How many protons are present in 5.6 L of oxygen at N.T.P.using O-16 isotope only ?

Explanation

No. of `O_(2)` molecules in 5.6 L of oxygen `=((6.022xx10^(10^(23))xx(5.6L))/((22.4L)))=1.506xx10^(23)`
No. of oxygen atoms present = `1.506xx10^(23)xx2`
No. of protons in each oxygen atom =8
Total no. of protons present `=2xx1.506xx10^(23)xx8=2.41xx10^(24)`.