$0.2 \%(\mathrm{w} / \mathrm{v})$ solution of NaOH is measured to have resistivity $870.0 \mathrm{~m} \Omega \mathrm{~m}$. The molar conductivity of the solution will be \_\_\_\_ $\times 10^{2} \mathrm{mS} \mathrm{dm}^{2} \mathrm{mol}^{-1}$. (Nearest integer)

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Redox Reactions and Electrochemistry

0.2% (w / v)

solution of NaOH is measured to have resistivity

870.0 m Ω m

. The molar conductivity of the solution will be ____

\times 1 0^{2} mS dm^{2} mol^{- 1}

. (Nearest integer)

Answer: 23

0.2% (w/v)

solution contains 2 g of NaOH per litre, so the molarity is

2/40 = 0.05 mol L^{- 1}

. The resistivity is

ρ = 870.0 m Ω m = 0.870 Ω m

, so the conductivity is

κ = 1/ ρ = 1.149 S m^{- 1}

. With concentration

c = 0.05 mol L^{- 1} = 50 mol m^{- 3}

, the molar conductivity is

Λ_{m} = κ / c = 1.149/50 = 0.0230 S m^{2} mol^{- 1}

. Converting, this equals

22.99 \times 1 0^{2} mS dm^{2} mol^{- 1}

, which to the nearest integer is

23 \times 1 0^{2} mS dm^{2} mol^{- 1}

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Integer answer