4g of Substance A dissolved in 100g $H_{2} O$ depressed the freezing point of water by $0 . 1^{o} c$ . While 4g of another substance B depressed the freezing point by $0 . 2^{o} c$ . What is the relation between molecular weights of A and B.

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$M_{A} = 4 M_{B}$

$M_{A} = M_{B}$

$M_{A} = 0.5 M_{B}$

$M_{A} = 2 M_{B}$

answer is D.

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Detailed Solution

$∆ T_{f} = K_{f} m 0.1 = K_{f} \times \frac{4}{M_{A}} \times \frac{1000}{100} \frac{0.2 \times M_{B}}{40} = \frac{0.1 \times M_{A}}{40} \Rightarrow M_{A} = 2 M_{B}$

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