Q136CP - Chemistry: Molecular Nature Of Matter And Change

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}}$}\right.$ )?

Step-by-Step Solution

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}$}\right.$.

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,

$\begin{array}{rcl}C& =& \frac{\text{0.65 mg}}{\text{1 L}}\times \frac{\text{1 g}}{\text{1000 mg}}\times \frac{\text{1 mol}}{\text{97 g}}\\ & =& 6.071\times {10}^{-6}\raisebox{1ex}{${\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 person can 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 $\raisebox{1ex}{$\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}$}\right.}}{1\text{ L}}\times \frac{\text{97 g}}{\text{1 mol}}\times \frac{{\text{10}}^{\text{9}}\text{ ng}}{\text{1 g}}\\ & =& 5.19\times {10}^5\raisebox{1ex}{${\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}$}\right.$.