Calculate the volume of occupied by 66.0g of carbon dioxide of `(CO_(2))` under N.T.P conditions.

Explanation

Gram molecular mass of `CO_(2)=12+2xx16=44g`.
`"44 g of "CO_(2) "correspond to mass"="1 gram mol"`
`"66g of "CO_(2) "correspond to mass"=(("1 gram mol")xx(66g))/((44g))=1.5"gram mol"`
`"Now, 1 gram mole of "CO_(2)"under N.T.P. conditions occupy"=22.4L`
1.5gram mole of `CO_(2)` under N.T.P conditions occupy `=((22.4L))/((1.0"gram mol"))xx(1.5"gram mol")`
=33.6L