# A 66.5*g mass of a solute whose formula mass is 80.05*g*mol^-1 is used to prepare a solution of 315*mL volume. What is its molarity?

$\text{Molarity} \cong 2.7 \cdot m o l \cdot {L}^{-} 1$
We want the quotient, $\text{moles of stuff (mol)"/"volume of solution (L)}$ to give the $\text{molarity}$.
And thus $\frac{\frac{66.5 \cdot g}{80.05 \cdot g \cdot m o {l}^{-} 1}}{315 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1} \cong 2.7 \cdot m o l \cdot {L}^{-} 1$