Question

Calculate molality of a `"1-L"` solution of `93%` `"H"_2"SO"_4` mass by volume. The density of the solution is `"1.84 g/mL"` ?

Answer 1

Here's what I got.

Explanation:

For starters, you know that you're dealing with a `93%` `"m/v"` sulfuric acid solution, which implies that every `"100 mL"` of this solution contain `"93 g"` of sulfuric acid, the solute.

To make the calculations easier, pick a sample of this solution that has a volume of `"100 mL"`. This sample will contain `"93 g"` of sulfuric acid and have a mass of

`100 color(red)(cancel(color(black)("mL solution"))) * "1.84 g"/(1color(red)(cancel(color(black)("mL solution")))) = "184 g"`

Now, in order to find the molality of the solution, you need to figure out the number of moles of sulfuric acid present for every `"1 kg"` of water, the solvent.

This sample contains

`"184 g " - " 93 g" = "91 g"`

of water, which implies that a sample that contains `"1 kg" = 10^3` `"g"` of water would contain

`10^3 color(red)(cancel(color(black)("g water"))) * ("93 g H"_2"SO"_4)/(91 color(red)(cancel(color(black)("g water")))) = "1021.98 g H"_2"SO"_4`

To convert the mass of sulfuric acid to moles, use the molar mass of the compound.

`1021.98 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"SO"_4)/(98.079 color(red)(cancel(color(black)("g")))) = "10.42 moles H"_2"SO"_4`

Since this represents the number of moles of sulfuric acid present for every `"1 kg"` of this solution, you can say that the molality of the solution is equal to

`color(darkgreen)(ul(color(black)("molality = 10.4 mol kg"^(-1))))`

I'll leave the answer rounded to three sig figs, but keep in mind that the value you have for the concentration of the solution justifies only two sig figs for the answer.

SIDE NOTE: Notice that the fact that the solution given to you has a volume of `"1 L"` is irrelevant to the calculations, meaning that as long as you know the percent concentration of the solution and its density, you can determine its molality.