Question

How many mole of CO2 are present in 55.5 l

Answer 1

Given,

Volume=55.5 l

At S.T.P

Pressure=1 atm

Temperature=298 K

R=0.083 l atm/K mol

We know,

PV=nRT

n=PV/RT

n=1×55.5/0.083×298

n=2.001 mol

Answer 2

Answer:

A: 22.4 L = 1mol

1mol

= 55.5 L x 22.4 L

= 2.48 mol of CO2.(Ans)