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

Answer 1

The correct option is B

$1.52S$
$K_{\alpha }=\frac{c{\alpha }^2}{1-\alpha }\Rightarrow 1.6\times {10}^{-5}=\frac{0.01\times {\alpha }^2}{1-\alpha }\alpha =\sqrt{\frac{1.6\times {10}^{-5}}{0.01}}\Rightarrow 0.04\alpha =\frac{{\Lambda }_{m.}}{{\Lambda }_m^{\infty }};{\Lambda }_m=0.04\times 380\times {10}^{-4}=15.2\times {10}^{-4}S{m}^{2}mo{l}^{-1}k={\Lambda }_m\times C=15.2\times {10}^{-4}\times {10}^3\times {10}^{-2}(\because 1m^3=1000litre)k=G.G^{\ast }and{G}^{\ast }=0.01m^{-1},$