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.........

Related questions
See all questions in Molarity
Impact of this question
1602 views around the world
You can reuse this answer
Creative Commons License