrix}$$

$\therefore x = \frac{5}{4} \times 100 = 125 \text{ cm.}^3$

Hence, volume of oxygen required = 125 cm³

Given, 20% of air contains oxygen

∴ 20% of 1000 cm³ = $\frac{20}{100}$ x 1000 = 200 cm³

[By Lussac's law]

$$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 &\longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} & : & 4\text{ vol.} \end{matrix}$$

To calculate the volume of propane consumed :

$$\begin{matrix}\text{O}_2 & : & \text{C}_3\text{H}_8 \\ 5 \text{ vol.} & : & 1 \text{ vol.} \\ 200 \text{ cm.}^3 & : & \text{x} \end{matrix}$$

$\therefore x = \frac{1}{5} \times 200 = 40 \text{ cm.}^3$

Hence, volume of propane is consumed is 40 cm³

[By Lussac's law]

$$\begin{matrix} 2\text{C}_2\text{H}_6 & + & 7\text{O}_2 &\longrightarrow & 4\text{CO}_2 & + & 6\text{H}_2\text{O} \\ 2\text{ vol.} & : & 7 \text{ vol.} & \longrightarrow & 4\text{ vol.} \end{matrix}$$

(i) To calculate the volume of CO₂ formed :

$$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{CO}_2 \\ 2 \text{ vol.} & : & 4 \text{ vol.} \\ 300 \text{ cc} & : & \text{x} \end{matrix}$$

∴

$x = \frac{4}{2} \times 300 = 600 \text{ cc}$

Hence, volume of carbon dioxide formed = 600 cc

(ii) (i) To calculate the volume of unused O₂ :

$$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{O}_2 \\ 2 \text{ vol.} & : & 7 \text{ vol.} \\ 300 \text{ cc} & : & \text{x} \end{matrix}$$

∴

$x = \frac{7}{2} \times 300 = 1050 \text{ cc}$

Unused oxygen = 1250 - 1050 = 200 cc.

Hence, volume of unused oxygen = 200 cc

[By Lussac's law]

$$\begin{matrix} 2\text{CO}& + & \text{O}_2 &\longrightarrow & 2\text{CO}_2 \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$$

To calculate the amount of CO required :

$$\begin{matrix}\text{CO}_2 & : & \text{CO} & \\ 2 \text{ vol.} & : & 2 \text{ vol.} \\ 750 \text{ ml} & : & \text{x} \end{matrix}$$

$\therefore x = \frac{2}{2} \times 750 = 750 \text{ ml}$

To calculate the amount of O₂ required :

$$\begin{matrix}\text{CO}_2 & : & \text{O}_2 & \\ \text{ 2 vol.} & : & 1 \text{ vol.} \\ 750 \text{ cm.}^3 & : & \text{x} \end{matrix}$$

$\therefore x = \frac{1}{2} \times 750 = 375 \text{ ml}$

Hence, CO required is 750 ml and O₂ required = 375 ml

Gay Lussac's law is associated in the question.

Volume is directly proportional to the number of molecules, hence gas D will have maximum no. of molecules as its volume is maximum.

If number of molecules of gas A is doubled, the volume will also be doubled.

(ii) 1 mole contains 6 x 10²³ number of molecules and occupies 22.4 lit. vol.

Given, volume of 'A' is 5.6 dm³ at s.t.p.

∴ vol. of D will be 4 × 5.6 = 22.4 lit.
No. of molecules in 22.4 lit. of D = 6 x 10²³ (Avogadro no.)

As D is 1 mole hence, mass of 1 mole of D (N₂O) = 2(14) + 16 = 28 + 16 = 44 g

Hence, mass of N₂O = 44 g

Gram molecular mass of C₂H₄
= 2(12) + 4(1)
= 24 + 4 = 28 g

As,

28 g of C₂H₄ = 1 mole
∴ 1.4 g of C₂H₄ = $\frac{1}{28}$ x 1.4 = 0.05 moles

1 mole = 6 × 10²³ molecules
∴ 0.05 moles = 6 × 10²³ x 0.05 = 3 x 10²² molecules

Hence, no. of moles is 0.05 and no. of molecules is 3 x 10²².

Vol. occupied by 1 mole = 22.4 lit
∴ Vol. occupied by 0.05 moles = 22.4 x 0.05 = 1.12 lit.

$\text{Vapour density} = \frac{\text{Molecular weight}}{2} = \frac{28}{2} = 14$

Hence, vol. occupied is 1.12 lit and vapour density is 14

(i) 2 moles of octane produce 16 moles of carbon dioxide

∴ 1 mole of octane will produce $\frac{16}{2}$ = 8 moles of carbon dioxide.

(ii) 1 moles occupies 22.4 lit. of vol.

∴ 8 mole will occupy $\frac{22.4}{1}$ x 8 = 179.2 lit. of vol.

