Q.Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Showing the 12 most recent of 20 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Resistance of 0.2 M solution of an electrolyte is $50 \Omega$. The conductivity of the solution is $1.3 \mathrm{Sm}^{-1}$. If the resistance of 0.4 M solution of the same electrolyte is $260 \Omega$, its molar conductivity is: (A) $6.25 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$ (B) $62.5 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$ (C) $6.25 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$ (D) $625 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$

   ›Reveal solutionSolution

   The key is to first find the cell constant from the 0.2 M data, then use it to find the conductivity of the 0.4 M solution, and finally compute its molar conductivity. The result matches option (C).

   Concept and intuition

   Molar conductivity ($\Lambda_m$) is the conductivity of a solution divided by its molar concentration: $\Lambda_m = \frac{\kappa}{c}$. But we cannot directly measure conductivity — we measure resistance $R$ of a solution in a cell with a fixed geometry. The cell constant $G^*$ (units m⁻¹) relates conductivity $\kappa$ to measured conductance $1/R$:

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

   So the plan: use the first solution’s known $\kappa$ and $R$ to find $G^*$, then use that $G^*$ with the second solution’s $R$ to find its $\kappa$, and finally compute $\Lambda_m$ for the second solution.

   ---

   Step-by-step solution

   1. Find the cell constant from the 0.2 M data For the 0.2 M solution:

   $\kappa_1 = 1.3\ \mathrm{S\,m^{-1}}, \quad R_1 = 50\ \Omega$

   Using $\kappa = G^* / R$:

   $G^* = \kappa_1 \cdot R_1 = 1.3 \times 50 = 65\ \mathrm{m^{-1}}$

   This cell constant is fixed for the same conductivity cell.

   2. Find the conductivity of the 0.4 M solution For the 0.4 M solution: $R_2 = 260\ \Omega$.

   $\kappa_2 = \frac{G^*}{R_2} = \frac{65}{260} = 0.25\ \mathrm{S\,m^{-1}}$

   3. Compute the molar conductivity of the 0.4 M solution Molar conductivity $\Lambda_m$ is defined as:

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

   where $c$ is concentration in mol/m³. …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Identify the correct mathematical expression which represents the variation in molar conductivity of a weak acid having concentration C and ionisation constant $\mathrm{K}_{\mathrm{a}}$ ( $\lambda_m^{\infty}=$ molar conductivity at infinite dilution, $\lambda_{\mathrm{m}}=$ molar conductivity at concentration C ) (A) $K_a=\lambda_m^2 C / \lambda_m^{\infty}\left(\lambda_m^{\infty}-\lambda_m\right)$ (B) $K_a=\lambda_m \lambda_m^{\infty}-\left(\lambda_m^{\infty}\right)^2+\lambda_m^2 C$ (C) $\lambda_m+\lambda_m^{\infty}+K_a C^{\frac{1}{2}}=0$ (D) $K_a=\lambda_m^2 C / \lambda_m^{\infty}\left(\lambda_m^{\infty}+\lambda_m\right)$

   ›Reveal solutionSolution

   The key idea is to combine Ostwald’s dilution law with the relation between molar conductivity and degree of dissociation. The correct expression is option (A).

   We start from the fundamental relationship for a weak electrolyte: the degree of dissociation $\alpha$ is given by the ratio of molar conductivity at concentration $C$ to that at infinite dilution:

   $\alpha = \frac{\lambda_m}{\lambda_m^\infty}$

   This works because $\lambda_m$ increases with dissociation, and at infinite dilution the electrolyte is fully dissociated.

   For a weak acid HA dissociating as:

   $HA \rightleftharpoons H^+ + A^-$

   Ostwald’s dilution law gives the ionization constant:

   $K_a = \frac{C \alpha^2}{1 - \alpha}$

   Now substitute $\alpha = \lambda_m / \lambda_m^\infty$:

   $K_a = \frac{C \left( \frac{\lambda_m}{\lambda_m^\infty} \right)^2}{1 - \frac{\lambda_m}{\lambda_m^\infty}} = \frac{C \lambda_m^2 / (\lambda_m^\infty)^2}{(\lambda_m^\infty - \lambda_m)/\lambda_m^\infty} = \frac{C \lambda_m^2}{\lambda_m^\infty (\lambda_m^\infty - \lambda_m)}$

