Question

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

• A. $31.5\text{u}$
• B. $6.3\text{u}$
• C. $3.15\text{u}$
• D. $63\text{u}$

Answer 1

The correct option is D $63\text{u}$

Isotonic solution means osmotic pressure of the solutions is same. Let the osmotic pressure be ${\pi }_1$ and ${\pi }_2$ respectively.
So, ${\pi }_1={\pi }_2$
$\Rightarrow C_{1}RT=C_{2}RT$
$\Rightarrow C_1=C_2$
Where, $C_1$ and $C_2$ are the concentrations

For very dilute solution molality = molarity.

molarity of glucose solution = $\frac{3\times 1000}{100\times 180}=\frac{30}{180}$
Let the molecular mass of the compound be $M$
According to the question,
$\frac{1.05\times 1000}{M\times 100}=\frac{30}{180}$
$\Rightarrow M=\frac{10.5\times 180}{30}$
$=63\text{u}$