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: Understanding the Concept
Formic acid partly ionises in water, so the number of particles is a little more than the number of moles dissolved. The van't Hoff factor \(i\) measures this ratio.

Step 2: Key Formula or Approach
\[ \Delta T_f = i\,K_f\,m \quad\Rightarrow\quad i = \frac{\Delta T_f}{K_f\,m} \]

Step 3: Detailed Explanation
Freezing point of solution is \(-0.045^{\circ}\)C, so \(\Delta T_f = 0 - (-0.045) = 0.045\) K.
\[ i = \frac{0.045}{1.86\times0.02} = \frac{0.045}{0.0372} = 1.21 \]
A value above 1 but well below 2 fits a weak acid that ionises only slightly.

Final Answer:
The van't Hoff factor is 1.21, option (A). \[ \boxed{1.21\ \text{(A)}} \]