   This matches option (A) exactly. …

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3. COMEDK 2026Set 2026-M1 markMCQ
   Q.Two statements, one Assertion and the other Reason are given. Choose the right option. Assertion: The Molar conductivity of KCl increases very slowly with dilution and approaches a limiting value when dilution is infinite. Reason: In case of KCl there is an increase in the number of ions on dilution due to complete ionisation at infinite dilution. (A) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion (B) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion (C) Assertion is incorrect but Reason is correct (D) Assertion is correct but Reason is incorrect

   ›Reveal solutionSolution

   Molar conductivity increases with dilution due to reduced ion-ion interactions, not because more ions are formed — KCl is already fully dissociated. The Assertion is correct, but the Reason is wrong.

   The key concept here is Kohlrausch’s law and the distinction between strong and weak electrolytes. KCl is a strong electrolyte — it dissociates completely in water at all dilutions. So the number of ions per formula unit does not increase as you dilute it; it’s already 100% ionised.

   The Assertion says molar conductivity of KCl increases slowly with dilution and approaches a limiting value. That is true. But why does it increase? Not because more ions appear — they were already there. The increase happens because ion-ion interactions weaken as the solution gets more dilute. Ions move more freely, so conductivity per mole (molar conductivity) rises. At infinite dilution, ions are independent and molar conductivity reaches its maximum, the limiting molar conductivity $\Lambda_m^\infty$.

   The Reason claims that dilution causes an increase in the number of ions due to complete ionisation at infinite dilution. That is false for KCl — it is already completely ionised even at moderate concentrations. The number of ions per mole of KCl does not change with dilution. So the Reason is incorrect.

   Let’s walk through it step by step.

   1. Identify the electrolyte type. KCl is a salt of a strong acid (HCl) and a strong base (KOH). In water, it dissociates fully: $\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-$. This is true at any dilution — there is no undissociated KCl present. So the degree of ionisation $\alpha = 1$ always.

   2. Understand molar conductivity. Molar conductivity $\Lambda_m$ is defined as $\Lambda_m = \frac{\kappa}{c}$, where $\kappa$ is conductivity and $c$ is concentration. For a strong electrolyte, $\Lambda_m$ increases as $c$ decreases, but the increase is gradual and follows the Kohlrausch square-root law: $\Lambda_m = \Lambda_m^\infty - A\sqrt{c}$. The limiting value $\Lambda_m^\infty$ is approached as $c \to 0$.

   3. Why does $\Lambda_m$ increase? At higher concentrations, ions are close together and their mutual electrostatic attractions (ion-ion interactions) slow them down. Dilution separates the ions, reducing these interactions, so each ion moves faster under the applied field. The number of charge carriers per mole is constant — it’s always 2 moles of ions per mole of KCl — but their mobility increases. That’s the sole reason for the rise in $\Lambda_m$. …

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4. KCET 2026Set D31 markMCQ
   Q.$\Lambda_m^o$(NH$_4$OH) is equal to (A) $\Lambda_m^o$(NH$_4$OH) + $\Lambda_m^o$(NH$_4$Cl) - $\Lambda_m^o$(HCl) (B) $\Lambda_m^o$(NH$_4$Cl) + $\Lambda_m^o$(NaOH) - $\Lambda_m^o$(NaCl) (C) $\Lambda_m^o$(NH$_4$Cl) + $\Lambda_m^o$(NaCl) - $\Lambda_m^o$(NaOH) (D) $\Lambda_m^o$(NaOH) + $\Lambda_m^o$(NaCl) - $\Lambda_m^o$(NH$_4$Cl)

   ›Reveal solutionSolution

   Apply Kohlrausch's law of independent ionic migration: combine strong electrolytes sharing NH$_4^+$ and OH$^-$ so the extra ions (Na$^+$, Cl$^-$) cancel out.

