Assuming equal concentrations, rank these aqueous solutions by their freezing point: low to high
NH4Cl, Sn(ClO3)4, Na2SO4, and AlBr3.
delta T = i*Kf*m
f.p. = 0 C - delta T.
So m and Kf are constant; the only variable is i = van't Hoff factor.
i = 2 for NH4Cl
i = 5 for Sn(ClO3)4
i = 3 for Na2SO4
i = 4 for AlBr3.
To rank these aqueous solutions by their freezing point, we need to consider their colligative properties. Colligative properties depend on the number of solute particles present in a solution rather than the type of solute particles.
NH4Cl dissociates into two ions: NH4+ and Cl-, so it will have more particles in solution compared to the other solutes. Sn(ClO3)4 dissociates into five ions: Sn4+, four ClO3-, while Na2SO4 dissociates into three ions: two Na+, and one SO4 2-. Finally, AlBr3 dissociates into four ions: Al3+, and three Br-.
Based on the number of particles each solute dissociates into, we can rank the solutions from low to high freezing point:
1. NH4Cl - It forms the greatest number of ions and will have the lowest freezing point.
2. AlBr3 - It forms the second greatest number of ions and will have the second lowest freezing point.
3. Sn(ClO3)4 - It forms the third greatest number of ions and will have the third lowest freezing point.
4. Na2SO4 - It forms the lowest number of ions and will have the highest freezing point.
Therefore, the ranking from low to high freezing point is NH4Cl, AlBr3, Sn(ClO3)4, and Na2SO4.
To rank the aqueous solutions by their freezing points, we need to consider the number of particles each compound will dissociate into when dissolved in water.
The general rule is that the more particles a compound dissociates into, the lower its freezing point will be. This is because the dissociated particles disrupt the crystal lattice formation, making it more difficult for the solvent to freeze.
Let's analyze each compound:
1. NH4Cl: This compound dissociates into two particles when dissolved in water: NH4+ and Cl-. Therefore, it will have a higher freezing point than compounds that dissociate into more particles.
2. Sn(ClO3)4: This compound contains Sn4+ cations and four ClO3- anions. However, each ClO3- anion already carries a -1 charge and does not dissociate further when dissolved in water. So, Sn(ClO3)4 also dissociates into two particles: Sn4+ and four ClO3-.
3. Na2SO4: This compound dissociates into three particles when dissolved in water: two Na+ cations and one SO4^2- anion. It will have a lower freezing point than NH4Cl and Sn(ClO3)4 due to the higher number of dissociated particles.
4. AlBr3: This compound dissociates into four particles when dissolved in water: three Br- anions and one Al3+ cation. As it has the largest number of dissociated particles among the given compounds, it should have the lowest freezing point.
Therefore, ranking the aqueous solutions by their freezing points from low to high would be:
AlBr3 < Na2SO4 < Sn(ClO3)4 < NH4Cl