Question

The chlorofluorocarbon compound, CCl₃F, is a substance for which no natural source is known and is considered an anthropogenic compound. As of 1971, CCl₃F was discovered in quantities of 40 ppt, while, as of 1994, it was found at 261 ppt. Compute the concentrations of CCl₃F in 1971 and 1994 under conditions of 200 K and 0.05 atm.

Answer 1

Solution.

Ppt is a part per trillion, or 1 ng/kg.

Assuming that we have 1 kg of the air, the mass of CCl₃F is 40 ng in 1971, and 261 ng in 1994.

The amount of substance is related to a mass by a formula n = m/M;

the molar mass of CCl₃F is M = 137.37 g/mol, so the amounts of substance are

in 1971: 40/137.37 = 0.291 nmol;

in 1994: 261/137.37 = 1.9 nmol.

To compute the molar concentration, the volume of a system is required.

The composition of the air by a molar fraction is 21% O₂ and 79% N₂.

The amounts of oxygen and nitrogen in 1 kg of the air are

n(O₂)M(O₂)+n(N₂)M(N₂) = 1000;

n(O₂)/n(N₂) = 0.21/0.79;

n(O₂) = 7.28 mol;

n(N₂) = 27.4 mol;

The volume of oxygen is (V = nRT/P):

Converting pressure to SI units.
P = 5066.25 Pa
T = 200 K
n = 7.28
R = 8.314 J/(mol*K)
Computing volume using the selected equation.
V = 2.389 m³
Converting volume to litres.
V = 2389 L;

The volume of nitrogen is (V = nRT/P):

Converting pressure to SI units.
P = 5066.25 Pa
T = 200 K
n = 27.4
R = 8.314 J/(mol*K)
Computing volume using the selected equation.
V = 8.992 m³
Converting volume to litre.
V = 8992 L;

The total volume of air is 8992+2389 = 11380 L;

The molar concentration is (c = n/V):

In 1971: 0.291 nmol/11380 L = 2.557*10^-14 mol/L;

In 1994: 1.9 nmol / 11380 L = 1.67*10^-13 mol/L.