   Step 1 — Express each electrolyte's limiting conductivity in terms of ions

   $\Lambda_m^o(\text{NH}_4\text{Cl}) = \lambda^o_{\text{NH}_4^+} + \lambda^o_{\text{Cl}^-}$

   $\Lambda_m^o(\text{NaOH}) = \lambda^o_{\text{Na}^+} + \lambda^o_{\text{OH}^-}$

   $\Lambda_m^o(\text{NaCl}) = \lambda^o_{\text{Na}^+} + \lambda^o_{\text{Cl}^-}$

   Step 2 — Combine to isolate NH$_4^+$ and OH$^-$

   Adding the first two and subtracting the third:

   $\Lambda_m^o(\text{NH}_4\text{Cl}) + \Lambda_m^o(\text{NaOH}) - \Lambda_m^o(\text{NaCl}) = (\lambda^o_{\text{NH}_4^+} + \lambda^o_{\text{Cl}^-}) + (\lambda^o_{\text{Na}^+} + \lambda^o_{\text{OH}^-}) - (\lambda^o_{\text{Na}^+} + \lambda^o_{\text{Cl}^-})$ …

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5. COMEDK 2025Set 2025-A1 markMCQ
   Q.Two statements, one Assertion and the other Reason are given. Choose the correct option. Assertion: For strong electrolytes the plot of Molar conductivity versus Concentration gives a straight line with slope equal to +A and intercept equal to $\lambda_{\mathrm{m}}$ Reason: For strong electrolytes, $\lambda_{\mathrm{m}}$ increases slowly with dilution due to increase in the distance between the ions and increase in ionic mobility (A) Assertion is correct but Reason is incorrect. (B) Both Assertion and Reason are incorrect. (C) Assertion is incorrect but Reason is correct. (D) Both Assertion and Reason are correct.

   ›Reveal solutionSolution

   The assertion incorrectly states the slope is positive (+A) when it is actually negative for strong electrolytes; the reason correctly describes the trend but misattributes it to "slow increase" rather than a small, gradual decrease. The correct option is (C).

   The key concept here is Kohlrausch’s law for strong electrolytes. For strong electrolytes (fully dissociated salts like NaCl, KCl), molar conductivity $\Lambda_m$ decreases linearly with the square root of concentration $\sqrt{c}$, not increases. The slope is negative because as concentration increases, ion-ion interactions impede mobility. The reason, while describing the correct physical trend (conductivity changes with dilution), gets the direction wrong—it says "increases slowly" when in fact it decreases slightly with increasing concentration.

   Let’s break it down step by step.

   1. Understand the Assertion

      The assertion says: For strong electrolytes, the plot of Molar conductivity versus Concentration gives a straight line with slope equal to +A and intercept equal to $\lambda_m$.

      • Kohlrausch’s empirical law states: $\Lambda_m = \Lambda_m^0 - A\sqrt{c}$, where $\Lambda_m^0$ is the limiting molar conductivity (intercept) and $A$ is a positive constant.
      • This is a straight line when $\Lambda_m$ is plotted against $\sqrt{c}$, not against $c$ directly. The slope is $-A$ (negative), not $+A$.
      • Therefore, the assertion is incorrect because it gives the wrong sign for the slope and implies a plot against $c$ (not $\sqrt{c}$).

   2. Understand the Reason

      The reason says: For strong electrolytes, $\lambda_m$ increases slowly with dilution due to increase in the distance between the ions and increase in ionic mobility.

      • As we dilute a strong electrolyte, the distance between ions increases, reducing interionic attractions. This does increase ionic mobility, so $\Lambda_m$ increases.
      • However, the phrase "increases slowly" is misleading: for strong electrolytes, $\Lambda_m$ increases sharply at low concentrations and then approaches a constant $\Lambda_m^0$; the change is not "slow" in the sense of a small slope. More importantly, the reason correctly identifies the cause (greater distance → less hindrance → higher mobility), but it misstates the rate as "slow". …

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6. COMEDK 2025Set 2025-E1 markMCQ
   Q.When $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ of KCl was filled in a Conductivity cell the resistance was 80 ohms at 298 K . (Conductivity of 0.1 M KCl at 298 K is $1.29 \mathrm{~S} \mathrm{~m}^{-1}$. The same cell when filled with an unknown electrolyte of concentration 0.025 M , had a resistance of 92 ohms. What is the Molar conductivity of the electrolyte at the given concentration? (A) $220.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ (B) $0.2206 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ (C) $448 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ (D) $0.449 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

   ›Reveal solutionSolution

   The key is to first find the cell constant from the known KCl data, then use it to get the conductivity of the unknown solution, and finally compute its molar conductivity. The result is $0.448 \, \mathrm{S \, m^2 \, mol^{-1}}$, which matches option (C) when converted to $\mathrm{S \, cm^2 \, mol^{-1}}$.

