Question

What is molar conductivity at zero concentration in $\Omega^{-1}$ cm$^2$ mol$^{-1}$ for aluminium sulphate, if molar ionic conductivities at zero concentration of Al$^{3+}$ and SO$_4^{2-}$ are 189 $\Omega^{-1}$ cm$^2$ mol$^{-1}$ and 50.1 $\Omega^{-1}$ cm$^2$ mol$^{-1}$ respectively?

• A. 239.1
• B. 428.1
• C. 478.2
• D. 528.3

Hint:

Always write out the proper chemical formula of the salt first! A common mistake is simply adding the two ionic conductivities together without accounting for the stoichiometric multipliers (2 and 3 in this case).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to calculate the limiting molar conductivity ($\Lambda^\circ_m$) for a salt (aluminium sulphate) given the individual limiting ionic conductivities ($\lambda^\circ$) of its constituent cation and anion.

Step 2: Key Formula or Approach:
According to Kohlrausch's Law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of the cation and anion, each multiplied by the number of ions present in one formula unit of the electrolyte.
$$\Lambda^\circ_m (A_xB_y) = x \lambda^\circ (A^{y+}) + y \lambda^\circ (B^{x-})$$

Step 3: Detailed Explanation:
The chemical formula for aluminium sulphate is $Al_2(SO_4)_3$.
Upon complete dissociation, one formula unit produces 2 aluminium ions ($Al^{3+}$) and 3 sulphate ions ($SO_4^{2-}$).
Apply Kohlrausch's Law:
$$\Lambda^\circ_m [Al_2(SO_4)_3] = 2 \times \lambda^\circ (Al^{3+}) + 3 \times \lambda^\circ (SO_4^{2-})$$
Substitute the given values:
$$\Lambda^\circ_m = 2 \times (189) + 3 \times (50.1)$$
$$\Lambda^\circ_m = 378 + 150.3$$
$$\Lambda^\circ_m = 528.3 \text{ } \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}$$

Step 4: Final Answer:
The molar conductivity is 528.3, matching option (d).