Correct Answer - 3
15 ppm means `15 g` of `CHCl_(3)` in `10^(6) g (or 1000 kg)` of solution. Since `CHCl_(3)` is present in trace amounts we can (approximately) consider mass of solvent equal to the mass of solvent, i.e., `1000 kg H_(2)O` contains `15 g CHCl_(3) ("or "15 g//119.5 g mol^(-1)=0.1255 mol CHCl_(3))`. Thus, 1 kg of the sample will contain.
`0.1255/1000 mol CHCl_(3)`
`=1.255xx10^(-4) mol CHCl_(3)`
Thus, the molality of the `CHCl_(3)` in the aqueous solution is `1.255xx10^(-4) m`.