Resistance of 0.2 M solution of an electrolyte is 50 ohm. The specific conductance of the solution i

Q: Why does the cell constant remain the same for both solutions?

The cell constant is defined as the ratio of the distance between the electrodes ($l$) to their cross-sectional area ($A$). Because you are using the exact same physical container (the conductivity cell) for both solutions, these physical dimensions do not change.

Q: Why did we convert Molarity to $\text{mol m}^{-3}$ instead of just using the standard formula with $1000$?

The standard formula $\Lambda_m = (\kappa \times 1000) / M$ is specifically designed for when $\kappa$ is in $\text{S cm}^{-1}$. In this problem, $\kappa$ is given in SI units ($\text{S m}^{-1}$). To maintain dimensional consistency without memorizing arbitrary conversion factors, it is much safer to convert Molarity ($\text{mol L}^{-1}$) directly into SI units ($\text{mol m}^{-3}$).

Q: What is the difference between specific conductance and molar conductivity?

Specific conductance ($\kappa$) is the conductance of a $1 \text{ m}^3$ (or $1 \text{ cm}^3$) volume of the solution. Molar conductivity ($\Lambda_m$) is the conducting power of all the ions produced by dissolving exactly $1 \text{ mole}$ of the electrolyte in the solution. It normalizes the conductance by the concentration.

Q: Can I solve this problem by converting everything to CGS units ($\text{cm}$) first?

Absolutely! You can convert $\kappa_1 = 1.4 \text{ S m}^{-1}$ to $1.4 \times 10^{-2} \text{ S cm}^{-1}$, find $G^*$ in $\text{cm}^{-1}$, find $\kappa_2$ in $\text{S cm}^{-1}$, and then use the standard formula $\Lambda_m = (\kappa \times 1000) / M$. Finally, you would convert the result back to $\text{S m}^2 \text{ mol}^{-1}$. However, sticking to SI units throughout is often faster and less prone to conversion errors.

Q: Why did the resistance increase from $50 \,\Omega$ to $280 \,\Omega$ when the concentration increased from $0.2 \text{ M}$ to $0.5 \text{ M}$?

Normally, for a strong electrolyte, increasing concentration increases the number of ions, which should decrease resistance. The data provided in this specific problem shows an increase in resistance with concentration, which implies the specific conductance dropped significantly. This behavior is characteristic of certain weak electrolytes where the degree of dissociation drops sharply at higher concentrations, or it may simply be hypothetical data designed to test your mathematical application of the formulas.

Question

Resistance of $0.2 	ext{ M}$ solution of an electrolyte is $50\Omega$. The specific conductance of the solution is $1.4 	ext{ S m}^{-1}$. The resistance of $0.5 	ext{ M}$ solution of the same electrolyte is $280\Omega$. The molar conductivity of $0.5 	ext{ M}$ solution of the electrolyte in $	ext{S m}^2 	ext{ mol}^{-1}$ is

Options:

(A) $5 	imes 10^{-4}$

(B) $5 	imes 10^{-3}$

(C) $5 	imes 10^3$

(D) $5 	imes 10^2$

Sigma Percentile · Accepted Answer

### The Anatomy of a Conductivity Cell

Imagine you are holding a conductivity cell in your hands. It consists of two parallel electrodes separated by a fixed distance $l$, and each electrode has a specific cross-sectional area $A$. Because these physical dimensions are rigidly built into the device, their ratio, known as the **cell constant** ($G^* = \frac{l}{A}$), is a permanent physical signature of that specific cell.

It does not matter whether you fill the cell with pure water, a highly concentrated acid, or a weak salt solution; the cell constant remains exactly the same. This is the fundamental secret to solving this problem: we can use the first solution to "calibrate" the cell and find its constant, and then use that same constant to unlock the properties of the second solution.

### Decoding the First Solution

Let's look at the first set of data provided. We are given a $0.2 	ext{ M}$ solution with a resistance $R_1 = 50 \,\Omega$ and a specific conductance $\kappa_1 = 1.4 	ext{ S m}^{-1}$.

We know the relationship between specific conductance, resistance, and the cell constant is given by:

$$ \kappa = \frac{1}{R} 	imes G^* $$

By rearranging this equation, we can isolate the cell constant:

$$ G^* = \kappa 	imes R $$

Substituting our known values:

$$ G^* = 1.4 	ext{ S m}^{-1} 	imes 50 \,\Omega = 70 	ext{ m}^{-1} $$

We have successfully calibrated our cell! The cell constant is $70 	ext{ m}^{-1}$.

### The Second Solution: A New Mystery

Now, imagine we empty the cell, rinse it, and fill it with a new $0.5 	ext{ M}$ solution of the same electrolyte. The resistance meter now reads $R_2 = 280 \,\Omega$.

Because we are using the exact same physical cell, our cell constant $G^*$ is still $70 	ext{ m}^{-1}$. We can use this to find the specific conductance ($\kappa_2$) of this new solution:

$$ \kappa_2 = \frac{G^*}{R_2} $$

$$ \kappa_2 = \frac{70 	ext{ m}^{-1}}{280 \,\Omega} = 0.25 	ext{ S m}^{-1} $$

### The Unit Trap: Molar Conductivity

The question ultimately asks for the **molar conductivity** ($\Lambda_m$) of the second solution. The formula is straightforward:

$$ \Lambda_m = \frac{\kappa}{C} $$

However, there is a massive trap here that catches many students. Our specific conductance $\kappa$ is in SI units ($	ext{S m}^{-1}$), but our concentration $C$ is given in Molarity ($	ext{mol L}^{-1}$). If you divide these directly, your units will clash, and your answer will be off by a factor of $1000$!

We must convert the concentration into SI units ($	ext{mol m}^{-3}$). Remember that $1 	ext{ L} = 10^{-3} 	ext{ m}^3$.

$$ C_2 = 0.5 	ext{ mol L}^{-1} = \frac{0.5 	ext{ mol}}{10^{-3} 	ext{ m}^3} = 500 	ext{ mol m}^{-3} $$

### The Final Calculation

Now that our units are perfectly aligned, we can execute the final division:

$$ \Lambda_m = \frac{0.25 	ext{ S m}^{-1}}{500 	ext{ mol m}^{-3}} $$

To make the math easier without a calculator, let's use scientific notation:

$$ \Lambda_m = \frac{25 	imes 10^{-2}}{5 	imes 10^2} $$

$$ \Lambda_m = 5 	imes 10^{-4} 	ext{ S m}^2 	ext{ mol}^{-1} $$

And there we have it! By carefully tracking our cell constant and rigorously managing our units, we arrive at the correct answer flawlessly.

Select Answer:

Visualized Solution

\text{The Conductivity Cell}

\text{Formula for Cell Constant}

\text{Case 1: } 0.2 \text{ M Solution}

\text{Calculating } G^*

\text{Case 2: } 0.5 \text{ M Solution}

\text{Calculating } \kappa_2

\text{Molar Conductivity Formula}

\text{Converting Concentration Units}

\text{Calculating } \Lambda_m

The Sigma Insight: Electrolytic Conduction

The Anatomy of a Conductivity Cell

Decoding the First Solution

The Second Solution

The Unit Trap

The Final Calculation

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