SolutionMedium
Question
The mass of KCl required to depress the freezing point of 500 g water by 2 K is (Kf = 1.86, K = 39)
Options
A.10.01 g
B.40.05 g
C.7.45 g
D.20.03 g
Solution
$\Delta T_{f} = K_{f}.m \Rightarrow 2 = 1.86 \times \frac{(w/74.5) \times 2}{500/1000} \Rightarrow w = 20.03\text{ gm}$
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