Answer
$n_{atoms}=1.3\times10^{27}\ atoms$
Work Step by Step
The volume of the room is $V=4.0m\times5.0m\times2.4m=48\ m^3\times\dfrac{1\times10^3\ L}{1\ m^3}=48\times10^3 L$
Since this room is entirely filled with argon, with a density of $\rho=1.78\ g/L$:
$m=\rho\times V=8.5\times10^4\ g$
With the atomic weight of $M= 39.948\ g/mol$
$n=m/M=2.1\times10^3\ mol$
From Avogadro's number: $N=6.022\times10^{23}\ atoms/mol$
$n_{atoms}=n\times N=1.3\times10^{27}\ atoms$
