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If 40.0 mL of a 3.0x10^(-5) solution is diluted to 100 mL, what is the resulting concentration?

Aug 26, 2016

${M}_{f} = 1.2 \times {10}^{- 5} M$

Explanation:

When a solution is diluted, the number of mole of particles in the solution does not change.

The relationship between the number of mole $n$ and the molarity $M$ is the following:

$M = \frac{n}{V} \implies n = M \times V$

where, $V$ is the volume of the solution.

If we consider that initially the volume and the molarity are:

${M}_{i} = 3.0 \times {10}^{- 5} M$

${V}_{i} = 40 m L$

What would be the final molarity M_f=? when the final volume is ${V}_{f} = 100 m L$?

since the number of mole does not change, we can write:

${M}_{i} \times {V}_{i} = {M}_{f} \times {V}_{f}$

$\implies {M}_{f} = \frac{{M}_{i} \times {V}_{i}}{{V}_{f}} = \frac{3.0 \times {10}^{- 5} M \times 40 \cancel{m L}}{100 \cancel{m L}} = 1.2 \times {10}^{- 5} M$

Here is a video that discusses the solution preparation and dilution:
Lab Demonstration | Solution Preparation & Dilution.

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