Question

Question #e16a6

Answer 1

Here's what I got.

Explanation:

The key here is the difference between the ebullioscopic constant, `K_b`, and the cryoscopic constant, `K_f`, of water. the solvent of the solution.

You know that you have--see here for the source

• `K_b = 0.512^@"C kg mol"^(-1)`
• `K_f = 1.86^@"C kg mol"^(-1)`

Now, the boiling-point elevation is calculated using

`DeltaT_"b" = i * K_b * b`

Here

• `i` is the van't Hoff factor
• `b` is the molality of the solution

Similarly, the freezing-point depression is calculated using

`DeltaT_"f" = i * K_f * b`

Since you're dealing with a single solution here, i.e. the same molality and van't Hoff factor, you can divide the two equations to get

`(DeltaT_"b")/(DeltaT_"f") = (color(red)(cancel(color(black)(i))) * K_b * color(red)(cancel(color(black)(b))))/(color(red)(cancel(color(black)(i))) * K_f * color(red)(cancel(color(black)(b))))`

This means that you have

`DeltaT_"f" = K_"f"/K_"b" * DeltaT_"b"`

Now, water has a normal boiling point of `100^@"C"`. In your case, the boiling point of the solution is said to be equal to `100.52^@"C"`, which means that the boiling-point elevation is equal to

`DeltaT_"b" = 100.52^@"C" - 100^@"C" = 0.52^@"C"`

Plug in your values to find the freezing-point depression

`DeltaT_"f" = (1.86 color(red)(cancel(color(black)(""^@"C kg mol"^(-1)))))/(0.512color(red)(cancel(color(black)(""^@"C kg mol"^(-1))))) * 0.52^@"C"`

`DeltaT_"f" = 1.89^@"C"`

Water has a normal freezing point of `0^@"C"`. The freezing-point depression is equal to `1.89^@"C"`, which means that the freezing point of the solution will be `1.89^@"C"` lower than the normal freezing point of pure water.

You thus have

`T_"f sol" = 0^@"C" - 1.89^@"C" = color(darkgreen)(ul(color(black)(-1.89^@"C")))`

SIDE NOTE The difference between this value and the value given to you in (a) is most likely caused by different values used for `K_"b"` and/or `K_"f"`.