Question

Calculate van't Hoff factor of aqueous solution of 0.18 m electrolyte that freezes at \(-0.54^{\circ} \mathrm{C}\).

\(\left(\mathrm{K}_{\mathrm{f}}\right.\) for solvent \(\left.=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\)

• A. 1.126
• B. 2.449
• C. 1.612
• D. 2.150

Hint:

Use the depression in freezing point formula with the van't Hoff factor and solve for i.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For an electrolyte, \(\Delta T_f=i\,K_f\,m\), where \(i\) is the van't Hoff factor.

Step 2: Data
Solution freezes at \(-0.54^{\circ}\)C, so \(\Delta T_f=0.54\) K.
\(K_f=1.86\) K kg mol\(^{-1}\) and \(m=0.18\) mol/kg.

Step 3: Calculation
\(i=\dfrac{\Delta T_f}{K_f\,m}=\dfrac{0.54}{1.86\times0.18}=\dfrac{0.54}{0.3348}\approx1.612\)

Final Answer:
The van't Hoff factor is about 1.612. \[ \boxed{\text{(C) }1.612} \]