What is the new boiling point for an aqueous "0.743 molal" solution of "KCl"? K_b = 0.512^@ "C/m"

1 Answer
Aug 9, 2017

As you should have seen in your book,

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

where:

  • DeltaT_b is the change in boiling point in ""^@ "C", from that of the pure solvent, T_b^"*", to that of the solution, T_b.
  • i is the van't Hoff factor, i.e. the effective number of solute particles in solution.
  • K_b = 0.512^@ "C/m" is the boiling point elevation constant of water.
  • m is the molality of the solution... "mol solute/kg solvent". Is the solute volatile or nonvolatile?

Assuming 100% dissociation...

"KCl"(aq) -> "K"^(+)(aq) + "Cl"^(-)(aq)

and 1 + 1 = 2 ~~ i, so...

color(blue)(T_b) = T_b^"*" + iK_bm

= 100^@ "C" + 2 cdot 0.512^@ "C/m" cdot "0.743 m"

= color(blue)ul(100.761^@ "C")

What was the change in boiling point?