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Q.

Moles of $K_{2} {SO}_{4}$ to be dissolved in $12$ mol water to lower its vapour pressure by $10 m m H g$ at a temperature at which vapour pressure of pure water is $50 m m$ is :

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a

$2 m o l$

b

$1 m o l$

c

$0.5 m o l$

d

$3 m o l$

answer is C.

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Detailed Solution

$\frac{Δ p}{p} = \frac{{in}_{B}}{{in}_{B} + n_{A}}$

Here $i = 3$

$\frac{10}{50} = \frac{3 n_{B}}{3 n_{B} + 12}$

$3 n_{B} + 12 = 15 n_{B}$

$12 n_{B} = 12$

$n_{B} = 1$

Hence, option C is correct.

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Moles of K2SO4 to be dissolved in 12 mol water to lower its vapour pressure by 10 mm Hg at a temperature at which vapour pressure of pure water is 50 mm is :