Question

Can someone help me with this molar mass problem?[It’s the last one]

Answer 1

Answer:

$54.18 \times 10^{23} \ moles$ in 3 mole of  $Al_2(SO_4)_3$

Explanation:

It is clear that in the given $1\ mole$ of $Al_2(SO_4)_3$ have $3\ ions$ of $SO_4^2^-$

Therefore 3 moles of $Al_2(SO_4)_3$ will have $3\times3=9 \ ions$ of   $SO_4^2^-$

Since 1 ion of anything is equivalent to $6.02\times10^{23} \ moles$

Therefore 3 moles of $Al_2(SO_4)_3$ will have $3\times3=9 \ ions$ of   $SO_4^2^-$

Which is equivalent to $9 \times6.02\times10^{23}=54.18\times10^{23} \ moles$

Thus 3 moles of  $Al_2(SO_4)_3$ gives $54.18\times10^{23} \ moles$ of  $SO_4^2^-$.