Calculate the number of molecules of oxygen in 150 ml of it at `20^(@)` C and 750 mm pressure

Explanation

Volume of oxygen at N.T.P.
`p_(1) = 750 mm, p_(2)=760`mm
`V_(1)=150 mL, V_(2)`= ?
`T_(1) = 273 + 20 = 293 K, T_(2)=273 K`
Applying, `(p_(1)V_(1))/T_(1) = (p_(2)V_(2))/T_(2)` or `V_(2) = (p_(1)V_(1)T_(2))/(T_(1)p_(2))`
`therefore V_(2) = (750 xx 150 xx 273)/(293 xx 760) = 137.9` mL
22400 mL of oxygen at N.T.P. has molecules.
`=6.022 xx 10^(23)`
137.9 mL of oxygen at N.T.P. has molecule.
`=(6.022 xx 10^(23) xx 137.9)/(22400)`
`=3.71 xx 10^(21)`