Question

380 mL of a gas at 27°C, 800 mm of Hg weights 0.455 g. The molecular weight of the gas is _____

Answer 1

Answer:

The standard way to do this problem is to use the gas law equation:

PV = nRT

P = pressure = 800/760 = 1.053 atm

V = volume = 0.380L

n = number of moles

R = gas law constant = 0.082057

T = temp in K = 27+273 = 300

Substitute:

1.053* 0.380 = n * 0.082057*300

n = ( 1.053*0.380) / ( 0.082057*300)

n = 0.400/ 24.6171

n = 0.0162 mol

0.455g = 0.0162 mol

1mol = 0.455/ 0.0162 = 28.00g/mol