# Question #7233e

##### 1 Answer

#### Explanation:

In order to find a solution's **molality**, **number of moles of solute**, which in your case is sulfuric acid, **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

#color(blue)(ul(color(black)("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 *weight by volume percent concentration*, **for every** **of solution** you get

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

#color(blue)(ul(color(black)(m_"solution" = m_"solute" + m_"solvent")))#

will be

#m_"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

#color(darkgreen)(ul(color(black)(b ="8.2 mol kg"^(-1)))) -># you get#8.2# molesof solute perkilogramof solvent

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