# What is the molality of a solution prepared by dissolving 19.9 g of KCI in 750.0 mL of water?

$\text{Molality}$ $=$ $\text{moles of solute"/"kilograms of solvent}$
We assume (reasonably) that we have a mass of $750.0 \cdot g$ of water.
And thus $\text{molality}$ $=$
$\frac{19.9 \cdot g}{74.55 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{0.750 \cdot k g}$ $\cong$ $0.4 \cdot m o l \cdot k {g}^{-} 1$