   ---

   Concept & Intuition

   Conductivity cells have a fixed geometry — the distance between electrodes and their area — captured by the cell constant $G^* = \frac{\ell}{A}$ (units: $\mathrm{m^{-1}}$).

   We cannot measure $\ell$ and $A$ directly, but we can find $G^*$ using a standard solution of known conductivity.

   Once we know $G^*$, any unknown solution’s conductivity is simply $\kappa = G^* / R$.

   Then molar conductivity $\Lambda_m = \frac{\kappa}{c}$ (with careful unit conversion) gives the answer.

   ---

   Step-by-step

   1. Find the cell constant from the KCl data For the KCl solution:

   $\kappa_{\text{KCl}} = 1.29 \, \mathrm{S \, m^{-1}}, \quad R_{\text{KCl}} = 80 \, \Omega$

   The cell constant is:

   $G^* = \kappa_{\text{KCl}} \times R_{\text{KCl}} = 1.29 \times 80 = 103.2 \, \mathrm{m^{-1}}$

   2. Use the cell constant to find the conductivity of the unknown For the unknown solution:

   $R_{\text{unknown}} = 92 \, \Omega$

   $\kappa_{\text{unknown}} = \frac{G^*}{R_{\text{unknown}}} = \frac{103.2}{92} \approx 1.12174 \, \mathrm{S \, m^{-1}}$

   3. Convert concentration to SI units The concentration is $0.025 \, \mathrm{M} = 0.025 \, \mathrm{mol \, L^{-1}}$. Since $1 \, \mathrm{L} = 10^{-3} \, \mathrm{m^3}$: $$$$

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7. COMEDK 2025Set 2025-M1 markMCQ
   Q.Arrange the following compounds in the decreasing order of the molar conductivities of their aqueous solutions. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} A B C D $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$ $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]$ $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$ $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$ (A) $\mathrm{B}>\mathrm{C}>\mathrm{A}>\mathrm{D}$ (B) $\mathrm{D}>\mathrm{A}>\mathrm{C}>\mathrm{B}$ (C) $\mathrm{B}>\mathrm{A}>\mathrm{C}>\mathrm{D}$ (D) $\mathrm{A}>\mathrm{B}>\mathrm{D}>\mathrm{C}$

   ›Reveal solutionSolution

   Molar conductivity depends on the number of ions produced per formula unit in solution. The more ions, the higher the conductivity. The correct order is D > A > C > B, which corresponds to option (B).

   The key concept here is Kohlrausch’s law of independent migration of ions: the molar conductivity of an electrolyte at infinite dilution is the sum of the conductivities of its constituent ions. For a given concentration (here, aqueous solutions at comparable dilution), the compound that dissociates into more ions will have a higher molar conductivity. So we simply count the number of ions each coordination compound releases when dissolved in water.

   ---

   1. Identify the number of ions per formula unit for each compound.

      Coordination compounds in water typically dissociate into the complex ion and the counter ions outside the coordination sphere. The ligands inside the brackets are tightly bound and do not dissociate.

      • A: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$

        The complex ion is $[\mathrm{Co(NH_3)_5Cl}]^{2+}$ and there are two $\mathrm{Cl}^-$ ions outside.

        → Total ions = 1 complex cation + 2 chloride ions = 3 ions.

      • B: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]$

        All three chlorines are inside the coordination sphere; no counter ions outside.

        → This is a neutral complex, so it does not dissociate into ions.

        → Total ions = 0 ions (or effectively 1 molecule, but conductivity is negligible).

      • C: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$

        The complex ion is $[\mathrm{Co(NH_3)_4Cl_2}]^{+}$ and one $\mathrm{Cl}^-$ outside.

        → Total ions = 1 complex cation + 1 chloride ion = 2 ions.

