A sample of drinking water was found to be severely contaminated with chloroform, `CHCl_(3)`, supposed to be carcinogen. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molality of chloroform in the water sample.

Options

A. `2.317xx10^(-4)m`
B. `0.7864xx10^(-4)`
C. `1.255xx10^(-4)m`
D. `0.5555xx10^(-4)m`

Answer

3

Explanation

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`.