(iii) To calculate the mass of CO₂ produced :

$$\begin{matrix} \text{C}_8 \text{H}_{18} & : & \text{CO}_2 & \\ \text{ 2 moles} & : & 16 \text{ moles} \\ \end{matrix}$$

Mass of 1 mole of CO₂ = 44 g

∴ Mass of 16 moles of CO₂ = 44 x 16 = 704 g

Hence, 8 moles of carbon dioxide are produced, vol occupied is 179.2 lit. and mass of carbon dioxide produced is 704 g

Mole — A mole is the amount of substance which contains the same number of units as the number of atoms in 12,000 g of carbon - 12 [₆C¹²]

(i) Gram molecular mass of N₂ = 2(14) = 28 g

6 × 10²³ molecules of N₂ weighs 28 g

∴ 24 × 10²⁴ molecules will weigh $\frac{28}{6 \times 10^{23}}$ x 24 × 10²⁴ = 1120 g

(ii) 6 × 10²³ molecules of N₂ occupies 22.4 dm³

∴ 24 × 10²⁴ molecules of N₂ will occupy $\frac{22.4}{6 \times 10^{23}}$ x 24 × 10²⁴ = 896 dm³

Hence, Mass of nitrogen gas = 1120 g and vol. occupied is 896 dm³

100 cm³ weighs 0.5 g .

22400 cm³ will weigh $\frac{0.5}{100}$ x 22400 = 112 g

Hence, relative molecular mass = 112 g.

CaCO₃ + 2HCl ⟶ CaCl₂ + H₂O + CO₂

(i) Given,

1 mole of CaCO₃ = molecular mass of CaCO₃ = 100 g

∴ 4.5 moles of CaCO₃ weighs $\frac{100}{1}$ x 4.5 = 450 g

(ii) 1 mole of CaCO₃ produces 1 mole of CO₂ and 1 mole occupies 22.4 l of volume.

∴ 4.5 moles of CaCO₃ will produce 4.5 moles of CO₂ and 4.5 moles will occupy 22.4 x 4.5 = 100.8 l

(iii) 1 mole CaCO₃ produces 111 g. CaCl₂

∴ 4.5 moles of CaCO₃ will produce = 111 x 4.5 = 499.5 g.

(iv)

$$\begin{matrix}\text{CaCO}_3 & : & \text{HCl} & \\ \text{ 1 mole} & : & 2 \text{ moles} \\ \text{ 4.5 mole} & : & x \text{ moles} \\ \end{matrix}$$

∴ number of moles of HCl used = 2 x 4.5 = 9 moles.

Hence, mass of CaCO₃ = 450 g, vol. of CO₂ liberated = 100.8 lits., mass of Cl₂ formed = 499.5 g, moles of HCl used = 9 moles.

(i) gram molecular mass of S = 32

6 × 10²³ atoms weigh = 32 g

10²² atoms will weigh = $\frac{32}{6 \times 10^{23}}$ x 10²² = 0.533 g

(ii) 1 mole of carbon dioxide weighs = C + 2(O) = 12 + 2(16) = 12 + 32 = 44 g

∴ 0.1 mole of carbon dioxide weighs = 44 x 0.1 = 4.4 g.

Hence, mass of 10²² atoms of sulphur = 0.533 g. and mass of 0.1 mole of carbon dioxide = 4.4 g.

gram molecular mass of SO₂ = 32 + 2(16) = 32 + 32 = 64 g

64 g of SO₂ occupy 22.4 lit of vol.

320 g of SO₂ will occupy = $\frac{22.4}{64}$ x 320 = 112 lit.

Hence, volume of 320 g of SO₂ = 112 lit.

5.6 dm³ weighs 12 g

∴ 22.4 dm³ weighs $\frac{12}{5.6}$ x 22.4 = 48 g

Hence, relative molecular mass of the gas = 48 g.

Given, V.D. = 8

Gram molecular mass = V.D. x 2 = 8 x 2 = 16 g.

16 g occupies 22.4 lit.

∴ 24 g. will occupy = $\frac{22.4}{16}$ x 24 = 33.6 lit.

Hence, volume occupied by gas = 33.6 lit.

1 mole occupies 22.4 lit.

0.01 mole will occupy = 22.4 x 0.01 = 0.224 lit.

Hence, vol. occupied is 0.224 lit.

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

(i) Gram molecular mass of ammonia = N + 3(H) = 14 + 3 = 17 g.

17 g. occupies 22.4 lit. of vol.

∴ 68 g will occupy = $\frac{22.4}{17}$ x 68 = 89.6 lit.

Hence, volume occupied by this gas = 89.6 lit.

(ii) 17 g = 1 mole

∴ 68 g = $\frac{1}{17}$ x 68 = 4 moles.

1 mole = 6 × 10²³

∴ 4 moles = 4 x 6.023 × 10²³ molecules.

Hence, Moles = 4 and molecules = 4 x 6.023 × 10²³ molecules

D : 6.02 x 10²³ atoms of carbon.

Reason
Gram molecular mass of carbon is least.

1 mole of Sulphur weighs 32 g and contains 6.02 x 10²³ atoms

∴ 3.2 g of Sulphur will contain = $\frac{6.02 \times 10^{23}}{32}$ x 3.2 = 6.02 x 10²² atoms.

6.02 x 10²³ atoms of Ca weighs = 40 g

∴ 6.02 x 10²² atoms of Ca will weigh = $\frac{40}{ 6.02 \times 10^{23}}$ x 6.02 x 10²² = 4 g.

Hence, mass = 4g.

(C) 1 : 1
Reason
1 mole of any substance contains same (Avogardo No.) of molecules. Hence, ratio is same.

20 lit. of gas weighs = 32 g.

