Question

Calculate the number of molecules present in 44.8 cm3 of oxygen gas at 273 k and 2 atmospheric pressure.

Answer 1

Pressure=2 atm
Temp.=273 K
Volume=44.8 cm3=44.8 litre
So,
By PV=nRT
n=PV/RT
mole=(2 x 44.8)/(0.0821 x 273)
So, no. of mole=3.998=4
So, no. of molecule=4 x 6.023 x 10^(23)
=24.08 x 10^(23)

Answer 2

Answer:

Step-by-step explanation:Pressure=2 atm

Temp.=273 K

Volume=44.8 cm3=44.8 litre

So,

By PV=nRT

n=PV/RT

mole=(2 x 44.8)/(0.0821 x 273)

So, no. of mole=3.998=4

So, no. of molecule=4 x 6.023 x 10^(23)

=24.08 x 10^(23)