The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H₂C₂O₄.2H₂O) in 250 mL of water in mol L⁻¹ is x × 10⁻². The value of x is _____________. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0]

Solution

Chemistry ma’am · 3 steps

Take the molar mass of the whole crystal, water of crystallisation included, then divide by litres.

1. Molar mass. The solid you weigh is the dihydrate, so its two waters count. H₂C₂O₄.2H₂O has 2 + 24 + 64 + 36 = 126 g mol⁻¹.
2. Moles of acid. n = 6.3/126 = 0.05 mol. Each mole of dihydrate gives one mole of oxalic acid in solution.

   M = n/(V (L)) = 0.05/0.250 = 0.20 mol L⁻¹

3. So. 0.20 mol L⁻¹ = 20 × 10⁻² mol L⁻¹, so x = 20.

The common slip: 28. Using the anhydrous molar mass of 90 g mol⁻¹ gives 0.07 mol and 0.28 mol L⁻¹, so x = 28. The 2H₂O must be counted.

The idea to keep. For a hydrated solid, the mass you weigh includes the water of crystallisation, so use the full hydrate formula for the molar mass. Always convert mL to L before dividing.

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