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

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 = $\frac{P}{K_{h}}$

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 $O_{2}$ = $7.9 \times 10^{2} bar/mol$

    $K_{h}$ for $N_{2}$ = $1.6 \times 10^{3} bar /mol$

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.

      $C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L$  

      $C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L$

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

         $n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3$ mol

         $n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3$ mol

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

Hence, mass of oxygen will be as follows.

         Mass of $O_2 = 32 \times 1.995 \times 10^-3 \times 1000$

                           = 63.84 mg

As the molar mass of $N_{2}$ = 28

       Mass of $N_{2} = 28 \times 3.66 \times 10^-3$ = 102.5 mg

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

Answer 2

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

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