Chemistry College

Calculate the solubility of nitrogen (in M) when the gas is at a pressure of
a) 2.00 atm
b) 688 mmHg

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Answers

Answer 1

A.) The solubility of nitrogen at a pressure of 2.00 atm is [tex]1.36 \times 10^{(-3)} M.[/tex]

B.) The solubility of nitrogen at a pressure of 688 mmHg is [tex]6.17 \times 10^{(-4)} M.[/tex]

To calculate the solubility of nitrogen (N2) in M (molarity) at different pressures, we need to use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation for Henry's Law is:

C = k * P

Where:

C is the solubility of the gas in M (molarity)

k is the Henry's Law constant

P is the partial pressure of the gas

For nitrogen, the Henry's Law constant (k) is approximately 6.8 x 10^(-4) M/atm.

a) To calculate the solubility of nitrogen at a pressure of 2.00 atm:

C = (6.8 x 10^(-4) M/atm) * (2.00 atm)

C = 1.36 x 10^(-3) M

Therefore, the solubility of nitrogen at a pressure of 2.00 atm is 1.36 x 10^(-3) M.

b) To calculate the solubility of nitrogen at a pressure of 688 mmHg:

First, we need to convert mmHg to atm by dividing by 760 (since 1 atm = 760 mmHg).

P = 688 mmHg / 760 mmHg/atm

P = 0.905 atm

C = (6.8 x 10^(-4) M/atm) * (0.905 atm)

C = 6.17 x 10^(-4) M

Therefore, the solubility of nitrogen at a pressure of 688 mmHg is 6.17 x 10^(-4) M.

It's important to note that the solubility of a gas can also depend on temperature, so these calculations assume a constant temperature. Additionally, Henry's Law is an approximation and may not hold true for all gas-liquid systems, especially at high pressures or when there are significant intermolecular interactions between the gas and liquid.

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