JEE Main 2025 Jan 28 Shift 1, Chemistry Q21: Solutions

JEE Main2025Jan 28, Shift 1Chemistry

Given below is the plot of the molar conductivity vs $concentration$ for KCl in aqueous solution.

If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100 $Ω$ , then the resistance of the same cell with the dilute solution is ' $x$ ' $Ω$ .

The value of $x$ is _________ (Nearest integer)

Your answer

Solution

From the plot, the higher concentration corresponds to $C = 0.15$ ( $C = 0.0225$ ) with $Λ_{m} = 100$ S cm $^{2}$ mol $^{- 1}$ , and the dilute solution corresponds to $C = 0.1$ ( $C = 0.01$ ) with $Λ_{m} = 150$ S cm $^{2}$ mol $^{- 1}$ .

For the same cell, the cell constant $G^{*} = ℓ / A$ is fixed. Using $κ = \frac{G ^{*}}{R}$ and $Λ_{m} = \frac{κ \cdot 1000}{C}$ :

$\frac{κ _{c}}{κ _{d}} = \frac{R _{d}}{R _{c}} = \frac{Λ _{m, c} C _{c}}{Λ _{m, d} C _{d}}$

$\frac{R _{d}}{100} = \frac{100 \times ( 0.15 ) ^{2}}{150 \times ( 0.1 ) ^{2}} = \frac{100 \times 0.0225}{150 \times 0.01} = \frac{2.25}{1.5} = 1.5$

$∴ R_{d} = 150 Ω$ .

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