How many atoms are in 1.75 moles of zinc?

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How many atoms are in 1.75 moles of zinc?

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Answer 1 · 2016-06-02T17:45:23

How many atoms in $1.5$ $1.5$ $dozen$ $"dozen"$ zinc atoms? Clearly, there are $18$ $18$ .

Explanation:

We know that $1$ $1$ $m o l$ $mol$ of stuff specifies $Avogadro's Number, N_{A}$ $"Avogadro's Number, "N_A$ individual items of stuff. Avogadro's Number $=$ $=$ $6.022 \times 10^{23} \cdot m o l^{- 1}$ $6.022xx10^23*mol^-1$ .

So you have $1.75 \times N_{A}$ $1.75xxN_A$ atoms of zinc. You do the arithmetic, and tell us what is the mass of this quantity, this number, of zinc atoms?

The use of $Avogadro's number$ $"Avogadro's number"$ as we use any other collective number, e.g. score, dozen, gross, is entirely legitimate. Why do use such an absurdly large number? It turns out that $N_{A}$ $N_A$ $^{1} H$ $""^1H$ atoms have a mass of $1 \cdot g$ $1*g$ precisely. $Avogadro's number$ $"Avogadro's number"$ is thus the link between atoms and molecules, the which we can't see, to the macro world of grams and litres, that which we can conveniently measure in the laboratory.

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How many atoms are in 1.75 moles of zinc?

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