# A 5.00 mL sample of an 18.2 M solution is diluted to 0.1100 L. Then 3.00 mL of this solution is...

## Question:

A 5.00 mL sample of an 18.2 M solution is diluted to 0.1100 L. Then 3.00 mL of this solution is diluted to 0.700 L. What is the final concentration?

## Dilution:

Increasing volume of dissolving medium cause dilution. Dilution always results in a diluted solution starting from a concentrated one. Since, moles are not affected by dilution, it is possible to find the concentration of prepared diluted solution.

Given Data:

• The initial molarity of solution 1 is 18.2 M.
• The initial volume of solution 1 is 5 mL.
• The final volume of solution 2 is 0.1100 L.
• The volume of solution 2 taken for second dilution is 3.00 mL.
• The final volume of solution 3 is 0.700 L.

The concentration of solution 2 obtained after diluting solution 1 is calculated by the formula shown below.

{eq}{M_1}{V_1} = {M_2}{V_2} {/eq}

Where,

• {eq}{M_1} {/eq} is the molarity of solution 1.
• {eq}{M_2} {/eq} is the molarity of solution 2.
• {eq}{V_1} {/eq} is the volume of solution 1.
• {eq}{V_2} {/eq} is the volume of solution 2.

Substitute the values in above formula.

{eq}\begin{align*}18.2\;{\rm{M}}\; \times 5 \times {10^{ - 3}}\;{\rm{L}} &= {M_2} \times 0.1100\;{\rm{L}}\\{M_2} &= \dfrac{{18.2\;{\rm{M}} \times 5\; \times {{10}^{ - 3}}{\rm{L}}}}{{0.1100\;{\rm{L}}}}\\ &= \;0.827\;{\rm{M}}\end{align*} {/eq}

Hence, the molarity of solution 2 obtained after first dilution is 0.827 M.

The concentration of solution 3 obtained after diluting solution 2 is calculated by the formula shown below.

{eq}{M_2}{V_2} = {M_3}{V_3} {/eq}

Where,

• {eq}{M_3} {/eq} is the molarity of solution 3.
• {eq}{V_3} {/eq} is the volume of solution 3.

Substitute the values in above formula.

{eq}\begin{align*}0.827\;{\rm{M}}\; \times 3 \times {10^{ - 3}}\;{\rm{L}} &= {M_2} \times 0.700\;{\rm{L}}\\{M_2} &= \dfrac{{0.827\;{\rm{M}}\; \times 3 \times {{10}^{ - 3}}\;{\rm{L}}}}{{0.700\;{\rm{L}}}}\\ &= \;3.544 \times {10^{ - 3}}\;{\rm{M}}\end{align*} {/eq}

Hence, the molarity of solution 3 obtained after second dilution is {eq}\boxed{3.544 \times {10^{ - 3}}\;{\rm{M}}} {/eq}.