Question

Question #7233e

Answer 1

`"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, `"H"_2"SO"_4`, present in `"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

`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 `80%` weight by volume percent concentration, `"w/v"`. That means that for every `"100 mL"` of solution you get `"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

`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` moles of solute per kilogram of solvent

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