A 259.6-L cylinder contains 63.5% He(g) and 36.5% Kr(g) by mass at 22.0°C and 1.25 atm total pressure. What is the mass of He in this container?

Question

Quizplus · Accepted Answer

1. Calculate the moles of He and Kr:
- Assume a total mass of 100 g, then there would be 63.5 g He and 36.5 g Kr.
- Calculate the number of moles:
n(He) = (63.5 g)/(4.003 g/mol) = 15.85 mol
n(Kr) = (36.5 g)/(83.80 g/mol) = 0.435 mol

2. Calculate the partial pressure of He:
- Use the ideal gas law: PV = nRT
- Rearrange to solve for P:
P = nRT/V
- Calculate the partial pressure of He:
P(He) = (15.85 mol)(0.0821 L·atm/mol·K)(295 K)/(259.6 L) = 1.25 atm x (63.5/100) = 0.793 atm

3. Calculate the mass of He:
- Use the ideal gas law again to find the number of moles of He in the whole cylinder:
n(He) = PV/RT = (1.25 atm x (259.6 L))/((0.0821 L·atm/mol·K)(295 K)) = 15.7 mol
- Calculate the mass of He:
mass(He) = n(He) x molar mass(He) = (15.7 mol)(4.003 g/mol) = 62.8 g

Therefore, the answer is B.

A 2596-L Cylinder Contains 63