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#?

This happens because salt completely dissociates in aqueous solution, while sugar does not. This aspect is important when dealing with colligative properties because the number of particles that you place in solution will differ for the aforementioned compounds.

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.

Sugar, on the other hand, does not dissociate in aqueous solution, which means that 1 mole of sugar will produce, well, 1 mole of sugar in solution.

This is important because of the equation we use to determine freezing point depression

#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.