Calculate the molality of a solution made by dissolving a compound in camphor if the freezing point of the solution is 162.33 degrees C: the melting point of pure camphor was measured to be 178.43 degrees C. K sub f for camphor is 37.7 degrees C/m sub c?

Please explain the equation used for this problem and how to get to the answer. The answer isn't needed to be provided.

1 Answer
May 2, 2017

Answer:

#0.42" molal"#

Explanation:

To make any sense of this question, let's first write down what we are given.

Given
- #color(green)("Pure camphor freezing point" = 178.43^@C)#
- #color(green)(K_(f) = 37.7^@C/m)#
- #color(green)("Solution's freezing point" = 162.33^@C)#

#color(white)(aaaaaa)"Now, what is this problem really saying and asking?"#

Well first off, we should understand the concept being tested - colligative properties.

Colligative properties describe the physical properties of solutions that depend only on the amount of solutes dissolved in solutions and not the type dissolved. Here, melting point/freezing point is being affected. When you add solutes to a pure solution, the freezing point of the solution decreases.

Clearly, the pure camphor and the camphor with the added compound have different freezing points.

#color(white)(aaaaaaaaaaaaaa)color(blue)"pure camphor" = 178.43^@C#
#color(white)(aaaaa)color(red)"camphor with compound" = 162.33^@C#

#color(white)(aaaaaaa)DeltaT = 178.43^@C-162.33^@C = 16.10^@C#

#---------------------#

To find the molality of the compound we need the following formula

#color(white)(aaaaaaaaaaaaaaa)color(magenta)(DeltaT = iK_(f)m)#

Where

  • #DeltaT = "change in the temperature" (color(white)(a)^@C)#
  • #i = "vant Hoff factor"#
  • #K_f = "molal freezing point constant" ((color(white)(a)^@C)/(m))#
  • #m = "molality" ("moles of solute"/"1 kg of solvent" or m)#

The vant Hoff factor of a nonelectrolyte solute is 1.

#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaa)#

Plugin, rearrange and solve (ignored units for simplicity sake)

  • #color(magenta)(DeltaT = iK_(f)m)#

  • #16.10 = (1)(37.7)(m)#

  • #(16.10)/[(1)(37.7)] = m#

  • #"molality" = 0.42" m, where m equals molal"#

#"Answer": "molality" = 0.42" m#