At 298 K,the equivalent conductance of 0.1 | vedclass

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(B) The degree of dissociation $\alpha$ is given by the formula: $\alpha = \frac{\lambda^{c}}{\lambda^{\infty}}$$(i)$ Calculate the limiting molar con

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$1.3$

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(B) The degree of dissociation $\alpha$ is given by the formula: $\alpha = \frac{\lambda^{c}}{\lambda^{\infty}}$$(i)$ Calculate the limiting molar conductivity of acetic acid:$\lambda^{\infty} (CH_{3}COOH) = \lambda^{\infty} (CH_{3}COO^{-}) + \lambda^{\infty} (H^{+})$$\lambda^{\infty} (CH_{3}COOH) = 40.9 + 349.8 = 390.7 \ S \ cm^{2} \ eq.^{-1}$$(ii)$ Calculate the degree of dissociation $\alpha$:$\alpha = \frac{5.20}{390.7} \approx 0.0133$$(iii)$ Convert to percentage:$\alpha \% = 0.0133 	imes 100 = 1.33 \% \approx 1.3 \%$

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