The concentration of glucose solution in molality which boils at 100.01∘CKb for water is 0.5∘C kg mol− is

A
0.01
B
0.02
C
0.03
D
0.04

Solution:

It is known that:

${ΔT}_{b} = K_{b} \times m . . (1)$

m=molality of solution

${ΔT}_{b} = 100 .01 - 100 = 0 {.01}^{\circ} C$

Calculating m:

$\begin{array}{l} K_{b} = 0 {.5}^{\circ} C kg {mol}^{- 1} \\ 0 {.01}^{\circ} C = 0 {.5}^{\circ} C kg {mol}^{-} \times m (substitute values in (1)) \\ m = \frac{0 .01}{0 .5} = 0 .02 mol {kg}^{-} \end{array}$

The concentration of glucose solution in molality which boils at $100 {.01}^{\circ} C (K_{b} for water is 0 {.5}^{\circ} C kg {mol}^{-})$ is

Solution:

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