Question

A river is contaminated with $\mathbf{\text{0.65}}\raisebox{1ex}{${\displaystyle \mathbf{\text{mg}}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{L}}$}\right.$of dichloroethylene $\mathbf{\left(}{\mathbf{\text{C}}}_{\mathbf{\text{2}}}{\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}\mathbf{\right)}$. What is the concentration (in $\raisebox{1ex}{${\displaystyle \mathbf{\text{ng}}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{L}}$}\right.$) of dichloroethylene at$\mathbf{\text{21}}\mathbf{°}\mathbf{\text{C}}$in the air breathed by a person swimming in the river (${\mathbf{\text{k}}}_{\mathbf{\text{H}}}{\mathbf{\text{C}}}_{\mathbf{\text{2}}}{\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}$for in water is $\mathbf{0}\mathbf{.}\mathbf{033}\raisebox{1ex}{${\displaystyle \mathbf{\text{mol}}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{L}}\mathbf{\cdot }\mathbf{\text{atm}}$}\right.$ )?

Short answer

The concentration of dichloroethylene in the air breathed by a person swimming in river is $5.19\times {10}^5\raisebox{1ex}{${\displaystyle \text{ng}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$.

Step-by-step solution

A constant:

Pressure of gas can be calculated by using ideal gas equation.

$\mathit{P}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$

Here, $\mathit{n}$is the number of moles of gas, $\mathit{P}$is pressure, $\mathit{V}$ is volume, $\mathit{R}$ is the gas constant, and $\mathit{T}$ is temperature.

It can write it as;

$\begin{array}{rcl}\mathit{P}& \mathbf{=}& \frac{\mathbf{n}}{\mathbf{V}}\mathit{R}\mathit{T}\\ & \mathbf{=}& \mathit{C}\mathit{R}\mathit{T}\end{array}$

Where, $\mathit{C}$ is the concentration of gas.

Pressure of dichloroethylene gas:

Consider the given data as below.

The value of concentration, $C=0.65\raisebox{1ex}{${\displaystyle \text{mg}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$

Converting this value into $\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$ in the following way.

Molar mass of dichloroethylene is $97\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$. Therefore,

$\begin{array}{rcl}C& =& \frac{\text{0.65mg}}{\text{1L}}\times \frac{\text{1g}}{\text{1000mg}}\times \frac{\text{1mol}}{\text{97g}}\\ & =& 6.071\times {10}^{-6}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$

Value of gas constant, $R=0.082\raisebox{1ex}{${\displaystyle \text{L}\cdot \text{atm}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\cdot \text{K}$}\right.$

The temperature,$T=21°\text{C}=(21+273)\text{K}=294\text{K}$

Thus, pressure of dichloroethylene gas can be calculated as,

$\begin{array}{rcl}{P}_{gas}& =& CRT\\ & =& 6.071\times {10}^{-6}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\times 0.082\raisebox{1ex}{${\displaystyle \text{L}\cdot \text{atm}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\cdot \text{K}$}\right.\times 294\text{K}\\ & =& 1.62\times {10}^{-4}\text{atm}\end{array}$

Define the Concentration of dichloromethane:

Determine the Concentration of dichloromethane in the air breathed by a personcan be represented as,

$C_{gas}=k_H\times P_{gas}$

Where,$k_H$is Henry’s constant.

Substitute known values in the above equation, and you have

$\begin{array}{rcl}{C}_{gas}& =& 0.033\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}\cdot \text{atm}$}\right.\times 1.62\times {10}^{-4}\text{atm}\\ & =& 5.35\times {10}^{-6}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$

Thus, the concentration of gas is $5.35\times {10}^{-6}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$.

Now, converting this concentration into role="math" localid="1663660749388" $\raisebox{1ex}{$\text{ng}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$.

$1\text{g}={10}^9\text{ng}$

Therefore, the concentration will be,

$\begin{array}{rcl}{C}_{gas}& =& \frac{5.35\times {10}^{-6}{\displaystyle \raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.}}{1\text{L}}\times \frac{\text{97g}}{\text{1mol}}\times \frac{{\text{10}}^{\text{9}}\text{ng}}{\text{1g}}\\ & =& 5.19\times {10}^5\raisebox{1ex}{${\displaystyle \text{ng}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$

Hence, the concentration of dichloroethylene in the air breathed by a person swimming in river is $5.19\times {10}^5\raisebox{1ex}{${\displaystyle \text{ng}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$.