Question #1a27e

1 Answer
Feb 18, 2016


#"8 mol kg"^(-1)#


In order to find a solution's molality, you need to know two things

  • how many moles of solute you have present
  • the mass of the solvent expressed in kilograms

Notice that the problem provides you with the solution's molarity, which as you know tells you how many moles of solute you get per liter of solution.

This means that you can use the solution's molarity to determine how many moles of sulfuric acid are present in solution.

Since molarity is given per liter of solution, you can pick a #"1.0-L"# sample of this #"6-M"# solution to make the calculations easier.

For a #"1.0-L"# sample, molarity and number of moles of solute are interchangeable. This means that your sample will contain #6# 8moles* of sulfuric acid

#1color(red)(cancel(color(black)("L"))) * "6 mol" color(red)(cancel(color(black)("L"^(-1)))) = "6 moles H"_2"SO"_4#

Use the density of the solution to determine the mass of the sample

#1 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1.34 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("given density")) = "1340 g"#

You can calculate how much water is present in the sample by calculating how much sulfuric acid is present in the sample. Use sulfuric acid's molar mass to convert the number of moles to grams

#6color(red)(cancel(color(black)("moles H"_2"SO"_4))) * "98.08 g"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "588.5 g H"_2"SO"_4#

This means that your sample will contain

#m_"sample" = m_"water" + m_(H_2SO_4)#

#m_"water" = "1340 g" - "588.5 g" = "751.5 g water"#

The molaity of the solution will thus be

#color(blue)(b = n_"solute"/m_"solvent")#

#b = "6 moles"/(751.5 * 10^(-3)"kg") = color(green)("8 mol kg"^(-1))#

The answer can only have one significant figure, since that's how many sig figs you have for the molarity of the solution.