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Standard XII

Chemistry

Elevation in Boiling Point

A solution co...

Question

A solution containing 0.2563 g of naphthalene (molecular mass = 128) in 50 g of carbon tetrachloride yields a boiling point elevation of ${0.201}^{o} C$ while a solution of 0.6216 g of an unknown solute in the same mass of the solvent gives a boiling point elevation of ${0.647}^{o} C .$ Find the molecular mass of the unknown solute.

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Solution

We know that,
$K_{b} = \frac{△ T_{b} \times W \times m}{1000 \times w}$
For $C C l_{4}$ , $K_{b} = \frac{0.201 \times 50 \times 128}{1000 \times 0.2563} = 5.019$
$K_{b}$ is now used in the second part of the problem.
$m = \frac{1000 \times K_{b} \times w}{△ T_{b} \times W}$
$= \frac{1000 \times 5.019 \times 0.6216}{0.647 \times 50} = 96.44$

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We know that,Kb=△Tb×W×m1000×wFor CCl4, Kb=0.201×50×1281000×0.2563=5.019Kb is now used in the second part of the problem.m=1000×Kb×w△Tb×W =1000×5.019×0.62160.647×50=96.44