Why does adding 1 mole of salt, #NaCl#, lower the freezing point of a sample of water more than adding 1 mole of sucrose, #C_12H_22O_11#?

Question

Naomi Aronson · Accepted Answer

This occurs because sugar and salt do not completely dissociate in aqueous solution. This is a significant difference when discussing colligative properties because the amount of particles in solution for the two compounds mentioned above will vary. Sodium chloride is an ionic compound that completely dissociates into #"Na"^(+)# and #"Cl"^(-)# ions in aqueous solution. What that means is that 1 mole of sodium chloride will produce 1 mole of #Na^(+)# cations and 1 mole of #Cl^(-)# anions.  However, sugar does not dissociate in aqueous solution, meaning that one mole of sugar in solution will yield one mole of sugar itself. Because of the formula we use to calculate freezing point depression, this is significant. #DeltaT_("freezing") = i * K_f * b#, where #K_f# - the cryoscopic constant - depends on the solvent;
#b# - the molality of the solution; 
#i# - the van't Hoff factor - the number of ions per individual molecule of solute. 
#DeltaT_("freezing")# - the freezing point depression - is defined as 
#T_(F("solvent")) - T_(F("solution"))#.  As you can see, #DeltaT_("freezing")# for two identical solutions that have different van't Hoff factors will be bigger in favor of the one that dissociates into more ions.  In your case, salt has a van't Hoff factor of #i=2# (1 mole of salt produces 2 moles of ions), while sugar has a van't Hoff factor of #i=1# (1 mole of sugar produces 1 mole of sugar).  As a comparison, 1 mole of calcium chloride (#CaCl_2#) will lower the freezing point of a sample of water even more because it dissociates into  #CaCl_(2(aq)) -> Ca_((aq))^(2+) + 2Cl_((aq))^(-)# In this case, the van't Hoff factor will be #i=3# #-># 1 mole of calcium chloride will produce 1 mole of #Ca^(2+)# cations and 2 moles of #Cl^(-)# anions.

Luna Chen · Answer

undefined Adding 1 mole of salt (NaCl) lowers the freezing point of water more than adding 1 mole of sucrose (C12H22O11) because salt dissociates into two ions (Na+ and Cl-) when dissolved in water, which increases the number of solute particles and thus lowers the freezing point more significantly compared to sucrose, which remains as individual molecules in solution.