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  4. Moles of K2SO4 to be dissolved in 12 moles of water to lower its vapour pressure by 10 mm Hgat a temperature at which vapour pressure of pure water is 50 mmHg is

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Q. Moles of $K_2SO_4$ to be dissolved in 12 moles of water to lower its vapour pressure by 10 mm Hgat a temperature at which vapour pressure of pure water is 50 mmHg is

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Solution:

Lowering ofV.P.is colligative property thus, $i_{K_2} SO _4=1+(y-1) x=1+2 x=3$
$\therefore \text { If } \frac{\varnothing_p}{P^*}=\frac{n_1 i}{n_1 i+n_2} $
$\frac{10}{50}=\frac{3 n_1}{3 n_1+12}=\frac{n_1}{n_1+4} n_1=1$