Q1: The vapor pressures of benzene and toluene are 3 and 4/3 atmospheres respectively. A liquid feed of 0.4 moles of benzene and 0.6 moles of toluene is vapourised. Assuming that the products are in equilibrium, the vapor phase mole fraction of benzene is

ANS:B - 0.6

To find the vapor phase mole fraction of benzene (𝑋BenzeneXBenzene​), we can use Raoult's law, which states that the partial pressure of a component in a mixture of ideal liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. Mathematically, for a binary mixture of benzene (B) and toluene (T): 𝑃Benzene=𝑋Benzene×𝑃Benzene∘PBenzene​=XBenzene​×PBenzene∘​ 𝑃Toluene=𝑋Toluene×𝑃Toluene∘PToluene​=XToluene​×PToluene∘​ Given that the vapor pressures (𝑃∘P∘) of benzene and toluene are 3 atm and 4334​ atm respectively, we can write: 𝑃Benzene=𝑋Benzene×3PBenzene​=XBenzene​×3 𝑃Toluene=𝑋Toluene×43PToluene​=XToluene​×34​ Since the sum of the mole fractions of all components in a mixture is 1, we have: 𝑋Benzene+𝑋Toluene=1XBenzene​+XToluene​=1 Given that the liquid feed contains 0.4 moles of benzene and 0.6 moles of toluene, the total moles of the mixture is 1: Total moles=0.4+0.6=1Total moles=0.4+0.6=1 Now, we can find the mole fractions: 𝑋Benzene=moles of benzenetotal moles=0.41=0.4XBenzene​=total molesmoles of benzene​=10.4​=0.4 𝑋Toluene=moles of toluenetotal moles=0.61=0.6XToluene​=total molesmoles of toluene​=10.6​=0.6 Now, using the equations for partial pressures: 𝑃Benzene=0.4×3=1.2 atmPBenzene​=0.4×3=1.2atm 𝑃Toluene=0.6×43=0.8 atmPToluene​=0.6×34​=0.8atm Since the products are in equilibrium, the sum of partial pressures equals the total pressure: 𝑃Benzene+𝑃Toluene=1.2+0.8=2 atmPBenzene​+PToluene​=1.2+0.8=2atm Now, to find the vapor phase mole fraction of benzene (𝑋BenzeneXBenzene​): 𝑋Benzene=𝑃Benzene𝑃Benzene+𝑃Toluene=1.22=0.6XBenzene​=PBenzene​+PToluene​PBenzene​​=21.2​=0.6 So, the vapor phase mole fraction of benzene is 0.60.6.