Question

A $5.0\,m \,mol\, dm ^{-3}$ aqueous solution of $KCl$ has a conductance of $0.55\,mS$ when measured in a cell constant $1.3\,cm ^{-1}$. The molar conductivity of this solution is________ $mSm ^{2} mol ^{-1}$ (Round off to the Nearest Integer)

Answer 1

Given conc $^{ n }$ of $KCl =\frac{ m \cdot mol }{ L }$

Conductance $( G )=0.55\,mS$

Cell constant $\left(\frac{\ell}{ A }\right)=1.3\,cm ^{-1}$

To Calculate : Molar conductivity

$\left(\lambda_{ m }\right)$ of sol. $\rightarrow$ Since $\lambda_{ m }=\frac{1}{1000} \times \frac{ k }{ m }$

$\rightarrow$ Molarity $=5 \times 10^{-3} \frac{ mol }{ L }$

$\rightarrow$ Conductivity $= G \times\left(\frac{\ell}{ A }\right)=0.55 mS \times \frac{1.3}{\frac{1}{100}} m ^{-1}$

$=55 \times 1.3 mSm ^{-1}$

$eq ^{ n }(1) \lambda_{ m }=\frac{1}{1000} \times \frac{55 \times 1.3}{\left(\frac{5}{1000}\right)} \frac{\,m\,Sm ^{2}}{ mol }$

$\Rightarrow \lambda_{ m }=14.3 \frac{ mSm ^{2}}{ mol }$