Q133CP - Chemistry: Molecular Nature Of Matter And Change

Question

Water treatment plants commonly use chlorination to destroy bacteria. A by-product is chloroform  $\mathbf{\left(}{\mathbf{\text{CHCl}}}_{\mathbf{\text{3}}}\mathbf{\right)}$, a suspected carcinogen, produced when $\mathbf{\text{HOCl}}$, formed by reaction of ${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}$ and water, reacts with dissolved organic matter. The United States, Canada, and the World Health Organization have set a limit of $\mathbf{\text{100 ppb}}$ of ${\mathbf{\text{CHCl}}}_{\mathbf{\text{3}}}$ in drinking water. Convert this concentration into molarity, molality, mole fraction, and mass percent.

Step-by-Step Solution

Answer

Thee molarity of the solution is $8.37\times {10}^{-7}M$.

The molality of a solution is $8.37\times {10}^{-7}m$.

Mole fractions of chloroform is $1.15\times {10}^{-7}$.

The mass percent of a compound is $9.99\times {10}^{-2}\%$.

A concept:

Molarity is defined as the number of moles dissolved in per liter of solution.

Molality is defined as the moles of solute per kilogram of a solvent.

Molarity:

Molarity can be calculated as,

$\text{molarity}=\frac{\text{wt.\hspace{0.17em}of\hspace{0.17em}solute}\times 1000}{\text{volume\hspace{0.17em}of\hspace{0.17em}solution}\times \text{mol.\hspace{0.17em}wt.\hspace{0.17em}of\hspace{0.17em}solute}}$ 

 

Now, the formula of $\text{ppb}$ is,

$$\begin{gathered} \text{ppb}=\frac{\text{wt.\hspace{0.17em}of\hspace{0.17em}solute}}{\text{volume\hspace{0.17em}of\hspace{0.17em}solution}}\times {10}^9 \\ \frac{\text{ppb}}{{10}^9}=\frac{\text{wt.\hspace{0.17em}of\hspace{0.17em}solute}}{\text{volume\hspace{0.17em}of\hspace{0.17em}solution}} \end{gathered}$$

 

The molar mass of chloroform is $119.38\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

So, calculate the molarity from $100\text{ ppb}$ is as follows.

$\begin{array}{rcl}\text{molarity}& =& \frac{100\text{ ppb}}{{10}^9}\times \frac{1000}{119.38{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 8.37\times {10}^{-7}\text{ M}\end{array}$ 

 

Hence, the molarity of solution is $8.37\times {10}^{-7}\text{\hspace{0.17em}M}$.

Molality:

Since the solution is very dilute we can assume that one litre of water is as similar as one kilogram of water.

Hence, the molality is equal to the molarity and value of it is $8.37\times {10}^{-7}\text{ M}$.

Mole fraction:

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components is termed a mole fraction.

To calculate the mole fraction, first, you have to calculate the moles of chloroform and water.

Consider the $1\text{ kg}$ or $\text{1 liter}$ of water and calculate the moles of chloroform from molarity as follow;

$\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}chloroform}& =& \text{molarity}\times \text{volume\hspace{0.17em}of\hspace{0.17em}water }\left(\text{L}\right)\\ & =& 8.37\times {10}^{-7}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\times 1\text{ L}\\ & =& 8.37\times {10}^{-7}\text{ mol}\end{array}$ 

 

Now, to calculate the moles of water, divide its mass by molar mass.

Assume, the mass of water is $1000\text{ g}$.

The molar mass of water is $18\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

$\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}water}& =& \frac{\text{mass}}{\text{molar\hspace{0.17em}mass}}\\ & =& \frac{1000\text{ g}}{18{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 55.55\text{ mol}\end{array}$ 

 

The total moles present in the solution

$\begin{array}{rcl}\text{Total moles}& =& \text{moles\hspace{0.17em}of\hspace{0.17em}chloroform}+\text{moles\hspace{0.17em}of\hspace{0.17em}water}\\ & =& 8.37\times {10}^{-7}\text{ mol}+55.55\text{ mol}\end{array}$ 

 

Thus, the mole fraction of chloroform can be calculated as,

$\begin{array}{rcl}{\text{mole\hspace{0.17em}fraction\hspace{0.17em}of\hspace{0.17em}CHCl}}_{\text{3}}& =& \frac{\text{moles\hspace{0.17em}of\hspace{0.17em}chloroform}}{\text{total\hspace{0.17em}moles}}\\ & =& \frac{8.37\times {10}^{-7}\text{ mol}}{(8.37\times {10}^{-7}+55.55)\text{ mol}}\end{array}$ 

Write it as follows:

$\begin{array}{rcl}{\text{mole\hspace{0.17em}fraction\hspace{0.17em}of\hspace{0.17em}CHCl}}_{\text{3}}& =& \frac{8.37\times {10}^{-7}\text{ mol}}{8.37\times {10}^{-7}\text{ mol}}+\frac{8.37\times {10}^{-7}\text{ mol}}{55.55\text{ mol}}\\ & =& 1+0.150\times {10}^{-7}\\ & =& 1.15\times {10}^{-7}\end{array}$

 

Hence. The mole fractions of chloroform is $1.15\times {10}^{-7}$.

Mass percent:

The molar mass of chloroform is $119.38\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

Moles of chloroform is $8.37\times {10}^{-7}\text{ mol}$.

 

Formula:

$\text{number\hspace{0.17em}of\hspace{0.17em}moles}=\frac{\text{mass}}{\text{molar\hspace{0.17em}mass}}$ 

$\begin{array}{rcl}\text{mass\hspace{0.17em}of\hspace{0.17em}chloroform}& =& \text{number\hspace{0.17em}of\hspace{0.17em}moles}\times \text{molar\hspace{0.17em}mass}\\ & =& 119.38\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 8.37\times {10}^{-7}\text{ mol}\\ & =& 0.9992\times {10}^{-4}\text{ g}\\ & \simeq & 1\times {10}^{-4}\text{ g}\end{array}$

 

The molar mass of water is $18\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

Moles of water is $55.55\text{ moles}$.

 $\begin{array}{rcl}\text{mass\hspace{0.17em}of\hspace{0.17em}water}& =& \text{moles\hspace{0.17em}of\hspace{0.17em}water}\times \text{molar\hspace{0.17em}mass}\\ & =& 18\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 55.55\text{ moles}\\ & =& 999.9\text{ g}\end{array}$

 

Now, it can calculate the mass percent of a compound by dividing the mass of chloroform by the total mass of the compound multiply with  $100$.

$\begin{array}{rcl}\text{mass\hspace{0.17em}percent}& =& \frac{\text{mass\hspace{0.17em}of\hspace{0.17em}chloroform}}{(\text{mass\hspace{0.17em}of\hspace{0.17em}chloroform}+\text{mass\hspace{0.17em}of\hspace{0.17em}water})}\times 100\%\\ & =& \frac{1\times {10}^{-4}\text{ g}}{(1\times {10}^{-4}+999.9)\text{ g}}\times 100\%\\ & =& 9.99\times {10}^{-2}\%\end{array}$

 

Hence, the mass percent of a compound is $\begin{array}{rcl}& & 9.99\times {10}^{-2}\%\end{array}$.