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Van't Hoff Factor

A 0.4 molal a...

Question

A 0.4 molal aqueous solution of $M_{x} A$ has freezing point ${3.72}^{o} C$ . The $K_{f}$ of $H_{2} O$ is $1.86 K m o l a l i t y^{- 1}$ . The value of x is :

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Solution

The expression for the depression in the freezing point is $Δ T_{f} = i \times K_{f} \times m$
Hence, $i = \frac{Δ T_{f}}{K_{f} \times m}$
Substitute values in the above expression.
$i = \frac{3.72 K}{1.86 K / m \times 0.4 m} = 5$
Thus, the Van't Hoff factor is 5.
This indicates 1 molecule of $M_{x} A$ dissociates to give 5 ions.
Hence, the molecule can be represented as $M_{4} A$ .
$M_{4} A ⇌ 4 M^{+} + X^{4 -}$ .
Thus, the value of X is 4.

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