Question

A sample of drinking water severely contaminated with chloroform , CHCl3 ,is supposed to be carcinogenic in nature . The level of combination was 15 ppm (by mass) .What is the molality of chloroform in the water sample ?

Answer 1

`"15 ppm" ~=` `(15xx10^-3*g)/(119.37*g*mol^-1)xx1/(1*kg)` `=` `??*mol*kg^-1`

Explanation:

`"1 ppm"` `=` `1*mg*L^-1`, by definition.

If there is a `"15 ppm"` solution of chloroform in water, this represents, `(15xx10^-3*g)/(119.37*g*mol^-1)xx1/(1*kg)` `=` `??*mol*kg^-1`, with respect to chloroform.

You might object to this treatment in that the chloroform solution will have a slightly different density than pure water, but it will have a negligible effect. This is quite probably concentrated enough to taste.

Answer 2

The molality of chloroform is `1.2 × 10^"-4"color(white)(l) "mol/kg"`.

Explanation:

The formula for molality is

`color(blue)(|bar(ul(color(white)(a/a) "Molality" = "moles of solute"/"kilograms of solute"color(white)(a/a)|)))" "`

Assume that you have 1 L of water. Then the mass of water is 1000 g.

`"Mass of CHCl"_3 = 1000 color(red)(cancel(color(black)("g water"))) × "15 g CHCl"_3/(10^6 color(red)(cancel(color(black)("g water")))) = "0.015 g CHCl"_3`

`"Moles of CHCl"_3 = 0.015 color(red)(cancel(color(black)("g CHCl"_3))) × "1 mol CHCl"_3/(119.38 color(red)(cancel(color(black)("g CHCl"_3))) ) = 1.2 × 10^"-4"color(white)(l) "mol CHCl"_3`

`"Molality" = (1.2 × 10^"-4" color(white)(l)"mol")/"1 kg" = 1.2 × 10^"-4" color(white)(l)"mol/kg"`