#(i)# What mass of anhydrous #MgI_2# is required to prepare a #400*mL# volume of #0.0250*mol*L^-1# concentration?

#(ii)# Given #8.0*mol*L^-1# #HClO_4(aq)#, what volume is required to prepare #4.00*L# of the acid at #0.250*mol*L^-1# concentration?

1 Answer
Jun 5, 2017

Answer:

Well, in both instances we use the relationship is:

#"Concentration"="Moles of solute"/"Volume of solution"#

Explanation:

WE use the quotient to determine the number of moles of solute we need.......

i.e. if the solution is #0.0250*mol*L^-1# with respect to #MgI_2#, we needs a #400xx10^-3Lxx0.0250*mol*L^-1=0.01*mol# quantity of the anhydrous salt.

And this represents a mass of #0.01*molxx278.11*g*mol^-1=2.78*g#.

And to prepare this solution, we take an accurately measured mass of magnesium iodide (if your salt is the hydrate, we change the mass appropriately), and dissolve it up in a #400*mL# volume of water.

The second solution REQUIRES that we add the ACID to water and NOT WATER to ACID (important! so get it right!).

We require #4.00*Lxx0.250*mol*L^-1=1.0*mol#. And thus we START with #125*mL# of the #8.0*mol*L^-1# solution of #HClO_4(aq)#, and we add this to a #3.875*L# volume of water.