Respuesta :
Answer: The concentration of dichloroethylene in the air breather by a person swimming in the river is [tex]5.19\times 10^{5}ng/L[/tex]
Explanation:
We are given:
Concentration of dichloroethylene = 0.65 mg/L
Converting this into mol/L, we use the conversion factor:
1 g = 1000 mg
Molar mass of dichloroethylene = 97 g/mol
Converting the concentration, we get:
[tex]\Rightarrow (\frac{0.65mg}{1L})\times (\frac{1g}{1000mg})\times (\frac{1mol}{97g})=6.701\times 10^{-6}mol/L[/tex]
To calculate the pressure of the gas, we use ideal gas equation:
[tex]PV=nRT\\\\P=\frac{n}{V}RT\\\\P=CRT[/tex]
where,
P = pressure of the gas
C = concentration of gas = [tex]6.701\times 10^{-6}M[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]21^oC=[21+273]K=294K[/tex]
Putting values in above equation, we get:
[tex]P=6.701\times 10^{-6}mol/L\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 294K\\\\P=1.62\times 10^{-4}atm[/tex]
To calculate the concentration of gas we use the equation given by Henry's law, which is:
[tex]C_{C_2H_2Cl_2}=K_H\times p_{C_2H_2Cl_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = 0.033 mol/L.atm
[tex]C_{C_2H_2Cl_2}[/tex] = molar solubility of dichloroethylene gas = ?
[tex]p_{C_2H_2Cl_2}[/tex] = partial pressure of dichloroethylene gas = [tex]1.62\times 10^{-4}atm[/tex]
Putting values in above equation, we get:
[tex]C_{C_2H_2Cl_2}=0.033mol/L.atm\times 1.62\times 10^{-4}atm\\\\C_{C_2H_2Cl_2}=5.35\times 10^{-6}mol/L[/tex]
Converting this into ng/L, we use the conversion factor:
[tex]1g=10^9ng[/tex]
Molar mass of dichloroethylene = 97 g/mol
Converting the concentration, we get:
[tex]\Rightarrow (\frac{5.35\times 10^{-6}mol}{1L})\times (\frac{97g}{1mol})\times (\frac{10^9ng}{1g})=5.19\times 10^{5}ng/L[/tex]
Hence, the concentration of dichloroethylene in the air breather by a person swimming in the river is [tex]5.19\times 10^{5}ng/L[/tex]
