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Elevation in Boiling Point

18 g of gluco...

Question

$18 g$ of glucose, $C_{6} H_{12} O_{6}$ is dissolved in $1 k g$ of water in a saucepan. At what temperature will the water boil
under atmospheric pressure? $K_{b}$ for water is $0.52 K k g m o l^{- 1}$

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Solution

Boiling point elevation of a solution is given by,
$△ T_{b} = K_{b} \times m$
where,
$△ T_{b}$ is the elevation in boiling point.
$K_{b}$ is molal elevation constant.
m is molality of solution.
$Molality = \frac{△ T_{b}}{K_{b}} = \frac{△ T_{b}}{0.52}$
Given,
Mass of the glucose = $18 g$
No. of moles of glucose $= \frac{18}{180}$
$Molality (m) = \frac{n_{solute} \times 1000}{W_{solvent} in gram}$
$∴$
$Molality = \frac{18}{180} \times \frac{1000}{1000} m o l / k g$
$∴$
$△ T_{b} = 0.52 \times \frac{18 \times 1000}{180 \times 1000}$
$Boiling point of pure water, T = 373 K$
$△ T_{b} = 0.052 T - 373 K = 0.052$
Thus, the temperature at which the given sugar solution boil is at $373.052 K$

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