Question

A weak monobasic acid is $5\%$ dissociated in $0.01$ mol $d{m}^{-3}$ solution. Limiting molar conductivity of acid at infinite dilution is $4\times {10}^{-2}$ $oh{m}^{-1}$ $m^2$ $mo{l}^{-1}$. What will be the conductivity of $0.05$ mol $d{m}^{-3}$ solution of the acid?

• A. $8.94\times {10}^{-6}$ $oh{m}^{-1}$ $c{m}^2$ $mo{l}^{-1}$
• B. $8.92\times {10}^{-4}$ $oh{m}^{-1}$ $c{m}^2$ $mo{l}^{-1}$
• C. $4.46\times {10}^{-6}$ $oh{m}^{-1}$ $c{m}^2$ $mo{l}^{-1}$
• D. $2.23\times {10}^{-5}$ $oh{m}^{-1}$ $c{m}^2$ $mo{l}^{-1}$

Answer 1

Answer: (B)

$8.92\times {10}^{-4}$ $oh{m}^{-1}$ $c{m}^2$ $mo{l}^{-1}$

Given,

$\alpha =5\%$

$C=0.01$ $mol$ $d{m}^{-3}$

We have to keep in mind that $K_{\alpha }$ remains constant for a reaction

We know that,

$K_{\alpha }=C{\alpha }^2=0.01\times (0.05{)}^2=2.5\times {10}^{-5}$

Now,

$K_{\alpha }=C{\alpha }^2$

$⟹2.5\times {10}^{-5}=0.05\times {\alpha }^2⟹\alpha =0.0223$

Also,

$\alpha =\frac{{\Lambda }_m^c}{{\Lambda }_m^{\infty }}$

$⟹{\Lambda }_m^c=0.0233\times 4\times {10}^{-2}=8.92\times {10}^{-4}oh{m}^{-1}c{m}^{2}mo{l}^{-1}$