Question

A 5% solution of canesugar (mol. wt. =342) is isotonic with 1% solution of a substance $X$. The molecular weight of $X$ is.

• A. 34.2
• B. 171.2
• C. 68.4
• D. 136.8

Answer 1

Answer: (C)

68.4

Answer 2

Molar concentration of cane sugar

$= \frac{5}{342} \times \frac{1000}{100} = \frac{50}{342}$

Molar concentration of X = $\frac{1}{m} \times \frac{1000}{100} = \frac{10}{m}$

$\frac{10}{m} = \frac{50}{342}$ or $m = 68.4$.