Question

A sample of drinking water was found to be severely contaminated with chloroform, ${\text{CHC}}{{\text{l}}_3}$, supposed to be carcinogenic in nature. The level of contamination was $15$ppm (by mass).
(i) Express this percent by mass.
(ii) Determine the molality of chloroform in the water sample.

Answer 1

Hint: We can determine the percent by mass by dividing the percent amount of chloroform from the total amount of the mixture. We can determine the mole of chloroform by using the mole formula. Mole formula gives the relation in mass, molar mass, and moles of a substance. After determining the moles of chloroform, the molality can be determined.

Complete Step by step answer:(i)
The given amount of chloroform is $15$ ppm (by mass). So, $15$ gram chloroform is present in ${10^6}$ gram of solution
The formula of percent by mass is as follows:
${\text{percent}}\,{\text{mass}}\,{\text{ = }}\,\dfrac{{{\text{mass}}\,{\text{of}}\,{\text{chloroform}}}}{{{\text{mass}}\,{\text{of}}\,{\text{solution}}}}{{ \times 100}}$
Substitute $15$gram for mass of chloroform and ${10^6}$gram for mass of solution.
${\text{percent}}\,{\text{mass}}\,{\text{ = }}\,\dfrac{{{\text{15}}}}{{{{10}^6}}}{{ \times 100}}$
${\text{percent}}\,{\text{mass}}\,{\text{ = }}\,{\text{1}}{\text{.5}}\,{{ \times 1}}{{\text{0}}^{ - 3}}$
Therefore, the percent by mass is ${\text{1}}{\text{.5}}\,{{ \times 1}}{{\text{0}}^{ - 3}}$%.

(ii)
We will use the mole formula to determine the mole of chloroform, present in $15$ gram.
The mole formula is as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
The molar mass of the chloroform is$119.5\,{\text{g/mol}}$.
Substitute $119.5\,{\text{g/mol}}$ for molar mass and$15$ gram for mass of iron.
${\text{mole}}\,\,{\text{ = }}\,\dfrac{{{\text{15}}\,{\text{g}}}}{{119.5\,{\text{g/mol}}}}$
${\text{mole}}\,\,{\text{ = }}\,0.1255$
So, $15$ grams of chloroform have $0.1255$ mole chloroform.
We will determine the mass of solvent by subtracting the mass of solute (chloroform) from mass of solution.
${\text{mass}}\,{\text{of}}\,{\text{solvent}}\,{\text{ = }}\,{\text{1}}{{\text{0}}^{\text{6}}}\,{\text{g}}\, - \,{\text{15}}\,{\text{g}}$
$\Rightarrow {\text{mass}}\,{\text{of}}\,{\text{solvent}}\,{\text{ = }}\,{\text{999985}}\,{\text{g}}$
So, the mass of solvent is ${\text{999985}}\,{\text{g}}$.
Convert the mass of solvent form gram to kg as follows:
${\text{1000}}\,{\text{g}}\,{\text{ = }}\,{\text{1}}\,{\text{kg}}$
$\Rightarrow {\text{999985}}\,{\text{g}}\,{\text{ = }}\,{\text{999}}{\text{.985}}\,{\text{kg}}$
Use the molality formula to determine the molality of chloroform solution as follows:
${\text{Molality}}\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{kgof}}\,{\text{solvent}}}}$
Substitute ${\text{999}}{\text{.985}}\,{\text{kg}}$ for kg of solvent and $0.1255$ for mole of solute.
${\text{Molality}}\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.1255}}\,{\text{mol}}}}{{{\text{999}}{\text{.985}}\,{\text{kg}}}}$
$\Rightarrow {\text{Molality}}\,{\text{ = }}\,1.255\, \times {10^{ - 4}}\,{\text{m}}$
Therefore, the molality of chloroform in the water sample $1.255\, \times {10^{ - 4}}\,{\text{m}}$.

Note: Full form of ppm is parts per million. One ppm is equal to ${10^6}$. Mass of the solute that is $15$ gram is very-very less than the mass of solvent that is ${10^6}$ gram so, the mass of solute can be neglected. The mass of solution can be taken as the mass of solvent. Molality is defined as the moles of solute present in kg of the solvent.