Question

A 1% (w/v) aqueous solution of a certain solute is isotonic with a 3% (w/v) solution of glucose (molar mass 180 g mol$^{-1}$). The molar mass of solute (in g mol$^{-1}$) is

• A. 120
• B. 60
• C. 360
• D. 80

Hint:

For isotonic solutions, use the relation $\frac{w_1}{M_1} = \frac{w_2}{M_2}$ when temperature and volume are same. Always compare the percentage weights and molar masses directly.

Answer 1

The Correct Option is B

Step 1: Use the concept of isotonic solutions.
Two solutions are isotonic when they have the same osmotic pressure. For non-electrolytes: \[ \pi = \frac{w \cdot R \cdot T}{M \cdot V} \] Where $\pi$ is osmotic pressure, $w$ is mass, $M$ is molar mass, and $V$ is volume. Since $R$ and $T$ are same and $\pi$ is equal for both: \[ \frac{w_1}{M_1} = \frac{w_2}{M_2} \] Step 2: Set up the relation from given data.
Given: - 1% (w/v) = 1 g per 100 mL for solute (unknown molar mass $M$) - 3% (w/v) = 3 g per 100 mL for glucose (molar mass = 180 g/mol) \[ \frac{1}{M} = \frac{3}{180} \Rightarrow M = \frac{1 \times 180}{3} = 60 \text{ g/mol} \] Step 3: Final Answer
Molar mass of solute = 60 g mol$^{-1$}