Question

A sample of drinking water was found to be severely contaminated with chloroform, $CHC{l}_3$, supposed to be carcinogenic in nature. The level of contamination was $15ppm\text{mass}$.


$A$: Express this in percent by mass.


$B$: Determine the molality of chloroform in the water sample.

Answer 1

$A$:

$1ppm$ means in million ${10}^6$ parts.

So, $\text{percent by mass}=\frac{15}{{10}^6}\times 100$
$=15\times {10}^{-4}$
$1.5\times {10}^{-3}\%$

$B$:


$\text{molality}=\frac{\text{mole of solute (mole)}}{\text{mass of solvent(kg)}}$

$\text{molality}\left(m\right)=\frac{\text{mass of solute}\left(CHC{l}_3\right)}{\text{molar mass}CHC{l}_3\times \text{mass of solvent}\left(kg\right)}$

Molar mass of Cholroform$\left(CHC{l}_3\right)=119.5gmo{l}^{-1}$

So, we have calculated in part 1 that percentage by mass of chloroform is $1.5\times {10}^{-3}$
Hence in $100g$ sample, there will be $1.5\times {10}^{-3}g$ of cholorform

Therefore $1000g\left(1kg\right)$ of the sample will contain choloform $1.5\times {10}^{-2}g$

So, molality $=\frac{1.5\times {10}^{-2}}{119.5}=1.25\times {10}^{-4}m$

Final Answer: Molality of choloform in water sample is $1.25\times {10}^{-4}m$