The elevation in the boiling point of one molar aqueous solution of glucose is:
(density = 1.2 g/mL)

1 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 D $0.98 K_{b}$
We know,
$Δ T_{b} = i K_{b} m$
for glucose $i = 1$
1 molar aqueous glucose solution = 1 mole of glucose present in 1000 mL of solution.
Mass of 1000 mL solution = 1200 g (from given density)
we know, molality $= \frac{m o l o f s o l u t e}{k g o f s o l v e n t}$
So, weight of solvent = $(1200 - 180) g$ (since 1 mole of glucose = 180 g of glucose)
$Δ T_{b} = 1 \times K_{b} \times \frac{1}{\frac{1200 - 180}{1000}}$
$= 1 \times K_{b} \times \frac{1000}{1200 - 180}$
$= 0.98 K_{b}$

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