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A solution is prepared from 0.05 mol of an ionic salt and 45 g of water. The vapour pressure of water above this solution is found to be 276 mmHg at 348.0 K. The vapour pressure of pure water at this temperature is 290 bar. If the salt is 73% dissociated, the number of ions present per formula unit of the salt is_____.

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answer is 3.

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Detailed Solution

$\frac{p^{0} - p}{p} = i x_{s o l u t e} \Rightarrow \frac{290 - 276}{290} = i \frac{0.05}{0.05 + 2.5} \Rightarrow i = 2.46 \begin{array}{l} A \overset{}{\to} n i o n s \\ 1 - 0.73 0.73 n \\ \Rightarrow 1 - 0.73 + 0.73 n = 2.46 \Rightarrow n = 3 \end{array}$

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