Answer the question on basis of passage given below:
\(G = \frac{\kappa A}{l} = \kappa \) (both\(A\) and \(l\) are unity in their appropriate units in m or cm)
Molar conductivity of a solution at a given concentration is the conductance of the volume \(V\) of solution containing one mole of electrolyte kept between two electrodes with area of cross section \(A\) and distance of unit length \(l\). Therefore,
\(\Lambda_m = \frac{\kappa A}{l}\)
Since \(l = 1\) and \(A = V\) (volume containing 1 mole of electrolyte) \(\Lambda _m =\kappa V\)
Calculate \(\Lambda ^o _m\) for aluminium sulphate. When \(\Lambda ^o_m\) for aluminium chloride, sulphuric acid and hydrochloric acid are \(236.2\), \(459.8\) and \(121.4\, \ Scm^2mol^{-1}\) respectively:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1123.4\, \ Scm^2mol^{-1}\)
By Kohlrausch law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the contributions of its ions, each multiplied by the number of ions in the formula: Λ°m = ν+λ°+ + ν−λ°−. So Λ°m of any electrolyte can be built up by adding and subtracting the Λ°m values of other electrolytes, as long as the ions that are not wanted cancel.
\( \ce{Al2(SO4)3} \) → \( \ce{2Al^{3+}} \) + \( \ce{3SO4^{2-}} \), so Λ°m[Al2(SO4)3] = 2λ°(\( \ce{Al^{3+}} \)) + 3λ°(\( \ce{SO4^{2-}} \)).
2Λ°(\( \ce{AlCl3} \)) = 2λ°(\( \ce{Al^{3+}} \)) + 6λ°(\( \ce{Cl^{-}} \))
3Λ°(\( \ce{H2SO4} \)) = 6λ°(\( \ce{H^{+}} \)) + 3λ°(\( \ce{SO4^{2-}} \))
6Λ°(HCl) = 6λ°(\( \ce{H^{+}} \)) + 6λ°(\( \ce{Cl^{-}} \))
Adding the first two and subtracting the third removes the 6 \( \ce{H^{+}} \) and 6 \( \ce{Cl^{-}} \):
Λ°m[Al2(SO4)3] = 2Λ°(\( \ce{AlCl3} \)) + 3Λ°(\( \ce{H2SO4} \)) − 6Λ°(HCl)
= 2(236.2) + 3(459.8) − 6(121.4)
= 472.4 + 1379.4 − 728.4
= 1123.4 S cm2 mol−1
NCERT (2026-27 edition):
Applications of Kohlrausch Law
