we are close to STP here where one mol is 22.4 liters
58.4/22.4 is
around 2.61 moles
10.3/2.61 = 3.95 grams/mol
Helium maybe ?
A 10.3g sample of an elemental gas has a volume of 58.4L at 758torr when the temperature is 2.5 C. Determine the identity of the gas.
2 answers
Damon's estimate is quite good (and reasonable) and the answer is correct. If you want to do it without assuming it is 22.4 L/mol, it is done this way.
Use PV = nRT and solve for n = number of mols of the gas.
(758*58.4)/(760*0.08206*275.6) = 0.257 = n.
Then mols = grams/molar mass or
molar mass = g/mol = 10.3/0.257 = 4.01. Indeed that is He.
Use PV = nRT and solve for n = number of mols of the gas.
(758*58.4)/(760*0.08206*275.6) = 0.257 = n.
Then mols = grams/molar mass or
molar mass = g/mol = 10.3/0.257 = 4.01. Indeed that is He.