Q1: Saturated solution of benzene in water is in equilibrium with a mixture of air and vapours of benzene and water at room temperature and pressure. Mole fraction of benzene in liquid is x_B and the vapour pressures of benzene and water at these conditions are p_v^B and p_v^w respectively. The partial pressure of benzene in air-vapour mixture is

ANS:B - x_B.P_v^B

In this scenario, benzene is dissolved in water to form a saturated solution. At equilibrium, the benzene molecules will exert a vapor pressure (PvB) while in the vapor phase. According to Raoult's Law, the partial pressure of a component in the vapor phase above a solution is proportional to its mole fraction in the solution. Raoult's Law states: 𝑃vap=𝑥⋅𝑃vappurePvap​=x⋅Pvappure​ Where:

• 𝑃vapPvap​ is the vapor pressure of the component in the solution,
• 𝑥x is the mole fraction of the component in the liquid phase,
• 𝑃vappurePvappure​ is the vapor pressure of the pure component.

In this case, the partial pressure of benzene in the air-vapor mixture will be 𝑥𝐵×𝑃vappurexB​×Pvappure​. Since benzene is in equilibrium with the air-vapor mixture, the vapor pressure of benzene (𝑃vappurePvappure​) is equal to the partial pressure of benzene in the air-vapor mixture. So, the correct expression for the partial pressure of benzene in the air-vapor mixture is 𝑥𝐵⋅𝑃𝑣𝐵xB​⋅PvB​.