The solubility of carbon dioxide at 20°C is 0.06 g of CO2 per kilogram of water when the pressure of CO2 is 760 mmHg. What would be the solubility if the pressure of CO2 is increased to 890 mmHg and the temperature left unchanged?

3g/L

Explanation:

it is simply using the idea gas law to calculate the new molar density (i. e. molar solubility) at a different pressure,

P V = nRT

where:

P is the pressure in atm

V is the volume in L.

n is the mols of gas in mol s.

R = 0.082057 L ⋅atm/mol. K is the universal gas constant in the appropriate units.

T is the temperature in K.

Solving for molar density, we get:

n V = PRT

Since we have two solubility to consider, we must have two states.

n ₁/v₁ = p₁/RT

n ₂/v₂ = p₂/RT

Therefore, we have:

RT = P₁V ₁/n₁ = P₂V ₂/n₂

Solving for the new molar density, we get:

n₂/v₂ = n₁/v₁ (p₂/p₁)

we can now convert the mass-based solubility to mols. We assume that the

CO₂ was as dissolved into the water as possible for the initial pressure, and that it doesn't change the volume of the water.

n₁v₁ = {0.06g CO ₂ * (1 mol CO ₂/44.009g CO ₂) }/0.100 L water

0.06g CO ₂ * (0.022722) }/0.100 L water = 0.001363/0.1 = 0.0136335mol/L

Therefore, the new solubility is: 0.0136335mol/L * 5.50/1.00 atm = 0.06816 mol/L

In the original units, we have:

0.06816 mol CO ₂/L solution * 44.009 g CO ₂/mol CO ₂ = 3g/L