Question #45e44

1 Answer
Jun 27, 2016

Answer:

#"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.

http://acidsandbasesfordummieschem.weebly.com/molarity.html

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"#.