MHT CET · Chemistry · Solutions
Calculate the percent dissociation of 0.02 m solution if its freezing point
depression is 0.046 K
\(\left[\mathrm{K}_{\mathrm{f}}\right.\) for water \(\left.=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ; \mathrm{n}=2\right]\)
- A \(12.3 \%\)
- B \(23.6 \%\)
- C \(35.00 \%\)
- D \(48.1 \%\)
Answer & Solution
Correct Answer
(B) \(23.6 \%\)
Step-by-step Solution
Detailed explanation
\(i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.046}{1.86 \cdot 0.02} = 1.2366\) \(\text{Percent dissociation} = \frac{i-1}{n-1} \cdot 100 = \frac{1.2366-1}{2-1} \cdot 100 = 23.66\% \approx 23.6\%\)
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