Question

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

• A. $1.01$
• B. $10$
• C. $10.1$
• D. $100$

Answer 1

The correct option is D $100$

We know,

$\Delta T_b=\frac{K_b\times w\times 1000}{M\times W}$

$\Rightarrow M=\frac{K_b\times w\times 1000}{\Delta T_b\times W}=\frac{2.16\times 0.15\times 1000}{0.216\times 15}=100.$