Question

calculate the normality of aqueous solution containing 31.5g.of oxalic acid in 1.5 litres of solutions. (molecular weight of oxalic acid is 126​

Answer 1

Answer:

1/3 hope this is correct

Answer 2

1. Question⇒ calculate the normality of aqueous solution containing 31.5g.of oxalic acid in 1.5 litres of solutions. (molecular weight of oxalic acid is 126​?

Explanation:

• Molar mass of the Oxalic acid = 126 g/mole.

• Mass of the Oxalic acid = 3.15 g.

• ∴ No. of moles of the Oxalic Acid = 3.15/126

• = 0.025

• Volume of the Oxalic acid solution. = 250 mL.

• = 0.25 L.

• For Molarity,

• ∵ Molarity = No. of moles of Solute/Volume of the Solution in L.

• ∴ Molarity = 0.025/0.25

• = 0.1 M.

• For Normality,

• Equivalent Mass = 126/2 [∵ Basicity of Oxalic Acid is 2.]

• = 63

• No. of grams equivalent = Mass/Equivalent Mass

• = 3.15/63

• = 0.05

• ∵ Normality = No. of grams Equivalent/Volume of the Solution in L.

• ∴ Normality = 0.05/0.25

• = 0.2 N.

Hope it helps.