Chemistry: A river is contaminated with 0.65 mg/L of dichloroethylene (C2H2Cl2). What is the concentration (in ng/L) of dichloroethylene at 21°C in the air breathed by a person swimming in the river (kH for C2H2Cl2 in water is 0.033 mol/L•atm)?

A river is contaminated with 0.65 mg/L of dichloroethylene

jennarosebrand3518

21.01.2022

The concentration of dichloroethylene in the air breather by a person swimming in the river is $5.19\times 10^{5}ng/L$

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:

$\Rightarrow (\frac{0.65mg}{1L})\times (\frac{1g}{1000mg})\times (\frac{1mol}{97g})=6.701\times 10^{-6}mol/L$

To calculate the pressure of the gas, we use ideal gas equation:

$PV=nRT\\\\P=\frac{n}{V}RT\\\\P=CRT$

where,

P = pressure of the gas

C = concentration of gas = $6.701\times 10^{-6}M$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

T = temperature of the gas = 21^oC=[21+273]K=294K

Putting values in above equation, we get:

$P=6.701\times 10^{-6}mol/L\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 294K\\\\P=1.62\times 10^{-4}atm$

To calculate the concentration of gas we use the equation given by Henry's law, which is:

$C_{C_2H_2Cl_2}=K_H\times p_{C_2H_2Cl_2}$

where,

K_H = Henry's constant = 0.033 mol/L.atm

$C_{C_2H_2Cl_2}$ = molar solubility of dichloroethylene gas = ?

$p_{C_2H_2Cl_2}$ = partial pressure of dichloroethylene gas = $1.62\times 10^{-4}atm$

Putting values in above equation, we get:

$C_{C_2H_2Cl_2}=0.033mol/L.atm\times 1.62\times 10^{-4}atm\\\\C_{C_2H_2Cl_2}=5.35\times 10^{-6}mol/L$

Converting this into ng/L, we use the conversion factor:

1g=10^9ng

Molar mass of dichloroethylene = 97 g/mol

Converting the concentration, we get:

$\Rightarrow (\frac{5.35\times 10^{-6}mol}{1L})\times (\frac{97g}{1mol})\times (\frac{10^9ng}{1g})=5.19\times 10^{5}ng/L$

Hence, the concentration of dichloroethylene in the air breather by a person swimming in the river is $5.19\times 10^{5}ng/L$

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