# What NaCl concentration results when 204 mLof a 0.760 M NaCI solution is mixed with 417 mL of a 0.430 M NaCl solution?

Aug 30, 2016

$\left[N a C l\right]$ $\cong$ $0.5 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$\text{Solution 1}$, $\text{moles of NaCl}$ $=$ $204 \times {10}^{-} 3 \cdot L \times 0.760 \cdot m o l \cdot {L}^{-} 1 = 0.155 \cdot m o l$.

$\text{Solution 2}$, $\text{moles of NaCl}$ $=$ $417 \times {10}^{-} 3 \cdot L \times 0.430 \cdot m o l \cdot {L}^{-} 1 = 0.179 \cdot m o l$.

Since we must consider the volumes to be additive when these solutions are mixed, we have the following concentration with respect to $N a C l$.

$\text{Concentration}$ $=$ $\frac{\left(0.155 + 0.179\right) \cdot m o l}{\left(204 + 417\right) \times {10}^{-} 3 \cdot L}$

$\cong$ $0.5 \cdot m o l \cdot {L}^{-} 1$ with respect to $N a C l$.