JEE Main 2026 · 6 Apr · Shift 1 · Chemistry Q58

← JEE Main 2026 · 6 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 58 of 75

Consider the following data.

Electrolyte	$\Lambda^\circ_m$ (S cm² mol $^{-1}$ )
$\ce{BaCl2}$	$x_1$
$\ce{H2SO4}$	$x_2$
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 $\Lambda_m=\Lambda^\circ_m$ )

\dfrac{10^6x^2}{\alpha^2\left(x_1+x_2-2x_3\right)^2}

\dfrac{x^2}{\left(x_1+x_2-2x_3\right)^2}

\dfrac{\alpha^2\left(x_1+x_2-2x_3\right)^2}{10^6x^2}

\dfrac{x^2}{\left(x_1+x_2+2x_3\right)^2}

Answer and worked solution

Official answer: A

Limiting molar conductivity.
$\Lambda^\circ(\ce{BaSO4})=\Lambda^\circ(\ce{BaCl2})+\Lambda^\circ(\ce{H2SO4})-2\Lambda^\circ(\ce{HCl})=x_1+x_2-2x_3.$

Solubility. $s=\dfrac{1000x}{x_1+x_2-2x_3}$ mol L $^{-1}$ .

Solubility product.
$K_{sp}=s^2=\frac{10^6x^2}{(x_1+x_2-2x_3)^2}.$

This is option (A). Its α² is not defined in the question; read it as 1.

(A)

Topics:Conductance and Kohlrausch's law Solubility product

← Previous question Next question →

JEE Main 2026 · 6 Apr · Shift 1, Chemistry question 58