What is the molecular weight of a gas if a 21.0 g sample has a pressure of at in a flask?
(R= 0.0821 L atm/ mol K)

Question

What is the molecular weight of a gas if a 21.0 g sample has a pressure of  at  in a  flask? 
(R= 0.0821 L atm/ mol K)

Quizplus · Accepted Answer

We can use the ideal gas law: PV = nRT, where P = 11.2 atm, V = 0.500 L, T = 298 K, and R = 0.0821 L atm/ mol K. 
Rearranging the equation to solve for n, the number of moles: 
n = PV/RT = (11.2 atm * 0.500 L)/(0.0821 L atm/ mol K * 298 K) = 0.949 mol. 
The molecular weight (MW) is given by MW = m/n, where m is the mass of the gas. 
Since the sample has a mass of 21.0 g, MW = 21.0 g/0.949 mol = 22.1 g/mol. 
Converting to atomic mass units (amu), we have MW = 22.1 g/mol * (1 amu/1.6605e-24 g) = 1.33e5 amu. 
However, this value is clearly too high for a gas, so we must check our calculations. 
We made an error in our conversion from grams to moles - we should have divided by the molar mass of the gas, not the number of moles. 
The correct MW is MW = mRT/PV = (21.0 g * 0.0821 L atm/ mol K * 298 K)/(11.2 atm * 0.500 L) = 234 amu. 
Therefore, the answer is C.

What Is the Molecular Weight of a Gas If a 21.0