If the boiling point elevation of an aqueous solution containing a non-volatile nonelectrolyte is 0.48◦C, what is the molality of the solution?

Answer in units of molality.

1 Answer
Truong-Son N.
Jan 15, 2018

I got #"0.94 molal"# to two sig figs.

Boiling point elevation is given by:

#DeltaT_b = T_b - T_b^"*" = iK_bm#,

where:

  • #T_b# is the boiling point of the solvent in the context of the solution (in #""^@ "C"#), and #"*"# indicates pure solvent.
  • #i# is the van't Hoff factor, i.e. the effective number of solute particles that have dissociated from the original solute particles.
  • #K_b = 0.512^@ "C"cdot"kg/mol"# is the boiling point elevation constant of water.
  • #m# is the molality of the solution in #"mols solute/kg solvent"#.

The solute is a nonelectrolyte, so #i = 1#. It also is nonvolatile, so it does not alter the vapor pressure other than through means of concentration at the surface.

Therefore,

#DeltaT_b = 0.48^@ "C" = (1)(0.512^@ "C"cdot"kg/mol")cdot m#

#=> color(blue)(m) = (0.48^@ "C")/((1)(0.512^@ "C"cdot"kg/mol"))#

#=# #color(blue)(0.93_8 " mols solute/kg solvent")#

Related questions
Impact of this question
5475 views around the world
You can reuse this answer
Creative Commons License