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?

Answer 1

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 = K_Hx

`=>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 = K_Hx

= `"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.