# Question #5f823

##### 1 Answer

#### Explanation:

The idea here is that you need to use the *underlying principle* of a **dilution**, which tells you that the concentration of a solution can be **decreased** by

keeping thenumber of moles of soluteconstantincreasingthe volume of the solution

Mathematically, this can be expressed by using the *molarities* and *volumes* of the stock solution, which is the solution you're starting with, and the of the target solution

#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

So, your starting solution has a volume of

#{(V_1 = "100.0 mL"), (c_1 = "1.35 M"):}#

The volume of the **target solution** will include the added water

#V_2 = "100.0 mL" + "650.0 mL" = "750.0 mL"#

Rearrange the above equation to solve for *diluted* solution.

#c_1V_1 = c_2V_2 implies c_2 = V_1/V_2 * c_1#

Plug in your values to find

#c_2 = (100.0 color(red)(cancel(color(black)("mL"))))/(750.0color(red)(cancel(color(black)("mL")))) * "1.35 M" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.180 M")color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.