      • D: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$

        The complex ion is $[\mathrm{Co(NH_3)_6}]^{3+}$ and three $\mathrm{Cl}^-$ outside. …

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8. COMEDK 2025Set 2025-M1 markMCQ
   Q.The conductivity of 0.01 M solution of $\mathrm{CH}_3 \mathrm{COOH}$ at 298 K is $1.65 \times 10^{-4} \mathrm{Scm}^{-1}$ What is the $\mathrm{pK}_{\mathrm{a}}$ value of the acid if $\lambda^0\left(\mathrm{H}^{+}\right)$and $\lambda^0\left(\mathrm{CH}_3 \mathrm{COO}\right)^{-1}$ are $349.1 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$ and $40.9 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$ respectively? (A) 4.73 (B) 1.87 (C) 3.47 (D) 2.95

   ›Reveal solutionSolution

   The pKa is found by first calculating the molar conductivity and degree of dissociation from the given conductivity and limiting conductivities, then using the dissociation constant expression for a weak acid. The result is approximately 4.73, corresponding to option (A).

   Concept & Intuition

   This problem connects conductivity measurements to acid dissociation. For a weak acid like acetic acid, the molar conductivity at a given concentration is less than the limiting molar conductivity because the acid is only partially dissociated. The ratio of these gives the degree of dissociation ($\alpha$). Once $\alpha$ is known, the dissociation constant $K_a$ follows from the equilibrium expression, and then $\mathrm{p}K_a = -\log K_a$.

   ---

   1. Calculate the molar conductivity ($\Lambda_m$) of the solution. Molar conductivity is given by $\Lambda_m = \frac{\kappa}{c}$, where $\kappa$ is the conductivity and $c$ is the concentration in mol/cm³ (since units must match). Here, $\kappa = 1.65 \times 10^{-4} \, \mathrm{S\,cm^{-1}}$ and $c = 0.01 \, \mathrm{mol\,L^{-1}} = 0.01 \times 10^{-3} \, \mathrm{mol\,cm^{-3}} = 10^{-5} \, \mathrm{mol\,cm^{-3}}$.

   $\Lambda_m = \frac{1.65 \times 10^{-4}}{10^{-5}} = 16.5 \, \mathrm{S\,cm^2\,mol^{-1}}$

   2. Find the limiting molar conductivity ($\Lambda_m^0$) of acetic acid. Using Kohlrausch’s law:

   $\Lambda_m^0(\mathrm{CH_3COOH}) = \lambda^0(\mathrm{H^+}) + \lambda^0(\mathrm{CH_3COO^-})$

   Given $\lambda^0(\mathrm{H^+}) = 349.1$ and $\lambda^0(\mathrm{CH_3COO^-}) = 40.9$ (both in $\mathrm{S\,cm^2\,mol^{-1}}$):

   $\Lambda_m^0 = 349.1 + 40.9 = 390.0 \, \mathrm{S\,cm^2\,mol^{-1}}$

   3. Determine the degree of dissociation ($\alpha$). For a weak electrolyte, $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$.

   $\alpha = \frac{16.5}{390.0} \approx 0.04231$

   4. Write the dissociation equilibrium and find $K_a$. For $\mathrm{CH_3COOH} \rightleftharpoons \mathrm{H^+} + \mathrm{CH_3COO^-}$, initial concentration $c = 0.01 \, \mathrm{M}$. At equilibrium: $$[\mathrm{H^+}] = [\mathrm{CH_3COO^-}] = c\alpha, \quad [\mathrm{CH_3COOH}] = c(1-\alpha) …$$

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9. KCET 2024Set B-21 markMCQ
   Q.The value of ‘A’ in the equation $\lambda_m = \lambda_m^0 - A\sqrt{C}$ is same for the pair : (A) NaCl and CaCl$_2$ (B) CaCl$_2$ and MgSO$_4$ (C) NaCl and KBr (D) MgCl$_2$ and NaCl

   ›Reveal solutionSolution

   The slope $A$ of the Debye–Hückel–Onsager plot is fixed by the electrolyte's charge type (1:1, 1:2, 2:2 …), so the pair with the same charge type shares the same $A$.

   Step 1 — What the equation says.

   For a strong electrolyte, molar conductivity falls with concentration as

   $\Lambda_m = \Lambda_m^{0} - A\sqrt{C}.$

   The $\sqrt{C}$ dependence comes from the ionic atmosphere (relaxation + electrophoretic effects). The Debye–Hückel–Onsager theory shows the constant $A$ depends on:

   • the nature of the solvent (dielectric constant, viscosity),
   • the temperature, and
   • the charge type of the electrolyte (i.e. the valencies $z_+$, $z_-$).

   It does not depend on which particular ions of that charge type are present.

