Question

Question #421fb

Answer 1

You're on the right track here.

Explanation:

You are indeed correct, the concentration of solution `"B"` is

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

That is the case because you've performed a `1:10` dilution of solution `"A"`. This implies that the concentration of the diluted solution, i.e. solution `"B"`, is `10` times lower than the concentration of the stock solution, i.e. solution `"A"`.

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 `"DF"` is the 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")))`