What is the boiling point of an aqueous solution of #"CaCl"_2# with a molal concentration of 1.56 mol/kg?

1 Answer
Ernest Z.
Apr 21, 2017

The boiling point of the solution is 108.705 °C.

Explanation:

The formula for boiling point elevation #ΔT_"b"# is

#color(blue)(bar(ul(|color(white)(a/a)ΔT_"b" = iK_"b"mcolor(white)(a/a)|)))" "#

where

  • #i# is the van't Hoff #i# factor
  • #K_"b"# is the molal boiling point elevation constant
  • #b# is the molality of the solution

In this problem,

#i = 3#, because 1 mol of #"CaCl"_2# gives 3 mol of ions in solution.
#K_text(b) = "1.86 °C·mol·kg"^"-1"#
#b = "1.56 mol·kg"^"-1"#

#ΔT_"b" = iK_"b"b = 3 × "1.86 °C·"color(red)(cancel(color(black)("kg·mol"^"-1"))) × 1.56 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "8.705 °C"#

#T_"b" = T_"b"^@ + ΔT_"b" = "100 °C + 8.705 °C" = "108.705 °C"#

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