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Solve the following :

The solubility of ethane at 25 °C is 0.92 × 10-3 g dm-3 at 1000 mm Hg pressure. Calculate Henry’s law constant.

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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

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