At \(25^{\circ} \mathrm{C}\), the vapor in equilibrium with a solution containing carbon disulfide and acetonitrile has a total pressure of 263 torr and is \(85.5\) mole percent carbon disulfide. What is the mole fraction of carbon disulfide in the solution? At \(25^{\circ} \mathrm{C}\), the vapor pressure of carbon disulfide is 375 torr. Assume the solution and vapor exhibit ideal behavior.

First, we need to convert the mole percent of carbon disulfide in the vapor to mole fraction. Mole percent is given by: Mole percent = (moles of carbon disulfide / total moles) × 100 Let the mole fraction of carbon disulfide be X_CS2 and mole fraction of acetonitrile be X_AN. Since there are only two components in the solution, their mole fractions add up to 1: X_CS2 + X_AN = 1 From the given information, 85.5 mole percent of carbon disulfide = (moles of carbon disulfide / total moles) ×100 Divide by 100: X_CS2 = 0.855 Now we can find the mole fraction of acetonitrile: X_AN = 1 - X_CS2 = 1- 0.855 = 0.145

Using the mole fractions, we can calculate the partial pressures of carbon disulfide (P_CS2) and acetonitrile (P_AN) in the vapor: Total pressure (P_total) = P_CS2 + P_AN The given total pressure is 263 torr. We can write: P_CS2 = X_CS2 × P_total P_CS2 = 0.855 × 263 torr = 224.865 torr Now we can find the partial pressure of acetonitrile: P_AN = P_total - P_CS2 = 263 torr - 224.865 torr = 38.135 torr

Raoult's law states that the partial pressure of each component in the vapor is equal to the product of its mole fraction in the solution and its vapor pressure: P_i = x_i × P_i^0 Where P_i is the partial pressure of component i in the vapor, x_i is the mole fraction of component i in the solution, and P_i^0 is the vapor pressure of the pure component i. We are given P_CS2 = 224.865 torr and P_CS2^0 (vapor pressure of pure carbon disulfide) = 375 torr. We need to find the mole fraction of carbon disulfide in the solution, x_CS2_solution. Using Raoult's law: 224.865 torr = x_CS2_solution × 375 torr To find x_CS2_solution, divide both sides by 375 torr: x_CS2_solution = 224.865 torr / 375 torr x_CS2_solution = 0.5998 Thus, the mole fraction of carbon disulfide in the solution is approximately 0.5998.