What is the freezing point of a solution that contains 3.5 mol of the ionic solid BaCl2 dissolved in 500 grams of water?

1 Answer
Junaid Mirza
Nov 28, 2017

#-39.06 °C#

Explanation:

#"Molality" = "Moles of solute"/"Mass of solvent (in kg)"#

Molality of given solution#= n_("BaCl"_2)/"Mass of water" = "3.5 mol"/"0.5 kg" = "7 m"#

#DeltaT_f = T_f - T_i = -iK_fm#

  • #i# of #"BaCl"_2# is 3
  • #K_f# of water is #(1.86 °C)/"m"#
  • #m# is Molality of solution
  • Freezing point of pure water (#T_i#) is #0°C#

#T_f - 0°C = -3 × (1.86 °C)/cancel"m" × 7 cancel"m" = -39.06°C#

#T_f = -39.06°C#

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