Question

Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 Torr. Assume that air is 21% oxygen and 78% nitrogen by volume.

Answer 1

Full Question:

Calculate the masses of oxygen and nitrogen that are dissolved in 6.0 L of aqueous solution in equilibrium with air at 25 °C and 760 Torr. Assume that air is 21% oxygen and 78% nitrogen by volume.

Gas constants are

O2 7.9x10^2 bar/M

N2 1.6x 10^3 bar/M

Step-by-step explanation:

Pressure = 760 torr = 1 atm

Temperature = 25 °C

Henry's law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

Henry's Law states that;

c = p/k

where p = partial pressure and k = Henry's constant for the gas.

Since Partial pressure is proportional to volume fraction;

Partial pressure of O2 = 0.21atm

Partial pressure of N = 0.78atm

From the question;

k for O2 = 769.2 L-atm/mol

k for N2 = 1639 L-atm/mol

Inserting the vlues of P and K, solving for C;

c(O2) = 0.21/769.2 = 2.73*10^-4 mol/L

c(N2) = 0.78/1639 = 4.76*10^-4 mol/L

Voulume of solutiion = 6.0L, so the no of moles is

n(O2) = 6 * 2.73*10^-4 = 1.64*10^-3 mol

n(N2) = 6 * 4.76*10^-4 = 2.86*10^-3 mol

Mass = Number of moles * Molar mass

molar mass of O2 = 32

Mass of O2 = 32 * 1.64 * 10^-3 = 52 mg of O2

molar mass of N2 = 28

Mass of N2 = 28 * 2.86 * 10^-3 = 80 mg of N2

Answer 2

Step-by-step explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

C =
`(P)/(K_(h))`

where, pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at
`25^(o)C` are as follows.

`p(O_(2))` = 0.21 atm = 0.21 bar

`p(N_(2))` = 0.78 atm = 0.78 bar

`K_(h)` for
`O_(2)` =
`7.9 * 10^(2) bar/mol`

`K_(h)` for
`N_(2)` =
`1.6 * 10^(3) bar /mol`

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

`C(O_(2)) = (0.21)/(7.9 * 10^(2)) = 2.66 * 10^-4 mol/L`

`C(N_2) = (0.78)/(1.6 * 10^3) = 4.875 * 10^-4 mol/L`

Now, we will calculate the number of moles as follows.

`n(O_(2)) = 7.5 * 2.66 * 10^-4 = 1.995 * 10^-3` mol

`n(N_2) = 7.5 * 4.875 * 10^-4 = 3.66 * 10^-3` mol

As the molar mass of
`O_2` = 32 g/mol

Hence, mass of oxygen will be as follows.

Mass of
`O_2 = 32 * 1.995 * 10^-3 * 1000`

= 63.84 mg

As the molar mass of
`N_(2)` = 28

Mass of
`N_(2) = 28 * 3.66 * 10^-3` = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg and nitrogen is 102.5 mg.