A solution containing $30 g m$ of non-volatile solute exactly in $90 g m$ of water has a vapour pressure of $2.8 k P a$ at $298 K$ . Further, $18 g m$ of water is then added to the solution and the new vapour pressure becomes $2.9 k P a$ at $298 K$ .
Calculate: (i) molar mass of the solute (ii) vapour pressure of water at $298 K$ .

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Solution

Weight of solute, $W_{B} = 30$ g,
Weight of water, $W_{A} = 90$ g

Vapour pressure of solution $P_{A} = 2.8$ kPa

According to Raoult's law, $\frac{P_{A}^{o} - P_{A}}{P_{A}^{o}} = X_{B} ≃ \frac{W_{B} M_{A}}{M_{B} W_{A}}$

$\frac{P_{A}^{o} - 2.8}{P_{A}^{o}} = \frac{30 \times 18}{M_{B} \times 90}$

$\frac{2.8}{P_{A}^{o}} = \frac{M_{B} - 6}{M_{B}}$ $. . . . . . (1)$
After adding water-
Weight of solute $W_{B} = 30$ g,Weight of water $W_{A} = 90 + 18 = 108$ g

Vapour pressure of solution $P_{A} = 2.9$ kPa

According to Raoult's law, $\frac{P_{A}^{o} - P_{A}}{P_{A}^{o}} = X_{B} ≃ \frac{W_{B} M_{A}}{M_{B} W_{A}}$

$\frac{P_{A}^{o} - 2.9}{P_{A}^{o}} = \frac{30 \times 18}{M_{B} \times 108}$

$\frac{2.9}{P_{A}^{o}} = \frac{M_{B} - 5}{M_{B}}$ $. . . . . . (2)$
Divide equation $(1)$ by equation $(2)$ , we get

$\frac{2.8}{2.9} = \frac{M_{B} - 6}{M_{B} - 5}$

$M_{B} = 34$ g/mol

Substituting the values of $M_{B}$ in equation $(1)$ , we get
$\frac{2.8}{P_{A}^{o}} = \frac{34 - 6}{34}$
$P_{A}^{0} = 3.4$ kPa

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