1 mole = 22.4 lit. of gas will weigh = $\frac{32}{20}$ x 22.4 = 35.84 g.

Hence, gram mol. weight of the gas = 35.84 g.

Gram molecular mass of oxygen = 32 g = 1 mole and contains 6 x 10²³ molecules i.e., 6 x 10²³ molecules weigh 32 g

∴ 12 x 10²⁴ molecules will weigh = $\frac{32}{ 6\times 10^{23}}$ x 12 x 10²⁴ = 640 g

(ii) 6 x 10²³ molecules occupy 22.4 lit. of vol.

∴ 12 x 10²⁴ molecules will occupy = $\frac{22.4}{ 6\times 10^{23}}$ x 12 x 10²⁴ = 448 lit.

Hence, mass of O₂ present in the cylinder = 640 g and volume of O₂ at S.T.P. present in the cylinder = 448 lit.

23 g of Na = 1 g. atom

∴ 4.6 g. of Na = $\frac{1}{23}$ x 4.6 = 0.2 g

Hence, 0.2 g. atoms

11.2 lit. weighs 24 g

∴ 22.4 lit. will weigh = $\frac{24}{11.2}$ x 22.4 = 48 g

Hence, gram molecular mass of the gas = 48 g.

(i) Gram molecular mass of oxygen = 32 g = 1 mole

32 g = 1 mole

∴ 12 g = $\frac{1}{32}$ x 12 = 0.375 moles.

(ii) 6 x 10²³ atoms of carbon weigh = 12 g

∴ 10²² atoms of carbon will weigh = $\frac{12}{ 6\times 10^{23}}$ x 10²² = 0.2 g.

Hence, number of moles = 0.375 moles and weight of 10²² atoms of carbon = 0.2 g.

Gram molecular mass of CO₂ = 12 + 32 = 44 g and occupies 22.4 lit. vol.

∴ 80 g. will occupy = $\frac{22.4}{44}$ x 80 = 40.73 lit.

Hence, volume occupied = 40.73 lits.

Gram molecular mass of CO₂ = 12 + 32 = 44 g and contains 6.023 x 10²³ molecules.

∴ 4.4 gm. of CO₂ will contain = $\frac{6.023 \times 10^{23}}{44}$ x 4.4 = 6.023 x 10²² molecules.

Hence, number of molecules = 6.023 x 10²² molecules.

The number of moles in 11 g. of nitrogen gas is 0.39

Working

Gram molecular mass of nitrogen gas (N₂) = 2(14) = 28

28 g of nitrogen gas = 1 mole

∴ 11 g of nitrogen gas = $\frac{1}{28}$ x 11 = 0.39

(i) Gram molecular mass of methane = V.D. x 2 = 8 x 2 = 16 g

16 g of methane occupies 22.4 lit.

∴ 40 g of methane will occupy = $\frac{22.4}{16}$ x 40 = 56 lits.

(ii) Gram molecular mass of NaOH = 23 + 16 + 1 = 40 g

40 g of NaOH = 1 mole

∴ 160 g of NaOH = $\frac{160}{40}$ = 4 moles.

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

∴ If gases under the same conditions have same number of molecules then they must have the same volume.

(i) So, X molecules of N₂ occupy V litres.

(ii) 3X molecules of CO will occupies 3V litres.

(iii) Molar mass of CO₂ = C + 2(O) = 12 + 2(16) = 44 g

1 mole of O₂ weighs 32 g and occupy 22.4 lit. volume

∴ 8 g of O₂ will occupy $\frac{22.4}{32}$ x 8 = 5.6 lit. vol. = Molar volume

If 8 g of O₂ occupies 5.6 lit. vol.

And 1 mole of CO₂ occupy 22.4 lit and weighs = 44 g at s.t.p.

∴ 5.6 lit. vol. of CO₂ will weigh = $\frac{44}{22.4}$ x 5.6 = 11 g

Hence, mass of CO₂ = 11 g

(iv) Avogadro's Law is used above.

Atomic weight is the number of times one atom of an element is heavier than 1⁄12 the mass of an atom of carbon (C¹²)

Avogadro's Law

Equal volumes of all gases under identical conditions contain the same number of molecules.

According to Avogadro's law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Hence, number of molecules of N₂ = Number of molecules of H₂ = X

(i) 2 g of hydrogen gas = 1 mole

∴ 1000 g of hydrogen gas = $\frac{1000}{2}$ = 500 moles.

(ii) Molar mass of CO₂ = C + 2(O) = 12 + 2(16) = 44 g

1 mole of carbon dioxide = 44 g

∴ 500 moles of carbon dioxide = 44 x 500 = 22,000 g = 22 kg.

Hence, Weight of carbon dioxide in cylinder = 22 kg

(iii) According to Avogadro's law — Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

∴ molecules in the cylinder of carbon dioxide = X.

(iv) Avogadro's law — Equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

According to Avogadro’s law : equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Given,

150 cc of a gas A contains X molecules

Therefore, 150 cc of B will also contain X molecule.

Hence, 75 cc of B will contain X/2 molecules.

The problem is based on Avogadro's law.

