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The boiling point of a solution of $0.1050 g m$ of a substance in $15.84 g r a m$ of ether was found to be $0.100 ° C$ higher than that of pure ether. What is the molecular weight of the substance [Molecular elevation constant of ether per $100 g = 2.16$ ]

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$144.50$

$143.18$

$140.28$

$146.66$

answer is B.

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Detailed Solution

$m = \frac{K_{b} \times w \times 1000}{Δ T_{b} \times W} = \frac{2.16 \times 0.1050 \times 1000}{0.100 \times 15.84} = 143.18$

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