# What is the molarity of a solution prepared by dissolving 0.450 of HNO_3 in 300 mL of water?

Moles of nitric acid: $\frac{0.450 \cdot g}{63.01 \cdot g \cdot m o {l}^{- 1}}$. This is dissolved in $300$ $m L$ of water. I have assumed that you mean $0.450 \cdot g$ of nitric acid.
$\text{Molarity}$ $=$ $\left(\text{moles of solute")/("volume of solution}\right)$.
$\frac{\frac{0.450 \cdot g}{63.01 \cdot g \cdot m o {l}^{- 1}}}{0.300 \cdot L}$ $=$ ?? $m o l \cdot {L}^{- 1}$
Please do the math, and report the answer in $m o l \cdot {L}^{- 1}$.