# A 1.5*dm^3 volume of 0.30*mol*dm^-3 NaCl(aq) was mixed with a 2.5*dm^3 volume of 0.70*mol*dm^-3 NaCl(aq). What is the final concentration of NaCl(aq)?

Oct 27, 2016

$\left[N a C l \left(a q\right)\right] = 0.55 \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

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

We have to calculate the quantity of all the ions in solution, and then divide this by the total volume.

$\text{Moles of NaCl(i)}$ $=$ $1.50 \cdot {\mathrm{dm}}^{3} \times 0.30 \cdot m o l \cdot {\mathrm{dm}}^{-} 3 = 0.45 \cdot m o l$

$\text{Moles of NaCl(ii)}$ $=$ $2.50 \cdot {\mathrm{dm}}^{3} \times 0.70 \cdot m o l \cdot {\mathrm{dm}}^{-} 3 = 1.75 \cdot m o l$

And thus $\left[N a C l\right]$ $=$ $\frac{0.45 \cdot m o l + 1.75 \cdot m o l}{4.0 \cdot {\mathrm{dm}}^{3}}$

$=$ $0.55 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$.

Note that $1 \cdot {\mathrm{dm}}^{3}$ $=$ ${\left({10}^{-} 1 m\right)}^{3}$ $=$ ${10}^{-} 3 {m}^{3} = 1 L$. We have (reasonably) assumed the volumes of solution to be additive.