QJEEAdv22Chab11: JEE Advanced 2022 Chemistry - Class 12 JEEAdv22Chab11 Solution: Consider the strong electrolytes and . Limiting molar…

Given,

$\Lambda_{\mathrm{V}_{\mathrm{m}} \mathrm{Y}_{\mathrm{p}}}^{\circ}=250 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

$\Lambda_{\mathrm{V}_m \mathrm{x}_n}=440 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

It is also given that

$\begin{aligned}& \lambda^{\circ} \mathrm{Z}^{n+}=50 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\& \lambda_{\mathrm{V} p+}^0=25.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\& \lambda_{\mathrm{V}^{n+}}^{\circ}=100.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\& \lambda_{\mathrm{X}^{m-}}^{\circ}=80.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\& \lambda_{\mathrm{Y}^{m-}}^{\circ}=100.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\\end{aligned}$
Now, $\Lambda_{\mathrm{V}_m \mathrm{Y}_p}=m \lambda_{\mathrm{V}^{+}}^{\circ}+p \lambda^{\circ}{ }_{\mathrm{Y}}$

$250=25 m+100 p$

$10=m+4p$ ..........(1)

Also, $\Lambda_{\mathrm{V}_m \mathrm{x}_n}=m \lambda_{\mathrm{v}^{+}}^{\circ}+n \lambda^{\circ} \mathrm{X}^{-}$

$440=100 m+80 n$

$22=5 m+4 n$ ........(2)

In the question, a graph of $(\Lambda)$ of $Z_m X_n$ Vs $C^{1 / 2}$ is given,

We know, $\lambda_m=\lambda_m^0-A \sqrt{C}$

For electrolyte $Z_m X_n$ and from given curve

$\begin{aligned}& \lambda_m\left(Z_m X_n\right)=\lambda_m^0\left(Z_m X_n\right)-A \sqrt{C} \\\\& -A=\frac{336-339}{0.04-0.01}=-\frac{3}{0.03} \\\\& \Rightarrow A=100 \\\\& \therefore \text { For } \lambda_m=336 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\& \Rightarrow 336=\lambda_m^{\circ}\left(Z_m X_n\right)-100 \times 0.04 \\\\& \lambda_m^{\circ}=336+4=340 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1}\end{aligned}$

So,

$\begin{aligned}\Lambda_{Z_m x_n} & =m \lambda_{\mathrm{Z}^{+}}+n \lambda_{\mathrm{X}^{-}} \\\\340 & =50 m+80 n \\\\34 & =5 m+8 n ........(3)\end{aligned}$

Solving eqn (2) and eqn (3),

$n=\frac{12}{4}=3$

$\Rightarrow$ $n=3$

Substituting the value of $n$ in eqn (2), we get,

$\begin{aligned}22 & =5 m+4(3) \\\\\Rightarrow 22 & =5 m+12 \\\\\Rightarrow 5 m & =22-12=10 \\\\\Rightarrow m & =\frac{10}{5}=2 \\\\\Rightarrow m & =2\end{aligned}$

Now, substituting the value of $m$ is eqn $(1)$ , we get

$\begin{aligned}10 & =m+4 p \\\\\Rightarrow 10 & =2+4 p \\\\\Rightarrow 8 & =4 p \\\\\Rightarrow p & =\frac{8}{4}=2 \\\\\Rightarrow p & =22\end{aligned}$

So, $m+n+p=2+3+2$

$\Rightarrow m+n+p=7$

Hence, the required value of $m+n+p$ is 7 .