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