Given,

Using the gas equation,

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$$\begin{aligned}\frac{380 \times 360}{360} = \frac{760 \times x}{273} \\[0.5em] x = \frac{38 \times 273}{76 } \\[0.5em] x = \frac{10,374}{76 } \\[0.5em] x = 136.5 \text{ cm}^3 = 0.1365 \text{ lit.}\end{aligned}$$

At s.t.p. 0.1365 lit weighs 0.546 g

Therefore, 1 lit of gas weighs = $\frac{0.546}{0.1365}$ x 1 = 4 g

Vapour density of gas =

$$\begin{aligned}\frac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \frac{4}{0.09} \\[0.5em] = 44.444\end{aligned}$$

Molecular weight = 2 x Vapour density
= 2 x 44.444 = 88.888 = 88.89 g

Hence, relative molecular mass of gas 88.89 g

Vapour Density.

2KMnO₄ ⟶ K₂MnO₄ + MnO₂ + O₂

Loss in mass = 1.32 g = 1 lit of oxygen

Vapour density of gas =

$$\begin{aligned}\frac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \frac{1.32}{0.0825} \\[0.5em] = 16 \text{ g}\end{aligned}$$

Molecular weight = 2 x Vapour density
= 2 x 16 = 32 g

Hence, relative molecular mass of oxygen is 32 g

10

Working

Vapour density of gas =

$$\begin{aligned}\frac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \frac{10}{2} \\[0.5em] = 5 \text{ g}\end{aligned}$$

Molecular weight = 2 x Vapour density
= 2 x 5 = 10 g

Hence, relative molecular mass of gas is 10 g

22

Working

Molecular weight of carbon dioxide = 12 + 2(16) = 12 + 32 = 44 g

Vapour density of gas =

$$\begin{aligned}\frac{\text{Molecular weight}}{2} \\[0.5em] = \frac{44}{2} \\[0.5em] = 22 \text{ g}\end{aligned}$$

Hence, vapour density of carbon dioxide is 22 g

Vapour density.

Molecular weight of magnesium nitrate Mg(NO₃)₂. 6H₂O

= 24 + 2[14 + 3(16)] + 6[2(1) + 16]
= 24 + 2[14 + 48] + 6(18)
= 24 + 2(62) + 6(18)
= 24 + 124 + 108
= 256 g

Mass of oxygen in Mg(NO₃)₂. 6H₂O
= 6(16) + 6(16)
= 96 + 96 = 192

256 g of magnesium nitrate crystals contains 192 g of oxygen

∴ 100 g of magnesium nitrate crystals contains = $\frac{192}{256}$ x 100 = 75%

Hence, total percentage of Oxygen in magnesium nitrate crystals is 75%

Molecular weight of urea [CO(NH₂)₂]
= 12 + 16 + 2[14 + 2(1)]
= 12 + 16 + 2[16] = 28 + 32
= 60 g

Converting given weight of urea in grams,
1000 kg of urea = 1000 x 10³ g of urea

60 g of urea contains 28 g of nitrogen

∴ 1000 x 10³ g of urea will contain = $\frac{28}{60}$ x 1000 x 10³ = 466.66 x 10³ g ≈ 467 x 10³ g = 467 kg of nitrogen.

Hence, mass of nitrogen is 467 kg

Molecular weight of borax Na₂B₄O₇.10H₂O
= 2(23) + 4(11) + 7(16) + 10[2(1)+ 16]
= 46 + 44 + 112 + 180 = 382 g

382 g of borax contains 44 g of boron

∴ 100 g of borax will contain = $\frac{44}{382}$ x 100 = 11.5%

Hence, Percentage of boron is 11.5%

Molecular weight of Ca(NO₃)₂
= 40 + 2[14 + 3(16)]
= 40 + 2[14 + 48]
= 40 + 2(62)
= 40 + 124 = 164 g

Total nitrogen that has to be replaced in 10 hectares of land = 20 x 10 = 200 kg

28 g of N is present in 164 kg of calcium nitrate

∴ 200 kg of N is present in = $\frac{164}{28}$ x 200 = 1171 kg

Hence, 1171 kg of fertilizer will be required

Molecular weight of Ca(H₂PO₄)₂
= 40 + 2[2(1) + 31 + 4(16)]
= 40 + 2[2 + 31 + 64]
= 40 + 2[97]
= 40 + 194
= 234 g

234 g of Ca(H₂PO₄)₂ contains 62 g of P

∴ 100 g of Ca(H₂PO₄)₂ will contain = $\frac{62}{234}$ x 100 = 26.49% = 26.5%

Hence, 26.5% phosphorous is present in superphosphate Ca(H₂PO₄)₂

Molecular weight of ammonium chloroplatinate (NH₄)₂PtCl₆
= 2[14 + 4(1)] + 195 + 6(35.5)
= 2[18] + 195 + 213
= 444 g

444 g of ammonium chloroplatinate contains 195 g of Pt

∴ 100 g of ammonium chloroplatinate will contain = $\frac{195}{444}$ x 100 = 43.91% = 44%

Hence, 44% percentage of platinum is present in ammonium chloroplatinate

Molecular weight of aluminium nitride AlN = 27 + 14
= 41 g

41 g of aluminium nitride contains 14 g of N

∴ 100 g of aluminium nitride will contain = $\frac{14}{41}$ x 100 = 34.15%

Hence, percentage of nitrogen in aluminium nitride is 34.15%

Molecular weight of sodium aluminium fluoride (Na₃AlF₆)
= 3(23) + 27 + 6(19)
= 69 + 27 + 114
= 210 g

