# What is the molarity of 40.0 grams of sodium hydroxide in 1.50 L of solution?

$\text{Molarity}$ $=$ $\text{Amount of substance (in moles)"/"Volume of solution (in litres)}$
So, $\frac{40.0 \cdot \cancel{g}}{40 \cdot \cancel{g} \cdot m o {l}^{-} 1} / \left(1.50 \cdot L\right)$ $=$ $0.667$ $m o l \cdot {L}^{-} 1$ with respect to sodium hydroxide.