ChemistryJEE Main 6 Apr 2026, Shift 1·Q8·MCQ·Moderate

Electrochemistry · Kohlrausch law and solubility product

Consider the following data :

Electrolyte	$\boldsymbol{\Lambda^\circ_\mathrm{m}}$ (S cm $^{\mathbf{2}}$ mol $^{\mathbf{-1}}$ )
$\ce{BaCl2}$	$x_1$
$\ce{H2SO4}$	$x_2$
HCl	$x_3$

$\ce{BaSO4}$ is sparingly soluble in water. If the conductivity of the saturated $\ce{BaSO4}$ solution is $x$ S cm $^{-1}$ then the solubility product of $\ce{BaSO4}$ can be given as.

(Here $\Lambda_\mathrm{m} = \Lambda^\circ_\mathrm{m}$ )

(1) $\dfrac{10^{6}x^{2}}{\alpha^{2}(x_1+x_2-2x_3)^{2}}$
(2) $\dfrac{x^{2}}{(x_1+x_2-2x_3)^{2}}$
(3) $\dfrac{\alpha^{2}(x_1+x_2-2x_3)^{2}}{10^{6}x^{2}}$
(4) $\dfrac{x^{2}}{(x_1+x_2+2x_3)^{2}}$

We corrected a printing error in the original paper

Source printed “(Here $\wedge_\mathrm{m} = \wedge^\circ_\mathrm{m}$ )” (logical-and sign for the Greek $\Lambda$ of the table).

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