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The boiling point of a solution of 0.1050 g of a substance in 15.84 g of ether was found to be $100^{o} C$ higher than that of pure ether. What is the molecular weight of the substance [Molecular elevation constant of ether per 100 g = 21.6]

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144.50

143.18

140.28

146.66

answer is B.

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Detailed Solution

$m = \frac{K_{b} \times w \times 1000}{{ΔT}_{b} \times W} m = \frac{21.6 \times 0.1050 \times 1000}{100 \times 15.84} m = 143 .18$

NOTE: There is a error in question

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