# Question #bca11

May 6, 2017

$\text{1 M}$

#### Explanation:

This is a classic example of a dilution problem.

You know that molarity, $c$, is defined as the number of moles of solute present per liter of solution. This can be expressed as

$c = \frac{n}{V}$

where

• $n$ is the number of moles of solute
• $V$ is the volume of the solution

Now, your initial solution has a concentration ${c}_{1}$, contains $n$ moles of solute, and has a volume

${V}_{1} = {\text{250 cm}}^{3}$

Your mixing this solution with ${\text{250 cm}}^{3}$ of water, which implies that the solution will still contain $n$ moles of solute, but this time, the volume will be

${V}_{2} = {\text{250 cm"^3 + "250 cm"^3 = "500 cm}}^{3} = \textcolor{red}{2} \cdot {V}_{1}$

This means that the molarity of the solution, ${c}_{2}$, will be

${c}_{2} = \frac{n}{V} _ 2$

which is equal to

${c}_{2} = \frac{n}{\textcolor{red}{2} \cdot {V}_{1}} = \frac{1}{\textcolor{red}{2}} \cdot \frac{n}{V} _ 1 = \frac{1}{\textcolor{red}{2}} \cdot {c}_{1}$

Therefore, you can say that

${c}_{2} = \frac{1}{\textcolor{red}{2}} \cdot \text{2 M" = "1 M}$

Long story short, doubling the volume of the solution while keeping the number of moles of solute, i.e. diluting the solution by a dilution factor of $2$, constant is equivalent to halving the concentration of the solution.