1 g of a gas occupies 300ml at 250C and 99 KPa.

Question

1 g of a gas occupies 300ml at 250C and 99 KPa. Find the following:

1. Number of moles
2. The molecular mass

Answer 1

The Ideal Gas Law says PV = nRT

P = 99 KPa

V = 0.300 L (300/1000)

T = 523.15 K (250 + 273.15)

R = (from a reference book or google) 8.314 L•KPa/(mol•K)

So n = PV/(RT) = (99)(0.300)/(8.314•523.25) = 29.7/4349.47 = 0.00683 moles (or rounded, n = 0.0068 moles)

The molecular mass is grams/mole so it would be (1 gram)/(0.00683 moles) = 146 g/mol (or rounded, = 150 g/mol)