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Chemistry Question 72 – JEE-MAIN 2025

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0.2\% (w/v) solution of NaOH is measured to have resistivity 870.0 mΩ m. The molar conductivity of the solution will be _______ ×102 mS dm2 mol−1. (Nearest integer)

Molar conductivity is a measure of the conducting power of all the ions produced by one mole of an electrolyte dissolved in a solution.

Step 1: Calculate Conductivity (κ)✦ Active

The resistivity ρ is given as 870.0 mΩ m. We convert this to Ω m:

ρ=870.0×10−3 Ω m=0.870 Ω m

Conductivity κ is the reciprocal of resistivity:

κ=1ρ=10.870 Ω m≈1.1494 S m−1
Step 2: Calculate Molar Concentration (C)○ Expand

A 0.2\% (w/v) solution of NaOH means 0.2 g of NaOH is present in 100 mL of solution. The molar mass of NaOH is 23+16+1=40 g/mol.

Moles of NaOH=0.2 g40 g/mol=0.005 mol

The volume of the solution is 100 mL, which is 0.1 L. So, the molarity is:

C=0.005 mol0.1 L=0.05 mol L−1

To use in SI units for molar conductivity, convert molarity to mol m−3:

C=0.05 mol L−1×1000 L/m3=50 mol m−3
Step 3: Calculate Molar Conductivity (Λm) and Convert Units○ Expand

Now, calculate the molar conductivity Λm:

Λm=κC=1.1494 S m−150 mol m−3≈0.022988 S m2 mol−1

The question asks for the value in mS dm2 mol−1 and in the format ×102. We use the conversion factors: 1 S=103 mS and 1 m2=100 dm2.

Λm=0.022988 S m2 mol−1×(103 mS1 S)×(100 dm21 m2)
Λm=0.022988×103×100 mS dm2 mol−1=2298.8 mS dm2 mol−1

To express this in the format ×102 mS dm2 mol−1:

Λm=2298.8100×102 mS dm2 mol−1=22.988×102 mS dm2 mol−1

Rounding to the nearest integer, the value is 23.

💡 Teacher's Secret Hint

Pay close attention to unit conversions, especially for resistivity, conductivity, and molar conductivity, to avoid common errors.

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