# What is the molarity of a solution of KNO_3 that contains 404 grams of KNO_3 in 2.00 liters of solution?

Apr 21, 2016

Approx. $2$ $m o l \cdot {L}^{-} 1$.
$\text{Concentration "= "Moles of solute"/"Volume of solution}$
$\text{Concentration } = \frac{404 \cdot \cancel{g}}{101.10 \cdot \cancel{g} \cdot m o {l}^{-} 1} \times \left(\frac{1}{2.00 \cdot L}\right)$ $=$ ??mol*L^-1.