Q1: In a mixture of benzene vapor and nitrogen gas at a total pressure of 900 mm Hg, if the absolute humidity of benzene is 0.2 kg benzene/kg nitrogen, the partial pressure of benzene in mm Hg is

ANS:B - 60.3

To find the partial pressure of benzene in the mixture, we can use Dalton's law of partial pressures, which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each individual gas. Given:

• Total pressure of the mixture = 900 mm Hg
• Absolute humidity of benzene = 0.2 kg benzene/kg nitrogen

We need to convert the absolute humidity of benzene to a partial pressure. The partial pressure of benzene (𝑃benzenePbenzene​) can be calculated using Raoult's law, which states that the partial pressure of a component in a non-ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. Let's assume the total moles of gas in the mixture is 1 mole for simplicity. Then, the mole fraction of benzene (𝑋benzeneXbenzene​) in the mixture is given by the ratio of the moles of benzene to the total moles of gas: 𝑋benzene=moles of benzenetotal moles of gasXbenzene​=total moles of gasmoles of benzene​ Given that the absolute humidity of benzene is 0.2 kg benzene/kg nitrogen, we can calculate the moles of benzene: moles of benzene=mass of benzenemolar mass of benzenemoles of benzene=molar mass of benzenemass of benzene​ The molar mass of benzene (C6H6) is approximately 78.11 g/mol. moles of benzene=0.2 kg0.2 kg/kg nitrogen×1000 g/kg×(1 mol/78.11 g)moles of benzene=0.2kg/kg nitrogen×1000g/kg×(1mol/78.11g)0.2kg​ moles of benzene≈0.278.11×103 molmoles of benzene≈78.11×1030.2​mol Now, we can calculate the mole fraction of benzene: 𝑋benzene=0.278.11×1031Xbenzene​=178.11×1030.2​​ 𝑋benzene≈0.278.11×103Xbenzene​≈78.11×1030.2​ Now, we can calculate the partial pressure of benzene using Raoult's law. The vapor pressure of benzene at room temperature is approximately 100 mm Hg. 𝑃benzene=𝑋benzene×𝑃vapor pressure of benzenePbenzene​=Xbenzene​×Pvapor pressure of benzene​ 𝑃benzene≈0.278.11×103×100 mm HgPbenzene​≈78.11×1030.2​×100mm Hg Calculating this gives us the partial pressure of benzene in the mixture. 𝑃benzene≈0.2×10078.11×103 mm HgPbenzene​≈78.11×1030.2×100​mm Hg