Question

A gas sample weighing 3.78 grams occupies a volume of 2280 mL at STP. What is the apparent molecular mass of the sample?

(Note: R = 8.314 kPa.dm3/K.mol or 0.0821 L.atm/K.mol or 62.4 L.torr/K.mol)

Answer 1

To calculate the apparent molecular mass of the gas sample, we can use the ideal gas law:

PV = nRT

where:
P is the pressure of the gas (at STP, P = 1 atm)
V is the volume of the gas (2280 mL = 2.280 L)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L.atm/K.mol)
T is the temperature of the gas (at STP, T = 273.15 K)

We can rearrange the ideal gas law to solve for n:

n = PV/RT

Let's plug in the given values and solve for n:

n = (1 atm) * (2.280 L) / (0.0821 L.atm/K.mol * 273.15 K)
n = 0.1002 mol

Now we can calculate the apparent molecular mass of the gas sample using the formula:

molecular mass = mass / moles

Let's plug in the given values and solve for the molecular mass:

molecular mass = 3.78 g / 0.1002 mol
molecular mass = 37.7 g/mol

Therefore, the apparent molecular mass of the gas sample is approximately 37.7 g/mol.