Q.i. Explain the effect of dilution of solution on conductivity ?

Showing the 12 most recent of 32 on this concept.

1. CBSE 2025Set ANNUAL1 mark
   Q.What is the SI unit of molar conductivity?

   ›Reveal solutionSolution

   Molar conductivity's SI unit is S m² mol⁻¹.

   Molar conductivity $\Lambda_m = \dfrac{\kappa}{C}$, where $\kappa$ (conductivity) has SI unit $S\,m^{-1}$ and concentration $C$ has SI unit $mol\,m^{-3}$.

   $\Lambda_m = \dfrac{S\,m^{-1}}{mol\,m^{-3}} = S\,m^2\,mol^{-1}$

   (In practice, chemists often quote it in $S\,cm^2\,mol^{-1}$, but the SI unit is $S\,m^2\,mol^{-1}$.)

   ✓Final answer

   S m² mol⁻¹

2. CBSE 2025Set ANNUAL1 mark
   Q.Define the following — Limiting molar conductivity

   ›Reveal solutionSolution

   Limiting molar conductivity is molar conductivity extrapolated to zero concentration.

   Limiting molar conductivity ($\Lambda_m^0$ or $\Lambda_m^\infty$) is the molar conductivity of an electrolyte solution when the concentration approaches zero (i.e. at infinite dilution). At infinite dilution, dissociation of the electrolyte is essentially complete and inter-ionic attractions vanish, so each ion conducts independently and to its maximum extent. For strong electrolytes, $\Lambda_m^0$ is obtained by extrapolating the $\Lambda_m$ vs $\sqrt{C}$ plot to $C \to 0$ (Kohlrausch's law); for weak electrolytes it is obtained using Kohlrausch's law of independent migration of ions.

   ✓Final answer

   Λm° = molar conductivity of an electrolyte at infinite dilution (C → 0)

3. CBSE 2025Set ANNUAL1 markMCQ
   Q.Equivalent conductances of sodium acetate, sodium chloride and hydrochloric acid at infinite dilution are 224, 38.2, 203 ohm^-1 cm^2 eqv^-1 respectively at 298K. So the (lambda)0 CH3COOH is:

   (a) 288.5 ohm^-1 cm^2 eqv.^-1

   (b) 288.8 ohm^-1 cm^2 eqv.^-1

   (c) 388.8 ohm^-1 cm^2 eqv.^-1

   (d) 59.2 ohm^-1 cm^2 eqv.^-1

   ›Reveal solutionSolution

   λ0(CH3COOH) = λ0(CH3COONa) + λ0(HCl) − λ0(NaCl) = 224 + 203 − 38.2 = 388.8 ohm^-1 cm^2 eqv^-1.

   CH3COOH is a weak electrolyte, so its limiting equivalent conductance cannot be found by direct extrapolation. Instead, Kohlrausch's law of independent migration of ions lets us combine the limiting conductances of related strong electrolytes.

   We want λ0(CH3COO-) + λ0(H+). Note that:

   λ0(CH3COONa) = λ0(CH3COO-) + λ0(Na+) = 224

   λ0(HCl) = λ0(H+) + λ0(Cl-) = 203

   λ0(NaCl) = λ0(Na+) + λ0(Cl-) = 38.2

   Adding the first two and subtracting the third cancels Na+ and Cl-, leaving exactly what we want:

   λ0(CH3COOH) = λ0(CH3COONa) + λ0(HCl) − λ0(NaCl)

   = 224 + 203 − 38.2 = 388.8 ohm^-1 cm^2 eqv^-1.

   ✓Final answer

   (c) 388.8 ohm^-1 cm^2 eqv^-1.

4. CBSE 2025Set ANNUAL1 mark
   Q.Fill in the blank: Molar conductivity ________ with decrease in concentration.

   ›Reveal solutionSolution

   Molar conductivity increases as concentration decreases (i.e., on dilution), reaching a maximum limiting value at infinite dilution.

   Molar conductivity is given by:

   Λm = κ x 1000 / M

   As a solution is diluted (concentration M decreases):

   • For weak electrolytes: the degree of dissociation increases sharply with dilution, so more ions are produced per mole, increasing Λm markedly.
   • For strong electrolytes: dissociation is already essentially complete, but the interionic (electrostatic) attractions between oppositely charged ions decrease on dilution, allowing ions to move more freely and conduct better, so Λm increases (though only slightly compared to weak electrolytes).

