Consider the strong electrolytes $Z_m{X}_n,U_m{Y}_p\text{and}{V}_m{X}_n.$ Limiting molar conductivity $\left({\Lambda }^{\circ }\right)\text{of}{U}_m{Y}_p\text{and}{V}_m{X}_n$
are $250$ and $440Sc{m}^{2}mo{l}^{-1}$, respectively. The value of $(m+n+p)$ is ______.

Ion	$Z^{n+}$	$U^{p+}$	$V^{n+}$	$X^{m-}$	$Y^{m-}$
${\lambda }^{\circ }\left(sc{m}^{2}mo{l}^{-1}\right)$	$50.0$	$25.0$	$100.0$	$80.0$	$100.0$

${\lambda }^{\circ }$ s the limiting molar conductivity of ions
The plot of molar conductivity
$\left(\Lambda \right)\text{of}{Z}_m{X}_{n}vs{c}^{1/2}$ is given below.

$\begin{array}{c}For{U}_m{Y}_p;{\lambda }_M^{\infty }=250Sc{m}^{2}mo{l}^{-1},\\ For{V}_m{X}_n;{\lambda }_M^{\infty }=440Sc{m}^{2}mo{l}^{-1},\end{array}$

${\lambda }_M^{\infty }\left(U_m{Y}_p\right)=m\times 25+p\left(100\right)=250$
$25m+100p=250$ ....(i)

${\lambda }_M^{\infty }\left(V_m{X}_n\right)=m\times 100+n\times 80=440$ ....(ii)

We know that;
${\lambda }_M={\lambda }_M^{\infty }$ - $b\sqrt{c}$
From graph:
$339={\lambda }_M^{\infty }$ - 0.01 b ..... (iii)
$336={\lambda }_M^{\infty }$ - 0.04 b ...... (iv)

From eqns (iii) and (iv) we get, ${\lambda }_M^{\infty }$ = 340 S$c{m}^{2}mo{l}^{-1}$
This could also be inferred by observing the graph directly.

${\lambda }_M^{\infty }\left(Z_m{X}_n\right)=m\times 50+n\times 80=340$ ...(v)

From equation (ii) & (v) $m=2$ & n = 3
and from equation (i) p = 2.

So
(m+n+p) = 7