Question

11. What pressure is exerted by 11.0 grams of carbon monoxide when contained in a 20.0 liter vessel at 27.0 C?

12. The weight of 151 ml of a certain vapor at 68.0 C and 768 mm is 1.384 grams. What is the molecular weight of the vapor?

Answer 1

11)

mass = 11.0 grams

Volume = V= 20.0L

T= 27C= 27+272= 300 K,           R= 0.0821 L-atm/mol-K

molar mass of CO = 28.0gram/mole

number of moles of CO = mass/molar mass = 11.0/28.0 = 0.393 moles

n= 0.393 moles

Ideal gas equation

PV=nRT

P = nRT/V = 0.393 x 0.0821 x 300/20.0 = 0.484 atm

Pressure = 0.484 atm.

12) Volume = V= 151 mL= 0.151L

T= 68C= 68+273= 341 K

Pressure = P= 768mm = 768.760 = 1.01atm

mass = 1.384 grams

PV=nRT

n = PV/RT = 1.01x 0.151/0.0821 x341 = 0.0054 moles

number of moles = mass/molar mass

molar mass of gas = mass/number of moles = 1.384/0.0054 = 256.296 = 256.3 grams

molecular weight of a vapour = 256.3 grams.