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__________× 10² mS dm² mol⁻¹. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Get conductivity and concentration into SI units first, divide, and only then switch to mS dm².

1. Concentration. A 0.2% (w/v) solution has 0.2 g of NaOH in 100 mL, which is 2 g L⁻¹. With M = 40 g mol⁻¹, c = 0.05 mol L⁻¹ = 50 mol m⁻³.
2. Conductivity. The resistivity is ρ = 870.0 mΩ m = 0.870 Ω m. Conductivity is its inverse: κ = 1/0.870 = 1.149 S m⁻¹.

   Λₘ = κ/c = (1.149 S m⁻¹)/(50 mol m⁻³) = 2.299 × 10⁻² S m² mol⁻¹

3. Change units. 1 S = 10³ mS and 1 m² = 10² dm², so 1 S m² = 10⁵ mS dm². That makes Λₘ = 2299 mS dm² mol⁻¹ = 22.99 × 10² mS dm² mol⁻¹.
4. So. The molar conductivity is about 22.99 × 10² mS dm² mol⁻¹, which rounds to 23.

The common slip: 230. Reading 0.2% (w/v) as 0.2 g per litre makes c ten times too small and gives 230.

The idea to keep. For Λₘ = κ / c, put κ in S m⁻¹ and c in mol m⁻³, then convert the answer to the units asked for. And w/v percent means grams per 100 mL.

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