Question #f97c6

1 Answer
anor277
Jun 22, 2017

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.........

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