Q23 / 160 · TS EAMCET 2024 (09 May Shift 2) · Chemistry
If the degree of dissociation of formic acid is $11.0 \%$, the molar conductivity of 0.02 M solution of it is (Given, $\lambda^{\circ}\left(\mathrm{H}^{+}\right)=349.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}, \lambda^{\circ}\left(\mathrm{HCOO}^{-}\right)=54.6$ $\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-1}$ )
- A$44.46 \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$
- Correct answerB$44.46 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
- C$22.23 \mathrm{~s} \mathrm{~m}^2 \mathrm{~mol}^{-1}$
- D$22.23 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
Solution
Given: $\lambda^0\left(\mathrm{H}^{+}\right)=349.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\begin{aligned}
& \lambda^{\circ}\left(\mathrm{HCOO}^{-}\right)=54.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\
& \alpha=11.0 \% \Rightarrow \frac{11}{100}=0.11 \\
& \lambda_{\mathrm{m}}^{\circ}(\mathrm{HCOOH})=\lambda^{\circ}\left(\mathrm{H}^{+}\right)+\lambda^{\circ}\left(\mathrm{HCOO}^{\ominus}\right) \\
& \lambda_{\mathrm{m}}^{\mathrm{m}}(\mathrm{HCOOH})=349.6+54.6 \\
& =404.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}$
For molar conductivity $\begin{aligned} & \lambda_{\mathrm{m}}=\alpha \times \lambda_{\mathrm{m}}^{\mathrm{o}} \\ & \lambda_{\mathrm{m}}=0.11 \times 404.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{m}}=44.46 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \end{aligned}$
For molar conductivity $\begin{aligned} & \lambda_{\mathrm{m}}=\alpha \times \lambda_{\mathrm{m}}^{\mathrm{o}} \\ & \lambda_{\mathrm{m}}=0.11 \times 404.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{m}}=44.46 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \end{aligned}$
Hence, the correct answer is option: 'B'.