Q. A certain substance ‘$A$’ tetramerises in water to the extent of 80%. A solution of 2.5 g of $A$ in 100 g of water lowers the freezing point by $0.3℃$. The molar mass of $A$ is
A) 31
B) 62
C) 122
D) 244
Explanation
(b)
$\alpha = \frac{1 - i}{1 - \frac{1}{n}}$
$0.8 = \frac{1 - i}{1 - \frac{1}{4}}$
$i = 0.4$
$\because\ \mathrm{\Delta}T = i \times k_{f} \times m$
$0.3 = 0.4 \times 1.86 \times \frac{w_{B} \times 1000}{m_{B} \times w_{A}}$
$0.3 = 0.4 \times 1.86 \times \frac{2.5 \times 1000}{m_{B} \times 100}$
$m_{B} = \frac{0.4 \times 1.86 \times 2.5 \times 1000}{0.3 \times 100} = 62$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.