Answer:
Explanation:
This problem is very similar to the other that you put before, so, we'll use the same principle here.
The ideal gas equation: PV = nRT
Where:
P: pressure in atm
V: volume in L
T: Temperature in K.
n: moles
R: Gas constant (In this case, we'll use 0.082 L atm/K)
to get the molar mass of the gas, we need to know the moles, and with the mass, we can know the molar mass. However we can put the ideal gas expression with the molar mass in this way:
we know that n is mole so:
n = g/MM
If we put this in the idea gases expression we have:
PV = gRT/MM
Solving for MM we have:
MM = gRT/PV
Now, let's convert the temperature and volume to K and L respectively:
T = 67 + 273 = 340 K
V = 350 / 1000 = 0.35 L
Now all we have to do is put all the data into the expression:
MM = 0.79 * 0.082 * 340 / 0.9 * 0.35
MM = 22.0252 / 0.315 = 69.92 g/mol rounded 70 g/mol
Now, the closest answer of your options would be 72 g/mol. This could be easily explained because we do not use all the significant figures of all numbers, including the gas constant of R. However, all the work, calculations and procedure is correct and fine, and we only have a minimum range of 2 units.
