Question #8956f
1 Answer
Explanation:
The trick here is to realize that you can use the volume of the diluted solution and the volume of the stock solution to determine the dilution factor.
As you know, in order to dilute a solution, you must increase its volume while keeping the *number of moles of solute * constant.
If you start from the molarity of the solution
#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 can use the fact that the number of moles of solute remains unchanged to write
#color(blue)(overbrace(c_1 xx V_1)^(color(green)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(green)("moles of solute in diluted solution"))#
Here
The dilution factor, which essentially tells you much concentrated the stock solution was compared with the dilute solution, will be equal to
#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_1/c_2 = V_2/V_1color(white)(a/a)|)))#
In your case, you know that the volume of stock solution was
#"D.F." = (2.50 color(red)(cancel(color(black)("L"))))/(250.0 * 10^(-3)color(red)(cancel(color(black)("L")))) = 10#
This tells you that the stock solution was
This means that the concentration of the dilute solution is
#"D.F." = c_1/c_2 implies c_2 = c_1/"D.F."#
which in your case is
#c_2 = "6.00 M"/10 = color(green)(|bar(ul(color(white)(a/a)"0.600 M"color(white)(a/a)|)))#
The answer is rounded to three sig figs.
So, by increasing the volume by a factor of