Question

Question #f97c6

Answer 1

Approx......`80*mL`

Explanation:

We need a molar quantity of `(22.0*g)/(213.9*g*mol^-1)=0.103*mol` with respect to `"aluminum nitrate"`.

Because `"concentration"="moles of solute"/"volume of solution (L)"`, we needs the quotient........

`"volume"="moles of solute"/"concentration"`

`=(0.103*mol)/(1.25*mol*L^-1xx10^-3*L*mL^-1)`

`=82.3*mL`, note that `1/(mL^-1)=1/(1/(mL))=mL` as required.........