B.P. of a solution is found to be $100.34$ on dissolving $12 g m$ glucose in $100 g m$ water. Calculate molal elevation constant for water?

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Solution

Boiling point of a solution is found to be $100.34^{o} C$ .Boiling point of pure
water is $100^{o} C$ .
The elevation in the boiling point $Δ T_{b} = 100.34 - 100 = 0.34^{o} C$ .
$12$ gm glucose (molecular weight $180$ g/mol) is dissolved in $100$ gm water.
The number of moles of glucose $= \frac{12 g}{180 g / m o l} = 0.0667 m o l$
Mass of water $= 100 g \times \frac{1 k g}{1000 g} = 0.100 k g$
Molality of solution $m = \frac{0.0667 m o l}{0.100 k g} = 0.667 m o l / k g$
The elevation in the boiling point $Δ T_{b} = K_{b} \times m$
$0.34^{o} C = K_{b} \times 0.667 m o l / k g$
$K_{b} = 0.51^{o} C k g / m o l$
The molal elevation constant for water is $0.51^{o} C k g / m o l$ .

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