StreakPeaked· Practice

ExamsJEE AdvancedChemistry

Calculate the molarity of a solution prepared by dissolving 6.3 g of oxalic acid dihydrate (H2C2O4.2H2O) in 250 mL of water. Express the answer as x * 10⁻² mol/L and give the nearest integer value of x. (Atomic masses: H = 1.0, C = 12.0, O = 16.0)

  1. 20
  2. 25
  3. 10
  4. 40

Correct answer: 20

Solution

Moles = 6.3/126 = 0.05 mol; molarity = 0.05/0.25 = 0.20 M = 20 * 10⁻², so x = 20.

Related JEE Advanced Chemistry questions

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →