Question #6469b

1 Answer
Jul 8, 2015

The molality of the solution is 2.29 molal.


There are a couple of things wrong with your question

  • you know that the solution's molarity is 2.05 M, so I presume that you're interested in its molality;
  • the density of acetic acid solutions doesn't come close to 1.20 g/mL (pure acetic acid has a density of 1.05 g/mL), so I presume that you meant to write 1.02 g/mL.

This being said, you need to focus on how to get the solution's molality, which is defined as moles of solute, in your case acetic acid, per kilograms of solvent, in your case water.

Let's assume a 1-L sample of your solution. This solution would contain

#C = n/V => n = C * V#

#n_"acetic acid" = "2.05 M" * "1 L" = "2.05 moles acetic acid"#

In order to get the mass of water, you need to use the solution's density.

#1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1.02 g"/(1cancel("mL")) = "1020 g"#

Since the solution contains 2.05 moles of acetic acid, you can use its molar mass to determine how many grams of acetic acid you have.

#2.05cancel("moles") * "60.05 g"/(1cancel("mole")) = "123.1 g acetic acid"#

This measn that the solution will contain

#m_"sol" = m_"water" + m_"acetic acid"#

#m_"water" = m_"sol" - m_"acetic acid" = 1020 - 123.1 = "896.9 g water"#

This means that the solution's molality will be

#b = n_"acetic acid"/m_"water"#

#b = "2.05 moles"/(869.9 * 10^(-3)"kg") = "2.2857 molal"#

Rounded to three sig figs, the answer will be

#b = color(green)("2.29 molal")#