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he purity percent of h2so4 with density 1.8 if five ml of it is neutralised competely by 84.6 ml of 2N NAOH solution

he purity percent of h2so4 with density 1.8 if five ml of it is neutralised competely by 84.6 ml of 2N NAOH solution

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
From N1V1 = N2V2 = constant we get,
N1*5 = 84.6*2
N1 = > 84.6*2/5 => 33.84 N
M = N/v.f => 33.84/2 => 16.92 M
that means 16.92 moles of H2SO4 is present in the 1000 ml of the soln.
mass of soln. => density of soln. x volme of soln. => 1.8*1000 => 1800 gms.
and mass of H2SO4 => 16.92*98 => 1658.16 gms.
Percentage purity of H2SO4 => (1658.16/1800)*100 = 92.12%
Henc ethe percentage purity for the H2SO4 in the soln is approx 92%.

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