Question

Equivalent conductivity ${\wedge }_{eq}$

• A. $\frac{k\times 1000}{m}oh{m}^{-1}c{m}^{2}E{q}^{-1},m-molality$
• B. $\frac{k\times 1000}{M}oh{m}^{-1}c{m}^{2}E{q}^{-1},M-molarity$
• C. $\frac{k\times 1000}{N}oh{m}^{-1}c{m}^{2}E{q}^{-1},N-normality$
• D. None of the above

Answer 1

The correct option is C $\frac{k\times 1000}{N}oh{m}^{-1}c{m}^{2}E{q}^{-1},N-normality$

Equivalent conductivity $\left({\wedge }_{eq}\right)=k\times V$
As equivalent conductivity is conductivity of 1 gm equivalent.
$\therefore {\wedge }_{eq}=k\times V$
$=k\times \frac{1000}{N}$ where N is normality.