How many milliliters (mL) of 5 M MgCl2 must be added to 25 mL of a 0.5 M MgCl2 solution to...

Firstly, we will find the molarity concentration of the mixture.

If we assume x mL of 5 M {eq}MgCl_2 {/eq} is added to 25 mL of 0.5 M {eq}MgCl_2 {/eq}, then the molarity concentration of the mixture is:

{eq}M \, = \, \frac{n_{total}}{v_{total}} \\ M \, = \, \frac{n_{MgCl_2 \,1} \, + \, n_{MgCl_2 \, 2}}{v_{MgCl_2 \,1} \, + \, v_{MgCl_2 \, 2}} \\ M \, = \, \frac{(M \, \times \, v)_{MgCl_2 \,1} \, + \,( M \, \times \, v)_{MgCl_2 \, 2} }{v_{MgCl_2 \,1} \, + \, v_{MgCl_2 \, 2}} \\ M \, = \, \frac{(5 \, M \, \times \,x \, mL) \, + \,( 0.5 \, M \, \times \, 25 \, mL) }{x \, mL \, + \,25 \, mL} {/eq}

Secondly, we will write the dilution equation and substitute the molarity of the mixture into the dilution equation.

After the solution is mixed, we dilute the mixture to 1500 mL of 1.75 M {eq}MgCl_2 {/eq}, then

{eq}\displaystyle \begin{align} (M \, \times \, v)_{before \, dilute} & = \, (M \, \times \, v)_{after \, dilute} \\ (\frac{(5 \, M \, \times \,x \, mL) \, + \,( 0.5 \, M \, \times \,25 \, mL) }{x \, mL \, + \,25 \, mL}) \, \times \, (x \, mL \, + \,25 \, mL) & = \, 1.75 \, M \, \times \, 1500 \, mL \\ (5 \, M \, \times \,x \, mL) \, + \,( 0.5 \, M \, \times \,25 \, mL) & = \, 1.75 \, M \, \times \, 1500 \, mL \\ (5 \, M \, \times \,x \, mL) \, + \,12.5 \, mmol & = \, 2625 \, mmol \\ (5 \, M \, \times \,x \, mL) & = \, 2625 \, mmol \, - \, 12.5 \, mmol \\ x \, mL & = \, \frac{2625 \, mmol \, - \, 12.5 \, mmol}{5 \, M} \\ x \, mL & = \, 522.5 \, mL \end{align} {/eq}

Since we assume the volume of 5 M {eq}MgCl_2 {/eq} as x, then the volume of 5 M {eq}MgCl_2 {/eq} is 522.5 mL.

Thirdly, we will find the volume of the water for dilution.

Since the volume of the mixture before dilution is 522.5 mL of 5 M {eq}MgCl_2 {/eq} and 25 mL of 0.5 M {eq}MgCl_2 {/eq}, and volume of the mixture after dilution is 1500 mL of 1.75 M {eq}MgCl_2 {/eq}, then the volume of water is:

{eq}\displaystyle \begin{align} v \ _{before\,dilution} & = \, v \ _{after\, dilution} \\ v \, _ {5 M \, of \, MgCl_2} \, + \, v \ _ {0.5 M \, of \, MgCl_2} \,+ \, v \ _{water} & = \, v \ _ {0.5 M \, of \, MgCl_2} \\ 522.5 \, mL \, + \, 25 \, mL \,+ \, v_{water} & = \, 1500 \, mL \\ v_{water} & = \, 1500 \, mL \, - \, (522.5 \, mL \, + \, 25 \, mL) \\ v_{water} & = \, 952.5 \, mL \end{align} {/eq}

Therefore, the volume of 5 M {eq}MgCl_2 {/eq} is 522.5 mL and the volume of water is 952.5 mL.