WBJEE · Chemistry · Electrochemistry
At \(25^{\circ} \mathrm{C}\), the molar conductance of \(0.007 \mathrm{M}\) hydrofluoric acid is 150 mho \(\mathrm{cm}^{2} \mathrm{mol}^{-1}\) and its \(\Lambda_{m}^{\circ}=500\)mho\(cm^2$$mol^{-1}\) The value of the dissociation constant of the acid at the given concentration at \(25^{\circ} \mathrm{C}\) is
- A \(7 \times 10^{-4} \mathrm{M}\)
- B \(7 \times 10^{-5} \mathrm{M}\)
- C \(9 \times 10^{-3} \mathrm{M}\)
- D \(9 \times 10^{-4} \mathrm{M}\)
Answer & Solution
Correct Answer
(D) \(9 \times 10^{-4} \mathrm{M}\)
Step-by-step Solution
Detailed explanation
Degree of dissociation, \(\alpha=\frac{\lambda^{\circ} c}{\lambda^{\circ}m}=\frac{150}{500}=0.3\) Given, \(C=0.007 \mathrm{M}\) Hydrofluoric acid dissociates in the following manner Dissociation constant,…
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