210 g of sodium aluminium fluoride contains 69 g of Na

∴ 100 g of sodium aluminium fluoride will contain = $\frac{69}{210}$ x 100 = 32.85% = 33%

Hence, percentage of sodium in sodium aluminium fluoride (Na₃AlF₆) is 33%

Molecular weight of ammonium nitrate NH₄NO₃ = 14 + 4(1) + 14 + 3(16)
= 32 + 48 = 80 g

80 g of ammonium nitrate contains 48 g of oxygen

∴ 100 g of ammonium nitrate will contain = $\frac{48}{80}$ x 100 = 60%

Hence, percentage of oxygen in ammonium nitrate is 60%

Given,

molecular mass of ammonium nitrate = 80

mass of nitrogen in 1 mole of ammonium nitrate (NH₄NO₃) is 2(14) = 28 g

percentage of nitrogen in ammonium nitrate = $\frac{28}{80}$ x 100 = 35%

mass of oxygen in 1 mole of ammonium nitrate (NH₄NO₃) is 3(16) = 48 g

percentage of oxygen in ammonium nitrate = $\frac{48}{80}$ x 100 = 60%

Hence, percentage of nitrogen is 35% and percentage of oxygen is 60%

Molecular weight of Mg(NO₃)₂.6H₂O = 24 + 2[14 + 3(16)] + 6[2(1) + 16]
= 24 + 2[14 + 48] + 6[18]
= 24 + 2(62) + 108
= 24 + 124 + 108 = 256 g

256 g of magnesium nitrate crystals contains 24 g of magnesium

∴ 100 g of magnesium nitrate crystals will contain = $\frac{24}{256}$ x 100 = 9.375% = 9.38%

Hence, percentage of magnesium in magnesium nitrate crystals is 9.38%

Molecular weight of CuSO₄.5H₂O = 64 + 32 + 4(16) + 5[2(1) + 16]
= 64 + 32 + 64 + 5(18) = 160 + 90 = 250 g

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

∴ 100 g of CuSO₄.5H₂O will contain = $\frac{90}{250}$ x 100 = 36%

Hence, percentage of water of crystallization is 36%

(i) Molecular weight of sodium aluminium fluoride (Na₃AlF₆)
= 3(23) + 27 + 6(19)
= 69 + 27 + 114
= 210 g

210 g of sodium aluminium fluoride contains 114 g of fluorine

∴ 100 g of sodium aluminium fluoride will contain = $\frac{114}{210}$ x 100 = 54.28% = 54%

(ii) 210 g of sodium aluminium fluoride contains 69 g of sodium

∴ 100 g of sodium aluminium fluoride will contain = $\frac{69}{210}$ x 100 = 32.85% = 33%

(iii) 210 g of sodium aluminium fluoride contains 27 g of aluminium

∴ 100 g of sodium aluminium fluoride will contain = $\frac{27}{210}$ x 100 = 12.85% = 13%

Hence, percentage of F = 54%, Na = 33%, Al = 13%

Molecular Formula of octane = C₈H₁₈ = (C₄H₉)2

Molecular formula = n[Empirical formula]

n = 2

Hence, Empirical formula = C₄H₉

Simplest ratio of whole numbers = C : H : Cl = 1 : 1 : 2

Hence, empirical formula is CHCl₂

Empirical formula weight = 12 + 1 + 2(35.5) = 84 g

Relative molecular mass = 168

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{168}{84} = 2\end{aligned}$$

Molecular formula = n[E.F.] = 2[CHCl₂] = C₂H₂Cl₄

∴ Molecular formula = C₂H₂Cl₄

Simplest ratio of whole numbers N : H = 1 : 2

Hence, empirical formula is NH₂

Empirical formula weight = 14 + 2(1) = 16

Relative molecular mass = 37

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{37}{16} = 2.31 = 2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[NH₂] = N₂H₄

Simplest ratio of whole numbers = C : H : Br = 1 : 2 : 1

Hence, empirical formula is CH₂Br

Empirical formula weight = 12 + 2(1) + 80 = 94

Vapour density (V.D.) = 94

Molecular weight = 2 x V.D. = 2 x 94

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{2 \times 94}{94} = 2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[CH₂Br] = C₂H₄Br₂

Empirical formula is X₂Y

Empirical formula weight = 2(10) + 5 = 25

Vapour density (V.D.) = 25

Molecular weight = 2 x V.D. = 2 x 25

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{2 \times 25}{25} = 2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[X₂Y] = X₄Y₂

Empirical formula is CH

Empirical formula weight = 12 + 1 = 13

Vapour density (V.D.) = 13

Molecular weight = 2 x V.D. = 2 x 13

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{2 \times 13}{13} = 2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[CH] = C₂H₂

Simplest ratio of whole numbers = C : H = 1 : 2.5 = 2 : 5

Hence, empirical formula is C₂H₅

Empirical formula weight = 2(12) + 5(1) = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{2 \times 29}{29} =2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[C₂H₅] = C₄H₁₀

Empirical formula = XY₂

Empirical formula weight = V.D.

Molecular weight = 2 x V.D.

