WBJEE · Chemistry · Solutions
The measured freezing point depression for a \(0.1 \mathrm{m}\) aqueous \(\mathrm{CH}_{3} \mathrm{COOH}\) solution is \(0.19^{\circ} \mathrm{C}\). The acid dissociation constant \(K_{a}\) at this concentration will be (Given, \(K_{f}\) the molal cryoscopic constant \(=1.86 \mathrm{K}\) \(\mathrm{kg} \mathrm{mol}^{-1}\) )
- A \(4.76 \times 10^{-5}\)
- B \(4 \times 10^{-5}\)
- C \(8 \times 10^{-5}\)
- D \(2 \times 10^{-5}\)
Answer & Solution
Correct Answer
(B) \(4 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \Delta T_{f} &=i \times K_{t} \times m \\ \therefore & i=\frac{\Delta T_{f}}{K_{t} \times m}=\frac{0.19}{1.86 \times 0.1}=1.02 \end{aligned}\) Again from, \(\alpha=\frac{i-1}{n-1}=\frac{1.02-1}{2-1}\) \(=0.02=2.0 \times 10^{-2}\)…
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