Gas question

[alanan84]

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What is the molar mass of an ideal gas if a 0.622g sample of this gas occupies a volume of 300mL at 35C and 789mmHg?

a) 44.8 g*mol-1
b) 48.9 g*mol-1
c) 50.5 g*mol-1
d) 54.5 g*mol-1

 

[jlt0530]

You should use the ideal gas law equation, which is PV=nRT. But, this can be also written as P(mm)/RT = mass/V. This is very often used when dealing with density. Here, you are solving for molar mass (mm) so set the equation as,

mm = mass x RT / VP Then, you just substitute the value you've got.
= 0.622g x 0.0821 x (273 + 35) / 0.3L x (789/760)atm
= 50.5 g/mol

So, the answer is (C).
Please correct me if I made any mistake! 🙂

 

[alanan84]

You should use the ideal gas law equation, which is PV=nRT. But, this can be also written as P(mm)/RT = mass/V. This is very often used when dealing with density. Here, you are solving for molar mass (mm) so set the equation as,

mm = mass x RT / VP Then, you just substitute the value you've got.
= 0.622g x 0.0821 x (273 + 35) / 0.3L x (789/760)atm
= 50.5 g/mol

So, the answer is (C).
Please correct me if I made any mistake! 🙂

You're exactly right! Thanks for the help, I forgot about that equation. As you can see, it's really been awhile since gen chem.

 

[Shingo123]

Knowing how to derive both that equation and the Density equation from the Ideal gas equation is a nice skill.

The trick is:

PV=nRT

The n which is moles can be changed to be mass/molar mass.

So it is:

PV= mass/molar mass x RT

From there it is just rearrangement:

M= mRT/PV

Density is similar:

Density is Mass/volume.

So D= m/V=MP/RT