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6g of non volatile solute is dissolved in 90g of water, such that the lowering of vapour pressure is 2% . The molecular weight of solute is

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detailed solution

Correct option is C

Wt of non volatile solute $W_{1}$ = 6g

Wt of solvent (Water)= $W_{2}$ =90g

GMW of solute $= M_{1} = ?$

GMW of solvent $(H_{2} O) = M_{2} = 18 g$

$R . L . V . P = \frac{∆ p}{p^{0}} = \frac{2}{100}$

$According to Raoult ’ s law$

$\frac{∆ p}{p^{0}} = \frac{W_{1}}{M_{1}} \times \frac{M_{2}}{W_{2}}$

$\frac{2}{100} = \frac{6}{M_{1}} \times \frac{18}{90}$

$M_{1} = \frac{6 \times 18}{2 \times 90} \times 100; M_{1} = 60$

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