Question

A $5.0{\text{mmol dm}}^{-3}$ aqueous solution of $KCl$ has a conductance of $0.55mS$ when measured in a cell of cell constant $1.3{\text{cm}}^{-1}$. The molar conductivity of this solution is $\left(mS\right)m^2{\text{mol}}^{-1}$. (Round off to the nearest integer).

Answer 1

$G_{KCl}=0.55\text{mS}=55\times {10}^{–5}S$
Cell constant $=\frac{l}{A}=1.3{\text{cm}}^{–1}$

$K=G\left(\frac{l}{A}\right)=55\times {10}^{-5}\times 1.3Sc{m}^{–1}$
${\lambda }_M=\frac{K\times 1000}{\text{Molarity}}=\frac{55\times 1.3\times {10}^{-5}\times 1000}{5\times {10}^{-3}}$
${\lambda }_M=11\times 1.3\times 10=11\times 13=143{\text{Scm}}^2{\text{mol}}^{-1}$
${\lambda }_M=\frac{143S}{({10}^{-2}m{)}^{-2}}{\text{mol}}^{-1}$
${\lambda }_M=143\times 1000\times {10}^{-4}\left(mS\right)m^2.mo{l}^{-1}$
${\lambda }_M=14.3\left(mS\right)m^2.{\text{mol}}^{-1}$
Nearest integer $=14$