Question

How much water would need to add to $500mL$ of a $2.4M\,KCL$ solution to make a $1.0M$ solution?

Answer 1

This type of question comes under the dilution calculations. To solve this question, first we have to find the molarity of the solution then find how many moles the solution contains. And, finally conclude the volume of water added to the solution.

Complete step by step solution:
According to the concept of Dilution, we can decrease the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.
Now, the molarity of a solution tells us the number of moles of solute present in $1litre$ of that solution:
$c = \dfrac{{{n_{solute}}}}{{{V_{solution}}}}$
In a dilution, we know that the number of moles of solute must remain constant. This means that we can use the molarity and volume of the initial solution to figure out how many moles it contains:
$c = \dfrac{n}{V} \\\ \Rightarrow n = c \times V \\\$
Now,
${n_{KCL}} = 2.4 \times {500.10^{ - 3}} = 1.2{\text{ }}moles\,KCL$
This is exactly how many moles of potassium chloride must be present in the target solution, which means that we have,
$c = \dfrac{n}{V} \\\ \Rightarrow V = \dfrac{n}{c} \\\$
Again, ${V_{t\arg et}} = \dfrac{{1.2}}{{1.0}} = 1.2L$
So, we need the volume of the target solution to be equal to
${V_{t\arg et}} = 1.2L = 1.2 \times {10^3}mL = 1200mL$
This means that we must add water:
${V_{added}} = 1200mL - 500mL = 700mL$
Hence, $700mL$ is added to our initial solution to get its molarity down from $2.4M$ to $1.0M$ .

Note:
The essential explanation you start with a concentrated arrangement and afterward weaken it to make a weakening is that it is unthinkable to precisely gauge solute to set up a weakened arrangement, so there would be a huge level of blunder in the concentration value.