   In both cases Λm increases with decreasing concentration, approaching a limiting molar conductivity Λm° at infinite dilution.

   ✓Final answer

   increases (with decrease in concentration / on dilution).

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.The unit of molar conductivity is

   (a) S cm^-2 mol^-1

   (b) S cm^2 mol^-1

   (c) S^-1 cm^2 mol^-1

   (d) S cm^2 mol

   ›Reveal solutionSolution

   Molar conductivity relates conductivity (S/cm) to concentration (mol/cm^3), giving the composite unit S cm^2 mol^-1.

   Molar conductivity is defined as:

   $\Lambda_m = \dfrac{\kappa \times 1000}{C}$

   where $\kappa$ (specific conductivity) has units S cm^-1 and $C$ (concentration) has units mol L^-1 (mol per 1000 cm^3).

   Working out the units: $\dfrac{\text{S cm}^{-1} \times \text{cm}^3\,\text{mol}^{-1}}{1} = \text{S cm}^2\,\text{mol}^{-1}$

   ✓Final answer

   (b) S cm^2 mol^-1.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The molar conductivity of a 0.1mol $L^{-1}$ solution of KCl with electrolytic conductivity 0.0129 S $cm^{-1}$ at 298 K is –

   (a) 12.9 S cm$^2$ mol$^{-1}$

   (b) 1.29 S cm$^2$ mol$^{-1}$

   (c) 0.0129 S cm$^2$ mol$^{-1}$

   (d) 129 S cm$^2$ mol$^{-1}$

   ›Reveal solutionSolution

   Converting conductivity (per cm) into molar conductivity requires dividing by the molar concentration expressed per cm³, giving a factor of 1000 in the standard formula.

   Molar conductivity is related to the specific conductivity (electrolytic conductivity, $\kappa$) and molar concentration $C$ (in $mol\,L^{-1}$) by:

   $\Lambda_m = \frac{\kappa \times 1000}{C}$

   (The factor of 1000 converts the concentration from $mol\,L^{-1}$ to $mol\,cm^{-3}$, since $\kappa$ is expressed per $cm$.)

   Substituting $\kappa = 0.0129\ S\,cm^{-1}$ and $C = 0.1\ mol\,L^{-1}$:

   $\Lambda_m = \frac{0.0129 \times 1000}{0.1} = \frac{12.9}{0.1} = 129\ S\,cm^2\,mol^{-1}$

   ✓Final answer

   (d) 129 S cm² mol⁻¹.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The formula used to calculate molar conductivity of an electrolyte is _____.

   (a) $\Lambda=\dfrac{1000c}{k}$

   (b) $c=\dfrac{1000\Lambda}{k}$

   (c) $\Lambda=\dfrac{1000k}{c}$

   (d) $k=\dfrac{1000}{\Lambda c}$

   ›Reveal solutionSolution

   $\Lambda_m = \dfrac{1000\,\kappa}{c}$.

   Molar conductivity ($\Lambda_m$) relates to specific conductivity ($\kappa$) and molar concentration $c$ (in $mol\,dm^{-3}$) by:

   $\Lambda_m = \dfrac{\kappa \times 1000}{c}$

   (the factor 1000 converts $\kappa$, usually in $S\,cm^{-1}$, to a per-litre basis consistent with $c$ in $mol\,dm^{-3} = mol\,L^{-1}$).

   ✓Final answer

   (c) $\Lambda=\dfrac{1000k}{c}$

8. CBSE 2025Set ANNUAL1 mark
   Q.What is the SI unit of molar conductivity? OR Write the relation between specific conductivity and molar conductivity.

   ›Reveal solutionSolution

   Molar conductivity's SI unit follows from $\Lambda_m = \kappa/C$: siemens metre-squared per mole.

   Molar conductivity is defined as $\Lambda_m = \dfrac{\kappa}{C}$, where $\kappa$ (specific/electrical conductivity) has SI unit $S\,m^{-1}$ and $C$ (molar concentration) has SI unit $mol\,m^{-3}$. Dividing, the SI unit of $\Lambda_m$ works out to

   $\frac{S\,m^{-1}}{mol\,m^{-3}} = S\,m^2\,mol^{-1}.$

   (In practical lab work, where $\kappa$ is often expressed in $S\,cm^{-1}$ and $C$ in $mol\,L^{-1}$, the commonly used relation is $\Lambda_m = \dfrac{1000\,\kappa}{C}$, giving the c.g.s.-style unit $S\,cm^2\,mol^{-1}$.)