   Step 2 — Classify each substance by charge type.

   Electrolyte	Ions	Charge type
   NaCl	Na$^+$, Cl$^-$	1 : 1
   KBr	K$^+$, Br$^-$	1 : 1

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10. COMEDK 2024Set 2024-A1 markMCQ
    Q.The limiting molar conductivity of $\mathrm{NH}_4 \mathrm{OH}$ is $238 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. At $25^{\circ} \mathrm{C}$, molar conductance of $0.1 \mathrm{M}$ aqueous solution of ammonium hydroxide is $9.54 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. The degree of ionisation of $\mathrm{NH}_4 \mathrm{OH}$ at the same concentration and temperature is: (A) 4.008% (B) 40.0% (C) 2.08% (D) 32.5%

    ›Reveal solutionSolution

    The degree of ionization is the ratio of molar conductivity at a given concentration to the limiting molar conductivity. For NH₄OH, this gives 9.54 / 238 ≈ 0.04008, which is 4.008%, so option (A) is correct.

    The key idea here is Kohlrausch’s law and the concept of degree of ionization for weak electrolytes. Ammonium hydroxide (NH₄OH) is a weak base; it does not fully dissociate in water. The molar conductivity at a given concentration (Λₘ) is less than the limiting molar conductivity (Λₘ⁰) because only a fraction of the molecules are ionized. For weak electrolytes, the degree of ionization (α) is simply:

    $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$

    This works because the mobility of ions is constant at infinite dilution, and the only reason Λₘ is smaller is that fewer ions are present.

    Now, let’s work through it step by step.

    1. Identify the given data

       • Limiting molar conductivity, $\Lambda_m^0 = 238 \, \text{S cm}^2 \text{mol}^{-1}$
       • Molar conductivity at 0.1 M, $\Lambda_m = 9.54 \, \text{S cm}^2 \text{mol}^{-1}$
       • Concentration = 0.1 M, temperature = 25°C

    2. Recall the formula for degree of ionization

       For a weak electrolyte, the degree of ionization is:

    $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$

    This is valid because at infinite dilution, the electrolyte is fully dissociated, so Λₘ⁰ represents the conductivity if all molecules were ions. At any finite concentration, the actual conductivity is proportional to the fraction dissociated.

    3. Plug in the numbers

    $\alpha = \frac{9.54}{238}$

    Calculate:

    $\alpha = 0.040084...$

    4. Convert to percentage

    $\alpha \times 100\% = 0.040084 \times 100\% = 4.0084\%$

    Rounded to three significant figures, this is 4.008%.

    5. Match with the options …

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11. COMEDK 2024Set 2024-E1 markMCQ
    Q.Given below are 2 statements: Assertion and Reason. Choose the correct option. Assertion: When Molar conductivity for a strong electrolyte is plotted versus $\sqrt{C}(\mathrm{~mol} / \mathrm{L})^{1 / 2}$, a straight line is obtained with intercept equal to Molar conductivity at infinite dilution for the electrolyte and Slope equal to $-\mathrm{A}$. All electrolytes of a given type have the same $\mathrm{A}$ value. Reason: At infinite dilution, strong electrolytes of the same type will have different number of ions due to incomplete dissociation. (A) Assertion is correct but Reason is incorrect statement. (B) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. (C) Both Assertion and Reason are incorrect statements. (D) Assertion is incorrect but Reason is correct statement.

    ›Reveal solutionSolution

    The assertion is correct (Kohlrausch’s law gives a straight line for strong electrolytes, with the same slope for a given type), but the reason is false (strong electrolytes dissociate completely at infinite dilution, so the number of ions is the same for a given type). Hence the correct option is (A).

    Concept and Intuition

    Kohlrausch discovered that for strong electrolytes, molar conductivity $\Lambda_m$ decreases linearly with the square root of concentration at low concentrations. This is described by $\Lambda_m = \Lambda_m^\infty - A\sqrt{C}$, where $\Lambda_m^\infty$ is the limiting molar conductivity and $A$ is a constant that depends only on the type of electrolyte (e.g., 1:1, 2:1, etc.) and the solvent, not on the specific ions. The reason given claims that at infinite dilution, strong electrolytes of the same type have different numbers of ions due to incomplete dissociation — but that’s wrong: strong electrolytes dissociate completely at all dilutions, and at infinite dilution they are fully dissociated, so all 1:1 electrolytes give exactly two ions per formula unit, all 2:1 give three, etc. The reason therefore does not explain the assertion and is itself false.

