# What the "millimolarities" with respect to sodium and chloride ions, of a 9.00*g mass of sodium chloride dissolved in a volume of 1*L of solution...?

Jun 25, 2017

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

And here, we have a concentration of $154 \cdot m m o l \cdot {L}^{-} 1$.

#### Explanation:

$\text{Molarity NaCl} = \frac{\frac{9.00 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1}}{1.000 \cdot L} = 0.154 \cdot m o l \cdot {L}^{-} 1$

But we were asked to supply $\text{millimolarity}$ of $N {a}^{+} \left(a q\right)$.

Because sodium chloride reacts in solution to give equimolar $N {a}^{+}$, and $C {l}^{-}$ as the aquated ions....

$\left[N {a}^{+}\right] = \left[N a C l \left(a q\right)\right] = 0.154 \cdot m o l \cdot {L}^{-} 1$

And..........

$\left[N {a}^{+}\right] = 0.154 \cdot m o l \cdot {L}^{-} 1 = 0.154 \cdot \cancel{m o l} \cdot {L}^{-} 1 \times {10}^{3} \cdot m m o l \cdot \cancel{m o {l}^{-} 1} = 154 \cdot m m o l \cdot {L}^{-} 1.$

And what is [Cl^-]??