# If I have 340 mL of a 0.5 M NaBr solution, what will the concentration be if l add 560 mL more water to it?

Apr 29, 2016

Approx. $0.20$ $m o l \cdot {L}^{-} 1$ with respect to $N a B r$

#### Explanation:

$\text{Concentration " = " Moles of Solute"/"Volume of Solution}$

$=$ $\frac{340 \times {10}^{-} 3 \cdot L \times 0.5 \cdot m o l \cdot {L}^{-} 1}{\left(340 + 560\right) \times {10}^{-} 3 L}$

The numerator gives the amount of $N a B r$ in moles in the initial volume. This is divided again by the new volume of the solution $900 \times {10}^{-} 3 \cdot L$.

$=$ ??*mol*L^-1?