Q25 / 150 · MHT CET 2026 (15 April Shift 2) · Chemistry
Calculate the concentration of silver nitrate solution if molar conductivity and conductivity of silver nitrate at $25^\circ$C are respectively 120 ohm$^{-1}$ cm$^2$ mol$^{-1}$ and 0.0024 ohm$^{-1}$ cm$^{-1}$.
- A$0.01$ M
- Correct answerB$0.02$ M
- C$0.03$ M
- D$0.04$ M
Solution
The relationship between molar conductivity $\Lambda_m$, conductivity $\kappa$, and concentration $C$ in molarity is given by:
$\Lambda_m = \dfrac{\kappa \times 1000}{C}$
Rearranging the formula to solve for concentration $C$:
$C = \dfrac{\kappa \times 1000}{\Lambda_m}$
Substituting the given values $\kappa = 0.0024 \text{ ohm}^{-1} \text{ cm}^{-1}$ and $\Lambda_m = 120 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1}$:
$C = \dfrac{0.0024 \times 1000}{120}$
$C = \dfrac{2.4}{120}$
$C = 0.02 \text{ M}$
Answer: $0.02$ M
$\Lambda_m = \dfrac{\kappa \times 1000}{C}$
Rearranging the formula to solve for concentration $C$:
$C = \dfrac{\kappa \times 1000}{\Lambda_m}$
Substituting the given values $\kappa = 0.0024 \text{ ohm}^{-1} \text{ cm}^{-1}$ and $\Lambda_m = 120 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1}$:
$C = \dfrac{0.0024 \times 1000}{120}$
$C = \dfrac{2.4}{120}$
$C = 0.02 \text{ M}$
Answer: $0.02$ M
Hence, the correct answer is option: 'B'.