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If $0.15 g$ of a solute dissolved in $15 g$ of solvent is boiled at a temperature higher by $0.216 ° C$ than that of the pure solvent. The molecular weight of the substance ( $molal$ elevation constant for the solvent is ${2.16}^{\circ} Ckg / mol)$ is

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$1.01$

$10$

$10.1$

$100$

answer is D.

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Detailed Solution

$m = \frac{K_{b} \times w \times 1000}{Δ T_{b} \times W} = \frac{2.16 \times 0.15 \times 1000}{0.216 \times 15} = 100$

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