Question

6.3 g oxalic acid dihydrate (C2H2O4·2H2O) was dissolved in 250 mL water. Molarity of the solution is

Answer 1

Answer:

Since oxalic acid is a dicarboxylic acid, 1 mole of acid reacts with 2 moles of NaOH. So by actual experiment 0.0179 x (1/2) = 0.00895 mol of acid was present. (. 98 g) / (90.0352 g H2C2O4/mol) = 0.01088 mol H2C2O4 is what was stared with, which compares roughly with what was determined by titration.

126.07 g/mol

Oxalic acid dihydrate, H2C2O4 * 2 H2O (molar mass = 126.07 g/mol) is often used as a primary standard for the standardization of an NaOH solution. If 0.147 g of oxalic acid dihydrate are neutralized by 23.64 mL of an NaOH solution, what is the molar concentration of the NaOH solution? Oxalic acid is a diprotic acid.now normality is 0.028/250 . 250 ml is 1/4th of a litre hence 0.028/0.25 which is 0.028 x 4 which gives answer to be 0.112N. But sometimes oxalic acid is taken as h2c204 along with 2h20. hence then molecular msass is 126 g and equivalent mass is 126/2 which is 63.prepare 1000 ml of 1 N Oxalic acid solution, the amount of oxalic acid required = 63 g.

Therefore, to prepare 1000 ml of 0.1 N Oxalic acid solution, the amount of oxalic acid required = 6.3 g.

Answer 2

Given:

Mass of oxalic acid dihydrate, w = 6.3 gm

The volume of water, V = 250 ml

To Find:

The molarity of the solution.

Calculation:

- The molar mass of oxalic acid dihydrate, M.wt = 126 gm/mol

- Molarity = (w × 1000)/(M.wt × V)

⇒ M = (6.3 × 1000)/(126 × 250)

⇒ M = 0.05 × 4

⇒ M = 0.2 M

- So, the molarity of the given solution is 0.2 M.