# For equimolar solutions of "sugar", "potassium sulfate", and "potassium hexaferricyanide", which will exert the LEAST vapour pressure?

We have sugar, which is molecular, ${K}_{2} S {O}_{4}$, 3 particles on dissolution, and the iron complex, ${K}_{3} \left[F e {\left(C \equiv N\right)}_{6}\right]$, which will likely yield 4 particles on dissolution. Because sugar does not speciate into solvated ions in solution, this will likely cause the least boiling point depression, i.e. will have the least diminution of vapour pressure.
Note that the iron(III) complex itself, ${\left[F e {\left(C \equiv N\right)}_{6}\right]}^{3 -}$, would be a fairly stable species in solution. Iron(II) salts (and this is not one!) are the old laboratory remedy to cyanide poisoning. If you get poisoned by cyanide in the lab, you take a solution of $F e \left(I I\right)$ salts. $F e \left(I I I\right)$ salts would be even more effective, given the ready formation and stability of ${K}_{3} \left[F e {\left(C \equiv N\right)}_{6}\right]$, but $F {e}^{3 +} \left(a q\right)$ is quite acidic, and would burn your insides.