Which aqeuous solution has a freezing point closest to that of 0.30M C12H22O11?
A) 0.075M AlCl3
B) 0.15M CuCl2
C) 0.30M NaCl
D) 0.60M C6H12O6
I know the answer but I don't know the reason or rationale behind it. Please explain.
I suggest you review Raoult's Law at
The freezing point lowering depends upon the solution molality and the number of ions the solute breaks up into, if any. The sugar molecule C12H22O11 does not dissociate. AgCl3 breaks up into 4 ions, so there will be the same number of ions in solution per unit volume as there are molecules of sugar, 0.30 moles per liter. So my guess is the answer is (A)
delta T = i*Kf * m
where i is the number of ions (particles in solution), K is the same constant for all, and m is molality.
Therefore, we simply multiply i*m to see which = 0.3
AlCl3 = 0.075 x 4 = ??
CuCl2 = 0.15 x 3 = ??
NaCl = 0.30 x 2 = ??
C6H12O6 = 0.6 x 1 = ??