Calculate relative lowering of vapour pressure in an aqueous solution of ${CaCl}_{2}$ having 1 mole of ${CaCl}_{2}$ dissolved in 324 g of water if degree of dissociation is 60%.

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$\frac{2.2}{19}$

$\frac{1}{19}$

$\frac{3}{19}$

$\frac{11}{101}$

answer is C.

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Detailed Solution

$\begin{array}{l} {CaCl}_{2} ⟶ {Ca}^{2 +} + 2 {Cl}^{-} \\ 1 - α α 2 α \\ i = 1 + 2 α = 1 + 2 \times 0.6 = 2.2 \\ R L V P = \frac{2.2 \times 1}{2.2 \times 1 + 18} = \frac{2.2}{20.2} = \frac{11}{101} \end{array}$

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