# Question #45e44

##### 1 Answer

#### 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,

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(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 **diluted solution** must contain **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 **moles** of solute, it follows that the *diluted solution* has a molarity of

#c_"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 *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

Rearrange to solve for

#c_1 * V_1 = c_2 * V_2 implies V_2 = c_1/c_2 * 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 = color(green)(|bar(ul(color(white)(a/a)color(black)("340. mL")color(white)(a/a)|)))#

So, you know that if you start with