Answer the following questions :
Calculate the molality of a sulphuric acid solution in which the mole fraction of water is $0.85$ .

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Solution

$\frac{n_{A}}{n_{A} + n_{B}} = 0.85$ ..........(i)

$∴ \frac{n_{B}}{n_{A} + n_{B}} = 1 - 0.85 = 0.15$ .........(ii)

Dividing (ii) by (i), we get

$\frac{n_{B}}{n_{A}} = \frac{0.15}{0.85}$ or $\frac{n_{B}}{1000 / 18} = \frac{0.15}{0.85}$

or $n_{B} = \frac{0.15}{0.85} \times \frac{1000}{18} = 9.8 m o l$

Hence, molality = $9.8 m$

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