# Question #2e57b

##### 1 Answer

#### Answer:

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 *for every*

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

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_"sol" = m_"NaOH" + m_"water"#

#m_"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

#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 = n/m_"water"#

#b = "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 = color(green)("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_"NaOH" = "86.96 g" " "# and#" "n_"NaOH" = "2.174 moles"#

*The molality in this case would once again be*

#b = "2.174 moles"/((1087 - 86.96) * 10^(-3)"kg water") = "2.17 molal"#