Question

What volume (in L) of a 2.46 M magnesium nitrate `Mg(NO_3)_2` solution would be needed to make 275 mL of a 0.758 M solution by dilution?

Answer 1

Under `0.1*L`

Explanation:

We need `275*mL` of a solution that is `0.758*mol*L^-1` with respect to magnesium nitrate.

`"Moles of solute"=0.275*Lxx0.758*mol*L^-1` `=` `0.209*mol`.

So we divide this molar quantity, by the molarity of the mother solution to gives us the volume:

`(0.209*mol)/(2.46*mol*L^-1)` `=` `85*mL`

So we would take this volume and dilute it with fresh solvent to give the required volume. Capisce?