Calculate the number of molecules present in (a) 1 kg oxygen (b) `1 dm^(3)` of hydrogen at N.T.P

Answer

(a) `1.88 xx 10^(25)` (b) `2.69 xx 10^(22)`

Explanation

(a) 32.0 grams of oxygen have molecules `= 6.022 xx 10^(23)`
1000.0 grams of oxygen have molecules `((1000.0g))/((32.0g))xx6.022xx10^(23)=1.88xx10^(25)`
(b) `22.4 dm^(3)` of hydrogen at N.T.P have molecules `= 6.022 xx 10^(23)`
`1.0 dm^(3)` of hydrogen at N.T.P have molecules `= (6.022)/(22.4)xx10^(23)=2.69xx10^(22)`.