Question cbbe3

Jan 19, 2016

color(blue)(1.25 %)
I have given both the shortcut method and also explained in detail what makes the shortcut work

Explanation:

$\textcolor{b l u e}{\text{Assumption 1}}$ The concentration is specified by percentage

$\textcolor{b l u e}{\text{Assumption 2}}$ The original solution is 100%

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The wording; "diluted to" is stating the $0.25$ forms part of the $20$

Thus, expressed as a fraction we have: $\frac{0.25}{20}$
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$\textcolor{b l u e}{\text{Changing this to a percentage}}$

$\textcolor{b r o w n}{\text{Shortcut method}}$

color(green)((0.25 -: 20) xx100-> 0.25/20xx100=1.25%)

$: : : : : : : : : : : : : : : : : : : : : : : : : : : : :$

$\textcolor{b r o w n}{\text{Introduction to first principles to demonstrate what percentage is}}$

What $\frac{0.25}{20}$ actually stating is that for every 20 of the whole we have 0.25 of what we are looking at.

To change this to percent we need to have $\frac{\text{some unknown value}}{100}$ which is the standardisation we call percent. That is how many have we got of what we are looking at for every 100 of the whole.

$\textcolor{b r o w n}{\text{Applying first principles to solve the problem}}$

For a moment lets think of the $\frac{0.25}{20}$ as a ratio

If we can retain this ratio but change the bottom number (denominator) to 100 we have our answer

Known: $20 \times 5 = 100$

So if we multiply $\frac{0.25}{20}$ by 1 but in the form of $\frac{5}{5}$ we change the way our number look without actually changing the intrinsic value of the whole.

color(blue)("As a percentage")color(brown)(-> 0.25/20xx5/5= 1.25/100=1.25%)

The %# is just another way of writing: how many out of a count of 100

$\textcolor{b r o w n}{\text{The shortcut just by passes the bit where I multiplied by 1 in the }}$$\textcolor{b r o w n}{\text{form of } \frac{5}{5}}$

Jan 21, 2016

$1 : 80$

Explanation:

Let me offer a more chemical-oriented approach to this problem.

As you know, a solution's molarity is defined as the number of moles of solute divided by the volume of the solution - expressed in liters.

$\textcolor{b l u e}{\text{molarity" = c = "moles of solute"/"liters of solution} = \frac{n}{V}}$

Now, the idea behind diluting a solution is that you can decrease its concentration by

• keeping the number of moles of solute constant
• increasing the volume of the solution by increasing the volume of the solvent

The underlying principle of a dilution

So, your initial sample has a total volume of $\text{0.25 mL}$ and an unknown concentration ${c}_{1}$.

The target solution has a total volume of $\text{20 mL}$ and an unknown concentration ${c}_{2}$.

The number of moles of solute present in these solutions will be

${n}_{1} = {c}_{1} \cdot {V}_{1} \text{ }$ and $\text{ } {n}_{2} = {c}_{2} \cdot {V}_{2}$

Since a dilution implies keeping the number of moles of solute constant

${n}_{1} = {n}_{2}$

you will get

$\textcolor{b l u e}{{c}_{1} {V}_{1} = {c}_{2} {V}_{2}} \to$ the equation for dilution calculations

Your goal is to determine the dilution factor, so you can rearrange the above equation to get

${c}_{1} / {c}_{2} = {V}_{2} / {V}_{1}$

Plug in your values to get

${c}_{1} / {c}_{2} = \left(20 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL"))))/(0.25color(red)(cancel(color(black)("mL}}}}\right) = 80$

Therefore, the initial solution was eighty times as concentrated as the target solution

${c}_{1} = 80 \times {c}_{2}$

And that's how you find the dilution factor. Simply divide the final volume of the solution by the initial volume of the sample

$\textcolor{b l u e}{\text{D"."F" = V_"final"/V_"initial}} \to$ the equation for finding the dilution factor

A dilution factor of $80$ will thus represent a $1 : 80$ dilution.