# What is the molality of a solution that contains 31.0 g HCl in 5.00 kg water?

Jan 22, 2016

$\text{Molality}$ $=$ $\left(\text{moles of solute")/("kg of solvent}\right)$

#### Explanation:

As you know, this unit is independent of volume, and depends solely on the amount of substance (in moles) with respect to mass of solvent.

$\text{Molality}$ $=$ $\left(\text{moles of solute")/("5 kg}\right)$
$=$ $\frac{\frac{31.0 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1}}{5.00 \cdot k g} \cong 0.2 \cdot m o l \cdot k {g}^{-} 1$
Assuming that the volume of solution changes marginally (a safe assumption) what is the estimated concentration in $m o l \cdot {L}^{-} 1$?