JEE Advanced 2022 — Paper 2 — CHEM-I solution

JEE Advanced2022Paper 2CHEM-I

Consider the strong electrolytes Z $_{m}$ X $_{n}$ , U $_{m}$ Y $_{p}$ and V $_{m}$ X $_{n}$ . Limiting molar conductivity ( $Λ^{0}$ ) 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 $^{m -}$
$λ^{0}$ (S cm $^{2}$ mol $^{- 1}$ )	50.0	25.0	100.0	80.0	100.0

$λ^{0}$ is the limiting molar conductivity of ions

The plot of molar conductivity ( $Λ$ ) of Z $_{m}$ X $_{n}$ vs $c^{1/2}$ is given below.

Your answer

Solution

Apply the Debye-Hückel-Onsager equation $Λ_{m} = Λ_{m}^{o} - A c$ to the Z $_{m}$ X $_{n}$ plot.

From the graph, slope $- A = \frac{336 - 339}{0.04 - 0.01} = - 100$ , so $A = 100$ .

Using the point $(0.04, 336)$ : $336 = Λ_{m}^{o} - 100 \times 0.04 \Rightarrow Λ_{m}^{o} (Z_{m} X_{n}) = 340$ S cm $^{2}$ mol $^{- 1}$ .

By Kohlrausch's law:

Z $_{m}$ X $_{n} \to m$ Z $^{n +} + n$ X $^{m -}$ : $50 m + 80 n = 340 \Rightarrow 5 m + 8 n = 34$ … (i)

U $_{m}$ Y $_{p} \to m$ U $^{p +} + p$ Y $^{m -}$ : $25 m + 100 p = 250 \Rightarrow m + 4 p = 10$ … (ii)

V $_{m}$ X $_{n} \to m$ V $^{n +} + n$ X $^{m -}$ : $100 m + 80 n = 440 \Rightarrow 5 m + 4 n = 22$ … (iii)

Subtracting (iii) from (i): $4 n = 12 \Rightarrow n = 3$ . Then $m = (22 - 12) /5 = 2$ . From (ii): $p = (10 - 2) /4 = 2$ .

$∴ m + n + p = 2 + 3 + 2 = 7$ .

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