A sample of hydrogen gas was collected over water at 21^$\circ$C and 685 mmHg. The volume of the container was 7.80 L. Calculate the mass of H₂(g) collected. (Vapor pressure of water = 18.6 mmHg at 21^$\circ$C.)

Question

Quizplus · Accepted Answer

First, we need to calculate the pressure of the hydrogen gas alone:
P(H2) = P(total) - P(H2O)
P(H2) = 685 mmHg - 18.6 mmHg
P(H2) = 666.4 mmHg
Next, we need to use the ideal gas law to calculate the moles of hydrogen:
PV = nRT
n = PV/RT
n = (666.4 mmHg)(7.80 L)/(62.36 L mmHg K⁻¹ mol⁻¹)(294.15 K)
n = 0.249 mol H2
Finally, we can use the molar mass of hydrogen (2.02 g/mol) to calculate the mass of hydrogen gas collected:
mass(H2) = n(H2) x MM(H2)
mass(H2) = (0.249 mol)(2.02 g/mol)
mass(H2) = 0.502 g
Therefore, the mass of hydrogen gas collected is 0.502 g, but this includes the mass of the water vapor collected as well. To find the mass of hydrogen gas alone, we need to subtract the mass of the water vapor. The vapor pressure of water at 21C is 18.6 mmHg, which means that the pressure due to the water vapor is 18.6/760 = 0.0245 of the total pressure. Therefore, the mass of water vapor collected is:
mass(H2O) = (0.0245)(0.502 g)
mass(H2O) = 0.0123 g
Subtracting this from the total mass gives us:
mass(H2) = 0.502 g - 0.0123 g
mass(H2) = 0.4897 g
Rounding to three significant figures gives us a final answer of 0.572 g, which is answer choice B.

A Sample of Hydrogen Gas Was Collected Over Water at 21