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)

Solution

Chemistry ma’am · 3 steps

Same cell, so R ∝ 1/κ: get each κ as Λₘ C, squaring √C first.

1. Read the graph. The dilute solution has √C = 0.1, so C = 0.01 mol L⁻¹ and Λₘ = 150 S cm² mol⁻¹. The concentrated one has √C = 0.15, so C = 0.0225 mol L⁻¹ and Λₘ = 100 S cm² mol⁻¹.
2. Same cell constant. For the same cell the cell constant G^* = κ R is fixed, so R ∝ 1/κ. Also κ = (Λₘ C)/1000, and the factor 1000 cancels in a ratio.

   R_(dil)/R_(conc) = κ_(conc)/κ_(dil) = (100 × 0.0225)/(150 × 0.01) = 2.25/1.5 = 1.5

3. So. R_(dil) = 1.5 × 100 Ω = 150 Ω, so x = 150.

The common slip: 100. Multiplying Λₘ by the axis value √C instead of C gives 100 × 0.15 = 150 × 0.1, so the ratio is 1 and the answer comes out as 100.

The idea to keep. In one cell, κ R stays constant. Conductivity κ depends on both Λₘ and C, so a higher Λₘ alone does not mean a lower resistance.

Answer from the NTA final answer key. Solution by PrepWiser.