Question

How many moles of oxygen ATOMS are present in a `27.3*g` mass of `SO_3(g)`?

Answer 1

Well what is the molar quantity of sulfur trioxide...?

Explanation:

`"Moles of sulfur trioxide"=(27.3*g)/(80.07*g*mol^-1)=0.341*mol.`

And thus `"moles of oxygen"=3xx0.341*mol=1.02*mol` WITH RESPECT TO OXYGEN ATOMS....

And since a mole specifies `6.022xx10^23` individual particles, we gots `6.022xx10^23*mol^-1xx1.02*mol=??"oxygen atoms."`

Answer 2

Sulpher tri oxide is expressed as `SO_3`

It's molecular weight is `(32+3*16)=80gm`

So,we can say, `3*6.023*10^23` atoms of oxygen are present in `80 gm` of `SO_3`

So,in `27.3 gm` ,number of atoms present will be `27.3 *(3/80)*6.023*10^23` i.e `1.02` i.e about `6.17*10^23`