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$K_{f}$ of 1,4-dioxane is 4.9 ${mol}^{- 1}$ for 1000 g. The depression in freezing point for a 0.001 m solution in dioxane is

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0.0049

4.9 + 0.001

4.9

0.49

answer is A.

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Detailed Solution

$ΔT = K_{f} \times Molality = 4.9 \times 0 .001=0.0049 K$

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