Question

The resistance of $N/10$ solution is found to be $2.5\times {10}^{3}ohm$. The equivalent conductance of the solution is (cell constant $=1.25c{m}^{-1})$.

• A. $2.5oh{m}^{-1}c{m}^{2}equi{v}^{-1}$
• B. $5oh{m}^{-1}c{m}^{2}equi{v}^{-1}$
• C. $2.5oh{m}^{-1}c{m}^{-2}equi{v}^{-1}$
• D. $5oh{m}^{-1}c{m}^{-2}equi{v}^{-1}$

Answer 1

Answer: (B)

$5oh{m}^{-1}c{m}^{2}equi{v}^{-1}$

Given Data : 1) Resistance = $R$ = $2.5\times {10}^3$

2) Cell constant = $k$ = $\frac{l}{a}$= $1.15$$c{m}^2$

3) Normality =$N$= $0.1N$

To Find : Equivalent conductance = $\Lambda$${}_e$${}_q$

The relation between $K$ ,$R$ and $k$ is,

$K$ = $\frac{1}{R}\times \frac{l}{a}$

where $K$ is specific conductance.

$\therefore$ Substituting the given values we get,

$K$ = $\frac{1}{2.5\times {10}^3}$ × $1.15c{m}^{-}$${}^1$

= $\frac{1.15}{2.5\times {10}^3}$ $ohm$${}^{-}$${}^1$$cm$${}^{-}$${}^1$

The relation between $\Lambda$${}_e$${}_q$ and $K$ is,

$\Lambda$${}_e$${}_q$ = $\frac{K\times 1000}{N}$

Substituting the value of $M$ and $K$ we get,

${\Lambda }_{eq}=\frac{1.15}{2.5\times {10}^3\times 0.1}\times 1000$

= $4.6$$oh{m}^{-}$${}^1$$cm$${}^2$$equi$${}^{-1}$ [Note:$\text{Normality= no. of equiv.}/c{m}^3$]

Here appproximation is taken,

$\approx$ $5$ $ohm$${}^{-}$${}^1$$cm$${}^2$$equi$${}^{-}$${}^1$

Hence the correct option is 'B'.