Question 457e2

Feb 15, 2016

Here's what I got.

Explanation:

All you really need to know here is what it means to dilute a solution. If you understand that concept, you don't need to remember any formula.

So, when you're performing a dilution, you are essentially doing one thing, and that is increasing the volume of a solution.

But the thing to remember here is that the number of moles of solute is always kept constant in a dilution. Always.

In other words, when you're diluting a solution you're adding more solvent, while keeping the amount of solute constant.

As you know, molarity is defined as moles of solute per liter of solution.

$\textcolor{b l u e}{\text{molarity" = "moles of solute"/"liter of solution}}$

Notice here that if the number of moles of solute is kept constant, increasing the volume of the solution by adding more solvent will decrease the concentration, i.e. make the solution more diluted.

If you start with a solution that has a molarity of ${c}_{1}$ and a volume of ${V}_{1}$, you can say that increasing the volume of the solution to ${V}_{2}$ will get you a new molarity ${c}_{2}$

$\textcolor{b l u e}{{\overbrace{{c}_{1} \times {V}_{1}}}^{\textcolor{p u r p \le}{\text{moles of solute before dilution")) = overbrace(c_2 xx V_2)^(color(purple)("moles of solute after the dilution}}}}$

That is the formula for dilution calculations.

Now, a very useful thing to remember is that the dilution factor can be determined by knowing concentrations or volumes of solution.

$\textcolor{b l u e}{\text{D.F.} = {V}_{2} / {V}_{1} = {c}_{1} / {c}_{2}}$

Simply put, to get the dilution factor you either have to know the initial and final concentration of the solution, or the final and initial volume of the solution.

Now let's take a few examples from your list.

• Example 1

Here you are told that the sample has a volume of $5$ drops. The volume of the diluent, which is what you add to dilute the solution, is also equal to $5$ drops.

The diluted volume is what you get after you add the diluent to the sample.

${V}_{\text{diluted" = "5 drops" + "5 drops" = "10 drops}}$

Now, you know that the initial concentration of the sample is $\text{10 M}$. The dilution factor will be

"D.F." = (10 color(red)(cancel(color(black)("drops"))))/(5color(red)(cancel(color(black)("drops")))) = 2

This tells you that you're performing a $1 : 2$ dilution. The final concentration will be

$\text{D.F." = c_1/c_2 implies c_2 = c_1/"D.F." = "10 M"/2 = "5 M}$

• Example 2

This time, you know the diluted volume and the initial volume of the sample. This means that the volume of diluent was

${V}_{\text{diluent" = V_"diluted" - V_"sample}}$

${V}_{\text{siluent" = "40 drops" - "2 drops" = "38 drops}}$

The dilution factor is

"D.F." = (40color(red)(cancel(color(black)("drops"))))/(2color(red)(cancel(color(black)("drops")))) = 20

This time, the final concentration will be

${c}_{2} = {c}_{1} / \text{D.F." = "0.3 M"/20 = "0.015 M}$

• Example 3

This time, you know the dilution factor and the initial concentration of the sample, which means that you can find its final concentration

${c}_{2} = {c}_{1} / \text{D.F." = "5 M"/10 = "0.5 M}$

Now you're going to use a bit of algebra to find the volume of the sample and the diluted volume.

If you take the volume of the sample to be $x$ milliliters, you can say that the diluted volume will be

${V}_{\text{diluted" = x + "5 mL}}$

But since you know the dilution factor, you will have

"D.F." = ((x + 5)color(red)(cancel(color(black)("mL"))))/(xcolor(red)(cancel(color(black)("mL")))) = 10#

This is equivalent to

$x + 5 = 10 x \implies x = \frac{5}{9} = 0.5556$

So, if you start with approximately $\text{0.56 mL}$ and add enough diluent to get the diluted volume to $\text{5.56 mL}$, you performed a $1 : 10$ dilution.

Since this is the exact same approach you will need for the other examples, I will leave them to you as practice.