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] \text{n} = \frac{\text{2 x V.D.}}{\text{V.D.}} \\[0.5em] \text{n} = 2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[XY₂] = X₂Y₄

Simplest ratio of whole numbers = N : H = 1 : 3

Hence, empirical formula is NH₃

Molecular formula of a compound = C₆H₁₈O₃ = 3[C₂H₆O]

As, molecular formula = n[E.F.] and n = 3

Hence, empirical formula = C₂H₆O

Empirical formula

Simplest ratio of whole numbers = C : H : N = 5 : 5 : 1

Hence, empirical formula is C₅H₅N

Empirical formula weight = 5(12) + 5(1) + 14 = 79

V.D. = 39.5

Molecular weight = 2 x V.D. = 2 x 39.5 = 79

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{79}{79} = 1\end{aligned}$$

∴ Molecular formula = n[E.F.] = 1[C₅H₅N] = C₅H₅N

$$\begin{matrix} \text{Na}_2\text{CO}_3.10\text{H}_2\text{O} & \xrightarrow{\Delta} & \text{Na}_2\text{CO}_3 & + & 10\text{H}_2\text{O} \\ 2(23) + 12 + 3 (16) & & 2(23) + 12 & & 10[2(1) \\ + 10[2(1) + 16] & & + 3(16) & & + 16] \\ = 46 + 12 + 48 & & = 46 + 12 & & = 180 \text{ g} \\ + 10(18) & & + 48 \\ = 286 \text{ g} & & = 106 \text{ g} \\ \end{matrix}$$

286 g of washing soda had 106 g of anhydrous sodium carbonate

∴ 57.2 g will have = $\frac{106}{286}$ x 57.2 = 21.2 g anhydrous sodium carbonate.

Hence, 21.2 g anhydrous sodium carbonate is left.

$$\begin{matrix} \text{Na}_2\text{SO}_4 & + & \text{Pb}(\text{NO}_3)_2 & \longrightarrow &\text{PbSO}_4 & + & 2\text{NaNO}_3 \\ 2(23) + 32 & & & & 207 + 32 \\ + 4(16) & & & & + 4(16) \\ = 46 + 32 & & & & = 207 + 32 \\ + 64 & & & & + 64 \\ = 142 \text{ g} & & & & = 303 \text{ g} \\ \end{matrix}$$

303 g of lead sulphate was obtained from 142 g of Na₂SO₄

∴ 15.15 g of lead sulphate will be precipitated from $\frac{142}{303}$ x 15.15 = 7.1 g.

Hence, 7.1 g of sodium sulphate was present in the original solution.

$$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & \xrightarrow{\Delta} & \text{Cr}_2\text{O}_3 & + & 4\text{H}_2\text{O} & + \text{ N}_2 \\ 2[14 + 4(1)] & & 2(52) & & & 1 \text{ mol.} \\ + 2(52) + 7(16) & & + 3(16) & & & \text{vol.} \\ = 36 + 104 + 112 & & = 104 + 48 & & & = 22.4 \text{ lit} \\ = 252 \text{ g} & & = 152 \text{ g} \\ \end{matrix}$$

252 g of ammonium dichromate gives 22.4 lit of nitrogen

∴ 63 g of ammonium dichromate will give $\frac{22.4}{252}$ x 63 = 5.6 lit of nitrogen.

Hence, 5.6 lit of nitrogen is evolved.

(ii) 252 g of ammonium dichromate gives 152 g Cr₂O₃

∴ 63 g of ammonium dichromate will give $\frac{152}{252}$ x 63 = 38 g

Hence, 38 g of Cr₂O₃ is formed.

$$\begin{matrix} \text{Na}_2\text{SO}_4 & + & \text{BaCl}_2 & \longrightarrow & \text{BaSO}_4 \downarrow & + & 2\text{NaCl} \\ 2(23) + 32 & & & & 137 + 32 \\ + 4(16) & & & & + 4(16) \\ = 46 + 32 & & & & = 137 + 32 \\ + 64 & & & & + 64 \\ = 142 \text{ g} & & & & = 233 \text{ g} \\ \end{matrix}$$

233 g of BaSO₄ is obtained from 142 g of Na₂SO₄

6.99 g of BaSO₄ will be obtained from $\frac{142}{233}$ x 6.99 = 4.26 g of Na₂SO₄.

∴ In 10 g mixture $\frac{4.26}{10}$ x 100 = 42.6% Na₂SO₄ is present.

Hence, 42.6% Na₂SO₄ is present in the original mixture

$$\begin{matrix} 2\text{KMnO}_4 & + 10 \text{FeSO}_4 & + & 8\text{H}_2\text{SO}_4 \\ 2[39 + 55 & 10[56+ 32 \\ + 4(16)] & + 4(16)] \\ = 2[39 + & 10[56 + \\ 55 + 64] & 32 + 64] \\ = 316 \text{ g} & = 1520\text{ g} \\ \end{matrix}$$

$\longrightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5 \text{Fe}_2(\text{SO}_4)_3 + 8\text{H}_2\text{O}$

316 g of potassium permanganate was used with 1520 g of iron (II) sulphate

∴ 15.8 g. of potassium permanganate will be used with $\frac{1520}{316}$ x 15.8 = 76 g of iron (II) sulphate

Hence, 76 g of iron (II) sulphate was used.

