# What is the molarity of a solution containing 12.0 g of NaOH in 250.0 mL of solution?

$\text{Molarity "="Moles of solute"/"Volume of solution}$ $=$ $\frac{12.0 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{250.0 \times {10}^{-} 3 \cdot L}$ $=$ ??mol*L^-1
$\text{Moles of sodium hydroxide}$ $=$ $\frac{12.0 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1}$.
$\text{Volume of solution}$ $=$ $250.0 \times {10}^{-} 3 \cdot L$