# 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"# ?

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

For starters, you know that you're dealing with a

To make the calculations easier, pick a sample of this solution that has a volume 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**

This sample contains

#"184 g " - " 93 g" = "91 g"#

of water, which implies that a sample that contains

#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 **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.*