# The molecular weight of Na is 58.44 grams/mole. To prepare a 0.50 M solution: how many grams of NaCI would you need in 1,000 mL of water?

$\text{Molarity}$ $=$ $\text{Moles of solute"/"Volume of solution}$, so here we need $29.22 \cdot g$ of $N a C l$
And thus, $\frac{29.22 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{1 \cdot L}$ $=$ $0.500 \cdot m o l \cdot {L}^{-} 1$.