Question

A 5% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol\(^{-1}\)) in the same solvent. If densities of both solutions are \(1 \, \text{g cm}^{-3}\), the molar mass of the substance is _ _ _ _ _ .

Hint:

For isotonic solutions with same density, directly equate molarity after converting % to molarity.

Answer 1

Correct Answer: 200

Step 1: Understanding the Concept:
Isotonic solutions are solutions that have the same osmotic pressure ($\pi$) at the same temperature. For non-electrolytes in the same solvent, this implies they have the same molar concentration ($C$).
Step 2: Key Formula or Approach:
Condition for isotonicity:
\[ \pi_{1} = \pi_{2} \implies C_{1}RT = C_{2}RT \implies C_{1} = C_{2} \]
Molarity $C = \frac{\text{Mass of solute}}{\text{Molar mass} \times \text{Volume of solution (L)}}$.
Step 3: Detailed Explanation:
1. Solution 1 (Substance): 5% solution means 5g of substance in 100mL of solution.
Concentration $C_{1} = \frac{5}{M \times 0.1}$ (where $M$ is the unknown molar mass).
2. Solution 2 (Urea): 1.5% solution means 1.5g of urea in 100mL of solution.
Concentration $C_{2} = \frac{1.5}{60 \times 0.1}$.
3. Equating concentrations:
\[ \frac{5}{M \times 0.1} = \frac{1.5}{60 \times 0.1} \]
\[ \frac{5}{M} = \frac{1.5}{60} \]
\[ M = \frac{5 \times 60}{1.5} \]
\[ M = \frac{300}{1.5} = 200\text{ g/mol} \]
Step 4: Final Answer:
The molar mass of the substance is 200 g/mol.