JEE Advanced 2022 · Paper 2 · Chemistry Q3

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ChemistryInteger+3 / -1Question 39 of 54

Consider the strong electrolytes $\mathrm{Z_mX_n}$ , $\mathrm{U_mY_p}$ and $\mathrm{V_mX_n}$ . Limiting molar conductivity $(\Lambda^0)$ of $\mathrm{U_mY_p}$ and $\mathrm{V_mX_n}$ are 250 and 440 S cm $^2$ mol $^{-1}$ , respectively. The value of $(m + n + p)$ is ________.

Given:

Ion	$\mathrm{Z^{n+}}$	$\mathrm{U^{p+}}$	$\mathrm{V^{n+}}$	$\mathrm{X^{m-}}$	$\mathrm{Y^{m-}}$
$\lambda^0$ (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 $\mathrm{Z_mX_n}$ vs $c^{1/2}$ is given below.

$Straight line of molar conductivity Λ of Z_mX_n (S cm^2 mol^-1, vertical axis) against c^{1/2} ((mol L^-1)^{1/2}, horizontal axis), decreasing linearly. Dashed guide lines: at c^{1/2} = 0.01 the line is at Λ = 339; at c^{1/2} = 0.04 it is at Λ = 336; intermediate gridlines at 0.02 and 0.03 fall on equally spaced values between (338 and 337).$

Your answer

Answer and worked solution

Official answer: 7

Limiting conductivity of $\mathrm{Z_mX_n}$ . The plot falls by 3 units over a change of 0.03 in $\sqrt c$ , a slope of $-100$ . Extrapolating from 339 at $\sqrt c=0.01$ gives $\Lambda^0=340$ .

Kohlrausch's law.

$\mathrm{Z_mX_n}$ : $50m+80n=340$ .
$\mathrm{V_mX_n}$ : $100m+80n=440$ .
$\mathrm{U_mY_p}$ : $25m+100p=250$ .

Solve. Subtracting the first two gives $50m=100$ , so $m=2$ . Then $n=3$ and $p=2$ .

Answer. $m+n+p=7$ .

Topics:Conductance and Kohlrausch's law

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