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03/01/2022 3:25 pm
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The partial pressure of ethane over a solution containing 6.56 x 10-3 g of ethane is 1 bar. If the solution contains 5.00 x 10-2 g of ethane, then what shall be the partial pressure of the gas?
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03/01/2022 4:36 pm
Molar mass of ethane (C2H6) = 2 × 12 + 6 × 1
= 30 g mol-1
Number of moles present in 6.56 × 10-3 g of ethane = \(\frac{6.56 \times 10^{-3}}{30}\)
= 2.187 × 10-4 mol
Let the number of moles of the solvent be x.
According to Henry's law,
ρ = KHx
⇒ 1 bar = KH.\(\frac{2.187 \times 10^{-4}}{2.187 \times 10^{-4} + x}\)
⇒ 1 bar = KH.\(\frac{2.187 \times 10^{-4}}{ x}\)(Since x >> 2.187 x 10-4)
⇒ KH = \(\frac{x}{2.187 \times 10^{-4}}\)bar