SolutionHard

Question

The mass of KCl required to depress the freezing point of 500 g water by 2 K is (K_f = 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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