The weight of a gaseous mixture containing 12.044xx10^(23) atoms of He and 3.011xx10^(23) molecules of hydrogen is _____________g.

The weight of a gaseous mixture containing 12.044xx10^(23) atoms of He

1.	The weight of a gaseous mixture containing 12.044xx10^(23) atoms of He and 3.011xx10^(23) molecules of hydrogen is _____________g.
Answer» SOLUTION :Helium weight = `(12.044xx10^(23))/(6.023xx10^(23))xx4=8g` Hydrogen weight = `(3.011xx10^(23))/(6.023xx10^(23))xx2=1g` Total = `8+1=9g`