JEE Advanced · Chemistry · 16. Solutions
When \(20 \mathrm{~g}\) of naphthoic acid \(\left(\mathrm{C}_{11} \mathrm{H}_8 \mathrm{O}_2\right)\) is dissolved in \(50 \mathrm{~g}\) of benzene \(\left(K_f=1.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\), a freezing point depression of \(2 \mathrm{~K}\) is observed. The van't Hoff factor \((i)\) is
- A \(0.5\)
- B 1
- C 2
- D 3
Answer & Solution
Correct Answer
(A) \(0.5\)
Step-by-step Solution
Detailed explanation
Actual molecular weight of napthoic acid \(\left(\mathrm{C}_{11} \mathrm{H}_8 \mathrm{O}_2\right)=172\)
\(\text {Molecular mass (calculated ) } =\frac{1000 . K_f \cdot w}{W . \Delta T_f} \)
\( =\frac{1000 \times 1.72 \times 20}{50 \times 2} \)
\( =344\)
\(\text {Molecular mass (calculated ) } =\frac{1000 . K_f \cdot w}{W . \Delta T_f} \)
\( =\frac{1000 \times 1.72 \times 20}{50 \times 2} \)
\( =344\)
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