The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temperature and pressure. In an experiment, \(8.00 \mathrm{~L}\) of argon gas is passed through $11.7872 \mathrm{~g}\( of liquid hexane \)\mathrm{C}_{6} \mathrm{H}_{14}$ at \(30.0^{\circ} \mathrm{C}\). The mass of the remaining liquid after the experiment is \(4.875 \mathrm{~g}\). Assuming that the gas becomes saturated with hexane vapor and that the total gas volume and temperature remain constant, what is the vapor pressure of hexane in atm?

Subtract the mass of the remaining liquid from the initial mass of the liquid hexane to find the mass of the hexane vapor. Mass of hexane vapor = Initial mass - Remaining mass = 11.7872 g - 4.875 g = 6.9122 g

To find the moles of hexane vapor, divide the mass of hexane vapor by its molar mass (86.18 g/mol, since hexane is \(\mathrm{C}_6\mathrm{H}_{14}\)). Moles of hexane vapor = Mass of hexane vapor / Molar mass of hexane = 6.9122 g / 86.18 g/mol = 0.0802 mol

First, calculate the moles of argon gas using the ideal gas law: PV = nRT Where P is pressure (in atm), V is volume (in L), n is the moles of gas, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature (in K). We're given that the volume of argon gas is 8.00 L and the temperature is 30°C, which is equal to 303.15 K. Assume the atmospheric pressure is 1 atm. 1 atm * 8.00 L = n * 0.0821 L·atm/mol·K * 303.15 K n (moles of argon) = 0.3283 mol Now, calculate the mole fractions of argon and hexane: Mole fraction of argon (X_Ar) = moles of argon / (moles of argon + moles of hexane) = 0.3283 mol / (0.3283 mol + 0.0802 mol) = 0.8036 Mole fraction of hexane (X_C6H14) = moles of hexane / (moles of argon + moles of hexane) = 0.0802 mol / (0.3283 mol + 0.0802 mol) = 0.1964

According to Dalton's Law of Partial Pressures, the partial pressure of each gas in a mixture is equal to its mole fraction times the total pressure. Partial pressure of hexane (P_C6H14) = Mole fraction of hexane (X_C6H14) * Total pressure = 0.1964 * 1 atm = 0.1964 atm Thus, the vapor pressure of hexane at 30.0°C is 0.1964 atm.