Question

An oxygen gas container has a volume of 20.0 L. How many grams of oxygen are in the container if the gas has a pressure of 883 mmHg at 24 ∘C? Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer : The mass of oxygen gas will be, 30.5 grams

Solution :

Using ideal gas equation,

`PV=nRT\\\\PV=(w)/(M)* RT`

where,

n = number of moles of gas

w = mass of gas = ?

P = pressure of the gas =
`883mmHg=(883)/(760)atm`

conversion used : (1 atm = 760 mmHg)

T = temperature of the gas =
`24^oC=273+24=297K`

M = molar mass of oxygen gas = 32 g/mole

R = gas constant = 0.0821 Latm/moleK

V = volume of gas = 20.0 L

Now put all the given values in the above equation, we get the mass of oxygen gas.

`((883)/(760)atm)* (20.0L)=(w)/(32g/mole)* (0.0821Latm/moleK)* (297K)`

`w=30.5g`

Therefore, the mass of oxygen gas will be, 30.5 grams