The vapor pressures of pure carbon disulfide and carbon tetrachloride are 360 and 99.8 Torr, respectively,At 296 K.What is the vapor pressure of a solution containing 50.0 g of each compound?

Question

Quizplus · Accepted Answer

To find the vapor pressure of the solution, we use Raoult's Law, which states that the vapor pressure of a solution is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent. First, we calculate the mole fractions of carbon disulfide (CS2) and carbon tetrachloride (CCl4). Molar mass of CS2 = 76.14 g/mol (32.07 for S * 2 + 12 for C)Moles of CS2 = 50.0 g / 76.14 g/mol = 0.657 molesMolar mass of CCl4 = 153.82 g/mol (35.45 for Cl * 4 + 12 for C)Moles of CCl4 = 50.0 g / 153.82 g/mol = 0.325 molesTotal moles = 0.657 + 0.325 = 0.982 molesMole fraction of CS2 = 0.657 / 0.982 = 0.669Mole fraction of CCl4 = 0.325 / 0.982 = 0.331Vapor pressure of the solution = (mole fraction of CS2 * vapor pressure of CS2) + (mole fraction of CCl4 * vapor pressure of CCl4)= (0.669 * 360 Torr) + (0.331 * 99.8 Torr)= 240.84 Torr + 33.06 Torr= 273.9 Torr, which rounds to 274 Torr.

The Vapor Pressures of Pure Carbon Disulfide and Carbon Tetrachloride