Question

The resistance of $1N$ solution of acetic acid is $250ohm.$ when measured in a cell of cell constant $1.15$ $c{m}^{-1}$. The equivalent conductance of $1N$ acetic acid (in $oh{m}^{-1}$ $c{m}^2$ $eq{v}^{-1}$) is:

• A. $4.6$
• B. $9.2$
• C. $18.4$
• D. $0.023$

Answer 1

Answer: (A)

$4.6$

The relationship between the specific conductance, resistance and the cell constant is $\kappa =\frac{1}{R}\times \frac{l}{a}$. Substitute $R=250ohm$ and $\frac{l}{a}=1.15{cm}^{-1}$.

Hence $\kappa =\frac{1}{250ohm}\times 1.15{cm}^{-1}$

$=\frac{1.15}{250}{ohm}^{-1}{cm}^{-1}$.

The relationship between the equivalent conductance and specifc conductance is ${\Lambda }_{eq}=\frac{\kappa \times 1000}{N}$.

Substitute $M=1N$ and $\kappa =\frac{1.15}{250}{ohm}^{-1}{cm}^{-1}$.

Hence, ${\Lambda }_{eq}=\frac{\frac{1.15}{250}\times 1000}{1}$

$=4.6{ohm}^{-1}{cm}^2{equiv}^{-1}$.

Hence, option $A$ is correct.