A company dissolves 'x' amount of CO₂ at 298 K in 1 litre of water to prepare soda water. X = __________ × 10⁻³ g. (nearest integer)
(Given : partial pressure of CO₂ at 298 K = 0.835 bar.
Henry's law constant for CO₂ at 298 K = 1.67 kbar.
Atomic mass of H, C and O is 1, 12, and 6 g mol⁻¹, respectively)

Solution

Chemistry ma’am · 4 steps

Henry's law gives the mole fraction. Turn it into moles using the moles of water, then into grams.

1. Mole fraction first. Henry's law is p = K_H x. Convert K_H to bar first: 1.67 kbar = 1670 bar.

   x_(CO₂) = p_(CO₂)/K_H = 0.835/1670 = 5 × 10⁻⁴

2. Moles of CO2. 1 L of water is 1000 g, which is 1000/18 ≈ 55.5 mol. The mole fraction is tiny, so n_(CO₂) ≈ x × n_(H₂O) = 5 × 10⁻⁴ × 55.5 = 2.775 × 10⁻² mol.
3. Convert to mass. The oxygen mass in the data is a misprint for 16, so M_(CO₂) = 12 + 2 × 16 = 44 g mol⁻¹. Mass = 2.775 × 10⁻² × 44 = 1.221 g.
4. So. x = 1.221 g = 1221 × 10⁻³ g, so the answer is 1221. If you carry 55.56 mol of water instead, you get 1222. The key uses 55.5.

The common slip: 666. Using the misprinted oxygen mass of 6 makes M_(CO₂) = 24, which gives 0.666 g, or 666. The molar mass of CO₂ is 44.

The idea to keep. For a gas dissolved in water, Henry's law gives the mole fraction. Since that mole fraction is tiny, moles of gas = mole fraction × moles of water. Always put p and K_H in the same pressure unit first.

Answer from the NTA final answer key. Solution by PrepWiser.