Answer:
[tex]54.18 \times 10^{23} \ moles[/tex] in 3 mole of [tex]Al_2(SO_4)_3[/tex]
Explanation:
It is clear that in the given [tex]1\ mole[/tex] of [tex]Al_2(SO_4)_3[/tex] have [tex]3\ ions[/tex] of [tex]SO_4^2^-[/tex]
Therefore 3 moles of [tex]Al_2(SO_4)_3[/tex] will have [tex]3\times3=9 \ ions[/tex] of [tex]SO_4^2^-[/tex]
Since 1 ion of anything is equivalent to [tex]6.02\times10^{23} \ moles[/tex]
Therefore 3 moles of [tex]Al_2(SO_4)_3[/tex] will have [tex]3\times3=9 \ ions[/tex] of [tex]SO_4^2^-[/tex]
Which is equivalent to [tex]9 \times6.02\times10^{23}=54.18\times10^{23} \ moles[/tex]
Thus 3 moles of [tex]Al_2(SO_4)_3[/tex] gives [tex]54.18\times10^{23} \ moles[/tex] of [tex]SO_4^2^-[/tex].
