# How would you calculate the molarity of the following solution: 0.900 mol of Na_2S in 1.10 L of solution?

$\text{Molarity}$ $=$ $\text{Moles of sodium sulfide"/"Volume of solution}$
$=$ $\frac{0.900 \cdot m o l}{1.10 \cdot L}$ $=$ $0.818 \cdot m o l \cdot {L}^{-} 1$ with respect to $\text{sodium sulfide}$. What is the concentration with respect to $N {a}^{+} \left(a q\right)$ ion?
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