    Step-by-step reasoning

    1. Understanding the assertion For a strong electrolyte, Kohlrausch’s law states:

    $\Lambda_m = \Lambda_m^\infty - A\sqrt{C}$

    This is a linear equation in $\sqrt{C}$. The intercept (at $\sqrt{C}=0$) is $\Lambda_m^\infty$, and the slope is $-A$. The constant $A$ depends on the stoichiometry (type) of the electrolyte and the solvent properties (viscosity, dielectric constant), but is the same for all electrolytes of a given type (e.g., all 1:1 electrolytes like NaCl, KCl, HCl have the same $A$ in the same solvent). So the assertion is correct.

    2. Examining the reason The reason says: “At infinite dilution, strong electrolytes of the same type will have different number of ions due to incomplete dissociation.”
       • Strong electrolytes (e.g., NaCl, KOH, HCl) dissociate completely in solution at all concentrations, including infinite dilution. …

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12. COMEDK 2024Set 2024-M1 markMCQ
    Q.$0.1 \mathrm{M}$ solution of $\mathrm{AgNO}_3$ is taken in a Conductivity cell and a potential difference of $40 \mathrm{~V}$ is applied across the ends of a column of this solution whose diameter is $4.0 \mathrm{~cm}$ and length of the column is $12 \mathrm{~cm}$. The current used is $0.4 \mathrm{~A}$. The Molar conductivity of the solution is _________. (A) $9.547 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$ (B) $95.5 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$ (C) $0.009546 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$ (D) $954.7 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}$

    ›Reveal solutionSolution

    The molar conductivity is found by first calculating the conductivity from the cell geometry and measured current/voltage, then dividing by the concentration. The result is $95.5 \, \mathrm{S\,cm^2\,mol^{-1}}$, which corresponds to option (B).

    We are given a conductivity cell with a solution of $0.1\,\mathrm{M}$ $\mathrm{AgNO}_3$. A voltage of $40\,\mathrm{V}$ is applied, producing a current of $0.4\,\mathrm{A}$. The column has diameter $4.0\,\mathrm{cm}$ and length $12\,\mathrm{cm}$. We need the molar conductivity $\Lambda_m$.

    Concept and intuition:

    Molar conductivity is the conductivity of a solution divided by its molar concentration. Conductivity $\kappa$ itself is the reciprocal of resistivity, and we can get it from the measured resistance $R = V/I$ and the cell constant $G^* = \ell / A$ (length over cross-sectional area). So the plan is:

    1. Find resistance from Ohm’s law.
    2. Compute the cell constant from the given dimensions.
    3. Conductivity $\kappa = \text{cell constant} / R$.
    4. Molar conductivity $\Lambda_m = \kappa / c$ (with careful unit conversion).

    Let’s go step by step.

    1. Find the resistance of the solution column. Ohm’s law: $R = V / I$.

    $R = \frac{40\,\mathrm{V}}{0.4\,\mathrm{A}} = 100\,\Omega.$

    2. Compute the cell constant $G^*$. The cell constant is $\ell / A$, where $\ell$ is the distance between electrodes (length of column) and $A$ is the cross-sectional area. Diameter $d = 4.0\,\mathrm{cm}$, so radius $r = 2.0\,\mathrm{cm}$. Area:

    $A = \pi r^2 = \pi (2.0)^2 = 4\pi \,\mathrm{cm}^2.$

    Length $\ell = 12\,\mathrm{cm}$.

    Hence:

    $G^* = \frac{\ell}{A} = \frac{12}{4\pi} = \frac{3}{\pi} \,\mathrm{cm}^{-1}.$

    3. Calculate the conductivity $\kappa$. Conductivity is related to resistance by:

    $\kappa = \frac{G^*}{R}.$

    So:

    $\kappa = \frac{3/\pi}{100} = \frac{3}{100\pi} \,\mathrm{S\,cm^{-1}}.$

    Numerically:

    $\kappa \approx \frac{3}{314.159} \approx 0.009549\,\mathrm{S\,cm^{-1}}.$

    4. Convert concentration to proper units. …

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