Question #7233e

1 Answer
Nov 14, 2016

Answer:

#"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.