# Question 45e44

Jun 27, 2016

$\text{340. mL}$

#### Explanation:

The first thing to do here is use the molarity and volume of the initial solution to determine how many moles of solute, which in your case is potassium iodide, $\text{KI}$, it contains.

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will have

n_("KI initial") = "5.00 mol" * color(red)(cancel(color(black)("L"^(-1)))) * overbrace(50.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

n_("KI initial") = "0.250 moles KI"

Now, you know that $\text{25 mL}$ of the diluted solution must contain $\text{3.05 g}$ of potassium iodide. Use the compound's molar mass to convert this to moles

3.05 color(red)(cancel(color(black)("g"))) * "1 mole KI"/(166.003 color(red)(cancel(color(black)("g")))) = "0.01837 moles KI"

Use the known volume of the dilute solution to figure out its molarity. Since $\text{25.0 mL}$ contains $0.01837$ moles of solute, it follows that the diluted solution has a molarity of

${c}_{\text{diluted" = "0.01837 moles"/(25.0 * 10^(-3)"L") = "0.7348 M}}$

You now know that the initial solution must be diluted to a final concentration of $\text{0.7348 M}$. As you know, diluting a solution implies keeping the number of moles of solute constant while increasing the volume of the solution.

The equation for dilution calculations looks like this

color(blue)(overbrace(c_1 xx V_1)^(color(darkgreen)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(darkgreen)("moles of solute in diluted solution"))

Here

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

Rearrange to solve for ${V}_{2}$, the volume of the diluted solution

${c}_{1} \cdot {V}_{1} = {c}_{2} \cdot {V}_{2} \implies {V}_{2} = {c}_{1} / {c}_{2} \cdot {V}_{1}$

Plug in your values to find

V_2 = (5.00 color(red)(cancel(color(black)("M"))))/(0.7348color(red)(cancel(color(black)("M")))) * "50.0 mL" = "340.22 mL"#

Rounded to three sig figs, the answer will be

${V}_{2} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{340. mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, you know that if you start with $\text{50.0 mL}$ of a $\text{5.00 M}$ potassium iodide solution and add enough water to get the volume of the solution to $\text{340. mL}$, you will make a target solution that contains $\text{3.05 g}$ of solute in $\text{25.0 mL}$.