Question

A sample of drinking water was found to be severely contaminated with chloroform (chcl3), supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass) a. Express this in percent by mass b. Determine the molality of chloroform in water sample.

Answer 1

Answer:Chloroform is present by 0.0014 % in a water sample.

Molality of chloroform in water sample is 0.000125 mol/kg.

Explanation:

The level of contamination = 15 ppm

This means that 15 mg of chloroform is present in 1 L of the water.

Density of water= 1g/mL

Mass of water = Density × Volume = 1 g/mL × 1000 mL = 1000 g = 1 kg

Mass of chloroform in the water = 15 mg = 0.015 g

Total mass of the solution = 1000 g+ 0.015 g = 1000.015 g

$Mass\%=\frac{0.015 g}{1000.015 g}\times 100=0.0014\%$

$Molality=\frac{0.015 g}{119.38 g/mol\times 1 kg}=0.000125 mol/kg$

Chloroform is present by 0.0014 % in a water sample.

Molality of chloroform in water sample is 0.000125 mol/kg.