JEE Advanced 2022 Paper 2 · Q03 · Electrolysis & Faraday's Laws

Question

Consider the strong electrolytes Z$_m$X$_n$, U$_m$Y$_p$ and V$_m$X$_n$. Limiting molar conductivity ($\Lambda^0_m$) of U$_m$Y$_p$ and V$_m$X$_n$ are 250 and 440 S cm$^2$ mol$^{-1}$, respectively. The value of $(m + n + p)$ is _____.

Given:

| Ion | Z$^{n+}$ | U$^{p+}$ | V$^{n+}$ | X$^{m-}$ | Y$^{m-}$ |
|---|---|---|---|---|---|
| $\lambda^0$ (S cm$^2$ mol$^{-1}$) | 50.0 | 25.0 | 100.0 | 80.0 | 100.0 |

$\lambda^0$ is the limiting molar conductivity of ions.

The plot of molar conductivity ($\Lambda$) of Z$_m$X$_n$ vs $c^{1/2}$ is given below (Debye-Hückel-Onsager plot): the intercept (at $c^{1/2} = 0$) is at $\Lambda = 339$ and the curve passes through ($c^{1/2} = 0.04$, $\Lambda = 336$).

SolveKar AI · Accepted Answer

Step 1 — limiting molar conductivity of Z$_m$X$_n$ from the Debye-Hückel-Onsager plot. $\Lambda = \Lambda^0_m - b c^{1/2}$. From the graph, at $c^{1/2} 	o 0$, $\Lambda 	o 339$ (intercept actually = $\Lambda^0_m$). Using two points (e.g. $\Lambda = 339$ at $c^{1/2} = 0.01$ and $\Lambda = 336$ at $c^{1/2} = 0.04$), $b = (339 - 336)/(0.04 - 0.01) = 100$, so $\Lambda^0_m (Z_m X_n) = 339 + 100 	imes 0.01 = 340$ S cm$^2$ mol$^{-1}$.

Step 2 — write the three Kohlrausch sums. For an electrolyte M$_a$N$_b$, $\Lambda^0_m = a\lambda^0_{M^+} + b\lambda^0_{N^-}$.
- Z$_m$X$_n$: $50m + 80n = 340$.
- U$_m$Y$_p$: $25m + 100p = 250$.
- V$_m$X$_n$: $100m + 80n = 440$.

Step 3 — solve. Subtract Z$_m$X$_n$ from V$_m$X$_n$: $(100 - 50)m = 440 - 340 \Rightarrow 50m = 100 \Rightarrow m = 2$. Then from Z$_m$X$_n$: $80n = 340 - 100 = 240 \Rightarrow n = 3$. From U$_m$Y$_p$: $100p = 250 - 50 = 200 \Rightarrow p = 2$.

$m + n + p = 2 + 3 + 2 = 7$.

JEE Advanced 2022 Paper 2 · Q03 · Electrolysis & Faraday's Laws

Solution