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:

\(1123.4\, \ Scm^2mol^{-1}\)

The correct answer is option 4. \(1123.4\, \ Scm^2mol^{-1}\).

The limiting molar conductance \((\Lambda _m^0 )\) for strong and weak electrolytes can be determined by using Kohlrauch's law which states that "the limiting molar conductance of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte."

The limiting molar conductance of the electrolytes given in the question can be represented as

\(\Lambda _{AlCl_3}^0 = \Lambda _{Al^{3+}}^0 + 3\Lambda _{Cl^-}^0\) -------\((i)\)

\(\Lambda _{H_2SO_4}^0 = 2\Lambda _{H^+}^0 + \Lambda _{SO_4^{2-}}^0\) -------\((ii)\)

\(\Lambda _{HCl}^0 = \Lambda _{H^+}^0 + \Lambda _{Cl^-}^0\) -------\((iii)\)

Also, the values of the limiting molar conductance are given in the question which are

\(\Lambda _{AlCl_3}^0 = 236.2\, \ Scm^2mol^{-1}\)

\(\Lambda _{H_2SO_4}^0 = 459.8\, \ Scm^2mol^{-1}\)

\(\Lambda _{HCl}^0 = 121.4\, \ Scm^2mol^{-1}\)

We have to find the limiting molar conductance of \(Al_2(SO_4)_3\) and this can be represented as :

\(\Lambda _{Al_2(SO_4)_3}^0 = 2\Lambda _{Al^{3+}}^0 + 3\Lambda _{SO_4^{2-}}^0\) -------\((iv)\)

Applying Kohlrausch's law  and equating as : \(2 × \text{eq(i)} + 3 × \text{eq(ii)} - 6 × \text{eq(i)}\)we can write

\(2 ×(\Lambda _{Al^{3+}}^0 + 3\Lambda _{Cl^-}^0) + 3 × (2\Lambda _{H^+}^0 + \Lambda _{SO_4^{2-}}^0) - 6 × (\Lambda _{H^+}^0 + \Lambda _{Cl^-}^0) = 2 × (\Lambda _{AlCl_3}^0) + 3 × (\Lambda _{H_2SO_4}^0 ) - 6 × (\Lambda _{HCl}^0)\)

Putting in all the values we can write as :

\(2\Lambda _{Al^{3+}}^0 + 6\Lambda _{Cl^-}^0 + 6\Lambda _{H^+}^0 + 3\Lambda _{SO_4^{2-}}^0 - 6\Lambda _{H^+}^0 - 6\Lambda _{Cl^-}^0= 2 × 236.8 + 3 × 459.8 - 6 × 121.4\)

Cancelling out the common terms we are left with

\(2\Lambda _{Al^{3+}}^0 + 3\Lambda _{SO_4^{2-}}^0 = 472.4 +1379.4 -728.4\)

or, \(\Lambda _{Al_2(SO_4)_3}^0 = 1851.8 - 728.4\)   [From equation \((iv)\)]

\(∴ \, \ \Lambda _{Al_2(SO_4)_3}^0 = 1123.4 \, \ Scm^2mol^{-1}\)