Question

Consider the strong electrolytes $Z_m X_n, U_m Y_p$ and $V_m X_n$. Limiting molar conductivity $\left(\Lambda^0\right)$ 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 (A) of $Z_{ m } X _{ n } v s c ^{1 / 2}$ is given below.

Answer 1

$ \Lambda^{\circ}\left(U_{ m } Y _{ p }\right) = m \times \lambda_{ u ^p}^0+ p \times \lambda_{ Y ^{+-}}^0=250 $
$ 25 m +100 p =250$
$ m +4 p =10....$(1)
$ \Lambda^{\circ}\left(V_{ m } X _{ n }\right)= m \times \lambda_{ v ^{2+}}+ n \times \lambda_{ x ^{m-}}^0=440$
$ 100 m +80 n =440$
$ 5 m +4 n =22....$(2)

From the extrapolation of curve
$\Lambda^{\circ}\left(Z_{ m } X _{ n }\right)=340 $
$ m \times \lambda_{ z ^{n+}}^{\circ}+ n \lambda_{ x ^{m-}}^{\circ}=340 $
$50 m +80 n =340 $
$ 5 m +8 n =34.....$(3)
(3) - (2) $\Rightarrow 4 n =12 \Rightarrow n =3$
Putting in (2) we get $m =2$
Putting in (1) we get $p =2$
$m + n + p =2+3+2=7$