# Question 2a118

Jul 21, 2016

Approx. $5 \cdot m m o l \cdot {L}^{-} 1$

#### Explanation:

"Concentration with respect to calcium hydroxide"= (100xx10^-3*Lxx0.0234*mol*L^-1)/(0.500*L)=??

Note that $m m o l$ $=$ ${10}^{-} 3 \cdot m o l$.

Jul 21, 2016

$\frac{1}{5} \cdot \text{0.0234 M}$

#### Explanation:

Here's an alternative approach to keep in mind when dealing with dilutions.

As you know, the underlying principle of a dilution is that you can decrease the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.

This implies that increasing the volume of the solution by a factor, let's say $\text{DF}$, will cause the concentration to decrease by the same factor $\text{DF}$.

This factor is called dilution factor and can be calculated like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you know that the initial volume of the solution, i.e. the volume of the concentrated solution, is equal to $\text{100 mL}$ and that the final volume of the solution, i.e. the volume of the diluted solution, is equal to

0.500 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "500 mL"

This means that the concentrated solution was diluted by a factor of

"DF" = (500 color(red)(cancel(color(black)("mL"))))/(100color(red)(cancel(color(black)("mL")))) = color(blue)(5)#

As a result, the concentration of the diluted solution will be $\frac{1}{\textcolor{b l u e}{5}} \text{th}$ the value of the concentration of the concentrated solution

${c}_{\text{diluted" = 1/color(blue)(5) * "0.0234 M}}$

Jul 21, 2016

I make it 0.00468 M.

#### Explanation:

I tend to think of it like this:

Work out how many moles of solute are contained in the original sample by multiplying the molarity by the number of litres (0.0234 x 0.1) = 0.00234 moles.

The number of moles in the final dilution will be the same, but the total volume will change from 100 ml to 0.5 l .

So the molarity at the end will be 0.00234 / 0.5 which is 0.00468 M.