A balloon which can be inflated upto 1 L is partially inflated with air and has a volume of 275 mL and contains 0.0120 mole of air. A piece of dry ice (solid `CO_(2)`) weighing 1.0 g is introduced into the vaporised, assuming that temperature and pressure remain constant. calculate the volume of balloon after addition of dry ice.

Explanation

1 g of `CO_(2)=(1)/(44)` mole =0.0227 mole (Molecular mass of `CO_(2)`=44)
Moles of air originally present=0.0120 mole
`:.` Total moles now present in the balloon =0.012+0.0227=0.0347
Thus, `" " n_(1)=0.012, n_(2)=0.03417`
Applying `" "` PV=nRT, as P and T remain constant
`(V_(1))/(V_(2))=(n_(1))/(n_(2))" "or" " V_(2)=V_(1)xx(n_(2))/(n_(1))=275xx(0.0347)/(0.0120)=795" mL"`