A dilute solution of glucose shows depression of freezing point by 1.86 K. k_f of water is 1.86km^-1. How many moles of glucose are present in 1L of solution and what is the mole fraction of glucose.

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1, 1.86

0.018, 1

1, 0.018

0.018, 1.86

answer is A.

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Detailed Solution

$\begin{array}{l} m = \frac{△ T_{f}}{k_{f}} = \frac{1 .86}{1 .86} = 1 i . e ., 1 mole is 1000 g solution \\ moles of glucose = 1 moles of water = \frac{1000}{18} = 55 .5 \\ x_{glucose} = \frac{1}{1 + 55 .5} = 0 .018 \end{array}$

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