JEE Main 2026 Apr 5 Shift 2, Chemistry Q24: Types Concentration And Solubility of Solutions

JEE Main2026Apr 5, Shift 2Chemistry

At 298 K, the molar conductivity of $x %$ (w/w) MX solution (aqueous) is 123.5 S cm $^{2}$ mol $^{- 1}$ . The conductance of the same solution is $1.9 \times 1 0^{- 3}$ S. The value of $x$ is ______ $\times 1 0^{- 2}$ .

(Given: cell constant $= 1.3$ cm $^{- 1}$ ; molar mass of MX is 75 g mol $^{- 1}$ ; density of aqueous solution of MX at 298 K is 1.0 g mL $^{- 1}$ .)

Your answer

Solution

Specific conductivity: $κ = G \times (cell constant) = 1.9 \times 1 0^{- 3} \times 1.3 = 2.47 \times 1 0^{- 3}$ S cm $^{- 1}$ .

$Λ_{m} = \frac{1000 κ}{c}$ (c in mol/L), so $c = \frac{1000 \times 2.47 \times 1 0 ^{- 3}}{123.5} = 0.02$ M.

Mass of MX in 1 L $= 0.02 \times 75 = 1.5$ g. Mass of solution (density 1.0 g/mL) $= 1000$ g.

$%$ w/w $= \frac{1.5}{1000} \times 100 = 0.15% = 15 \times 1 0^{- 2} %$ .

$∴ x = 15$ .

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JEE Main 2026 Apr 5 Shift 2, Chemistry Q24: Types Concentration And Solubility of Solutions

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