Given : Solubility of ethane = S
= 0.92 × 10-3 g dm-3
Pressure of ethane = P = 1000 mm Hg
Molar mass of ethane (C2H6) = 30 g mol-1
Henry’s law constant = KH = ?
S = 0.92 × 10-3 g dm-3
= \(\frac{0.92}{30}\) x 10-3
= 3.067 × 10-5 mol dm-3
P = 1000 mm = \(\frac{1000}{760}\) atm = 1.316 atm
By Henry’s law,
S = KH × P
∴ KH = \(\frac{S}{P}\) = \(\frac{3.067\times 10^{-5}}{1.3016}\)
= 2.33 × 10 mol-5 dm-3 atm-1
Henry’s law constant = KH
= 2.33 × 10-5 mol dm-3 atm-1