Question

Choose the correct statement.

• A. A solution formed by adding carbon disulphide to acetone forms a maximum boiling azeotrope
• B. Hypotonic solution is more concentrated with respect to the other solution separated by a semi-permeable membrane
• C. For a solvent, \( K_b = \frac{R \times M_1 \times T_b^2}{1000 \times \Delta H_{\text{vap}}} \), where \(R\) is gas constant, \(M_1\) is molar mass of solvent and \(T_b\) is boiling point of the solvent
• D. A 1.0 molal solution of glucose in water is more concentrated than 1.0 M glucose solution in the same solvent

Hint:

For colligative properties, remember the standard constants: \(K_b\) depends on boiling point, molar mass of solvent and enthalpy of vaporisation.

Answer 1

The Correct Option is C

Step 1: Analyse option (A).
Carbon disulphide and acetone show positive deviation from Raoult's law.
Positive deviation generally leads to minimum boiling azeotrope, not maximum boiling azeotrope.
Hence, option (A) is incorrect.

Step 2: Analyse option (B).
A hypotonic solution has lower solute concentration compared to the other solution separated by a semi-permeable membrane.
So, it is less concentrated, not more concentrated.
Hence, option (B) is incorrect.

Step 3: Analyse option (C).
The ebullioscopic constant \(K_b\) is given by:
\[ K_b = \frac{R M_1 T_b^2}{1000 \Delta H_{\text{vap}}} \]
where \(R\) is gas constant, \(M_1\) is molar mass of solvent, \(T_b\) is boiling point of solvent, and \(\Delta H_{\text{vap}}\) is enthalpy of vaporisation.
Thus, option (C) is correct.

Step 4: Analyse option (D).
Molality is moles of solute per kg of solvent, while molarity is moles of solute per litre of solution.
For aqueous glucose solution, 1.0 molal solution is not necessarily more concentrated than 1.0 M solution.
Therefore, option (D) is incorrect.

Step 5: Compare all statements.
Only option (C) gives the correct expression for ebullioscopic constant.

Step 6: Final conclusion.
\[ \boxed{K_b = \frac{R M_1 T_b^2}{1000 \Delta H_{\text{vap}}}} \]