A 5.0 m mol

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Question:

A $5.0 \mathrm{~m} \mathrm{~mol} \mathrm{dm}^{-3}$ 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:

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

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

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

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

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

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