A #0.795*g# mass of gas is confined to a #1*L# volume at a temperature of #89# #""^@C# under a pressure of #771*mm*Hg#. What is the molar mass of the gas?

1 Answer
Apr 29, 2017

Answer:

These questions ought NOT to be asked of students..........

Explanation:

Mercury has all but disappeared from modern laboratories. Aside from its toxicity, mercury spilled on a laboratory bench is a MAJOR AND DEMANDING clean up job, the which I doubt that the person who asked this question has ever performed. Pressures of over ONE atmosphere should thus be reported with units of atmospheres or kilopascals, and not some whack units of mercury............

This being said, we know that #1*atm# of pressure will support a column of mercury that is #760*mm# high.

And thus #"Pressure"-=(771*"mm Hg")/(760*"mm Hg"*atm^-1)=1.014*atm#

And thus from the Ideal Gas equation, #n=(PV)/(RT)#

#=(1.014*atmxx1.00*L)/(0.0821*(L*atm)/(K*mol)xx362*K)=3.41xx10^-2*mol#.

And thus #"molar mass"="Mass of gas (g)"/"Moles of gas (moles)"#

#=(0.795*g)/(3.41xx10^-2*mol)=23.3*g*mol^-1#

Again, this is a very poorly proposed question, in that we would think that we would get a molar mass of #28*g*mol^-1#, the which corresponds to a real gas, i.e. #CO# or #N_2#.......I cannot think of a real gas that has a molar mass of #23*g*mol^-1#, not methane, not ethane, not carbon monoxide, not ammonia, not acetylene. C'est la vie..........