JEE Main 2026 Apr 6 Shift 1, Chemistry Q8: Abnormal Molar Masses

JEE Main2026Apr 6, Shift 1Chemistry

Consider the following data:

Electrolyte	$Λ_{m}^{\circ}$ (S cm $^{2}$ mol $^{- 1}$ )
BaCl $_{2}$	$x_{1}$
H $_{2}$ SO $_{4}$	$x_{2}$
HCl	$x_{3}$

BaSO $_{4}$ is sparingly soluble in water. If the conductivity of the saturated BaSO $_{4}$ solution is $x$ S cm $^{- 1}$ then the solubility product of BaSO $_{4}$ can be given as.

(Here $Λ_{m} = α Λ_{m}^{\circ}$ )

A
$\frac{1 0 ^{6} x ^{2}}{α ^{2} ( x _{1} + x _{2} - 2 x _{3} ) ^{2}}$
B
$\frac{x ^{2}}{( x _{1} + x _{2} - 2 x _{3} ) ^{2}}$
C
$\frac{α ^{2} ( x _{1} + x _{2} - 2 x _{3} ) ^{2}}{1 0 ^{6} x ^{2}}$
D
$\frac{x ^{2}}{( x _{1} + x _{2} + 2 x _{3} ) ^{2}}$

Solution

Apply Kohlrausch's law to obtain the limiting molar conductivity of BaSO $_{4}$ :

$Λ_{m}^{\circ} (BaSO_{4}) = Λ_{m}^{\circ} (BaCl_{2}) + Λ_{m}^{\circ} (H_{2} SO_{4}) - 2 Λ_{m}^{\circ} (HCl) = x_{1} + x_{2} - 2 x_{3}$ .

Given $Λ_{m} = α Λ_{m}^{\circ}$ , and the standard relation $Λ_{m} = \frac{κ \cdot 1000}{S}$ (with $κ = x$ S cm $^{- 1}$ and $S$ in mol L $^{- 1}$ ):

$α (x_{1} + x_{2} - 2 x_{3}) = \frac{1000 x}{S} \Rightarrow S = \frac{1000 x}{α ( x _{1} + x _{2} - 2 x _{3} )}$ .

For BaSO $_{4} \to Ba^{2 +} + SO_{4}^{2 -}$ , $K_{s p} = S^{2} = \frac{1 0 ^{6} x ^{2}}{α ^{2} ( x _{1} + x _{2} - 2 x _{3} ) ^{2}}$ .

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JEE Main 2026 Apr 6 Shift 1, Chemistry Q8: Abnormal Molar Masses

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