Elevation of boiling point of 1 molar aqueous glucose solution (density = 1.2 g/ml) is

K_{b}

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1.20 K_{b}

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1.02 K_{b}

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0.98 K_{b}

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Solution

The correct option is B $0.98 K_{b}$
Here 1 molar solution contains 1 mole of glucose in $1000 c m^{3}$ of solution.
molar mass of glucose= $180.156 g (m o l)^{- 1}$
mass of glucose= $1 \times 180.56 = 180.56 g$
mass of solution $= d e n s i t y \times v o l u m e = 1.2 \times 1000 = 1200 g$
mass of water= $1200 g - 180.56 g = 1019.44 g = 1.01944 k g$
$Δ T_{b} = K_{b} \frac{n_{s o l u t e}}{mass of solvent(in kg)}$
$Δ T_{b} = K_{b} \frac{1 m o l}{1.01944 k g} = 0.98 K_{b}$

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