If 0.385 g of a gas occupies a volume of 200 mL at -73 degrees C and
750 torr, what is the molecular mass of the gas?
a. 21.4
b. 32.0
c. 44.0
d. 48.0

Hi!!

Again we will use the Ideal Gas Law:

P.V = n.R.T 

Then:

n = P.V / R.T

In this problem:
P = 750 torr = 0.986842 atm
V = 200 mL = 0.2 L
T = (-73+273)K = 200K
R = 0.082058 atm/(mol * K)

n = (0.986842 atm x 0.2 L) / (0.082058 atm/(mol * K) x 200K) =
  = (0.1973684 / 16.4116) mol = 0.012 mol

There are 0.012 moles of gas.

1 mole of a pure substance has a mass in grams equal to its molecular mass, then:

0.012 mol ---(weight)---> 0.385g
1 mol -----(weight)-----> Molecular Mass in grams

Then:
Molecular Mass in grams = 0.385g x 1mol/ 0.012mol = 32.083g

The correct answer is b.

Hope this helps.

Regards.
livioflores-ga