Question 2e57b

Oct 6, 2015

Here's how you could do that.

Explanation:

In order to be able to find this solution's molality, you need to determine how much solute and how much solvent you get in a random sample.

In this case, you know that the solution's percent concentration by mass is $\text{8% NaOH}$. This means that you get $\text{8 grams}$ of sodium hydroxide for every $\text{100 g}$ of solution.

Since molality is defined as moles of solute, in your case sodium hydroxide, divided by the mass of the solvent - expressed in kilograms, you can pick a sample of $\text{1.00 kg}$ of solution to work with.

You know that this much solution will contain

1.00color(red)(cancel(color(black)("kg solution"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "8 g NaOH"/(100color(red)(cancel(color(black)("g solution")))) = "80 g NaOH"

The mass of water in this sample will be

${M}_{\text{sol" = m_"NaOH" + m_"water}}$

${m}_{\text{water" = m_"sol" - m_"NaOH" = "1000 g" - "80 g" = "920 g water}}$

All you need now is the number of moles of sodium hydroxide you get in $\text{80 g}$ of the compound. To find that, use sodium hydroxide's molar mass

80color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "2.0 moles NaOH"

The molality of the solution will thus be

$b = \frac{n}{m} _ \text{water}$

$b = \text{2 moles"/(920 * 10^(-3)"kg") = "2.17 molal}$

If you round this to one sig fig, the number of sig figs you gave for the percent concentration by mass of the solution, you get

$b = \textcolor{g r e e n}{\text{2 molal}}$

SIDE NOTE If you start with a volume of solution, let's say 1.00 L, you need to use the solution's density to find its mass first, then use the approach I showed you here.

http://www.handymath.com/cgi-bin/naohtble3.cgi?submit=Entry

According to this table, an 8% w/w solution will ahve a density of approximately 1.087 g/mL at room temperature. This means that a 1.00-L sample will have a mass of

1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.087 g"/(1color(red)(cancel(color(black)("mL")))) = "1087 g"#

The molality MUST come out the same, regardless of the mass of the sample.

You would then go on to find the mass and number of moles of sodium hydroxide in this much solution

${m}_{\text{NaOH" = "86.96 g" " }}$ and $\text{ "n_"NaOH" = "2.174 moles}$

The molality in this case would once again be

$b = \text{2.174 moles"/((1087 - 86.96) * 10^(-3)"kg water") = "2.17 molal}$