A certain public water supply contained 0.10 parts per billion of chloroform `CHCl_(3)` How many molecules of `CHCl_(3)` would be contained in a 0.05 ml drop of this water ?

Explanation

Volume = 0.05 mL
Amount of `CHCl_(3)` in th drop `=(0.05 xx 0.1)/10^(5) = 5.0 xx 10^(-12)`g
Molecule mass of `CHCl_(3) = 12+1 + 3 xx 35.5`
`=119.5 g mol^(-1)`
Moles of `CHCl_(3)` present `=(5.0 xx 10^(-12))/119.5`
Molecules of `CHCl_(3)` present in the drop
`=(5.0 xx 10^(-12))/119.5 xx 6.022 xx 10^(23) = 2.5 xx 10^(19)`.