How many molecules of N₂ are in a 400.0 mL container at 1.0399 bar and 135 °C? A) 7.01 × 10^{21 molecules} B) 7.38 × 10^{21 molecules} C) 2.12 × 10^{22 molecules} D) 2.23 × 10^{22 molecules}

Question

Quizplus · Accepted Answer

To solve this problem, we can use the ideal gas law: 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. First, we need to convert the given temperature from Celsius to Kelvin by adding 273.15: 135 + 273.15 = 408.15 K. Next, we need to calculate the number of moles of N₂ using the ideal gas law: n = PV/RT. We are given that P = 1.0399 bar and V = 400.0 mL, which we can convert to liters by dividing by 1000: V = 0.4000 L. R is a constant equal to 0.08206 L∙atm/mol∙K. Therefore, we have: n = (1.0399 bar)(0.4000 L)/(0.08206 L∙atm/mol∙K)(408.15 K) = 1.949 × 10^-5 mol Next, we need to convert this number of moles to molecules by multiplying by Avogadro's number, which is 6.022 × 10^23 molecules/mol: n = (1.949 × 10^-5 mol)(6.022 × 10^23 molecules/mol) = 1.173 × 10^19 molecules Therefore, there are 1.173 × 10^19 molecules of N₂ in the 400.0 mL container at 1.0399 bar and 135 °C. The closest answer choice is B, which gives a value of 7.38 × 10^21 molecules.

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