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What weight of glycerol should be added to 600 g of water in order to lower its freezing point by $10^{\circ} C (K_{f} = {1.86}^{\circ} {Cm}^{- 1})$ (molecular weight of glycerol is 92 g)

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310 g

426 g

297 g

496 g

answer is B.

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Detailed Solution

$Δ T_{f} = K_{f} \times m = \frac{K_{f} \times W_{2} \times 1000}{M_{2} \times W_{1}}$

$W_{2} = \frac{10 \times 92 \times 600}{1.86 \times 1000} = 296.77 g$

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