Question

Calculate van't Hoff factor for aqueous solution of 0.02 m formic acid if it freezes at -0.045 \(^{\circ}\text{C}\).
[ \(K_f\) = 1.86 K kg \(\text{mol}^{-1}\) and freezing point of water = \(0 ^{\circ}\text{C}\) ]

• A. \(1.21\)
• B. \(1.46\)
• C. \(1.68\)
• D. \(1.05\)

Hint:

Use \(\Delta T_f = i K_f m\) with \(\Delta T_f = 0.045\) K.

Answer 1

The Correct Option is A

Step 1: Find the ideal depression
If formic acid did not ionise ($i=1$), $\Delta T_f = 1.86\times0.02 = 0.0372$ K.

Step 2: Compare with observed
Observed depression is 0.045 K, so $i = \dfrac{0.045}{0.0372} = 1.21$. This is option (A).

Final Answer:
The factor is $i=1.21$, option (A). \[ \boxed{1.21} \]