Q.Which one is the SI unit of molar 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 $mol\,m^{-3}$.

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

   …

   🔒 Start your 14-day free trial to see the full solution →
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 $\Lambda_m$ …

   🔒 Start your 14-day free trial to see the full solution →
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

   …

   🔒 Start your 14-day free trial to see the full solution →
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. …

   🔒 Start your 14-day free trial to see the full solution →
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).

   …

   🔒 Start your 14-day free trial to see the full solution →
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:

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

   …

   🔒 Start your 14-day free trial to see the full solution →
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:

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

   …

   🔒 Start your 14-day free trial to see the full solution →
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 $S\,cm^2\,mol^{-1}$.)

   …

   🔒 Start your 14-day free trial to see the full solution →
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 co …

   🔒 Start your 14-day free trial to see the full solution →
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 …

    🔒 Start your 14-day free trial to see the full solution →
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):

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

    🔒 Start your 14-day free trial to see the full solution →
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.

    …

    🔒 Start your 14-day free trial to see the full solution →