Question

A solution of ethanedioic acid was prepared by dissolving 1.01 g of the acid in water and making the volume up to 250 cm in a volumetric flask. A 25.0 cm sample of this solution was titrated against NaOH and required 18.40 cm NaOH for neutralisation. (COOH)₂ + 2NaOH —— (COONa)₂ + 2H₂O Calculate the number of moles of ethanedioic acid in 1.01 g.

Answer 1

To find the number of moles of ethanedioic acid in 1.01 g, divide the mass by the molecular weight of ethanedioic acid (126.07 g/mol) resulting in 0.00801 moles of ethanedioic acid.

Step-by-step explanation:

To calculate the number of moles of ethanedioic acid in 1.01 g, one must first need the molecular weight of ethanedioic acid (also known as oxalic acid), which is 126.07 g/mol. Using the mass of ethanedioic acid, the number of moles can be calculated using the formula:

moles = mass (g) ÷ molecular weight (g/mol).

Substituting the given values:

moles = 1.01 g ÷ 126.07 g/mol = 0.00801 mol.

Therefore, there are 0.00801 moles of ethanedioic acid in 1.01 g.