Question

What is the boiling point of a 0.527 m aqueous solution of LiBr?

Answer 1

`"boiling point" = 100.540^"o""C"`

Explanation:

We're asked to find the new boiling point of a solution with a molal concentration of aqueous `"LiBr"` of `0.527m`.

To do this, we can use the equation

`DeltaT_b = imK_b`

where

• `DeltaT_b` is the change in boiling point of the solution (not necessarily the actual boiling point!)

• `i` is the van't Hoff factor of the solute, which for these purposes is essentially the number of dissolved ions per unit of the solute.

Since `"LiBr"` splits into `"Li"^+` and `"Br"^-`, the van't Hoff factor here would be `color(orange)(2`.

• `m` is the molality of the solution, given as `color(purple)(0.527m`

• `K_b` is the molal boiling point elevation constant for the solvent; the solvent here is water ("aqueous")

`K_b` for water (at `25^"o""C"`) is `color(green)(0.512` `color(green)(""^"o""C/"m` (it may be useful to know this)

We therefore have

• `DeltaT_b = ?`

• `i = color(orange)(2`

• `m = color(purple)(0.527m`

• `K_b = color(green)(0.512` `""^"o""C/"m`

Plugging these into our equation, we have

`DeltaT_b = (color(orange)(2))(color(purple)(0.527)cancel(color(purple)(m)))(color(green)(0.512)(color(green)(""^"o""C"))/(cancel(color(green)(m)))) = color(red)(0.540^"o""C"`

This quantity represents by how much the boiling point increases (colligative properties), so to find the new boiling point, we simply add this to the normal boiling point of water (`100^"o""C"`):

`"boiling point" = 100^"o""C" + color(red)(0.540^"o""C") = color(blue)(100.540^"o""C"`