# What is the molarity of a solution prepared by 45.04 grams of potassium (K_2CO_3) in enough water to make 500.0 mL of solution?

Approx. $0.7 \cdot m o l \cdot {L}^{-} 1$
$\text{Molarity}$ $=$ $\text{Moles of solute"xx1/"Volume of solution}$ $=$
$\frac{45.04 \cdot g}{138.21 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{500.0 \times {10}^{-} 3 L}$ $=$ ??mol*L^-1