Astudent collects 350 ml of a vapor at a temperature of 67°c. the atmospheric pressure at the time of collection is 0.900 atm. the student determines that the mass of the vapor collected is 0.79 g. what is the molar mass of the gas?

36 g/mol
72 g/mol
88 g/mol
176 g/mol

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.

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