An aqueous solution is prepared by dissolving 0 . 1 mole of an ionic salt in 1 . 8kg of water at 35 ° C . The salt remains 90 % dissociated in the solution. The vapour pressure of the solution is 59 . 724mm of Hg .

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Number of ions present per formula unit of ionic salt $= x$
Van 't Hoff factor
\left(i\right) = 1 + \left(n - 1\right)\alpha = 1 + \left(x - 1\right) \times 0 . 9 = 0 . 1 + 0 . 9x $(Assuming$ 90 % $dissociation)$ ∴ $Relative lowering in vapour pressure$ = i \times $Mole fraction of solute$ ⇒ \dfrac{60 - 59 . 724}{60} = \dfrac{i \times 0 . 1}{\dfrac{1800}{18} + 0 . 1} $$ ⇒ 0 . 0046 = \dfrac{i \times 0 . 1}{100 + 0 . 1} $$ 0 . 0046 \approx \dfrac{\left(0 . 9x + 0 . 1\right) \times 0 . 1}{100} $$ ⇒ 0 . 9x + 0 . 1 = 4 . 6 $$ ⇒ x = \dfrac{4 . 5}{0 . 9} = 5 $$ x = 5

An aqueous solution is prepared by dissolving 0 . 1 mole of an...

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