A stock solution is prepared by dissolving an unknown mass w of oxalic acid dihydrate ( H ₂ C ₂ O ₄ 2 H ₂ O ) in water to make 250 mL of solution. A 25 mL aliquot of this stock solution is taken and diluted with water to a final volume of 200 mL . If the molarity of the resulting diluted solution is 0.25 M , what is the value of w (in grams)? Given: Atomic masses of H , C , and O are 1 , 12 , and 16 g mol ⁻¹ respecti

Step-by-step solution

Let the molarity of the stock solution be M₁ . Using the dilution equation M₁V₁ = M₂V₂ for the aliquot: M₁ 25 mL = 0.25 M 200 mL M₁ = 50 25 = 2.0 M The stock solution has a volume of 250 mL ( 0.25 L ). The number of moles of oxalic acid dihydrate in the stock solution is: n = M₁ V_ stock = 2.0 mol L ⁻¹ 0.25 L = 0.5 mol The molar mass of oxalic acid dihydrate ( H ₂ C ₂ O ₄ 2 H ₂ O ) is: 2(1) + 2(12) + 4(16) + 2(18) = 2 + 24 + 64 + 36 = 126 g mol ⁻¹ The mass w dissolved is: w = n Molar mass = 0.5 mol 126 g mol ⁻¹ = 6