Question

What is the mass associated with a numerical quantity of `9.033xx10^23` fluorine atoms?

Answer 1

Well we know that `6.022xx10^23` `""^19F` atoms have a mass of `19.0*g`....

Explanation:

Of course, fluorine normally occurs as the diatomic molecule, i.e. `F_2`, as do most of the gaseous elements..

And so we take the product...

`(9.033xx10^23*"atoms")/(6.022xx10^23*"atoms"*mol^-1)=1.5*mol`...

And so we gots a mass of .................

`1.5*molxx19.00*g*mol^-1=28.5*g`.