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Question
The partial pressure of ethane over a solution containing 6.56 × 10−3 g of ethane is 1 bar. If the solution contains 5.00 × 10−2 g of ethane, then what shall be the partial pressure of the gas?
Solution
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 = `(6.56xx10^(-3))/30`
= 2.187 × 10−3 mol
Let the number of moles of the solvent be x.
According to Henry’s law,
p = KHx
`=>1 "bar" = "K"_"H" (2.187xx10^(-3))/(2.187xx10^(-3)+"x")`
`=>1 "bar" = "K"_"H" (2.187xx10^(-3))/"x"` ...(Since x >> 2.187 × 10−3)
`=> "K"_"H" = "x"/(2.187xx10^(-3)) "bar"`
Number of moles present in 5.00 × 10−2 g of ethane = `(5.00xx10^(-2))/30` mol
= 1.67 × 10−3 mol
According to Henry’s law,
p = KHx
= `"x"/(2.187xx10^(-3))xx(1.67xx10^(-3))/(1.67xx10^(-3)+"x")`
= `"x"/(2.187xx10^(-3))xx(1.67xx10^(-3))/(1.67xx10^(-3)+"x")` ...(Since x >> 1.67 × 10−3)
= 0.764 bar
Hence, the partial pressure of the gas shall be 0.764 bar.
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