What is the freezing point of a 1molal and 0.432molal #Zn(H2O)_6 ^(+2)#?

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1 Answer
Mar 30, 2017

I'll do the 1 molal solution. You should then be able to do a 0.432 molal solution.

#DeltaT_f = T_f - T_f^"*" = -iK_fm#,

  • #T_f# is the freezing point, of course, and #T_f^"*"# is that of water.
  • #i# is the number of ions in solution. We ignore ion pairing for simplicity.
  • #K_f = 1.86^@ "C/m"#
  • #m# is the molality, traditionally in units of #"m"# or "molal".

Clearly, water is not an ion, and the hexahydrate acts as the simple cation in water. Thus, #i = 1#, and we simply have:

#DeltaT_f = T_f - 0^@ "C" = color(blue)(T_f)#

#= -(1)(1.86^@ "C/m")("1 m") = color(blue)(-1.86^@ "C")#