An aqueous solutions of diabasic acid (molecular mass $= 118$ ) containing $35.4$ g of acid per litre of the solution has density $1.0077 g m L^{- 1}$ . Find the mole fraction of the solvent.

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Solution

$M = \frac{35.4 \times 1000}{118 \times 1000} = 0.3$
$m = \frac{W_{2} \times 1000}{M w_{2} \times (V_{s o l} \times d_{s o l} - W_{2})}$
$= \frac{35.4 \times 1000}{118 (1000 \times 1.0077 - 35.4)}$
$= \frac{35.4 \times 1000}{118 \times 972.3} = 0.31$
$N = n \times M = 2 \times 0.3 = 0.6 N$
Weight of solvent $(W_{1})$ $=$ Weight of solution -Weight of solute
$= V_{s o l} \times d_{s o l} - W_{2}$ $= 1000 \times 1.0077 - 35.4 = 972.3$
The mole fraction of solute $χ_{2} = \frac{n}{n_{1} + n_{2}} = \frac{W_{2} / M w_{2}}{\frac{W_{1}}{M w_{1}} + \frac{W_{2}}{M w_{2}}}$
$= \frac{\frac{35.4}{118}}{\frac{972.3}{18} + \frac{35.4}{118}}$
$= \frac{0.3}{54 + 0.3} = 0.0055$
The mole fraction of solvent $χ_{1} = 1 - χ_{2} = 1 - 0.0055 = 0.9945$ .

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