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
#"KCl"(aq) -> "K"^(+)(aq) + "Cl"^(-)(aq)#
and
#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?
