Question

0.25 moles of $ \text{CH}_2\text{FCOOH} $ was dissolved in $ 0.5 \, \text{kg} $ of water. The depression in freezing point of the resultant solution was observed as $ 1^\circ \text{C} $. What is the van't Hoff factor? ($ K_f = 1.86 \, \text{K kg mol}^{-1} $)

• A. $ 0.93 $
• B. $ 1.07 $
• C. $ 1.25 $
• D. $ 1.50 $

Hint:

To calculate the van't Hoff factor ($ i $): 1. Use the formula $ \Delta T_f = i \cdot K_f \cdot m $. 2. Rearrange to solve for $ i $: $ i = \frac{\Delta T_f}{K_f \cdot m} $.

Answer 1

The Correct Option is B

\textbf{Step 1: Recall the Formula for Depression in Freezing Point} The depression in freezing point ($ \Delta T_f $) is given by: $$ \Delta T_f = i \cdot K_f \cdot m $$ where:
$ i $ is the van't Hoff factor,
$ K_f $ is the freezing point depression constant of the solvent ($ K_f = 1.86 \, \text{K kg mol}^{-1} $ for water),
$ m $ is the molality of the solution.
Step 2: Determine Molality ($ m $)
Molality ($ m $) is defined as: $$ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} $$ Given:
Moles of solute ($ \text{CH}_2\text{FCOOH} $): $ 0.25 \, \text{mol} $ Mass of solvent (water): $ 0.5 \, \text{kg} $ Substitute the values: $$ m = \frac{0.25}{0.5} = 0.5 \, \text{mol kg}^{-1} $$ Step 3: Use the Depression in Freezing Point Formula
Rearrange the formula to solve for $ i $: $$ i = \frac{\Delta T_f}{K_f \cdot m} $$ Given:
$ \Delta T_f = 1^\circ \text{C} $,
$ K_f = 1.86 \, \text{K kg mol}^{-1} $,
$ m = 0.5 \, \text{mol kg}^{-1} $. Substitute the values: $$ i = \frac{1}{1.86 \cdot 0.5} $$ $$ i = \frac{1}{0.93} \approx 1.07 $$ Step 4: Analyze the Options
Option (1): $ 0.93 $
Incorrect — does not match the calculated value. Option (2): $ 1.07 $
Correct — matches the calculated value. Option (3): $ 1.25 $
Incorrect — does not match the calculated value. Option (4): $ 1.50 $
Incorrect — does not match the calculated value.