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 is (molal elevation constant for the solvent is $2.16℃$)
A) 100
B) 102
C) 104
D) 1.02
Explanation
(a)
$m = \frac{k_{b} \times w \times 1000}{\mathrm{\Delta}T_{b} \times W} = \frac{2.16 \times 0.15 \times 1000}{0.216 \times 15} = 100$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.