Question

The ionization constant of a weak acid is $1.6\times {10}^{-5}$ and the molar conductivity at infinite dilution is $380\times {10}^{-4}S{m}^{2}mo{l}^{-1}$. If the cell constant is $0.01m^{-1}$ then conductance of $0.01M$ acid solution is:

• A. $1.52\times {10}^{-5}S$
• B. $1.52\times {10}^{-7}S$
• C. $1.52\times {10}^{-3}S$
• D. $1.52\times {10}^{-4}S$

Answer 1

The correct option is D $1.52\times {10}^{-4}S$

As we know,

$K=C{\alpha }^2;\Rightarrow \alpha =4\times {10}^{-2}$

Note: $K=1.6\times {10}^{-5};C=0.01M$

Also,

$\alpha ={\Lambda }_{eq}/{\Lambda }_{\infty };{\Lambda }_{eq}=4\times {10}^{-2}\times 380\times {10}^{-4}=1.52\times {10}^{-3}$

And ${\Lambda }_{eq}=k\times \left(1000/0.01\right)$

$k=1.52\times {10}^{-8}$ and

$k=G\times cellconstant;G=1.52\times {10}^{-4}$

[Note: cell constant $=0.01m^{-1}={10}^{-4}c{m}^{-1}$]

Hence, the correct option is $\text{B}$