Question

What would be the equivalent conductivity of a cell in which $0.5$ N salt solution offers a resistance of $40$ ohm whose electrodes are $2$ cm apart and $5$ $c{m}^2$ in area?

• A. $10$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$
• B. $20$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$
• C. $30$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$
• D. $25$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$

Answer 1

Answer: (B)

$20$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$

Given,

$R=40ohm,l=2cm,A=2c{m}^2$

We know the relation,

$\kappa =\frac{1}{R}\times \frac{l}{A}=\frac{1}{40}\times \frac{2}{5}=\frac{1}{100}oh{m}^{-1}c{m}^{-1}$

Now we also know the relation,

${\Lambda }_{eq}=\frac{\kappa \times 1000}{N}=\frac{1}{100}\times \frac{1000}{0.5}=20oh{m}^{-1}c{m}^{2}e{q}^{-1}$

Hence, option B is correct.