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?

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`