An aqueous solution of 6.3 g of oxalic acid dihydrate ( ) is made up to 250 mL. The volume of 0.1 N — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Molar mass of $H_2C_2O_4 \cdot 2H_2O = 126$ g/mol. Moles = 6.3 / 126 = 0.05. Molarity of 250 mL solution = 0.05 / 0.25 L = 0.2 M. Normality (n-factor for acid = 2) = $0.2 \times 2 = 0.4 N$ . By equivalence: $N_{acid} V_{acid} = N_{base} V_{base} \Rightarrow 0.4 \times 10 = 0.1 \times V_{base} \Rightarrow V_{base} = 40$ mL.