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

1 Answer
Jul 13, 2017

Answer:

#"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"#