# Suppose 85 mL of a 2.0 M HCl solution is diluted with pure water to a total volume of 0.15 L....

## Question:

Suppose 85 mL of a 2.0 M HCl solution is diluted with pure water to a total volume of 0.15 L. What is the final concentration of the solution?

## Dilution:

The term dilution is used when adding more amounts of solvent to a concentrated solution to make the solution less concentrated. For example, when 10 g of sugar is dissolved in 1 L of water, it has a specific sugar concentration. By adding 1 L of water, the sugar concentration is reduced.

Given data

• The initial concentration of HCl is 2.0 M.
• The initial volume of the solution is 85 mL.
• The final volume of the solution is 0.15 L.

The standard relationship between mL and L is as given below:

{eq}\rm 1\;L =\rm 1000\;mL {/eq}

The conversion of 85 mL to L is as given below:

{eq}\begin{align*} \rm 85mL &=\rm 85\;mL \times \dfrac{{1\;L}}{{1000\;mL}}\\ &=\rm 0.085\;L \end{align*} {/eq}

The final concentration of the solution is calculated by using the formula as given below:

{eq}\rm {C_1} \times {V_1} =\rm {C_2} \times {V_2} {/eq};

Where,

• {eq}\rm {C_1} {/eq} is the initial concentration of the solution, i.e., 2.0 M.
• {eq}\rm {V_1} {/eq} is the initial volume of the solution, i.e., 0.085 L.
• {eq}\rm {C_2} {/eq} is the final concentration of the solution.
• {eq}\rm {V_2} {/eq} is the final volume of the solution, i.e., 0.15 L.

On rearranging the above equation, we get

{eq}\rm {C_2} =\rm \dfrac{{{C_1} \times {V_1}}}{{{V_2}}} {/eq}

On substituting all the known values in the above equation, we get

{eq}\begin{align*} \rm {C_2} &=\rm \dfrac{{{C_1} \times {V_1}}}{{{V_2}}}\\ &=\rm \dfrac{{2.0\,M \times 0.085\,L}}{{0.15\;L}}\\ &=\rm \dfrac{{0.17\;M}}{{0.15}}\\ &=\rm 1.13\;M \end{align*} {/eq}

Thus, the final concentration of the solution is 1.13 M.