Question

The conductivity of an aqueous solution of a weak monoprotic acid is 0.000033 ohm${}^{-1}$ cm${}^{-1}$ at a concentration, 0.2M. If at this concentration the degree of dissociation is 0.015, calculate the value of ${\lambda }_0$ (ohm${}^{-1}$ cm${}^2$ /eqt).

• A. $11$
• B. $10$
• C. $12$
• D. $20$

Answer 1

The correct option is A $11$

We know,

$\Lambda =\frac{1000\kappa }{C}$

$\Lambda =\frac{1000\times 0.000033}{0.2}=0.165$

Again,

$\alpha =\frac{\Lambda }{{\Lambda }_0}$

$0.015=\frac{0.165}{{\Lambda }_0}$

${\Lambda }_0=\frac{0.165}{0.015}=11oh{m}^{-1}c{m}^{2}eqt$

Option A is correct.