Q27 / 180 · COMEDK 2024 (Morning Shift) · Chemistry
$0.1 \mathrm{M}$ solution of $\mathrm{AgNO}_3$ is taken in a Conductivity cell and a potential difference of $40 \mathrm{~V}$ is applied across the ends of a column of this solution whose diameter is $4.0 \mathrm{~cm}$ and length of the column is $12 \mathrm{~cm}$. The current used is $0.4 \mathrm{~A}$. The Molar conductivity of the solution is _________.
- A$0.009546 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$
- B$954.7 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$
- Correct answerC$95.5 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$
- D$9.547 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$
Solution
The resistance $R$ of the solution is given by Ohm's law: $R = \dfrac{V}{I} = \dfrac{40 \text{ V}}{0.4 \text{ A}} = 100 \text{ } \Omega$.
The area of cross-section $A$ of the column is $\pi r^2$. Given diameter $d = 4.0 \text{ cm}$, the radius $r = 2.0 \text{ cm}$. Thus, $A = \pi \times (2.0 \text{ cm})^2 = 4\pi \text{ cm}^2 \approx 12.566 \text{ cm}^2$.
The cell constant $G^{*}$ is given by $\dfrac{l}{A}$, where $l = 12 \text{ cm}$. So, $G^{*} = \dfrac{12}{4\pi} = \dfrac{3}{\pi} \text{ cm}^{-1} \approx 0.9549 \text{ cm}^{-1}$.
The conductivity $\kappa$ is given by $\kappa = \dfrac{G^{*}}{R} = \dfrac{3/\pi}{100} = \dfrac{3}{100\pi} \text{ S cm}^{-1} \approx 0.009549 \text{ S cm}^{-1}$.
The molar conductivity $\Lambda_m$ is calculated using the formula $\Lambda_m = \dfrac{1000 \times \kappa}{C}$, where $C = 0.1 \text{ M}$.
Substituting the values: $\Lambda_m = \dfrac{1000 \times (3 / (100\pi))}{0.1} = \dfrac{30}{\pi \times 0.1} = \dfrac{300}{\pi} \text{ S cm}^2 \text{ mol}^{-1}$.
Calculating the value: $\Lambda_m \approx \dfrac{300}{3.14159} \approx 95.493 \text{ S cm}^2 \text{ mol}^{-1}$.
Rounding to the nearest option, we get $95.5 \text{ S cm}^2 \text{ mol}^{-1}$.
Answer: $95.5 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$
The area of cross-section $A$ of the column is $\pi r^2$. Given diameter $d = 4.0 \text{ cm}$, the radius $r = 2.0 \text{ cm}$. Thus, $A = \pi \times (2.0 \text{ cm})^2 = 4\pi \text{ cm}^2 \approx 12.566 \text{ cm}^2$.
The cell constant $G^{*}$ is given by $\dfrac{l}{A}$, where $l = 12 \text{ cm}$. So, $G^{*} = \dfrac{12}{4\pi} = \dfrac{3}{\pi} \text{ cm}^{-1} \approx 0.9549 \text{ cm}^{-1}$.
The conductivity $\kappa$ is given by $\kappa = \dfrac{G^{*}}{R} = \dfrac{3/\pi}{100} = \dfrac{3}{100\pi} \text{ S cm}^{-1} \approx 0.009549 \text{ S cm}^{-1}$.
The molar conductivity $\Lambda_m$ is calculated using the formula $\Lambda_m = \dfrac{1000 \times \kappa}{C}$, where $C = 0.1 \text{ M}$.
Substituting the values: $\Lambda_m = \dfrac{1000 \times (3 / (100\pi))}{0.1} = \dfrac{30}{\pi \times 0.1} = \dfrac{300}{\pi} \text{ S cm}^2 \text{ mol}^{-1}$.
Calculating the value: $\Lambda_m \approx \dfrac{300}{3.14159} \approx 95.493 \text{ S cm}^2 \text{ mol}^{-1}$.
Rounding to the nearest option, we get $95.5 \text{ S cm}^2 \text{ mol}^{-1}$.
Answer: $95.5 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$
Hence, the correct answer is option: 'C'.