5.238 x 10^24 molecules of aluminum bromide to grams

Philip Stratford asked Nov 22, 2023

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Philip Stratford asked Nov 22, 2023

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24 votes

Answer

2319.698 g

Procedure

Use the Avogadro's number and then the molecular weight to convert into grams. The formula for aluminum bromide is AlBr₃.

$5.238*10^(24)\text{ molecules AlBr}_3\frac{1\text{ mole}}{6.022*10^(23)}\frac{266.69\text{ g}}{1\text{ mole}}=2319.698\text{ grams}$

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