# What volume should you dilute 124 mL of an 8.10 M CuCl_2 solution so that 52.0 mL of the diluted...

## Question:

What volume should you dilute {eq}124 \ mL {/eq} of an {eq}8.10 \ M \ CuCl_2 {/eq} solution so that {eq}52.0 \ mL {/eq} of the diluted solution contains {eq}5.9 \ g \ CuCl_2 {/eq} ?

## Dilution of Solution

The dilution process is commonly done by adding water to a concentrated solution to make a dilute solution. This is basically defined by an equation given as {eq}M_1V_1 = M_2V_2 {/eq}, where {eq}M_1{/eq} is the intial concentration, {eq}V_1{/eq} is the intial volume, {eq}M_2{/eq} is the final concentration, {eq}V_2{/eq} is the intial volume.

Solving the volume of the diluted solution using

{eq}M_1V_1 = M_2V_2 {/eq}

where,

• {eq}M_1 = 8.10\ M {/eq} is the intial concentration,
• {eq}V_1 = 124\ mL {/eq} is the intial volume
• {eq}M_2 = \dfrac{(5.9\ g)\cdot \left(\dfrac{1}{134.45\ \frac{g}{mol}}\right)}{(52.0\ mL)\cdot \left(\dfrac{1\ L}{1000\ mL}\right)} = 0.84\ M {/eq} is the final concentration
• {eq}V_2 {/eq} is the final volume.

Note that the molar mass of {eq}CuCl_2 {/eq} is {eq}134.45\ \frac{g}{mol} {/eq}.

Putting {eq}V_2 {/eq} on one side of the equation and substituting all given values. Therefore,

{eq}V_2 = \dfrac{M_1V_1}{M_2}\\ V_2 = \dfrac{(8.10\ M)(124\ mL)}{0.84\ M}\\ \boxed{V_2 = 1195.7\ mL} {/eq}