Question

`(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?

Answer 1

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.