# If water is added to 50 ml of a 0.04M solution so that it fills a 200 mL beaker, what is the final concentration?

Sep 28, 2016

Final concentration $=$ $0.01 \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution in litres}$

$=$ $\frac{50 \times {10}^{-} 3 \cdot L \times 0.04 \cdot m o l \cdot {L}^{-} 1}{200 \times {10}^{-} 3 L}$

$=$ $\frac{50 \times \cancel{{10}^{-} 3} \cdot \cancel{L} \times 0.04 \cdot m o l \cdot {L}^{-} 1}{200 \times \cancel{{10}^{-} 3 L}}$

$=$ $\frac{1 \times \cancel{{10}^{-} 3} \cdot \cancel{L} \times 0.04 \cdot m o l \cdot {L}^{-} 1}{4 \times \cancel{{10}^{-} 3 L}}$

$=$ $0.01 \cdot m o l \cdot {L}^{-} 1$