Question

Equivalent conductance of 0.01 N Na2SO4 solution is 112.4 ohm–1 cm2 eq–1. The equivalent conductance at infinite dilution is 129.9 ohm–1 cm2eq–1. What is the degree of dissociation in 0.01 N Na2SO4 solution?

Answer 1

Explanation:

$\alpha = \frac{molar \: conductance \: at \: given \: concentration}{limiting \: molar \: conductance}$

For NaCl,

Eq. Conductance = Molar Conductane (v.f. = 1)

Therefore,

$\alpha = \frac{112.4}{129.9}$

$\alpha = 0.865$

$\% \alpha = 86.5\%$

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