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: Plan
Rearrange the formula for molality and plug in numbers.

Step 2: Effective particles
The product $i K_f = 1.25 \times 1.86 = 2.325$ K kg/mol is the depression per molal of formula units.

Step 3: Divide
$m = 0.93/2.325 = 0.40$ mol/kg.

Step 4: Check
Back-calculate: $1.25 \times 1.86 \times 0.4 = 0.93$ K, which matches the given freezing point of $-0.93^\circ\text{C}$. Options 0.2, 0.3 and 0.5 give 0.465, 0.70 and 1.16 K.

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