Consider the statements S1 and S2 :
S1 : Conductivity always increases with decrease in the concentr

Q: Why does conductivity ($\kappa$) decrease with dilution?

When you dilute a solution, you are adding more solvent. The total number of ions remains the same, but they are now spread over a much larger volume. Therefore, the number of ions present in a specific *unit volume* (like $1 \text{ cm}^3$) decreases, leading to a drop in conductivity.

Q: If $\kappa$ decreases, shouldn't $\Lambda_m$ also decrease since $\Lambda_m = \kappa \times V$?

This is a classic trap! While $\kappa$ does decrease, the volume $V$ containing one mole of the electrolyte increases drastically upon dilution. The massive increase in volume completely overpowers the slight decrease in $\kappa$, causing the overall product ($\Lambda_m$) to increase.

Q: Does this rule apply to both strong and weak electrolytes?

Yes! For strong electrolytes, the increase in $\Lambda_m$ is gradual due to increased ionic mobility (less interionic attraction). For weak electrolytes, the increase is steep because dilution also increases the degree of dissociation ($\alpha$), producing more ions per mole.

Q: What is the difference between conductance and conductivity?

Conductance ($G$) is a property of the entire solution present between the electrodes, depending on the cell dimensions. Conductivity ($\kappa$) is a specific property, defined as the conductance of exactly one unit volume of the solution.

Q: What happens at infinite dilution?

At infinite dilution (concentration approaching zero), molar conductivity reaches its maximum limiting value, denoted by $\Lambda_m^\circ$. At this point, ions are so far apart that they don't interact with each other at all.

Question

Consider the statements $S_1$ and $S_2$ :
$S_1$ : Conductivity always increases with decrease in the concentration of electrolyte.
$S_2$ : Molar conductivity always increases with decrease in the concentration of electrolyte.
The correct option among the following is

Options:

(A) $S_1$ is correct and $S_2$ is wrong

(B) $S_1$ is wrong and $S_2$ is correct

(C) Both $S_1$ and $S_2$ are wrong

(D) Both $S_1$ and $S_2$ are correct

Sigma Percentile · Accepted Answer

### The Battle of Conductivities

When dealing with electrolytic solutions, two terms often cause immense confusion: **Conductivity** ($\kappa$) and **Molar Conductivity** ($\Lambda_m$). They sound similar, but they behave completely differently when you add water to the solution (dilution). Let's break down the physics behind these two statements and see why one is a classic trap.

### Decoding Statement $S_1$: The Unit Volume Trap

Imagine a crowded concert hall. If you draw a $1 	ext{ meter} 	imes 1 	ext{ meter}$ square on the floor, you might count 10 people standing inside it. This is exactly what **Conductivity** ($\kappa$) measures. It is the conductance of all the ions present in exactly **one unit volume** (like $1 	ext{ cm}^3$) of the solution.

Now, what happens when we dilute the solution? Dilution means we are adding more solvent, effectively decreasing the concentration. The total volume of the solution expands, and the ions spread out to occupy this new space.

If you look back at your fixed $1 	ext{ cm}^3$ box, the number of ions inside it has now decreased because they have drifted apart. Fewer ions in that specific volume mean less charge can be carried through it. Therefore, **conductivity ($\kappa$) strictly decreases with a decrease in concentration**.

Statement $S_1$ claims that conductivity *increases* with a decrease in concentration. This is a fundamental misunderstanding of the unit volume constraint. Thus, $S_1$ is absolutely wrong.

### Decoding Statement $S_2$: The Power of One Mole

Now let's shift our perspective to **Molar Conductivity** ($\Lambda_m$). Instead of restricting ourselves to a fixed volume, we are now tracking a fixed *amount* of substance. $\Lambda_m$ is the conducting power of all the ions produced by exactly **one mole** of the electrolyte, regardless of how much volume it occupies.

Mathematically, they are related by the master equation:

$$\Lambda_m = \frac{\kappa 	imes 1000}{M}$$

Here, $M$ is the molarity (concentration). When we dilute the solution, the concentration $M$ decreases. We already established that $\kappa$ also decreases. So, what happens to the overall fraction?

This is where the magic happens. The volume containing that one mole of electrolyte increases *drastically*. This massive expansion in volume completely overpowers the slight drop in $\kappa$. Because the denominator $M$ shrinks so significantly, the overall value of the fraction shoots up.

Therefore, **molar conductivity ($\Lambda_m$) always increases with a decrease in concentration**. Statement $S_2$ is perfectly correct.

### The Final Verdict

By understanding the physical constraints—$\kappa$ is bound by a fixed volume, while $\Lambda_m$ is bound by a fixed number of moles—we can confidently navigate this problem. Statement $S_1$ is wrong, and statement $S_2$ is correct. This leads us straight to option (b). Always visualize the ions before jumping into the math!

Select Answer:

Visualized Solution

$Visualizing Concentration$

$Understanding Conductivity (κ)$

$Effect of Dilution on κ$

$Evaluating Statement S_{1}$

$Understanding Molar Conductivity (Λ_{m})$

$Effect of Dilution on Λ_{m}$

$Evaluating Statement S_{2}$

$Final Conclusion$

The Sigma Insight: Electrolytic Conduction

The Battle of Conductivities

Decoding Statement $S_{1}$

Decoding Statement $S_{2}$

The Final Verdict

Similar Questions

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

Resistance of $0.2 M$ solution of an electrolyte is $50 Ω$ . The specific conductance of the solution is $1.3 S m^{- 1}$ . If resistance of the $0.4 M$ solution of the same electrolyte is $260 Ω$ , its molar conductivity is

Match List-I with List-II. $List-I (Parameter) A. Cell constant B. Molar conductivity C. Conductivity D. Degree of dissociation of electrolyte List-II (Unit) 1. S cm^{2} mol^{- 1} 2. Dimensionless 3. m^{- 1} 4. Ω^{- 1} m^{- 1}$ Choose the most appropriate answer from the options given below.

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The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution is shown in the given figure. The electrolyte X is

The equivalent conductance of NaCl at concentration $C$ and at infinite dilution are $λ_{c}$ and $λ_{\infty}$ , respectively. The correct relationship between $λ_{c}$ and $λ_{\infty}$ is given as

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing Concentration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Understanding Conductivity (κ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Effect of Dilution on κ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Statement S1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Understanding Molar Conductivity (Λm​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Effect of Dilution on Λm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Statement S2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion