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Q.

On dissolving 10.8 g of glucose (m.w. = 180) in 240 g of water, its boiling point increases by 0.13 $°$ . Then, the molal elevation constant of water is

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a

0.42 $°$

b

0.52 $°$

c

0.62 $°$

d

0.72 $°$

answer is B.

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Detailed Solution

The formula for elevation in boiling point of solution is

$∆ T_{b} = K_{b} \times \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}}$

$∴ K_{b} = \frac{∆ T_{b} \times M \times W_{A}}{W_{B} \times 1000}$

$= \frac{0.13 \times 18 \times 240}{10.8 \times 1000} = 0.52$

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On dissolving 10.8 g of glucose (m.w. = 180) in 240 g of water, its boiling point increases by 0.13° . Then, the molal elevation constant of water is