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

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%.