# What is the molality of a solution made made by dissolving 2 moles of NaOH in 6 kg of water?

Jan 26, 2016

$\text{0.3 molal}$

#### Explanation:

Molality is a measure of a solution's concentration in terms of amount of solute and amount of solvent present in said solution.

More specifically, molality is defined as the number of moles of solute divided by the mass of the solvent expressed in kilograms.

$\textcolor{b l u e}{\text{molality" = "moles of solute"/"kg of solvent}}$

In your case, sodium hydroxide, $\text{NaOH}$ is the solute and water is the solvent.

The problem provides you with all the information you need to find the solution's molality, so just plug in your values to get

$\textcolor{b l u e}{b = {n}_{\text{solute"/m_"solvent}}}$

$b = \text{2 moles"/"6 kg" = 0.333color(white)(a) overbrace("mol kg"^(-1))^(color(purple)("molal")) = "0.333 molal}$

Rounded to one significant figure, the answer will be

$b = \textcolor{g r e e n}{\text{0.3 molal}}$

Alternatively, you can say that because sodium hydroxide is a strong base that dissociates completely in aqueous solution, you will have a solution that is $\text{0.3 molal}$ in sodium cations, ${\text{Na}}^{+}$, and $\text{0.3 molal}$ in hydroxide anions, ${\text{OH}}^{-}$.