Elevation in b.p. of a molar glucose solution 4 $(d = 1.2 g m L^{- 1})$ is:

0.98 K_{b}

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K_{b}

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1.20 K_{b}

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1.02 K_{b}

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Solution

The correct option is A $0.98 K_{b}$
Let us consider the problem.
$D e n s i t y = 1.2 g / m l$
Assume 1 litre solution
$t o t a l m a s s o f s o l u t i o n = 16 \times d e n s i t y \times 1000 m l / L = 1200 g m o f s o l u t i o n$
Therefore the,
$1 M o f glucose = 1 \times 180 = 180 g m o f glucose$
$W e i g h t o f t h e s o l v e n t = 1200 - 180 = 1020 g m$
$M o l a l i t y = \frac{1 m o l o f s o l u t e}{1.02 k g o f s o l v e n t} = \frac{1}{1.02} = 0.98$

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