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

$\left(i i\right)$ Given $8.0 \cdot m o l \cdot {L}^{-} 1$ $H C l {O}_{4} \left(a q\right)$, what volume is required to prepare $4.00 \cdot L$ of the acid at $0.250 \cdot m o l \cdot {L}^{-} 1$ concentration?

Jun 5, 2017

Well, in both instances we use the relationship is:

$\text{Concentration"="Moles of solute"/"Volume of solution}$

Explanation:

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

i.e. if the solution is $0.0250 \cdot m o l \cdot {L}^{-} 1$ with respect to $M g {I}_{2}$, we needs a $400 \times {10}^{-} 3 L \times 0.0250 \cdot m o l \cdot {L}^{-} 1 = 0.01 \cdot m o l$ quantity of the anhydrous salt.

And this represents a mass of $0.01 \cdot m o l \times 278.11 \cdot g \cdot m o {l}^{-} 1 = 2.78 \cdot g$.

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 $400 \cdot m L$ 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 $4.00 \cdot L \times 0.250 \cdot m o l \cdot {L}^{-} 1 = 1.0 \cdot m o l$. And thus we START with $125 \cdot m L$ of the $8.0 \cdot m o l \cdot {L}^{-} 1$ solution of $H C l {O}_{4} \left(a q\right)$, and we add this to a $3.875 \cdot L$ volume of water.