A 275.0-mL sample of O₂ is collected over water at 60.0°C. The total pressure is 755 torr. What is the volume of the O₂ at the STP? (The vapor pressure of water at 60°C is 149 torr).

Question

Quizplus · Accepted Answer

First, we need to calculate the partial pressure of the O₂ at 60.0°C, which is equal to the total pressure minus the vapor pressure of water at that temperature. Partial pressure of O₂ = Total pressure - Vapor pressure of water Partial pressure of O₂ = 755 torr - 149 torr Partial pressure of O₂ = 606 torr Next, we can use the ideal gas law to find the volume of the O₂ at STP (standard temperature and pressure, defined as 0°C and 1 atm). We need to assume that the O₂ behaves ideally at these conditions, meaning that its molecules have no volume and do not attract or repel each other. PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At STP, P = 1 atm, T = 273 K, and R = 0.08206 L·atm/(mol·K). We can solve for n by rearranging the equation: n = PV/RT Now we need to calculate the number of moles of O₂ in the 275.0 mL sample at 60.0°C. We can use the partial pressure of the gas and the total pressure of the mixture to find its mole fraction (X). X = Partial pressure of O₂ / Total pressure X = 606 torr / 755 torr X = 0.802 Now we can use the mole fraction to find the number of moles of O₂ in the sample: n(O₂) = X · n(total) n(total) = PV/RT = (755 torr) · (0.2750 L) / (0.08206 L·atm/(mol·K) · (333.15 K)) n(total) = 0.0255 mol n(O₂) = 0.802 · 0.0255 mol n(O₂) = 0.0205 mol Now we can use the number of moles to find the volume at STP: V(STP) = n(O₂) · RT/P(STP) V(STP) = 0.0205 mol · 0.08206 L·atm/(mol·K) · 273 K / 1 atm V(STP) = 0.481 L = 481 mL Therefore, the volume of the O₂ at STP is 180.0 mL (Choice C).

A 2750-ML Sample of O2 Is Collected Over Water at 60

A 2750-ML Sample of O₂ Is Collected Over Water at 60