# What is the molality of a "H"_2"SO"_4 solution which is "96% w/w" ? M_M = "98 g/mol"

Feb 26, 2018

Here's what I got.

#### Explanation:

For starters, it's worth pointing out that the terms solute and solvent do not really apply here because sulfuric acid is miscible in water, meaning that the two liquids can be mixed in all proportions to form a homogeneous mixture, i.e. a solution.

So the classical representation of a solvent being the substance that is present in a greater amount does not really make sense here because sulfuric acid, which we will take as being the solute, is actually present in the greater amount in this solution than water, which we take as being the solvent.

So for the sake of simplicity, I will refer to sulfuric acid as the solute and to water as the solvent, but keep in mind that these terms are not relevant when dealing with miscible liquids.

$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

As you know, molality is defined as the number of moles of solute present for every $\text{1 kg" = 10^3 quad "g}$ of solvent.

This means that your first goal here will be to figure out the mass of sulfuric acid, the solute, present for every $\text{1 kg}$ of water, the solvent.

You know that the solution is $\text{96% m/m}$, which implies that it contains $\text{96 g}$ of sulfuric acid for every $\text{100 g}$ of the solution.

To make the calculations a little easier, let's pick a sample of this solution that has a mass of $\text{100 g}$. This sample will also contain $\text{96 g}$ of sulfuric acid, so you can say that the mass of water will be equal to

m_ ("H"_ 2"O") = "100 g" - "96 g"

m_ ("H"_ 2"O") = "4 g"

Now, you know that you get $\text{96 g}$ of sulfuric acid for every $\text{4 g}$ of water, so you can say that ${10}^{3}$ $\text{g}$ of water will contain

10^3 color(red)(cancel(color(black)("g H"_ 2"O"))) * ("96 g H"_ 2"SO"_ 4)/(4color(red)(cancel(color(black)("g H"_ 2"O")))) = "24,000 g H"_ 2"SO"_ 4

Next, use the molar mass of sulfuric acid to convert the number of grams to moles.

${\text{24,000" color(red)(cancel(color(black)("g"))) * ("1 mole H"_ 2"SO"_ 4)/(98color(red)(cancel(color(black)("g")))) = "244.9 moles H"_2"SO}}_{4}$

Since this represents the number of moles of solute present for every ${10}^{3} \quad \text{g" = "1 kg}$ of solvent, you can say that the molality of the solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molality = 240 mol kg}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the percent concentration.