A $2.72 \cdot g$ $2.72*g$ mass of aluminum constitutes WHAT molar quantity? And how many aluminum atoms does this molar quantity represent?

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A $2.72 \cdot g$ $2.72*g$ mass of aluminum constitutes WHAT molar quantity? And how many aluminum atoms does this molar quantity represent?

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Answer 1 · 2016-02-10T09:13:32

Avogadro's number of $A l$ $Al$ atoms has a mass of $26.98$ $26.98$ $g$ $g$ . So you tell us.

Explanation:

If there are $6.022$ $6.022$ $\times$ $xx$ $10^{23}$ $10^23$ individual aluminum atoms, there is a mass of $26.98$ $26.98$ $g$ $g$ of aluminum. $6.022$ $6.022$ $\times$ $xx$ $10^{23}$ $10^23$ individual aluminum atoms represents $1$ $1$ $m o l$ $mol$ of aluminum.

So, $(2.72*cancelg)/(26.98*cancelg*mol^-1)$ $=$ $??$ $mol$ ?

Whatever this molar quantity is, we multiply this by Avogadro's number, $6.022$ $xx$ $10^23$ $mol^-1$ , to get the number of aluminum atoms.

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A $2.72 \cdot g$ $2.72*g$ mass of aluminum constitutes WHAT molar quantity? And how many aluminum atoms does this molar quantity represent?

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A 2.72*g2.72⋅g2.72*g mass of aluminum constitutes WHAT molar quantity? And how many aluminum atoms does this molar quantity represent?

1 Answer

Explanation:

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A $2.72 \cdot g$ $2.72*g$ mass of aluminum constitutes WHAT molar quantity? And how many aluminum atoms does this molar quantity represent?