The equivalent conductivity of monobasic acid at infinite dilution is  348 ohm^{-1}cm^2eq^{-1}. If the resitivity of the solution containing 15g acid in 1 litre is 18.5 ohm cm , What is the degree of dissociation of acid and percentage of ionization?

Option 1)  45.9 %

Option 2)  40.2 %

Option 3)  60.4 %

Option 4)  50.7 %

Answers (1)

As we learnt in 

Application of Kohlrausch's law -

Determination of Degree of dissociation of weak electrolytes.

- wherein

Degree of dissociation

\alpha = \frac{\Lambda m}{\Lambda_{m}^{0} }

 

Conductivity= \frac{1}{resistivity}= \frac{1}{18.5}\, \, ohm^{-1}cm^{-1}

Degree of dissociation,\alpha =\frac{\Lambda ^{c}m}{\Lambda ^{\infty }m}

 

\Lambda ^{c}m=\frac{k \times 1000}{C_{m}}  k=conductivity ,  C_{m}=Molar concentration

C_{m}=\frac{15}{49\times 1}= 0.306m

\therefore \Lambda _{m}^{c}=\frac{1000}{18.5\times 0.306} =176.65\ ohm^{-1}cm^{2}mol^{-1}

\therefore degree of dissociation,\therefore \alpha =\frac{176.65}{348}=0.507

\therefore \% \text { of ionization }=50.7 \%

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