A solution containing 30 g of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mmHg at $25^{o} C$ . Further 18 g of water is then added to the solution. The resulting solution has a vapour pressure of 22.15 mmHg at $25^{o} C$ . Calculate the molecular weight of the solute

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74.2

78.7

75.6

67.83

answer is C.

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Detailed Solution

We have,

$\frac{p^{o} - 21 .85}{21 .85} = \frac{30 \times 18}{90 \times m}, for I case . .... (i)$

wt. of solvent $= 90 + 18 = 108 gm$

$\frac{p^{o} - 22 .15}{22 .15} = \frac{30 \times 18}{108 \times m}, for II case . .... (ii)$

By eq. (1) $p^{o} m - 21 .85 m = 21 .85 \times 6 = 131 .1$

By eq. (2) $p^{o} m - 22 .15 m = 22 .15 \times 5 = 110 .75$

On doing (1)-(2)

$0.30 m = 20.35$

$m = \frac{20 .35}{0 .30} = 67 .83$

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