Question

Calculate the molality of aqueous solution of electrolyte that freezes at \(-0.93\,^{\circ}\text{C}\) if \(K_f\) for water and van't Hoff factor respectively are \(1.86\,\text{K kg mol}^{-1}\) and \(1.25\). (Freezing point of water \(= 0\,^{\circ}\text{C}\) )

• A. \(0.2\) m
• B. \(0.5\) m
• C. \(0.3\) m
• D. \(0.4\) m

Hint:

Use depression in freezing point = i x Kf x m.

Answer 1

The Correct Option is D

Step 1: Understand the concept
For an electrolyte, the freezing point depression is \(\Delta T_f = i\,K_f\,m\), where \(i\) is the van't Hoff factor and \(m\) is the molality.

Step 2: Find the depression
Water freezes at \(0^\circ\text{C}\) and the solution freezes at \(-0.93^\circ\text{C}\), so \(\Delta T_f = 0.93\) K.

Step 3: Substitute
\[ m = \frac{\Delta T_f}{i\,K_f} = \frac{0.93}{1.25 \times 1.86} = \frac{0.93}{2.325} = 0.4\ \text{m} \]

Step 4: Check the options
If we had ignored \(i\), we would get \(0.93/1.86 = 0.5\) m, which is option (B), a trap. The correct use of \(i = 1.25\) gives 0.4 m.

Final Answer:
The molality is 0.4 m. This is option (D). \[ \boxed{\text{(D) }0.4\ \text{m}} \]