# Question d99e7

Jan 24, 2015

The solution's molality is $\text{5.0 molal}$.

So, you know that you are dealing with a $\text{20%}$ by weight solution of $\text{NaOH}$. In order to determine how much $\text{NaOH}$ you actually have, you must use the solution's density, which is not given to you.

You could assume that you have a solution with a density that's a little bigger than water's, maybe $\text{1.1 g/mL}$ or $\text{1.2 g/mL}$, or you could look up the density of a 20% by weight sodium hidroxide solution, which is listed as $\text{1.2191 g/mL}$.

Now, you know what the mass of the solution will be

rho = m/V => m_("solution") = rho * V = "1.2191 g/mL" * "550 mL"

m_("solution") = "670.5 g"

You can now find out how much $\text{NaOH}$ you have

"w/w%" = m_("NaOH")/(m_("solution")) * 100 => m_("NaOH") = ("w/w" * m_("solution"))/100

m_("NaOH") = (20 * 670.5)/100 = "134 g"#

Since molality is expressed in moles of solute divided by the mass of the solution - in kilograms, you'll need the number of $\text{NaOH}$ moles

$\text{134 g" * ("1 mole NaOH")/("40.0 g") = "3.35 moles}$

Thus, the solution's molality is

$\text{b" = ("3.35 moles")/("670.5" * 10^(-3)"kg") = "4.99 molal}$, or $\text{5.0 molal}$ - rounded to two sig figs.