# If 25.0 g of K_2CO_3, potassium carbonate, are dissolved in 450 cm^3 of solution, what is the molarity?

Approx. $0.4 \cdot m o l \cdot {L}^{-} 1$
$\text{Molarity}$ $=$ $\text{Moles of potassium carbonate (mol)"/"Volume of solution (L)}$
$= \frac{\frac{25 \cdot g}{138.21 \cdot g \cdot m o {l}^{-} 1}}{0.450 \cdot L} \cong 0.4 \cdot m o l \cdot {L}^{-} 1$
Note that when we do the calculation this wa we automatically get appropriate units. We seek the molarity, which should properly have units of $m o l \cdot {L}^{-} 1$, and lo and behold, the expression gives us such units as the grams cancel out, and the $m o {l}^{-} 1$ is inverted to give $\frac{1}{\frac{1}{m o l}} = m o l$.......