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What is the mass of 19.42 atoms of F?

1 Answer

Aug 26, 2016

Well $6.022 \times 10^{23}$ $6.022xx10^23$ atoms of fluorine (as the atom, not as $F_{2}$ $F_2$ ) have a mass of $19.00 \cdot g$ $19.00*g$ .

Explanation:

And thus $\frac{19}{6.022 \times 10^{23} \cdot m o l^{- 1}} \times 19.00 \cdot g \cdot m o l^{- 1}$ $19/(6.022xx10^23*mol^-1)xx19.00*g*mol^-1$

$=$ $=$ $not too many grams$ $"not too many grams"$ .

We cannot really speak of $0.42 \cdot atoms$ $0.42*"atoms"$ of fluorine. We can have the one atomic particle; we cannot really have 1/2 of an atom.

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