# Question 7233e

Nov 14, 2016

$\text{8.2 molal}$

#### Explanation:

In order to find a solution's molality, $b$, you need to know the number of moles of solute, which in your case is sulfuric acid, ${\text{H"_2"SO}}_{4}$, present in $\text{1 kg}$ of solvent.

The first thing to do here will be to convert the volume of the solution to mass by using its density. Since you know that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 L" = 10^3"mL}}}}$

you can say that your solution will have a mass of

1 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1.8 g"/(1color(red)(cancel(color(black)("mL")))))^(color(blue)("the given density")) = "1800 g"

Now, this solution is said to have a 80% weight by volume percent concentration, $\text{w/v}$. That means that for every $\text{100 mL}$ of solution you get $\text{80 g}$ of sulfuric acid.

You can thus say that this sample contains

10^3 color(red)(cancel(color(black)("mL solution"))) * ("80 g H"_2"SO"_4)/(100color(red)(cancel(color(black)("mL solution")))) = "800 g H"_2"SO"_4

This means that the mass of the solvent, which is given by

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{m}_{\text{solution" = m_"solute" + m_"solvent}}}}}$

will be

${m}_{\text{water" = overbrace("1800 g")^(color(purple)("mass of solution")) - overbrace("800 g")^(color(orange)("mass of solute")) = "1000 g}}$

Next, use the molar mass of sulfuric acid to calculate how many moles you have in your sample

800 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"SO"_4)/(98.08color(red)(cancel(color(black)("g")))) = "8.157 moles H"_2"SO"_4

Since you know that the solution contains

1000 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "1 kg"#

of water, you can say that its molality will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{b = {\text{8.2 mol kg}}^{- 1}}}} \to$ you get $8.2$ moles of solute per kilogram of solvent

I'll leave the answer rounded to two sig figs.