Question

The \(5%\) solution of cane sugar \((\text{molecular weight}=342)\) is isotonic with \(1%\) solution of substance \(X\). The molecular weight of \(X\) is:

• A. \(342\)
• B. \(171.2\)
• C. \(68.4\)
• D. \(136.8\)

Hint:

For isotonic non-electrolyte solutions: \[ \frac{w_1}{M_1}=\frac{w_2}{M_2} \] Lower molecular weight means larger number of solute particles for same mass percentage.

Answer 1

The Correct Option is C

Concept: Two solutions are said to be isotonic when they have the same osmotic pressure at the same temperature. For dilute solutions: \[ \pi = CRT \] For isotonic solutions: \[ C_1=C_2 \] Since concentration in percentage is given, we use: \[ \frac{w_1}{M_1}=\frac{w_2}{M_2} \] where:
• \(w\) = mass of solute in same volume of solution
• \(M\) = molecular weight

Step 1: Writing the given data.
For cane sugar: \[ w_1=5\,g \] \[ M_1=342 \] For substance \(X\): \[ w_2=1\,g \] \[ M_2=? \]

Step 2: Applying isotonic condition.
\[ \frac{w_1}{M_1}=\frac{w_2}{M_2} \] Substituting values: \[ \frac{5}{342}=\frac{1}{M_2} \] Cross multiplying: \[ M_2=\frac{342}{5} \] \[ M_2=68.4 \] Therefore, the molecular weight of substance \(X\) is: \[ 68.4 \]