Find the purity percentage of h2so4 with density 1.8, if 5ml of h2so4 is neutralised completely by 84.6 ml of 2n NaOH solution

N1V1 = N2V2

N1=?

V1=5

N2=2

V2=84.6

Apply into the equation

N1*5 = 84.6*2

N1 = 84.6*2/5 = 33.84 N

M = 33.84/2 = 16.92 M

16.92 moles of H2SO4 is present in the 1000 ml of the soln.

Density = mass/volume then,

mass = density x volume = 1.8*1000 = 1800g.

hence the mass of H2SO4 =16.92*98 = 1658.16g.

Percentage purity of H2SO4 = (1658.16/1800)*100 = 92.12%

therefore the percentage purity for the H2SO4 in the solution is 92% approximately

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