Question

At a certain temperature the vapor pressure of pure thiophene (CH S) is measured to be 0.39 atm. Suppose a solution is prepared by mixing 71.5 g of thiophene and 68.6 g of heptane (CGH16) Calculate the partial pressure of thiophene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal.

Answer 1

According to Raoult's law the vapour pressure of the component in a solution is equal to the product of its vapour pressure in the pure state and it's mole fraction.

Answer - heptane Thiophene (C7H16) (Culhus) mollector molecular mass molecular mats - (7x12.) +16x1) = (4x 12 ) +(4+1) + (32x1) = 10o amu = 84 amu moles (ha)= Given mass ling) Moles (ng) = Civen massling molecular mass A molecular mals nA = 71.5 no = 68.6 84 100 HA=0.85 no = 0.686 • Mole fraction of thiophene XA = A = 0.85 hatho a85+ 0.686 XA = 0.55 • According to raoult's lave PA = PÅ XA = 0.39 X 0.50 = 0.21 atm

P​​​​​​A = PA° × X​​​​A

P​​​​​​A = vapour pressure of component A in solution

P°A= vapour pressure of component A in pure form

X​​​​​​A = mole fraction of A