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?

Updated On: Sep 27, 2023
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Solution and Explanation

The correct answer is: \( \frac{x}{(2.187\times 10^{-4})}bar\)
Molar mass of ethane \((C_2H_6) = 2\times12 + 6\times1 = 30 \,g mol^{ - 1}\) 
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,
\(ρ = K_HX\)
\( ⇒1 bar = K_H. \frac{(2.187\times 10^{-4})}{(2.187\times10^{-4}+x)}\)
 \(⇒1 bar = KH. \frac{(2.187\times 10^{-4})}{x} \,\,\,\,\,(Since x>>2.187\times10^{-4})\)
 \(⇒K_H = \frac{x}{(2.187\times 10^{-4})}bar\)
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