# How many grams are in 4.5 x 1028 moles of sodium fluoride, NaF?

Feb 20, 2018

I get around $3139542 \setminus \text{g}$.

#### Explanation:

How many grams are in $4.5 \times {10}^{28}$ molecules of sodium fluoride, $N a F$?

That's because if you talk about $4.5 x 1028$ moles of sodium fluoride, you would get a really absurd number of grams.

So, I'd solve your problem assuming your question is How many grams are in $4.5 \times {10}^{28}$ molecules of sodium fluoride, $N a F$?.

In that, we first have to convert the amount of molecules into moles.

We know that $1 \setminus \text{mol} = 6.02 \cdot {10}^{23}$ molecules, so in here, we get

(4.5*10^28cancel"molecules")/(6.02*10^23cancel"molecules""/mol")=74750.8306~~74751 \ "mol"

Sodium fluoride has a molar mass of $42 \setminus \text{g/mol}$. So, the mass of $74751$ moles of sodium fluoride will be

$42 \text{g/"cancel"mol"*74751cancel"mol"=3139542 \ "g}$