# Question c92ff

May 30, 2017

Here's what I got.

#### Explanation:

Molality is defined as the number of moles of solute present for every $\text{1 kg}$ of solvent.

You know that your solution has a molarity of $\text{3 M}$, which implies that every $\text{1 L}$ of solution contains $3$ moles of sodium chloride, the solute.

To make the calculations easier, let's pick a $\text{1-L}$ sample of this solution. Use the molar mass of sodium chloride to calculate how many grams of solute it contains

3 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "175.32 g"

Now, use the density of the solution to determine its mass.

1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1.25 g"/(1color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the density of the solution")) = "1250 g"

To find the mass of water present in the sample, subtract the mass of the solute from the mass of the solution

${m}_{\text{water" = "1250 g" - "175.32 g}}$

${m}_{\text{water" = "1074.68 g}}$

This is equivalent to

1074.68 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "1.07468 kg"

So, if your solution contains $3$ moles of sodium chloride for every $\text{1.07468 kg}$ of water, the solvent, you can say that $\text{1 kg}$ of water will contain

1 color(red)(cancel(color(black)("kg water"))) * "3 moles NaCl"/(1.07468 color(red)(cancel(color(black)("kg water")))) = "2.79 moles NaCl"#

Therefore, the molality of the solution will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molality = 2.8 mol kg}}^{- 1}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the molarity of the solution.