$$\begin{matrix} 2\text{NaHCO}_3 & \longrightarrow & \text{Na}_2\text{CO}_3 & + & \text{H}_2\text{O} + \text{CO}_2 \\ 2 \text{mol.} & & 1 \text{mol.} \\ 2 \times 84 & & 106\text{ g.} \\ = 168\text{ g.} & & \end{matrix}$$

106 g of Na₂CO₃ is obtained from 168 g of NaHCO₃

∴ 21.2 g. of Na₂CO₃ is obtained from $\frac{168}{106}$ x 21.2 = 33.6 g of NaHCO₃

(ii)

$$\begin{matrix} \text{NaCl} & + & \text{NH}_3 & + & \text{CO}_2 & + & \text{H}_2\text{O} \\ & & & & 1\text{mol.} & \\ & & & & 22.4 \text{lit.} \\ \longrightarrow & \text{NaHCO}_3 & + & \text{NH}_4\text{Cl} \\ & 1 \text{mol.} \\ & 84\text{ g.} \end{matrix}$$

84 g of NaHCO₃ is obtained from 22.4 lit of CO₂

∴ 33.6 g. of NaHCO₃ is obtained from $\frac{22.4}{84}$ x 33.6 = 8.96 lit.

Hence, 8.96 lit of CO₂ is required.

$$\begin{matrix} 3\text{CuO} & + & 2\text{NH}_3 \\ 3\text{ mol.} & & 2 \text{ mol.} \\ = 3 \times 80 \text{ g.} & & = 2 \times 22.4 \text{ lit.} \\ = 240 \text{ g.} & & = 44.8\text{ lit.} \\ \end{matrix}$$

$\longrightarrow 3\text{Cu} + 3\text{H}_2\text{O} + \text{N}_2$

240 g of CuO is reduced by 44.8 lit of NH₃

∴ 120 g of CuO is reduced by $\frac{44.8}{240}$ x 120 = 22.4 lit.

Hence, 22.4 lit of NH₃ is required

$$\begin{matrix} \text{NH}_4\text{NO}_3 &\xrightarrow{\Delta} & \text{N}_2\text{O} & + & 2\text{H}_2\text{O} \\ 1\text{ vol.} = 80 \text{g} & & 1 \text{ vol.} & & 2\text{ vol.} \\ \end{matrix}$$

(i) Given,

1 vol. of di nitrogen produced at the same time as 8.96 litres of steam (2 vol)

Hence, Vol of di nitrogen = $\frac{8.96}{2}$ = 4.48 lit.

Hence, volume of di nitrogen oxide produced = 4.48 lit.

(ii) 2 vol = (2 x 22.4) lit steam is produced from 80 g NH₄NO₃

∴ 8.96 lit of steam will be produced by $\frac{80}{2 \times 22.4}$ x 8.96 = 16 g

Hence, 16 g of ammonium nitrate is required to be heated.

$\underset{12 \text{g}}{\text{C}} + \underset{2(98) = 196 \text{ g}}{2\text{H}_2\text{SO}_4} \longrightarrow \text{CO}_2 + 2\text{H}_2\text{O} + 2\text{SO}_2$

(i) 196 g of sulphuric acid oxidizes 12 g carbon

∴ 49 g of sulphuric acid will oxidize = $\frac{12}{196}$ x 49 = 3 g

Hence, 3 g of carbon is oxidized.

(ii) 12 g carbon liberates 2 vol = (2 x 22.4) lit of SO₂

∴ 3 g of carbon will liberate $\frac{2 \times 22.4}{12}$ x 3 = 11.2 lit of SO₂.

Hence, 11.2 lit of SO₂ is liberated.

$\underset{58 \text{g}}{\text{NaCl}} + \text{AgNO}_3 \longrightarrow \underset{143 \text{ g}}{\text{AgCl}} + \text{NaNO}_3$

(i) 143 g AgCl is formed by 58 g NaCl

∴ 14.3 g of AgCl will be formed by $\frac{58}{143}$ x 14.3 = 5.8 g

Percentage of NaCl = $\frac{5.8}{30}$ x 100 = 19.33%

Hence, percentage of NaCl is 19.33%

$$\begin{matrix} \text{C}_3\text{H}_8 & + &5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 3(12) + 8(1) & & 5 \text{ vol} \\ = 44 \text{ g}& & 5 (22.4) \text{ lit} \\ \end{matrix}$$

(i) 44 g propane requires 5 x 22.4 lit of oxygen

∴ 8.8 g of propane will require $\frac{5 \times 22.4}{44}$ x 8.8 = 22.4 lit.

Hence, 22.4 lit of Oxygen is required.

$$\begin{matrix} \text{P} & + & 5\text{HNO}_3 & \longrightarrow & \text{H}_3\text{PO}_4 & + & \text{H}_2\text{O} & + & 5\text{NO}_2 \\ 31 \text{ g} & & [conc.] & & 3(1) + 31 \\ & & & & + 4(16) \\ & & & & = 98 \text{ g} \\ \end{matrix}$$

(i) 31 g of P = 1 mole

∴ 9.3 g of P = $\frac{1}{31}$ x 9.3 = 0.3 moles.

Hence, 0.3 moles of phosphorous was taken for the reaction.

(ii) 31 g of P forms 98 g of phosphoric acid

∴ 9.3 g will form $\frac{ 98}{31}$ x 9.3 = 29.4 g.

