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Chemistry Question on Solutions

The partial pressure of ethane over a solution containing 6.56×10–3g6.56 × 10^{–3} g of ethane is 1 bar. If the solution contains 5.00×10–2g5.00 × 10^{–2} g of ethane, then what shall be the partial pressure of the gas?

Answer

The correct answer is: x(2.187×10−4)bar \frac{x}{(2.187\times 10^{-4})}bar
Molar mass of ethane (C2H6)=2×12+6×1=30 gmol−1(C_2H_6) = 2\times12 + 6\times1 = 30 \,g mol^{ - 1}
Number of moles present in 6.56×10−3g6.56 × 10^{ - 3} g of ethane =(6.56×10−3)30=\frac{(6.56\times 10^{-3})}{30}
=2.187×10−4mol=2.187 × 10^{- 4} mol
Let the number of moles of the solvent be x.x.
According to Henry's law,
ρ=KHXρ = K_HX
⇒1bar=KH.(2.187×10−4)(2.187×10−4+x)⇒1 bar = K_H. \frac{(2.187\times 10^{-4})}{(2.187\times10^{-4}+x)}
⇒1bar=KH.(2.187×10−4)x     (Sincex>>2.187×10−4)⇒1 bar = KH. \frac{(2.187\times 10^{-4})}{x} \,\,\,\,\,(Since x>>2.187\times10^{-4})
⇒KH=x(2.187×10−4)bar⇒K_H = \frac{x}{(2.187\times 10^{-4})}bar