Question

A sample of drinking water was found to be severely contaminated with chloroform $\left( {CHC{l_3}} \right)$ supposed to be a carcinogen. The level of contamination was $15ppm$ (by mass):
A. Express this in percent by mass
B. Determine the molality of chloroform in the water sample

Answer 1

Hint: The level of contamination is given in $ppm$ where $1ppm$ stands for $1$ part out of the one million parts i.e. ${10^6}$ parts. By using this information, one can calculate the mass percent by using the relevant formula. To calculate molality one can use the data that it depends on the mass of the solvent in grams. Here one needs to find out the molality of chloroform. The number of moles to be found out needs to be as per $1000g$ of solvent. One can put these values in the molality equation and find out the solution.

Complete step by step answer:
-First, we will analyze the level of contamination by mass is $15ppm$ which stands for $15$ parts per million. It can be written as $15$ parts per ${10^6}$ part.
- Now to calculate the percent by mass, means we need to find the percentage of the parts with respect to total parts which is shown below in mathematical form,
${\text{Mass percent of chloroform = }}\dfrac{{{\text{Parts of chloroform present}}}}{{{\text{Total parts }}}} \times 100$
${\text{Mass percent of chloroform = }}\dfrac{{15}}{{{{10}^6}}} \times 100 = 1.5 \times {10^{ - 3}}\% $
Hence, the mass percent of $15ppm$ chloroform is $1.5 \times {10^{ - 3}}\% $
- Now let’s calculate the molality of chloroform in the water sample, the formula for molality is,
${\text{Molality}} = \dfrac{{{\text{Number of moles of solute}}}}{{{\text{Mass of solvent in grams}}}} \times 1000$
Here, we need to find out the mass of chloroform,
${\text{Mass of chloroform(CHC}}{{\text{l}}_3}{\text{) = 12 + 1 + (3}} \times {\text{35}}{\text{.5) = 119}}{\text{.5 g/mol}}$
As in the above step, we calculated the percent of chloroform that means, $100g$ the sample has the $1.5 \times {10^{ - 3}}g$ chloroform.
Therefore, the $1000g$ sample will have $(1.5 \times {10^{ - 3}}) \times 1000 = 1.5 \times {10^{ - 2}}g$ of chloroform.
By putting the values in the molality equation we get,
${\text{Molality = }}\dfrac{{1.5 \times {{10}^{ - 3}}}}{{119.5}} \times 1000 = 1.25 \times {10^{ - 4}}mol/kg$
Therefore, the molality of chloroform in the water sample is $1.25 \times {10^{ - 4}}mol/kg$.

Note: The percent mass calculated is always taken $100g$ as the total mass. While calculating the molality one needs to remember that we use the mass of solvent in grams and not in liters or milliliters. Molality always has the gram factor in it at the denominator of the equation. The unit we used for molality, $mol/kg$ is the $SI$ unit, one can use simple $molal$ or $m$ as the unit for molality.