c
Given conc \(^{\text {n }}\) of \(KCl =\frac{ m . mol }{ L }\)

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

\(\Rightarrow \lambda_{ m }=14.3 \, \frac{ mSm ^{2}}{ mol }\)