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.
If the molar conductance values of \( Ca^{2+} \) and \( Cl^- \) at infinite dilution are, respectively, \( 118.8 × 10^{-4}\, ohm^{-1}\, m^2\, mol^{-1} \) and \( 77.33 × 10^{-4}\, ohm^{-1}\, m^2\, mol^{-1} \), then the molar conductance of \( CaCl_2 \) at infinite dilution will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(273.46\, ohm^{-1}\, m^2\, mol^{-1}\)
Kohlrausch’s law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the contributions of its cation and anion, each multiplied by the number of such ions in one formula unit:
\( \displaystyle \Lambda^{0}_m = \nu_{+}\lambda^{0}_{+} + \nu_{-}\lambda^{0}_{-} \)
Calcium chloride gives one \( \ce{Ca^2+} \) ion and two \( \ce{Cl-} \) ions:
\( \displaystyle \ce{CaCl2 -> Ca^2+ + 2Cl-} \)
So \( \nu_{+} = 1 \) and \( \nu_{-} = 2 \):
\( \displaystyle \Lambda^{0}_m(\ce{CaCl2}) = \lambda^{0}(\ce{Ca^2+}) + 2\,\lambda^{0}(\ce{Cl-}) \)
\( \displaystyle \Lambda^{0}_m = (118.8 + 2 \times 77.33) \times 10^{-4}\ \mathrm{ohm^{-1}\,m^{2}\,mol^{-1}} \)
\( \displaystyle \Lambda^{0}_m = 273.46 \times 10^{-4}\ \mathrm{ohm^{-1}\,m^{2}\,mol^{-1}} \)
The number 273.46 is the one given in Option (1).
Note: It seems that there is a mistake in the original NTA paper, which prints the options without the factor 10−4 (273.46 ohm−1 m2 mol−1) when it should read 273.46 × 10−4 ohm−1 m2 mol−1.