# Question aaf98

Feb 2, 2016

Here's what I got.

#### Explanation:

From what I can tell by looking at the problem, you don't really need to know the density of water.

Your strategy here will be to pick a sample of this $\text{20.0% w/w}$ sodium hydroxide solution and use its percent concentration by mass to figure out

• the mass of the solute, which in your case is sodium hydroxide
• the mass of the solvent, which in your case is water

So, a solution's percent concentration by mass is defined as the mass of the solute divided by the total mass of the solution, and multiplied by $100$

$\textcolor{b l u e}{\text{% w/w" = "mass of solute"/"mass of solution} \times 100}$

To make the calculations easier, let's pick a $\text{100.0-g}$ sample of this sodium hydroxide solution. This sample will contain

${m}_{N a O H} = \frac{{m}_{\text{solution" * "%w/w}}}{100}$

m_(NaOH) = ("100.0 g" * 20.0)/100 = "20.0 g NaOH"

This means that the sample will contain

${m}_{\text{solution" = m_"water}} + {m}_{N a O H}$

${m}_{\text{water" = "100.0 g" - "20.0 g" = "90.0 g water}}$

Use the molar masses of sodium hydroxide and water to figure out how many moles of each you have in the sample

20.0 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(39.997color(red)(cancel(color(black)("g")))) = "0.500 moles NaOH"

80.0 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "4.441 moles H"_2"O"

The mole fraction of sodium hydroxide is defined as the number of moles of sodium hydroxide divided by the total number of moles present in solution.

${n}_{\text{total" = n_(NaOH) + n_"water}}$

${n}_{\text{total" = "0.500 moles" + "4.441 moles" = "4.941 moles}}$

This means that you have

$\textcolor{b l u e}{{\chi}_{\text{NaOH" = n_(NaOH)/n_"total}}}$

${\chi}_{N a O H} = \left(0.500 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(4.941color(red)(cancel(color(black)("moles}}}}\right) = \textcolor{g r e e n}{0.101}$

To get the molality of the solution, you need to divide the number of moles of solute by the mass of the solvent expressed in kilograms.

$\textcolor{b l u e}{b = {n}_{N a O H} / {m}_{\text{water}}}$

Convert the mass of water from grams to kilograms

80.0 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 80.0 * 10^(-3)"kg"

The molality of the solution will thus be

b = "0.500 moles"/(80.0 * 10^(-3)"kg") = color(green)("6.25 mol kg"^(-1))#