Read the passage carefully and answer the Questions.
Molar conductivity (\( Λ_m \)) of a solution at a given concentration (c) is the conductance of volume, V of solution, containing one mole of electrolyte kept between the two electrodes with area of cross-section A and at a distance of unit length. It increases with the decrease in concentration and when the concentration approaches zero, the molar conductivity is called limiting molar conductivity (\( Λ^0_m \)). For a strong electrolyte, \( Λ_m \) increases linearly with dilution and is given by \( Λ_m = Λ^0_m -Ac^{1/2} \). The value of the constant A for a given solvent depends on the type of electrolyte along with temperature. According to Kohlrausch law, the value of \( Λ^0_m \) for an electrolyte is \( Λ^0_m=v_+λ^0_+ + v_-λ^0_- \), where \( v_+ \) and \( v_- \) are the number of cations and anions, respectively, per molecule of the electrolyte and \( λ^0_+ \) and \( λ^0_- \) are limiting molar conductivities of cation and anion, respectively. Kohlrausch law finds many applications, like determining the solubility of a sparingly soluble salt, determining the degree of dissociation (\( Λ_m/Λ^0_m \)) and the dissociation constant of a weak electrolyte.
The unit of constant A is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(S\, cm^2\, mol^{-3/2}\,L^{1/2}\)
For a strong electrolyte the molar conductivity falls with concentration according to
\( \displaystyle \Lambda_m = \Lambda^{0}_m - A\,c^{1/2} \)
Quantities can be added or subtracted only when they have the same unit. So the term \( A\,c^{1/2} \) must have the unit of molar conductivity, \( \mathrm{S\,cm^{2}\,mol^{-1}} \), and the unit of A follows by dividing by the unit of \( c^{1/2} \):
\( \displaystyle A = \frac{\Lambda^{0}_m - \Lambda_m}{c^{1/2}} \)
\( \displaystyle \text{unit of } A = \frac{\mathrm{S\,cm^{2}\,mol^{-1}}}{(\mathrm{mol\,L^{-1}})^{1/2}} = \frac{\mathrm{S\,cm^{2}\,mol^{-1}}}{\mathrm{mol^{1/2}\,L^{-1/2}}} \)
\( \displaystyle \text{unit of } A = \mathrm{S\,cm^{2}\,mol^{-1-1/2}\,L^{1/2}} = \mathrm{S\,cm^{2}\,mol^{-3/2}\,L^{1/2}} \)
So the unit of the constant A is S cm2 mol−3/2 L1/2.
Shortcut: Dividing by \( c^{1/2} \) adds −1/2 to the power of mol and +1/2 to the power of L. Only Option (3) has mol−3/2 together with L+1/2.