# Solve the following

Question:

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

Solution:

(14.3)

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

$\therefore$ Conductance $(\mathrm{G})=0.55 \mathrm{mS}$

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

To Calculate : Molar conductivity $\left(\lambda_{\mathrm{m}}\right)$ of sol.

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

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

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

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

$\Rightarrow \lambda_{\mathrm{m}}=14.3 \frac{\mathrm{m} \mathrm{Sm}^{2}}{\mathrm{~mol}}$