# Question #421fb

##### 1 Answer

#### Answer:

You're on the right track here.

#### Explanation:

You are indeed correct, the concentration of solution

#["B"] = 2.351 * 10^(-2)"M"#

That is the case because you've performed a **times lower** than the concentration of the stock solution, i.e. solution

The equation you have to work with

#color(blue)(ul(color(black)(M_1V_1 = M_2V_2)))#

where

#M_1# ,#V_1# represent the molarity and volume of the concentrated solution#M_2# ,#V_2# represent the molarity and volume of the diluted solution

expresses the *underlying concept* of a dilution -- the number of moles of solute **must remain constant**.

Since **molarity** is defined as moles of solute per liters of solution, you can say that

#overbrace(M_1 = n_1/V_1)^(color(blue)("concentrated solution"))" "# and#" " overbrace(M_2 = n_2/V_2)^(color(purple)("diluted solution"))#

In other words, you have

#overbrace(M_1V_1)^(color(blue)("moles of solute in concentrated solution")) = overbrace(M_2V_2)^(color(purple)("moles of solute in diluted solution"))#

Another cool thing to notice here is that you can rearrange this equation to get

#"DF" = M_1/M_2 = V_2/V_1#

Here **dilution factor**, which tells you how many times more concentrated the stock solution was compared to the diluted solution.

In this case, you have

#V_1 = "10.00 mL" " "# and#" "V_2 = "100.00 mL"#

which means that

#"DF" = (100.00 color(red)(cancel(color(black)("mL"))))/(10.00color(red)(cancel(color(black)("mL")))) = color(blue)(10)#

Consequently, you can say that

#"DF" = M_1/M_2 implies M_2 = M_1/"DF"#

This will once again get you

#M_2 = (2.351 * 10^(-1)"M")/color(blue)(10) = color(darkgreen)(ul(color(black)(2.351 * 10^(-2)"M")))#