If #k_f=1.86*""^(@)C*mol^-1*kg# for water, what is the freezing point of a solution composed of #255*kg# #CaCl_2#, and #33.8*L# of water?

Question

Caleb Adams · Accepted Answer

Well, #DeltaT_"fusion"=k_fxxcxxi# where.................

Where #k_f# is the #"molal freezing point depression constant"#, which you have kindly given, #c="molal concentration"# in #mol*kg^-1#, and #i#, the so-called #"van't Hoff factor"# which reflects the speciation of an ionic solid in a solvent. Clearly, here, #i=3#. Why? Because in solution, #"calcium chloride"# speciates to give three ions...... #CaCl_2(aq) stackrel(H_2O)rarrCa^(2+) + 2Cl^(-)# And so............... #Delta_"fusion"=1.86*""^(@)C*m^-1xx3xx((2.55xx10^3*g)/(110.98*g*mol^-1))/(33.8*kg)=3.80# #""^@C#. This reflects the depression of the fusion point (the melting point) with RESPECT to PURE SOLVENT. And thus #"fusion point"# #-=# #-3.80# #""^@C#. In practice, the melting point probably would not be depressed so far. And this is the province of measurement.

Luna Chen · Answer

undefined Use the formula: $\Delta T_f = i \cdot K_f \cdot m$
$i$ for CaCl₂ is 3 (calcium chloride dissociates into three ions)
$\Delta T_f = 3 \cdot 1.86 \, ^\circ C \cdot \frac{mol}{kg} \cdot 255 \, kg$
$\Delta T_f \approx 1409.1 \, ^\circ C$
Freezing point depression: $0 - \Delta T_f$
$0 - 1409.1 \approx -1409.1 \, ^\circ C$