Hence, 29.4 g. of phosphoric acid is formed

(iii) 31 g of P produces 5 vol = 5 x 22.4 lit.

∴ 9.3 g will produce = $\frac{5 \times 22.4}{31}$ x 9.3 = 33.6 lit.

$$\begin{matrix} 2\text{KClO}_3 & \xrightarrow{MnO_2} & 2\text{KCl} & + & 3\text{O}_2 \\ 2[39 + 35.5 & & 2[39 & & 3(22.4) \\ + 3(16)] & & + 35.5] & & = 67.2 \text{ lit.} \\ = 245 \text{ g} & & = 149 \text{g} \\ \end{matrix}$$

67.2 lit. of O₂ is produced by 245 g of KClO₃

∴ 6.72 lit of O₂ will be obtained from $\frac{245}{67.2}$ x 6.72 = 24.5 g.

(ii) 22.4 lit = 1 mole
∴ 6.72 lit = $\frac{1}{22.4}$ x 6.72 = 0.3 moles

1 mole = 6.023 x 10²³ molecules
∴ 0.3 moles = 0.3 x 6.023 x 10²³ molecules.

$$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & \xrightarrow{\Delta} & \text{N}_2 & + & 4\text{H}_2\text{O} & + & \text{Cr}_2\text{O}_3 \\ 2[14 + 4(1)] & & & & & & 2(52) \\ + 2(52) + 7(16) & & & & & & + 3(16) \\ = 36 + 104 & & & & & & = 104 + 48 \\ + 112 = 252 \text{ g} & & & & & & = 152 \text{ g} \\ \end{matrix}$$

(i) 252 g of (NH₄)₂Cr₂O₇ = 1 mole

∴ 63 g of (NH₄)₂Cr₂O₇ = $\frac{1}{252}$ x 63 = 0.25 moles

Hence, no. of moles = 0.25 moles

(ii)

$$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & : & \text{N}_2 \\ 1 \text{ mol.} & : & 1 \text{ mol.} \\ 0.25 \text{ mol.} & : & x \\ \end{matrix}$$

Hence, 0.25 moles of (NH₄)₂Cr₂O₇ will produce 0.25 moles of nitrogen.

(iii) 1 mole of N₂ occupies 22.4 lit.

∴ 0.25 moles of N₂ will occupy = 22.4 x 0.25 = 5.6 lit.

Hence, volume of N₂ evolved at s.t.p = 5.6 lit.

(iv) 1 mole of Cr₂O₇ = 152 g.

∴ 0.25 moles of Cr₂O₇ = 152 x 0.25 = 38 g.

Hence, mass in gms of Cr₂O₇ formed = 38 g.

$$\begin{matrix} 2\text{Ca}(\text{NO}_3)_2 & \longrightarrow & 2\text{CaO} & + & 4\text{NO}_2 & + & \text{O}_2 \\ 2[(40) + & & 2(40 & & 4[22.4] \\ 2(14 +3(16))] & & + 16) & & \text{ lit.} \\ = 328 \text{ g} & & 112 \text{g} \\ \end{matrix}$$

(i) 328 g of Ca(NO₃)₂ produces 112 g of calcium oxide.

∴ 82 g of Ca(NO₃)₂ will produce = $\frac{112}{328}$ x 82 = 27.99 g = 28 g

Hence, 28 g of calcium oxide is produced.

(ii) 328 g of Ca(NO₃)₂ produces 4(22.4) lit of nitrogen dioxide.

∴ 82 g of Ca(NO₃)₂ will produce = $\frac{4 \times 22.4}{328}$ x 82 = 22.4 lit.

Hence, vol of nitrogen dioxide evolved = 22.4 lit.

$$\begin{matrix} \text{Al}_4\text{C}_3 & + & 12\text{H}_2\text{O} & \longrightarrow & 4\text{Al(OH)}_3 & + & 3\text{CH}_4 \\ 144 \text{ g} & & & & 4(78) & & 3(22.4) \\ & & & & = 312 \text{ g} & & = 67.2 \text{ lit.} \\ \end{matrix}$$

144 g of aluminium carbide forms 312 g of aluminium hydroxide.

∴ 12 g of aluminium carbide will form $\frac{312}{144}$ x 12 = 26 g of aluminium hydroxide

Hence, 26 g of aluminium hydroxide is formed.

(ii) 144 g of aluminium carbide forms 67.2 lit of methane.

∴ 12 g of aluminium carbide will form $\frac{67.2}{144}$ x 12 = 5.6 lit.

Hence, vol. of methane obtained at s.t.p. = 5.6 lit.

$$\begin{matrix} 2\text{NaOH} & + & \text{CuSO}_4 & \longrightarrow & \text{Na}_2\text{SO}_4 & + & \text{Cu(OH)}_2\downarrow \\ 2[23 + 16 & & & & & & 64 + 2[40] \\ + 1] & & & & & & = 98 \text{g} \\ = 80 \text{ g} & & & & & & \\ \end{matrix}$$

80 g of sodium hydroxide precipitates 98 g of copper hydroxide.

∴ 200 g of sodium hydroxide will precipitate $\frac{98}{80}$ x 200 = 245 g of copper hydroxide.

Hence, 245 g. of copper hydroxide is precipitated.