What is the mass of a sample of O₂ gas, which has a pressure of 740.mmHg, at a temperature of 25 °C, in a volume of 250. mL?

Question

Quizplus · Accepted Answer

We can use the ideal gas law to solve for the mass of the gas: 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 in Kelvin. We can rearrange this equation to solve for the number of moles: n = PV/RT We are given P = 740 mmHg, V = 250 mL, T = 25 °C = 298 K, and we can look up R = 0.08206 L·atm/(mol·K). We need to convert the pressure to atm and the volume to L: P = 740 mmHg * 1 atm/760 mmHg = 0.974 atm V = 250 mL/1000 mL/L = 0.250 L Now we can plug in and solve: n = (0.974 atm * 0.250 L) / (0.08206 L·atm/(mol·K) * 298 K) n = 0.0126 mol Finally, we can use the molar mass of O₂ gas to convert from moles to grams: m = n * M where M is the molar mass. We can calculate the molar mass from the molecular formula: M = (1*15.999) + (1*32) + (12*12.011) + (1*32) + (1*15.999) + (12*12.011) + (1*32) + (1*15.999) + (12*12.011) + (1*15.999) + (1*32) + (1*15.999) + (12*12.011) + (1*32) + (1*15.999) + (1*32) + (1*15.999) + (12*12.011) + (1*32) + (1*15.999) + (12*12.011) + (1*15.999) + (1*32) + (1*15.999) M = 828.5 g/mol Now we can calculate the mass: m = 0.0126 mol * 828.5 g/mol = 10.47 g However, we need to round to the nearest hundredth because we only have two significant figures in the given pressure: m = 10.47 g ≈ 10.5 g ≈ 0.318 g

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What Is the Mass of a Sample of O₂ Gas