Question

How many moles of PBr3 contain 3.68 x 10^25 bromine atoms?

Answer 1

3.68 x 10²⁵ bromine atoms * 1mol/6.02*10²³ atoms=
 = 61.13 mol of bromine atoms

1 mol PBr3 ----- 3 mol Br
x mol PBr3 -----61.13 mol Br

x= 1*61.13/3 = 20.4 mol PBr3.


20.4 mol PBr3 contain 3.68 x 10^25 bromine atoms.

Answer 2

Answer:

20.4 moles of $PBr_{3}$ contain $3.68\times 10^{25}$ bromine atoms

Explanation:

1 molecule of $PBr_{3}[/tex contains one P atom and three Br atomsSo 1 mol of [tex]PBr_{3}[/tex contains 3 moles of Br.We know that 1 mol = [tex]6.023\times 10^{23}$ molecules

So, there are $(3\times 6.023\times 10^{23})$ atoms of Br in 1 mol of $PBr_{3}$

Hence $3.68\times 10^{25}$ bromine atoms are present in $\frac{3.68\times 10^{25}}{3\times 6.023\times 10^{23}}$ moles of $PBr_{3}$ or 20.4 moles of $PBr_{3}$