Q33 / 150 · MHT CET 2026 (19 April Shift 1) · Chemistry
What is the conductivity of $0.02\ \text{M}\ \text{AgNO}_3$ solution having cell constant $1.2\ \text{cm}^{-1}$ and resistance $95.0\ \text{ohms}$ ?
- A$1.140 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$
- Correct answerB$1.263 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$
- C$1.340 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$
- D$1.463 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$
Solution
Given:
Resistance, $R = 95.0\ \Omega$
Cell constant, $G^{*} = 1.2\ \text{cm}^{-1}$
The conductivity $\kappa$ is given by the relation:
$\kappa = \dfrac{1}{R} \times G^{*}$
Substituting the given values:
$\kappa = \dfrac{1}{95.0} \times 1.2$
$\kappa = 0.01263\ \Omega^{-1}\ \text{cm}^{-1}$
$\kappa = 1.263 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$
Answer: $1.263 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$
Resistance, $R = 95.0\ \Omega$
Cell constant, $G^{*} = 1.2\ \text{cm}^{-1}$
The conductivity $\kappa$ is given by the relation:
$\kappa = \dfrac{1}{R} \times G^{*}$
Substituting the given values:
$\kappa = \dfrac{1}{95.0} \times 1.2$
$\kappa = 0.01263\ \Omega^{-1}\ \text{cm}^{-1}$
$\kappa = 1.263 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$
Answer: $1.263 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$
Hence, the correct answer is option: 'B'.