Multiple choice

250 ml of a solution contains 6.3 grams of oxalic acid(mol. wt. =126). What is the volume (in litres) of water to be added to this solution to make it a 0.1 N solution?

  1. $750$
  2. $7.5$
  3. $0.075$
  4. $0.75$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

6.3g of oxalic acid (mol wt 126) is 0.05 moles. Since it is a dibasic acid, equivalent weight = 63. Equivalents = 6.3 / 63 = 0.1 eq. Normality = 0.1 eq / 0.25 L = 0.4 N. To make it 0.1 N, V2 = N1V1 / N2 = (0.4 * 0.25) / 0.1 = 1.0 L. Water to add = 1.0 - 0.25 = 0.75 L.

AI explanation

Since the equivalent weight of oxalic acid is half of its molecular weight, we calculate it as 126 divided by 2, which equals 63, allowing us to find the initial equivalents of the acid by dividing 6.3 grams by 63 to get 0.1 equivalents. For the final solution to have a normality of 0.1 N, the total volume in liters must be the number of equivalents divided by the normality, which is 0.1 divided by 0.1, meaning 1 liter. Subtracting the initial 0.25 liters from the required 1 liter gives the volume of water to be added as 0.75 liters.

Keep practicing — related questions