Question

A solution of a substance containing $1.05$ g per $100$ mL was found to be isotonic with $3\%$ glucose solution. The molecular mass of the substance is:

• A. $31.5$
• B. $6.3$
• C. $630$
• D. $63$

Answer 1

The correct option is D $63$

Given that the unknown solution

is isotonic with glucose solution

That means there osmotic

pressure is same.

$\pi =CRT=\frac{W_2}{M_{2}V}\times RT$

${(\frac{W_2}{M_{2}V}\times RT)}_{unknown}={(\frac{W_2}{M_{2}V}\times RT)}_{glucose}$

${(\frac{1.05}{M_2\times 0.1}\times RT)}_{unknown}={(\frac{3}{180\times 0.1}\times RT)}_{glucose}$

$M_2=\frac{180\times 0.1\times 1.05}{0.1\times 3}=63$

molecular mass of the unknown solute $=63g$