   OR (relation between specific and molar conductivity): Molar conductivity relates to specific conductivity by dividing by molar concentration: $\Lambda_m = \kappa / C$ (SI, C in $mol\,m^{-3}$), or equivalently $\Lambda_m = \dfrac{1000\,\kappa}{C}$ when $\kappa$ is in $S\,cm^{-1}$ and $C$ is in $mol\,L^{-1}$.

   ✓Final answer

   SI unit: $S\,m^2\,mol^{-1}$; relation: $\Lambda_m = \kappa/C$.

9. CBSE 2024Set A11 markMCQ
   Q.When the concentration of electrolytic solution approaches zero, the resulting molar conductivity is known as ;

   (a) specific conductance

   (b) resistivity

   (c) conductivity

   (d) limiting molar conductivity

   ›Reveal solutionSolution

   Molar conductivity at zero concentration (infinite dilution) is called the limiting molar conductivity — option (d).

   Molar conductivity $\Lambda_m$ increases as an electrolytic solution is diluted, because more of the electrolyte is present as free, effectively conducting ions. As the concentration approaches zero (infinite dilution), $\Lambda_m$ reaches a limiting maximum value denoted $\Lambda_m^\circ$, the limiting molar conductivity. Specific conductance/conductivity ($\kappa$) and resistivity ($\rho$) are different quantities, so (a), (b) and (c) are incorrect.

   ✓Final answer

   (d) limiting molar conductivity

10. CBSE 2024Set B1 mark
    Q.Answer in one word/sentence: Write the unit of Equivalence conductivity.

    ›Reveal solutionSolution

    Equivalent conductivity is conductivity per gram-equivalent of electrolyte per unit volume, giving units of S cm^2 eq^-1.

    Equivalent conductivity, $\Lambda_{eq}$, is defined as $\Lambda_{eq} = \kappa \times V$, where $\kappa$ (specific conductivity) has units of S cm^-1 (ohm^-1 cm^-1) and V is the volume in cm^3 containing one gram-equivalent of the electrolyte, so the overall unit works out to S cm^2 eq^-1 (equivalently, ohm^-1 cm^2 g-equiv^-1).

    ✓Final answer

    S cm^2 eq^-1 (ohm^-1 cm^2 equiv^-1).

11. CBSE 2024Set ANNUAL1 mark
    Q.The molar conductivity of $2.5 \times 10^{-2}$ M of methanoic acid is 46.1 S cm$^2$ mol$^{-1}$. Calculate the degree of dissociation. (Molar conductivity of H$^+$ and HCOO$^-$ at infinite dilutions are $\lambda°_{H^+} = 349.6$ S cm$^2$ mol$^{-1}$ and $\lambda°_{HCOO^-} = 54.6$ S cm$^2$ mol$^{-1}$).

    ›Reveal solutionSolution

    Adding the limiting ionic molar conductivities gives Λm°, and dividing the measured Λm by it gives the degree of dissociation.

    For a weak electrolyte, the degree of dissociation is given by:

    $\alpha = \frac{\Lambda_m}{\Lambda_m°}$

    Limiting molar conductivity of methanoic acid (using Kohlrausch's law of independent migration of ions):

    $\Lambda_m° = \lambda°_{H^+} + \lambda°_{HCOO^-} = 349.6 + 54.6 = 404.2\ S\,cm^2\,mol^{-1}$

    Degree of dissociation:

    $\alpha = \frac{\Lambda_m}{\Lambda_m°} = \frac{46.1}{404.2} = 0.1140$

    ✓Final answer

    $\alpha = \dfrac{46.1}{404.2} \approx \mathbf{0.114}$ (11.4% dissociated).

12. CBSE 2024Set ANNUAL1 mark
    Q.Write SI unit of molar conductivity.

    ›Reveal solutionSolution

    SI unit of molar conductivity is $S\,m^2\,mol^{-1}$.

    Molar conductivity $\Lambda_m$ is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte, measured between electrodes 1 m apart.

    Its SI unit is derived from conductivity ($S\,m^{-1}$) divided by molar concentration ($mol\,m^{-3}$): $\Lambda_m = \dfrac{\kappa}{C}$, giving units $\dfrac{S\,m^{-1}}{mol\,m^{-3}} = S\,m^2\,mol^{-1}$.

    ✓Final answer

    $S\,m^2\,mol^{-1}$ (siemens metre² per mole).