`15 g` sample of an alloy containing `Cu` and `Zn` reacts completely with `3M HNO_(3)` as,
`Cu+HNO_(3)rarrCu^(2+)+NO_(2(g))+H_(2)O`
`Zn+HNO_(3)rarrZn^(2+)+NH_(4)^(+)+H_(2)O`
The liberated `NO_(2(g))` was found to be `4.647 "litre"` at `1 atm` and `300 K`. Find the amount of zince ( to the closest value) in alloy. `(Cu-63.6, R=0.0821)`

Answer

9

Explanation

Eq. of `Cu=(w)/(63.6//2)`
`=Eq."of" NO_(2)=("mole of" NO_(2))/(1)`
`=(1xx4.647)/(0.0821xx300)=0.19`
`w_(Cu)=6.15`
`w_(Zn)=8.86approx9`