# 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 ?

May 17, 2016

$\text{15 ppm} \cong$ $\frac{15 \times {10}^{-} 3 \cdot g}{119.37 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{1 \cdot k g}$ $=$ ??*mol*kg^-1

#### Explanation:

$\text{1 ppm}$ $=$ $1 \cdot m g \cdot {L}^{-} 1$, by definition.

If there is a $\text{15 ppm}$ solution of chloroform in water, this represents, $\frac{15 \times {10}^{-} 3 \cdot g}{119.37 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{1 \cdot k g}$ $=$ ??*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.

May 17, 2016

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.

${\text{Mass of CHCl"_3 = 1000 color(red)(cancel(color(black)("g water"))) × "15 g CHCl"_3/(10^6 color(red)(cancel(color(black)("g water")))) = "0.015 g CHCl}}_{3}$

${\text{Moles of CHCl"_3 = 0.015 color(red)(cancel(color(black)("g CHCl"_3))) × "1 mol CHCl"_3/(119.38 color(red)(cancel(color(black)("g CHCl"_3))) ) = 1.2 × 10^"-4"color(white)(l) "mol CHCl}}_{3}$

$\text{Molality" = (1.2 × 10^"-4" color(white)(l)"mol")/"1 kg" = 1.2 × 10^"-4" color(white)(l)"mol/kg}$