Question

a) A sample of water was found to be severely contaminated by Chloroform$(CHC{l_3})$
supposed to be a carcinogen. The level of contamination was 15ppm (by mass).
i) Express this in percent by mass.
ii) Determine the molality of chloroform in the water sample.
b) What role does the molecular interaction play in a solution of alcohol and water?

Answer 1

Hint: Since the ppm (parts per million) amount of Chloroform is given, we can find out the
amount of Chloroform and can use it to find the percentage by mass and molality of the given solution.
For the second question, if we are aware of the proportionality of solubility and molecular weight, we
are able to answer the question.
Formula used:
i) ${\text{Percentage by mass = }}\dfrac{{{\text{Mass of
Chloroform}}}}{{{\text{Mass of solution}}}} \times 100$
ii) Molality
$m = \dfrac{{{\text{Mass of Chloroform}}}}{{{\text{Molar mass of chloroform x (Mass of
solution - Mass of chloroform)}}}} \times 100$

Complete step by step answer:
Let us now find the percentage by mass. For finding it we should need to know the mass of Chloroform
and the mass of solution. We may not have both of them now, but we have the ppm concentration of
Chloroform with which we are able to find both mass of chloroform and the solution.

Given, concentration of chloroform = $15ppm$
$1ppm$of any compound in a solution means $1g$ of the compound is present in ${10^6}g$ of the
given solution. Applying this to the problem there are $15g$ of Chloroform present in ${10^6}g$ of the
given solution.
Therefore, mass of chloroform = $15g$
Mass of the solution = ${10^6}g$
Now, we can find the percentage by mass, by applying the values in the formula,
$\dfrac{{{\text{Mass of Chloroform}}}}{{{\text{Mass of solution}}}} \times 100$
$ = \dfrac{{15}}{{{{10}^6}}} \times 100$
$ = 1.5 \times {10^{ - 3}}\% $
Now, let us find the molality of the given solution. Molality is the number of moles of a given solute per
kg of solvent.
Since we know the formula for molality, and the only parameter missing out is molar mass of
Chloroform. The formula of chloroform is $(CHC{l_3})$ hence the molar mass will be
$ = (12 \times 1) + (1 \times 1) + (35.5 \times 3)$ (Adding the individual molar weight of Carbon,
Hydrogen and Chlorine) whose value is $119.5gmo{l^{ - 1}}$.
We know that the equation for molality,
$m = \dfrac{{{\text{Mass of Chloroform}}}}{{{\text{Molar mass of chloroform }} \times {\text{ (Mass of
solution - Mass of chloroform)}}}} \times 100$
Substituting all the values,
$m = \dfrac{{15}}{{{\text{119}}{\text{.5 x (1}}{{\text{0}}^6}{\text{ - 15 )}}}} \times 100$
$m = 1.255 \times {10^{ - 4}}m$
Now, we have to find both the percentage by mass and molality.
Let us head for the second one.

(b) The lower members of alcohols are highly soluble in water but the solubility decreases with increase
in the molecular weight. The solubility of lower alcohols in water is due to the formation of hydrogen
bonds (Hydrogen bonding) between alcohols & water molecules.

However, as the size of alcohol molecule increases, the alkyl groups becomes larger & prevents the
formation of hydrogen bonds with water & hence the solubility goes on decreasing with increase in the
length of a carbon chain. Also, the interaction between the molecules of alcohol and water is weaker
than alcohol−alcohol and water−water interactions.
As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the
molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the
boiling point of the resulting solution.

Note:
You should always change the ppm into grams so that you will be able to simplify the question .
Always convert the symbols which are convenient for the solution. We can also change the grams back
to ppm if the net total concentration is required in ppm.