Step-by-Step Solution

1. Understand Kohlrausch's Law

   Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. For a general strong electrolyte $A_x B_y$, which dissociates into $x$ cations $A^{y+}$ and $y$ anions $B^{x-}$, the law is expressed as:

   $\Lambda^0(A_x B_y) = x \lambda^0(A^{y+}) + y \lambda^0(B^{x-})$

   where $\Lambda^0$ is the limiting molar conductivity of the electrolyte and $\lambda^0$ is the limiting molar conductivity of the individual ions.

2. Formulate Equations for $U_m Y_p$ and $V_m X_n$

   We are given the limiting molar conductivities for two electrolytes and the ionic conductivities in a table. We can apply Kohlrausch's law to them.

   • For the electrolyte $U_m Y_p$, the ions are $m$ moles of $U^{p+}$ and $p$ moles of $Y^{m-}$. $\Lambda^0(U_m Y_p) = m \lambda^0(U^{p+}) + p \lambda^0(Y^{m-})$ Substituting the given values: $250 = m(25.0) + p(100.0)$ Dividing the equation by 25, we get: $10 = m + 4p \quad \cdots(1)$

   • For the electrolyte $V_m X_n$, the ions are $m$ moles of $V^{n+}$ and $n$ moles of $X^{m-}$. $\Lambda^0(V_m X_n) = m \lambda^0(V^{n+}) + n \lambda^0(X^{m-})$ Substituting the given values: $440 = m(100.0) + n(80.0)$ Dividing the equation by 20, we get: $22 = 5m + 4n \quad \cdots(2)$

3. Determine $\Lambda^0$ for $Z_m X_n$ from the Graph

   The plot of molar conductivity ($\\/Lambda$) versus the square root of concentration ($c^{1/2}$) for a strong electrolyte is described by the Debye-Hückel-Onsager equation: $\\/Lambda = \Lambda^0 - A\sqrt{c}$, where $\Lambda^0$ is the limiting molar conductivity (the y-intercept).

   The graph shows a straight line with a negative slope. The equation for the line is provided on the graph as $\Lambda = 340 - 100 \sqrt{c}$. By comparing this with the Debye-Hückel-Onsager equation, we can find the y-intercept, which occurs at $\sqrt{c}=0$. $\Lambda^0 = 340 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ So, the limiting molar conductivity of $Z_m X_n$ is $340 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$.

4. Formulate Equation for $Z_m X_n$

   • For the electrolyte $Z_m X_n$, the ions are $m$ moles of $Z^{n+}$ and $n$ moles of $X^{m-}$. $\Lambda^0(Z_m X_n) = m \lambda^0(Z^{n+}) + n \lambda^0(X^{m-})$ Substituting the values we have: $340 = m(50.0) + n(80.0)$ Dividing the equation by 10, we get: $34 = 5m + 8n \quad \cdots(3)$

5. Solve the System of Linear Equations

   We now have a system of three linear equations with three variables ($m, n, p$): (1) $10 = m + 4p$ (2) $22 = 5m + 4n$ (3) $34 = 5m + 8n$

   Subtracting equation (2) from equation (3): $(5m + 8n) - (5m + 4n) = 34 - 22$ $4n = 12$ $n = 3$

   Substitute $n=3$ into equation (2): $22 = 5m + 4(3)$ $22 = 5m + 12$ $10 = 5m$ $m = 2$

   Substitute $m=2$ into equation (1): $10 = 2 + 4p$ $8 = 4p$ $p = 2$

   Thus, we have found the integer values: $m=2, n=3, p=2$.

6. Calculate the Final Value

   The question asks for the value of $(m+n+p)$. $m + n + p = 2 + 3 + 2 = 7$

   The final answer is 7.