Consider the strong electrolytes Z_mX_n, U_mY_p and V_mX_n. Limiting molar conductivity (Λ°) of U_mY_p and V_mX_n are 250 and 440 S cm² mol⁻¹, respectively. The value of (m + n + p) is_______.

Given:

Ion	Zn+	Up+	Vn+	Xm−	Ym−
λ° (S cm2 mol−1)	50.0	25.0	100.0	80.0	100.0

λ° is the limiting molar conductivity of ions

The plot of molar conductivity (Λ_m) of Z_mX_n vs c^1/2 is given below:

The correct answer is 7.

Step 1: Understanding Kohlrausch's Law of Independent Migration of Ions

According to Kohlrausch's law, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent cations and anions, each multiplied by the number of ions present per formula unit.

For an electrolyte A_xB_y containing x cations of charge +y and y anions of charge -x:

${\text{Λ}}_{\text{m}}^{°}=x·{\text{λ}}^{°}\left(\text{cation}\right)+y·{\text{λ}}^{°}\left(\text{anion}\right)$

Step 2: Set up equations using the given data for U_mY_p and V_mX_n

From the given table, the ionic limiting molar conductivities (in S cm² mol⁻¹) are:

• λ°(Zⁿ⁺) = 50.0

• λ°(U^p+) = 25.0

• λ°(Vⁿ⁺) = 100.0

• λ°(X^m−) = 80.0

• λ°(Y^m−) = 100.0

1. For the electrolyte U_mY_p:

The formula unit contains m U^p+ ions and p Y^m− ions.

${\text{Λ}}_{\text{m}}^{°}\left({\text{U}}_{\text{m}}{\text{Y}}_{\text{p}}\right)=m·{\text{λ}}^{°}\left({\text{U}}^{\text{p+}}\right)+p·{\text{λ}}^{°}\left({\text{Y}}^{\text{m−}}\right)$

Given Λ°(U_mY_p) = 250 S cm² mol⁻¹:

$250=25.0m+100.0p$

Dividing the entire equation by 25 gives:

$m+4p=10$    --- (Equation 1)

2. For the electrolyte V_mX_n:

The formula unit contains m Vⁿ⁺ ions and n X^m− ions.

${\text{Λ}}_{\text{m}}^{°}\left({\text{V}}_{\text{m}}{\text{X}}_{\text{n}}\right)=m·{\text{λ}}^{°}\left({\text{V}}^{\text{n+}}\right)+n·{\text{λ}}^{°}\left({\text{X}}^{\text{m−}}\right)$

Given Λ°(V_mX_n) = 440 S cm² mol⁻¹:

$440=100.0m+80.0n$

Dividing the entire equation by 20 gives:

$5m+4n=22$    --- (Equation 2)

Step 3: Determine the value of m from the plot of Z_mX_n

The graph shows a linear plot of molar conductivity (Λ_m) of Z_mX_n vs c^1/2:

From the plot, we can read two distinct coordinates (c^1/2, Λ_m):

• At c^1/2 = 0.01 (mol L⁻¹)^1/2, Λ_m = 339 S cm² mol⁻¹

• At c^1/2 = 0.04 (mol L⁻¹)^1/2, Λ_m = 336 S cm² mol⁻¹

According to Debye-Hückel-Onsager equation for a strong electrolyte:

${\text{Λ}}_{\text{m}}={\text{Λ}}_{\text{m}}^{°}-A\sqrt{c}$

We can find the limiting molar conductivity ${\text{Λ}}_{\text{m}}^{°}\left({\text{Z}}_{\text{m}}{\text{X}}_{\text{n}}\right)$ by finding the y-intercept (extrapolating to c^1/2 = 0):

$\text{Slope}(-A)=\frac{336-339}{0.04-0.01}=\frac{-3}{0.03}=-100$

Now using the point (0.01, 339):

$339={\text{Λ}}_{\text{m}}^{°}\left({\text{Z}}_{\text{m}}{\text{X}}_{\text{n}}\right)-100\left(0.01\right)$

${\text{Λ}}_{\text{m}}^{°}\left({\text{Z}}_{\text{m}}{\text{X}}_{\text{n}}\right)=339+1=340{\text{S cm}}^2{\text{mol}}^{-1}$

Now apply Kohlrausch's law for Z_mX_n:

${\text{Λ}}_{\text{m}}^{°}\left({\text{Z}}_{\text{m}}{\text{X}}_{\text{n}}\right)=m·{\text{λ}}^{°}\left({\text{Z}}^{\text{n+}}\right)+n·{\text{λ}}^{°}\left({\text{X}}^{\text{m−}}\right)$

$340=50.0m+80.0n$

Dividing by 10 gives:

$5m+8n=34$    --- (Equation 3)

Step 4: Solve for m, n, and p

Subtract Equation 2 from Equation 3:

$(5m+8n)-(5m+4n)=34-22$

$4n=12\Rightarrow n=3$

Substitute n = 3 into Equation 2:

$5m+4\left(3\right)=22$

$5m+12=22\Rightarrow 5m=10\Rightarrow m=2$

Substitute m = 2 into Equation 1:

$2+4p=10$

$4p=8\Rightarrow p=2$

Step 5: Calculate the value of (m + n + p)

$m+n+p=2+3+2=7$

Thus, the value of (m + n + p) is 7.