Question

Henrys law constant for $C{O}_2$ in water is $1.6\times {10}^8$ Pa at 298K.

Calculate quantity of $C{O}_2$ in 500 mL of soda water when packed under 2.5 atm $C{O}_2$ pressure at 298K.

Answer 1

According to Henry's law, $p_{C{O}_2}$ = K_Hx

Here x is the mole fraction of carbon dioxide in the carbon dioxide water mixture.

$p_{C{O}_2}$ = 2.5 atm = 2.5 $\times$ 1.01325 $\times$ 10⁵ Pa = 2.533125 $\times$ 10⁵ Pa

x = mole fraction = 2.533125 $\times$ 10⁵/ 1.67 $\times$ 10⁸ = 0.00152

x = moles of CO₂/(moles of CO₂ + moles of H₂O) ≈ moles of CO₂/moles of H₂O as mole of CO₂ are negligible in comparison to the number of moles of water

500 ml of water = 500g of water = 500/18 = 27.78 moles of water.

So, moles of CO₂/27.78 = 0.00152 or moles of CO₂ = 0.042 moles.
Hence the amount of CO₂ in 500 ml of soda water is 0.042 * 44 = 1.848 g.