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0.1 M KI and 0.2 M ${AgNO}_{3}$ are mixed in 3:1 volume ratio. The depression of freezing point of the resulting solution will be $[K_{f} (H_{2} O) = 1.86 Kg {mol}^{- 1}]$

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3.72 K

1.86 K

0.93 K

0.279 K

answer is D.

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Detailed Solution

$M_{R} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}}$

$= \frac{0.1 \times 3 + 0.2 \times 1}{4}$

= 0.125

$∆ T_{f} = K_{f} \times m$

$= 1.86 \times 0.125$

=0.2325

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