JEE Main · 2022 · Shift-IIeasyEC-105

The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is m1 and that of 20 moles of…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is Λm1\Lambda_{m1} and that of 20 moles of another identical cell having 80 mL NaCl solution is Λm2\Lambda_{m2}. The conductivities exhibited by these two cells are same. The relationship between Λm2\Lambda_{m2} and Λm1\Lambda_{m1} is

Options
  1. a

    Λm2=2Λm1\Lambda_{m2} = 2\Lambda_{m1}

    ✓
  2. b

    Λm2=Λm1/2\Lambda_{m2} = \Lambda_{m1}/2

  3. c

    Λm2=Λm1\Lambda_{m2} = \Lambda_{m1}

  4. d

    Λm2=4Λm1\Lambda_{m2} = 4\Lambda_{m1}

Correct Answera

Λm2=2Λm1\Lambda_{m2} = 2\Lambda_{m1}

Detailed Solution

Strategy: Use the relation Λm=(1000⋅κ)/C\Lambda_m = (1000 \cdot \kappa) / C. Sin\ce κ\kappa is constant for both cells, the ratio of molar conductivities is inversely proportional to the ratio of their concentrations.

Step 1: Calculate Concentrations C1=10 moles/0.020 L=500 MC_1 = 10 \text{ moles} / 0.020 \text{ L} = 500 \text{ M} C2=20 moles/0.080 L=250 MC_2 = 20 \text{ moles} / 0.080 \text{ L} = 250 \text{ M}

Step 2: Compare Λm\Lambda_m Λm1=1000κ500\Lambda_{m1} = \frac{1000 \kappa}{500} Λm2=1000κ250\Lambda_{m2} = \frac{1000 \kappa}{250}

Step 3: Find the relationship Λm2Λm1=500250=2  ⟹  Λm2=2Λm1\frac{\Lambda_{m2}}{\Lambda_{m1}} = \frac{500}{250} = 2 \implies \Lambda_{m2} = 2\Lambda_{m1}

Answer: (a) Λm2=2Λm1\boxed{\text{Answer: (a) } \Lambda_{m2} = 2\Lambda_{m1}}

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