Question

Calculate the number of molecules present in 44.8 $$ of oxygen gas at 273 K and 2 atmosphere pressure.

Answer 1

Pressure= 2 atm

temperature= 273K

volume= 44.8$c{m}^3$=44.* litre

so,By PV=nRT

n=$\frac{PV}{RT}$

mole=$\frac{2\times 44.8}{0.0821\times 273}$

so, no of moles= 3.998=4

hence, $4\times 6.023\times {10}^{23}$

=$24.08\times {10}^{23}$