A 0.795*g0.795g mass of gas is confined to a 1*L1L volume at a temperature of 8989 ""^@CC under a pressure of 771*mm*Hg771mmHg. What is the molar mass of the gas?

1 Answer
Apr 29, 2017

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*atm1atm of pressure will support a column of mercury that is 760*mm760mm high.

And thus "Pressure"-=(771*"mm Hg")/(760*"mm Hg"*atm^-1)=1.014*atmPressure771mm Hg760mm Hgatm1=1.014atm

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

=(1.014*atmxx1.00*L)/(0.0821*(L*atm)/(K*mol)xx362*K)=3.41xx10^-2*mol=1.014atm×1.00L0.0821LatmKmol×362K=3.41×102mol.

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

=(0.795*g)/(3.41xx10^-2*mol)=23.3*g*mol^-1=0.795g3.41×102mol=23.3gmol1

Again, this is a very poorly proposed question, in that we would think that we would get a molar mass of 28*g*mol^-128gmol1, the which corresponds to a real gas, i.e. COCO or N_2N2.......I cannot think of a real gas that has a molar mass of 23*g*mol^-123gmol1, not methane, not ethane, not carbon monoxide, not ammonia, not acetylene. C'est la vie..........