Consider the strong electrolytes ZmXn, UmYp and VmXn. Limiting molar conductivity of UmYp and VmXn a

Q: Why do we extrapolate the graph to zero concentration?

At zero concentration, the electrolyte is completely dissociated, and the ions do not interact with each other. This gives us the limiting molar conductivity ($\Lambda^0$), which is a fundamental property of the electrolyte used in Kohlrausch's law.

Q: How did we find the slope of the graph?

We used the two points given on the dashed lines: $(0.01, 339)$ and $(0.04, 336)$. The slope is simply the change in $y$ divided by the change in $x$, which is $(336 - 339) / (0.04 - 0.01) = -100$.

Q: Why did we subtract equation (3) from equation (1)?

Both equations had the term $5m$ (or $50m$ before simplification). Subtracting them eliminated the variable $m$, allowing us to easily solve for $n$. This is a standard algebraic technique for solving systems of linear equations.

Q: Can the stoichiometric coefficients be fractions?

In the context of a chemical formula like $Z_mX_n$, the coefficients $m, n$, and $p$ represent the number of ions in one formula unit of the electrolyte, so they must be positive integers. If you get a fraction, recheck your calculations!

Question

Consider the strong electrolytes $Z_mX_n$, $U_mY_p$ and $V_mX_n$. Limiting molar conductivity ($\Lambda^0$) of $U_mY_p$ and $V_mX_n$ are 250 and 440 S cm$^2$ mol$^{-1}$, respectively. The value of (m + n + p) is _______.
Given:
$\begin{array}{|c|c|c|c|c|c|} \hline 	ext{Ion} & Z^{n+} & U^{p+} & V^{n+} & X^{m-} & Y^{m-} \ \hline \lambda^0 (	ext{S cm}^2 	ext{ mol}^{-1}) & 50.0 & 25.0 & 100.0 & 80.0 & 100.0 \ \hline \end{array}$
$\lambda^0$ is the limiting molar conductivity of ions
The plot of molar conductivity ($\Lambda$) of $Z_mX_n$ vs $c^{1/2}$ is given below.

Sigma Percentile · Accepted Answer

### Decoding the Graph

Let's start by analyzing the graph provided in the question. It plots the molar conductivity ($\Lambda$) against the square root of concentration ($c^{1/2}$) for the strong electrolyte $Z_mX_n$. According to the Debye-Hückel-Onsager equation, strong electrolytes follow a linear relationship at low concentrations:

$$\Lambda = \Lambda^0 - A\sqrt{c}$$

To find the limiting molar conductivity ($\Lambda^0$), which is the y-intercept of this graph, we need to extrapolate the straight line to zero concentration ($c 	o 0$). Let's calculate the slope of this line using the two given points: $(0.01, 339)$ and $(0.04, 336)$.

$$	ext{Slope} = \frac{336 - 339}{0.04 - 0.01} = \frac{-3}{0.03} = -100$$

Now, using the point-slope form or simply substituting back into the equation, we can find the intercept. At $c^{1/2} = 0.01$, $\Lambda = 339$. Moving back to $c^{1/2} = 0$, we get:

$$\Lambda^0 = 339 - (-100)(0.01) = 340 	ext{ S cm}^2	ext{ mol}^{-1}$$

So, the limiting molar conductivity of $Z_mX_n$ is **$340 	ext{ S cm}^2	ext{ mol}^{-1}$**.

### Kohlrausch's Law of Independent Migration

Now, we bring in Kohlrausch's law of independent migration of ions. It states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its constituent ions, multiplied by their respective stoichiometric coefficients. Mathematically:

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

Let's apply this law to our three electrolytes using the ionic conductivities provided in the table. This will give us a system of three linear equations for the variables $m$, $n$, and $p$.

For $Z_mX_n$:
$$m(50.0) + n(80.0) = 340 \implies 5m + 8n = 34 \quad 	ext{--- (1)}$$

For $U_mY_p$:
$$m(25.0) + p(100.0) = 250 \implies m + 4p = 10 \quad 	ext{--- (2)}$$

For $V_mX_n$:
$$m(100.0) + n(80.0) = 440 \implies 10m + 8n = 44 \implies 5m + 4n = 22 \quad 	ext{--- (3)}$$

### Solving the Linear Equations

We now have a neat system of linear equations. Look closely at equations (1) and (3). They both contain a $5m$ term. This is a perfect opportunity to eliminate $m$ by subtracting equation (3) from equation (1):

$$(5m + 8n) - (5m + 4n) = 34 - 22$$
$$4n = 12 \implies \mathbf{n = 3}$$

Now that we have $n$, we can substitute it back into equation (3) to find $m$:

$$5m + 4(3) = 22$$
$$5m + 12 = 22 \implies 5m = 10 \implies \mathbf{m = 2}$$

Finally, substitute $m = 2$ into equation (2) to solve for $p$:

$$2 + 4p = 10$$
$$4p = 8 \implies \mathbf{p = 2}$$

### The Final Sum

We have successfully determined all the stoichiometric coefficients: $m = 2$, $n = 3$, and $p = 2$. The question asks for the sum of these three values.

$$m + n + p = 2 + 3 + 2 = \mathbf{7}$$

The final answer is 7. This problem beautifully connects graphical analysis with fundamental electrochemical laws and basic algebra.

Enter Numerical Value:

Visualized Solution

\text{Analyzing the Graph}

\text{Extrapolating for } \Lambda^0

\text{Calculating } \Lambda^0(Z_mX_n)

\text{Kohlrausch's Law of Independent Migration}

\text{Setting up Equations}

\text{Solving for } n

\text{Solving for } m \text{ and } p

\text{Final Calculation}

The Sigma Insight: Electrolytic Conduction

Decoding the Graph

Kohlrausch's Law of Independent Migration

Solving the Linear Equations

The Final Sum

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