Q80 / 180 · NEET 2026 (Re-NEET) · Chemistry
For a salt $\text{XY}$, which is a strong electrolyte, the plot of $\Lambda_m$ versus $\sqrt{c}$ has a slope of $-90.0$ S cm$^2$ mol$^{-3/2}$ L$^{1/2}$ at $298$ K. At $0.01$ M concentration of $\text{XY}$, the value of $\Lambda_m$ is $145.0$ S cm$^2$ mol$^{-1}$. The limiting molar conductivity of $\text{Y}^-$ ion ($\lambda^0_{Y^-}$, in S cm$^2$ mol$^{-1}$) at $298$ K will be
(Given : $\lambda^0_{X^+}=74.0$ S cm$^2$ mol$^{-1}$)
(Given : $\lambda^0_{X^+}=74.0$ S cm$^2$ mol$^{-1}$)
- A$76.0$
- Correct answerB$80.0$
- C$100.0$
- D$90.0$
Solution
According to the Debye-Huckel-Onsager equation for strong electrolytes:
$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$
The slope of the plot of $\Lambda_m$ versus $\sqrt{c}$ is $-A$.
Given slope $= -90.0$, we have $A = 90.0$.
At $c = 0.01$ M, $\sqrt{c} = 0.1$ M$^{1/2}$.
Substituting the given values into the equation:
$145.0 = \Lambda_m^0 - 90.0 \times 0.1$
$145.0 = \Lambda_m^0 - 9.0$
$\Lambda_m^0 = 154.0$ S cm$^2$ mol$^{-1}$
According to Kohlrausch's law of independent migration of ions:
$\Lambda_m^0(\text{XY}) = \lambda^0_{\text{X}^+} + \lambda^0_{\text{Y}^-}$
$154.0 = 74.0 + \lambda^0_{\text{Y}^-}$
$\lambda^0_{\text{Y}^-} = 154.0 - 74.0 = 80.0$ S cm$^2$ mol$^{-1}$
Answer: $80.0$
$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$
The slope of the plot of $\Lambda_m$ versus $\sqrt{c}$ is $-A$.
Given slope $= -90.0$, we have $A = 90.0$.
At $c = 0.01$ M, $\sqrt{c} = 0.1$ M$^{1/2}$.
Substituting the given values into the equation:
$145.0 = \Lambda_m^0 - 90.0 \times 0.1$
$145.0 = \Lambda_m^0 - 9.0$
$\Lambda_m^0 = 154.0$ S cm$^2$ mol$^{-1}$
According to Kohlrausch's law of independent migration of ions:
$\Lambda_m^0(\text{XY}) = \lambda^0_{\text{X}^+} + \lambda^0_{\text{Y}^-}$
$154.0 = 74.0 + \lambda^0_{\text{Y}^-}$
$\lambda^0_{\text{Y}^-} = 154.0 - 74.0 = 80.0$ S cm$^2$ mol$^{-1}$
Answer: $80.0$
Hence, the correct answer is option: 'B'.