Question

Question #45e44

Answer 1

`"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, `"KI"`, it contains.

`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 `"25 mL"` of the diluted solution must contain `"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 `"25.0 mL"` contains `0.01837` 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 `"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 * 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 `"50.0 mL"` of a `"5.00 M"` potassium iodide solution and add enough water to get the volume of the solution to `"340. mL"`, you will make a target solution that contains `"3.05 g"` of solute in `"25.0 mL"`.