If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be?

Feb 7, 2016

Approx. $0.10$ $m o l \cdot {L}^{-} 1$

Explanation:

$\text{Concentration}$ $=$ $\left(\text{MOLES of SOLUTE")/("Volume of SOLUTION}\right)$

So all we need to is to calculate the one quantity; $\text{Volume of SOLUTION}$ has been specified to be $150$ $m L$.

So, $\text{MOLES of SOLUTE}$ $=$ $0.15 \cdot m o l \cdot {L}^{-} 1$ $\times$ $100 \times {10}^{-} 3 L$ $=$ ??"mol"; this was our starting solution.

And final $\text{CONCENTRATION}$ $=$

$\frac{0.15 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 100 \times {10}^{-} 3 \cancel{L}}{150 \times {10}^{-} 3 \cdot L}$ $=$ ??mol*L^-1