Question

When 0.890 g of neon is added to an 550-cm3 bulb containing a sample of argon, the total pressure of the gases is found to be 2.51 atm at a temperature of 292 K .

Find the mass of the argon in the bulb.

Express your answer to two significant figures and include the appropriate units.

Answer 1

For Ne:

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 292K

P = pressure = ?

n = No . of moles of Ne = mass/molar mass = 0.890g/20.18(g/mol) = 0.0441mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = volume of the vessel = 550 cm³ x(10^-3 L/ cm³) = 0.550L

Plug the values we get P = (nRT) / V = 1.92 atm

Given total pressure , P' = 2.51 atm

So pressure exerted by Ar = total pressure - pressure exerted by Ne

= 2.51 - 1.92

= 0.59 atm

Caculation of number of moles of Ar present , n' = (PV)/(RT)

= (0.59 atm x 0.550 L) / ( 0.0821 L.atm/mol-K x292K)

= 0.0135 moles

So mass of Ar = number of moles x molar mass

= 0.0135 mol x 40.0 g/mol

= 0.54 g