3% H2O2 by mass solution 3g of H2O2 present in 100g of
solution
volume of solution = mass of solution/density
                        Â
= 100/1.01Â Â = 99ml
99ml of solution contains 3g of H2O2
18.95ml of solution contains = 3*18.95/99 = 0.574g
no of moles of H2O2 = W/G.M.Wt
                              Â
= 0.574/34Â Â = 0.0168moles
2H2O2(aq)→2H2O(l)+O2(g)
2 moles of H2O2 decomposes to gives 1 mole of O2
0.0168moles of H2O2 decomposes to gives = 1*0.0168/2Â Â
= 0.0084 moles of O2
P = 5 bar = 5*0.986923atm  = 4.934615atm
T = 27+273 = 300K
n = 0.0084moles
PV = nRT
VÂ Â = nRT/P
     =
0.0084*0.0821*300/4.934615Â Â = 0.042L
>>>answer
0.042L of O2 >>>>answer
