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Depression in Freezing Point

0.1 M KI and ...

Question

$0.1 M K I$ and $0.2 M$ $A g N O_{3}$ are mixed in $3 : 1$ volume ratio. The depression of freezing point of the resulting solution will be [ $k_{f} (H_{2} O) = 1.86 K k g m o l^{- 1}$ ]:

3.72 K

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1.86 K

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0.93 K

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0.279 K

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Solution

The correct option is C $0.279 K$
(d) $K I (a q) + A g N O_{3} (a q) + K N O_{3} (a q)$
$K I + A g N O_{3} \to K N O_{3} + A g I$
Initial : 0.3 mole 0.2 mole - -
Final : 0.1 mole - 0.2 0.2
Total moles of solute ions $= (0.1 + 0.2) \times 2 (∵ i = 2)$
[Solute] = $\frac{0.6}{4} M$
$Δ T_{f} = K_{f} \frac{0.6}{4}$
$= 1.86 \times \frac{0.6}{4} = 0.279 K$

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