Q. If 0.15 g of a solute, dissolved in 15 g of solvent, is boiled at a temperature higher by $0.216℃$ than that of the pure solvent. The molecular weight of the substance (molal elevation constant for the solvent is $2.16℃)$ is
A) 100
B) 10.1
C) 10
D) 1.001
Explanation
(a)
w =0.15 g, W=15 g,
$\mathrm{\Delta}T_{b}$ =0.216$℃$
$k_{b}$ = 2.16$℃$
$\therefore$ M=$\frac{k_{b} \times w \times 100}{\mathrm{\Delta}T_{b} \times W}\$
$= \frac{2.16 \times 0.15 \times 1000}{0.216 \times 15} = 100$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.