Partial pressure needed to dissolve 21 mg of CO₂ in 100 gr of water at 298K is (K_H for CO₂ is 2.937 KPa m³ mol^–1)

Given:

Mass of CO₂ (m): 21 mg = 0.021 g

Volume of water (V): 100 g (assuming the density of water is 1 g/mL, this equals 100 mL or 0.1 L)

Temperature (T): 298 K

Henry's Law constant for CO₂ (KH): 2.937 kPa·m³·mol⁻¹

Convert mass of CO₂ to moles:

The molar mass of CO₂ is 44 g/mol.

Moles of CO₂ (n) = mass / molar mass = 0.021 g / 44 g/mol = 4.773 × 10⁻⁴ mol

Calculate the molarity (C) of CO₂ in water:

Molarity (C) = moles of solute / volume of solvent in liters

C = 4.773 × 10⁻⁴ mol / 0.1 L = 4.773 × 10⁻³ mol/L

Apply Henry's Law to find the partial pressure (P):

Henry's Law is expressed as:

P = KH × C

Substituting the known values:

P = 2.937 kPa·m³·mol⁻¹ × 4.773 × 10⁻³ mol/L

Since 1 L = 0.001 m³, we convert mol/L to mol/m³:

4.773 × 10⁻³ mol/L = 4.773 mol/m³

Now, calculate the pressure:

P = 2.937 kPa·m³·mol⁻¹ × 4.773 mol/m³ = 14.01 kPa

Final Answer

The partial pressure needed to dissolve 21 mg of CO₂ in 100 g of water at 298 K is approximately 14 kPa.