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The partial pressure of ethane over a solution containing 6.56xx10^(-3)g of ethane is 1 bar. If the solution contains 5.00xx10^(-2)g of ethane, then what shall be the partial pressure of the gas ? |
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Answer» <P> Solution :Molar mass of ethane `(C_(2)H_(6))``= 2xx12+6xx1` = 30 G `mol^(-1)` `therefore` Number of molar present in `6.56xx10^(-3)` g of ethane `= (6.56xx10^(-2))/(1)` `= 2.187xx10^(-4)` Let the number of MOLES of the solvent be 55.55 assuming solvent is water. According to Henry.s low, `p = K_(H)X` 1 bar `= KH(2.187xx10^(-4)//55.455)` [Assuming DILUTION condition i.e., moles of solvent `gt gt` moles of solute] `K_(H)=1` bar / `(0.039xx10^(-4))` If mass of ethane = 0.05 g then moles of ethane = 0.05/30 mol = 0.00166 mol So `p=K_(H)X` `p=1//0.039xx10^(-4)xx(0.00166//55.55)` = 7.66 bar |
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