(a) A hydrate of calcium sulphate CaSO4.xH₂O contains 21% water of crystallisation. Find the value of x.

(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.

(c) How much volume will be occupied by 2g of dry oxygen at 27°C and 740 mm pressure?

(d) What would be the mass of CO₂ occupying a volume of 44 litres at 25°C and 750 mm pressure?

(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.

AgNO₃ (aq) + NaCl (aq) ⟶ AgCl (s) + NaNO₃

Calculate the percentage of NaCl in the mixture.

Relative molecular mass of CuSO₄.xH₂O

= 40 + 32 + (4×16) + [x(2+16)]

= 40 + 32 + 64 + 18x

= 136 + 18x

∴ 21% water of crystallization = 18 x

$\Rightarrow \dfrac{18x}{136 + 18x } = \dfrac{21}{100} \\[1em] 1800x = 21(136 + 18x) \\[1em] 1800x = 2856 + 378x \\[1em] 1800x - 378x = 2856 \\[1em] 1422x = 2856 \\[1em] x = \dfrac{2856}{1422} \\[1em] x = 2$

Hence, water of crystallization = 2

(b) Molar mass of H₂O = 2(1) + 16 = 18 g

For 18 g water, vol. of hydrogen needed = 22.4 litre

∴ For 1.8 g, vol. of H₂ needed = $\dfrac{22.4}{18}$ x 1.8 = 2.24 L

According to Gay Lussac's Law, 2 vol of hydrogen requires 1 vol of oxygen

When 2 vol of hydrogen in 1.8 g H₂O is 2.24 L, then one vol. of oxygen will be:

$\dfrac{2.24}{2}$ = 1.12 L

(c) Gram molecular mass of O₂ = 2 x 16 = 32 g

1 mole of O₂ weighs 32 g and occupies 22.4 lit. vol.

∴ 2 g of O₂ occupies = $\dfrac{22.4}{32} \times 2 = 1.4 \text{ lit.}$

Volume occupied by 2 g of O₂ gas at 27°C and 740 mm pressure:

Using the gas equation,

$\dfrac{P${1}V{1}}{T{1}} = \dfrac{P{2}V{2}}{T{2}}

Substituting the values we get,

$\dfrac{760 \times 1.4}{273} = \dfrac{740 \times x}{300} \\[0.5em] x = \dfrac{760 \times 1.4 \times 300}{740 \times 273 } \\[0.5em] x = \dfrac{3,19,200}{2,02,020} \\ \\[0.5em] x = 1.58 \text{ lit}$

Hence, the volume occupied by 2 g of O₂ gas at 27°C and 740 mm pressure is 1.58 lit.

(d) Gram molecular mass of CO₂ = 12 + 2(16) = 12 + 32 = 44 g

Using the gas equation,

$\dfrac{P${1}V{1}}{T{1}} = \dfrac{P{2}V{2}}{T{2}}

Substituting the values we get,

$\dfrac{750 \times 44}{298} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{750 \times 44 \times 273}{760 \times 298 } \\[0.5em] x = \dfrac{9,009,000}{226,480} \\ \\[0.5em] x = 39.78 \text{ lit}$

22.4 litre of CO₂ at S.T.P. has mass = 44 g

39.78 litre of CO₂ at S.T.P. has mass $\dfrac{44}{22.4}$ x 39.78

= 78.14 g

(e)

$\begin{matrix} \text{AgNO}$3 & + &\text{NaCl} & \longrightarrow & \text{AgCl} & + & \text{NaNO}3 \ && 23 + 35.5 && 108 + 35.5 \ && = 58.5 \text{g}&& = 143.5\text{g} \ \end{matrix}

(i) 143.5 g AgCl is formed by 58.5 g NaCl

∴ 1.435 g of AgCl will be formed by $\dfrac{58.5}{143.5}$ x 1.435 = 0.582 g

Percentage of NaCl = $\dfrac{0.582}{1}$ x 100 = 58.5%

Hence, percentage of NaCl is 58.5%