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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?
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Solution
According to Henry’s law,
m = kP
For the first case,
6.56 × 10−3 = k × 1 ...(i)
For the second case,
5 × 10−2 = k × p ...(ii)
Dividing eq. (i) by eq. (ii), we get
`(6.56 xx 10^-3)/(5 xx 10^-2) = (k xx 1)/(k xx p) = 1/p`
∴ p = `(5 xx 10^-2)/(6.56 xx 10^-3)`
= 7.62 bar
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