Question

A solution has been prepared by dissolving 0.49g of sulphuric acid in 450 ml of solution. Find the normality of solution?

Answer 1

Answer:

H - 2x1.00=2.00

S - 1x32.06=32.06

O - 4x16.00=64.00

64.00+32.06+2.00=98.06

so, 0.49grams/98.06grams per mol = 5*10^-3mol of H2SO4

and

5*10^-3 mol H2SO4/100mL H2O = 0.05 Molar

and since H2SO4 has two Hydrogen atoms

0.05M x 2N per M = 0.1 Normal

Hope it helps!

Pls mark as brainliest!

Answer 2

Explanation:

H - 2x1.00=2.00

S - 1x32.06=32.06

O - 4x16.00=64.00

64.00+32.06+2.00=98.06

so, 0.49grams/98.06grams per mol = 5*10^-3mol of H2SO4

and

5*10^-3 mol H2SO4/100mL H2O = 0.05 Molar

and since H2SO4 has two Hydrogen atoms

0.05M x 2N per M = 0.1 Normal

Hope it helps

Pls mark me a brainliest