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Van't Hoff Factor

When 3.24 g...

Question

When $3.24 g$ of mercuric nitrate $H g (N O_{3})_{2}$ dissolved in $1 k g$ of water, the freezing point of the solution is found to be $- {0.0558}^{\circ} C$ . When $10.84 g$ of mercuric chloride $H g C l_{2}$ is dissolved in $1 k g$ of water, the freezing point of the solution is $- {0.0744}^{\circ} C$ . $K_{f} = 1.86 m o l^{- 1} K k g$ . Will either of these dissociate into ions in an aqueous solution?

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Solution

We have,
in case of $H g (N O_{3})_{2}$
$Δ T_{f} = i K_{f} m$
$0.0558 = i * 1.86 * \frac{3.24}{324}$
$i = 3$
in case of $H g C l_{2}$
$Δ T_{f} = i K_{f} m$
$0.0744 = i * 1.86 * \frac{10.84}{271}$
$i = 1$
so, $H g (N O_{3})_{2}$ will dissociate into ions in an aqueous solution.

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H g (N O_{3})_{2}

- {0.0558}^{0} C

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K_{f}

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H g C l_{2} (l)

{1.24}^{o} C . K_{f}

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[H g - 200 g, C l - 35.5 g]

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[K_{f} = 186 K - k g m o 1^{- 1}]

C_{6} H_{6}

{1.28}^{o} C

{1.40}^{o} C

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C_{6} H_{6}

5.12 K {m o l}^{- 1} k g

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