Question

(5×22.4)/3 litres of a mixture of O2 and O3 weighs 64 grams at stp. Then volume of O2 will be

Answer 1

22.4 litres of any gas simply means one mole of that gas.
(as it is also said that 1 mole of any gas occupies 22.4 litres of volume)

it is given 5/3(22.4) which means 5/3 moles of a mixture of O2 and O3 is present at STP.

let us consider 'x' as the number of moles of oxygen ......hence '(5/3-x)' becomes number of moles of ozone.
molecular weights of O2 and O3 are 32 and 48 respectively.

now,
             32(x) +  48(5/3 - x)  =  64
             32x  +  80  -  48x   = 64
             16x  = 16
                 x  =  1

therefore 1 mole of oxygen is present in the mixture which is 22.4 litres of O2 at STP.