# Question 41e95

Nov 19, 2015

$\text{0.0433 M}$

#### Explanation:

The idea here is that the number of moles of potassium nitrate will remain constant after you dilute the initial solution.

This is what a dilution essentially means - keeping the number of moles of solute constant, while increasing the volume of the solution.

So, use the molarity and volume of the initial solution to figure out how many moles of potassium nitrate it contained

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{K N {O}_{3}} = {\text{0.866 M" * 25.0 * 10^(-3)"L" = "0.02165 moles KNO}}_{3}$

Since you're only adding water to this initial solution, this is exactly how many moles of potassium nitrate the target solution will contain.

If the volume of the solution is now equal to $\text{500.0 mL}$, then the concentration of the solution will be

c = "0.02165 moles"/(500.0 * 10^(-3)"L") = color(green)("0.0433 M")

This is what the formula for dilution calculations actually means. It's all about keeping the number of moles of solute constant

${\overbrace{\textcolor{b l u e}{{c}_{1} {V}_{1}}}}^{\textcolor{red}{\text{moles of solute in initial solution")) = overbrace(color(blue)(c_2 V_2))^(color(green)("moles of solute in target solution}}}$

If you use the formula, you will get the exact same result, since the same principle is at work

${c}_{1} {V}_{1} = {c}_{2} {V}_{2} \implies {c}_{2} = {V}_{1} / {V}_{2} \cdot {c}_{1}$

c_2 = (25.0color(red)(cancel(color(black)("mL"))))/(500.0color(red)(cancel(color(black)("mL")))) * "0.866 M"

c_2 = 1/20 * "0.866 M" = color(green)("0.0433 M")#