# Question f5c04

Dec 29, 2015

$\text{1070 mL}$

#### Explanation:

You're interested in finding the volume of water needed to reduce the concentration of $\text{800.0 mL}$ of water from $\text{1.75 M}$ to $\text{0.750 M}$.

As you know, molarity is defined as moles of solute divided by liters of solution.

$\textcolor{b l u e}{\text{molarity" = "moles of solute"/"liters of solution}}$

The idea here is that adding water to an aqueous solution will keep the number of moles of solute constant, but will increase the volume of the solution - this is known as diluting a solution.

The molarity of the solution will decrease because the same number of moles of solute will now be distributed in a larger volume.

So, how many moles of solute are present in your initial solution?

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

$n = \text{1.75 M" * 800.0 * 10^(-3)"L" = "1.40 moles}$

Now the question becomes what volume of solution is needed in order for $1.40$ moles of solute to correspond to a molarity of $\text{0.750 M}$.

Rearrange the above equation to solve for $V$, the volume of the target solution

$\textcolor{b l u e}{c = \frac{n}{V} \implies V = \frac{n}{c}}$

V = (1.40 color(red)(cancel(color(black)("moles"))))/(0.750color(red)(cancel(color(black)("moles")))/"L") = "1.8667 L"

So, the target solution must have a volume of $\text{1.8667 L}$. This means that you must add

$V = {V}_{\text{initial" + V_"water}}$

${V}_{\text{water" = "1.8667 L" - 800.0 * 10^(-3)"L" = "1.0667 L}}$

of water to the initial solution to get its concentration to drop from $\text{1.75 M}$ to $\text{0.750 M}$.

Rounded to three sig figs, and expressed in milliliters, the answer will be

V_"water" = color(green)("1070 mL")

Alternatively, you can use the formula for dilution calculations

$\textcolor{b l u e}{{c}_{1} \cdot {V}_{1} = {c}_{2} \cdot {V}_{2}} \text{ }$, where

${c}_{1}$, ${V}_{1}$ - the molarity and volume of the initial solution
${c}_{2}$, ${V}_{2}$ - the molarity and volume of the target solution

${V}_{2} = {c}_{1} / {c}_{2} \cdot {V}_{1}$
V_2 = (1.75 color(red)(cancel(color(black)("M"))))/(0.750 color(red)(cancel(color(black)("M")))) * "800.0 mL" = "1866.7 mL"
V_"water" = "1866.7 mL" - "800.0 mL" = color(green)("1070 mL")#