Question 421fb

Jan 28, 2017

You're on the right track here.

Explanation:

You are indeed correct, the concentration of solution $\text{B}$ is

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

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

The equation you have to work with

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{M}_{1} {V}_{1} = {M}_{2} {V}_{2}}}}$

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

${\overbrace{{M}_{1} {V}_{1}}}^{\textcolor{b l u e}{\text{moles of solute in concentrated solution")) = overbrace(M_2V_2)^(color(purple)("moles of solute in diluted solution}}}$

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

$\text{DF} = {M}_{1} / {M}_{2} = {V}_{2} / {V}_{1}$

Here $\text{DF}$ 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

${V}_{1} = \text{10.00 mL" " }$ and $\text{ "V_2 = "100.00 mL}$

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

$\text{DF" = M_1/M_2 implies M_2 = M_1/"DF}$

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")))#