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Question
If 1 mole of a non-volatile and non-electrolyte solute in 1000 g of water depresses the freezing point by 1.86°C, then what will be the freezing point of a solution of 1 mole of the solute in 500 g of water?
Options
A.−0.93°C
B.−1.86°C
C.3.72°C
D.−3.72 °C
Solution
$\Delta T_{f} = K_{f}.m$
$1.86 = K_{f}.\frac{1}{1} $$$\Delta T_{f} = K_{f}.\frac{1}{0.5}$$
$\therefore\text{ Required }\Delta T_{f} = 3.72^{o}C \Rightarrow \text{ Freezing point = } - 3.72^{o}C$
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