5.00 grams of a gas occupy 2.55 L at STP. What must be the molar mass of the gas?

We know that @ STP one mole of gas occupies 22.4 L and we have 2.55 L @STP. This means we can find the number of mol of the unknown gas we have by dividing 2.55 by 22.4:

2.55 L /22.4 L = .114 mol

To find the molar mass we divide the total number of grams we have by the number of moles, as follows:

5.00g / .114 mol = 43.9 g/mol

The molar mass of CO2 is 44.00 g/mol which is pretty close to 43.9g/mol. I'm fairly confident I approached that the correct way, if someone else could verify that would be great.

For the second part we can just use the Ideal Gas Law:

PV=nRT

P= 1atm

V= ?

n= 3.25 mol

R= .0821

T= 273K

(1atm) V = (3.25mol)(.0821)(273K)

V = 72.84L

Hope that helps. :)

If a fifteen.eighty 3 g pattern of a gasoline occupies 10.0 L at STP, what's the molar mass of the gasoline at one hundred twenty five stages Celsius? --------------------------------------... 15.eighty 3 g/10 L = x/22.4 L (15.eighty 3 g x 22.4 liters/mole)/10 liters = x = 35.6 grams/mole. x = molar mass, which does no longer exchange with temperature.