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Depression in Freezing Point

When 3.24 g o...

Question

When 3.24 g of mercuric nitrate, $H g (N O_{3})_{2}$ is dissolved in 1000 g of water, the freezing point of the solution is found to be $- {0.0558}^{0} C$ . But when 10.84 g of $H g C l_{2}$ is dissolved in 1000 g of water, the freezing point depression is ${0.0744}^{0} C$ . $K_{f}$ for water is ${1.86}^{0} C m o l a l^{- 1} .$ Now, select the correct statement(s). (Given:` Atomic mass of Hg = 200)

Mercuric nitrate dissociates to give one

H g^{2 +}

and two nitrate ions.

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

does not dissociate.

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H g C l_{2}

dissociates to give one

H g^{2 +}

and two chloride ions.

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H g C l_{2}

does not dissociate.

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Solution

The correct options are
A Mercuric nitrate dissociates to give one $H g^{2 +}$ and two nitrate ions.
D $H g C l_{2}$ does not dissociate.
The expression for the molality of $H g (N O_{3})_{2} solution is m = \frac{Δ T_{f}}{K_{f}} < (\frac{0.0558}{1.86} = 0.03 m)$ .
But 1000 g contains $= \frac{3.24}{324} = 0.01$ moles of $H g (N O_{3})_{2}$ .
The observed molality is thrice of the calculated value.
Thus, each molecule of $H g (N O_{3})_{2}$ dissociates to give three ions.
$H g (N O_{3})_{2} \to H g^{2 +} + 2 N O_{3}^{2 -}$
The molality of $H g C l_{2}$ solution is $\frac{0.0744}{1.86} = 0.04 m$ .
The number of moles of $H g C l_{2} = \frac{10.84}{271} = 0.04 m o l$ .
The observed value is equal to the calculated value. Hence, the molecules of $H g C l_{2}$ do not undergo dissociation.

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Similar questions

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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

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H g C l_{2}

is dissolved in

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K_{f} = 1.86 m o l^{- 1} K k g

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Q. A certain mass of a substance, when dissolved in 100 g of

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dissolved in 1000 g of water constitutes a solution having a freezing point of

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