100ml of 0.1 N H2SO4 by diluting from original H2SO4.

Question

how to solve this? 100ml of 0.1 N H2SO4 by diluting from original H2SO4.

Atomic mass of: O=16 , H=1 S=32

d H2SO4=1.08 purity%=98%

i need the solution.

Answer 1

Prepare 100. ml of 0.1 N H2S04 by diluting from original H2S04. Eq. wt. of H,S04 = 98/2=49 . dH2S04= 1.08 Purity =98%

1 N H2SO4 = 49 g H2SO4 / liter (1 grm equiv wt /L x 1 grm equiv. wt / 2 = 98 / 2 = 49 g)

0.1 N H2SO4 = 4.9 g / L

To make 100 mls (0.100 L) we would need 4.9 g / L x 0.1 L = 0.49 g H2SO4 / 100 mls

Original

density = 1.08 g / ml

with a purity of 98%, this converts to 1.08 g x 0.98 = 1.058 g /ml

0.49 g H2SO4 x 1 ml / 1.058 g = 0.46 mls of original H2SO4 needed

To make the desired solution

Take 0.46 mls of the original H2SO4 and dilute to a final volume of 100 mls.