# What is the final concentration of a 25.7*mL volume of KmnO_4(aq) that is diluted to a 50.00*mL volume?

Sep 21, 2016

Approx. $1.56 \cdot m o l \cdot {L}^{-} 1$, as you have diluted the starting solution by half.

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution (L)}$, and thus $\text{Concentration}$ has the units $m o l \cdot {L}^{-} 1$. With this relationship we can solve most problems of dilution.

We started with $25.7 \cdot m L$ of a $3.12 \cdot m o l \cdot {L}^{-} 1$ $K M n {O}_{4}$ solution.

$\text{Moles of } K M n {O}_{4}$ $=$ $25.7 \times {10}^{-} 3 \cdot L \times 3.12 \cdot m o l \cdot {L}^{-} 1$ $=$ ??*mol.

This molar quantity was then diluted to a $50.00 \cdot m L$ volume.

$\text{Final concentration}$ $=$ $\frac{25.7 \times {10}^{-} 3 \cdot \cancel{L} \times 3.12 \cdot m o l \cdot \cancel{{L}^{-} 1}}{50.00 \times {10}^{-} 3 \cdot L}$ $=$ ??mol*L^-1