Question

A river is contaminated with \(0.65 \mathrm{mg} / \mathrm{L}\) of dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right) .\) What is the concentration (in \(\mathrm{ng} / \mathrm{L}\) ) of dichloroethylene at \(21^{\circ} \mathrm{C}\) in the air breathed by a person swimming in the river \(\left(k_{\mathrm{H}}\right.\) for \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\) in water is \(\left.0.033 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{atm}\right) ?\)

Short answer

The concentration of dichloroethylene in air breathed is 816,000 ng/L.

Step-by-step solution

Convert concentration from mg/L to mol/L

First, convert the concentration of dichloroethylene in water from mg/L to mol/L. The molar mass of dichloroethylene \(\text{C}_2\text{H}_2\text{Cl}_2\) is calculated as follows: \(2 \times 12 + 2 \times 1 + 2 \times 35.5 = 96 \text{g/mol}\). To convert mg/L to \(\text{mol/L}\), use the formula: \[ \text{Concentration (mol/L)} = \frac{\text{Concentration (mg/L)}}{\text{Molar mass (mg/mol)}} \].So, the concentration in mol/L is \( \frac{0.65 \text{ mg}}{96 \text{ mg/mol}} = 6.77 \times 10^{-3} \text{ mol/L} \).

Calculate the partial pressure of dichloroethylene

Use Henry's Law formula \( c = k_H \times P \), where \( c \) is the concentration of the solute in mol/L, \( k_H \) is the Henry's Law constant \( (0.033 \text{mol/L} \cdot \text{atm})\), and \( P \) is the partial pressure of the gas. \[ P = \frac{c}{k_H} = \frac{6.77 \times 10^{-3} \text{mol/L}}{0.033 \text{mol/L} \cdot \text{atm}} = 0.205 \text{ atm} \].

Convert partial pressure to concentration in air

Use the ideal gas law to convert the partial pressure of the gas to concentration in air: \( P = \frac{nRT}{V}\), where \( n \) is the number of moles, \( R = 0.0821 \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}\), \( T = 294 \text{K} \ (21^{\circ}C + 273K) \), and \( V \) is the volume.Rearranging to solve for \( n/V = P / RT\), the concentration in mol/L is \[ \frac{P}{RT} = \frac{0.205 \text{ atm}}{0.0821 \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \times 294 \text{K}} = 8.5 \times 10^{-3} \text{ mol/L} \]

Convert concentration from mol/L to ng/L

Finally, convert the concentration from mol/L to ng/L. The concentration in ng/L can be calculated using the molar mass of dichloroethylene: \[ \text{Concentration} = \text{concentration (mol/L)} \times \text{molar mass (ng/mol)} \].Since 1 mol = 96g, and 1g = 1,000,000,000 ng, the molar mass in ng/mol is \[ 96 \times 10^9 \text{ ng/mol} \].Thus, the concentration in ng/L is \[ 8.5 \times 10^{-3} \text{mol/L} \times 96 \times 10^9 \text{ng/mol} = 8.16 \times 10^5 \text{ng/L} = 816,000 \text{ng/L} \].

Key concepts

Concentration Conversion

First, let's tackle concentration conversion, which is essential to understand the problem at hand. Concentration is often given in various units such as mg/L, mol/L, or ng/L.
To solve the exercise, we need to convert between these units.
First, the concentration of dichloroethylene (\text{C}_2\text{H}_2\text{Cl}_2\text{C}_2) in water is 0.65 mg/L.
To convert mg/L to mol/L, use the molecular weight of the substance. Here, the molecular weight of \text{C}_2\text{H}_2\text{Cl}_2 is 96 g/mol.
The formula to convert mg/L to mol/L is: \[ \text{Concentration (mol/L)} = \frac{\text{Concentration (mg/L)}}{\text{Molar mass (mg/mol)}} \]
Using the formula: \[ \frac{0.65 \text{ mg}}{96 \text{ mg/mol}} = 6.77 \times 10^{-3} \text{ mol/L} \]
Understanding conversions will help in all kinds of chemical calculations.

Henry's Law

Henry's Law is crucial in understanding how gases dissolve in liquids. It states that the concentration of a gas in a liquid is proportional to its partial pressure above the liquid.
The formula is: \[ c = k_H \times P \] Where:

• c is the concentration of the gas in the liquid (in mol/L)
• k_H is the Henry's Law constant for the gas (in mol/L·atm)
• P is the partial pressure of the gas (in atm)

For our problem: \[ P = \frac{c}{k_H} = \frac{6.77 \times 10^{-3} \text{mol/L}}{0.033 \text{ mol/L} \times \text{atm}} = 0.205 \text{ atm} \]
Hence, the partial pressure of dichloroethylene in air can be determined using Henry's Law.

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas. The formula is: \[ PV = nRT \] Rearanging for our purposes: \[ P = \frac{nRT}{V} \] Where:

• P is the pressure of the gas (in atm)
• V is the volume of the gas (in L)
• n is the number of moles of the gas
• R is the ideal gas constant (0.0821 L·atm/mol·K)
• T is the temperature (in K)

For the problem, T is given as 21˚C, converting to Kelvin: \[ 21 + 273 = 294 K \]
Thus, to find the concentration of the dichloroethylene in air: \[ \frac{P}{RT} = \frac{0.205 \text{atm}}{0.0821 \text{ L·atm/mol·K } \times 294 \text{K}} = 8.5 \times 10^{-3} \text{ mol/L} \]
This is the concentration of the gas in the air.

Units of Measurement

Finally, understanding units of measurement is vital to accurately communicate scientific concepts.
In this exercise, we've used units like mg/L, mol/L, atm, and ng/L.
Be aware of the context and required conversion between these units.
For the last step, converting the concentration of dichloroethylene from mol/L to ng/L involves: \[ \text{Concentration (ng/L)} = \text{Concentration (mol/L)} \times \text{Molar mass (ng/mol)} \]
Given the molar mass in ng/mol is: \[ 96 \times 10^9 \text{ ng/mol} \]
Hence the final concentration: \[ 8.5 \times 10^{-3} \text{mol/L} \times 96 \times 10^9 \text{ng/mol} = 816,000 \text{ ng/L} \]
This is the concentration in the air breathed by the person swimming in the river.
Always pay attention to converting units correctly to ensure precise results.