which of the following would be expected to have the lowest freezing temperature?
a. Ca(NO3)2
b. NaC2H3O2
c. CuSO4
d. HC2H3O2
e. C2H5OH
per what? per gram? per mol? Any could have the lowest freezing point if I put a cup of any one and a scant amount of the others. For equal molality, what you want is
delta T = Kb*i*molality.
Kb is a constant, IF it is equal molality the m is equal, and the secret then is to determine the number of particles (i) from the others.
To determine which of the given substances would have the lowest freezing temperature, we need to consider the concept of colligative properties, specifically freezing point depression.
The freezing point depression of a solution is proportional to the number of particles present in the solution, rather than the nature of the particles themselves. This property depends on the molality of the solution, which is the moles of solute per kilogram of solvent.
The formula to calculate the change in freezing temperature (delta T) is:
delta T = Kb * i * molality
In this equation, Kb is the cryoscopic constant, which is specific to the solvent being used, i is the van't Hoff factor, which represents the number of particles the solute dissociates into in the solution, and molality is the concentration of the solute expressed in moles per kilogram of solvent.
To determine which substance would have the lowest freezing temperature, we need to compare the values of i for each substance. The van't Hoff factor depends on the ability of a substance to dissociate into ions when dissolved in a solution.
In this case, let's analyze the substances:
a. Ca(NO3)2: This compound dissociates into three ions when dissolved in water, one calcium ion (Ca2+) and two nitrate ions (NO3-). The van't Hoff factor (i) for Ca(NO3)2 is 3.
b. NaC2H3O2: This compound dissociates into two ions in water, one sodium ion (Na+) and one acetate ion (C2H3O2-). The van't Hoff factor (i) for NaC2H3O2 is 2.
c. CuSO4: This compound dissociates into three ions in water, one copper ion (Cu2+) and four sulfate ions (SO4^2-). The van't Hoff factor (i) for CuSO4 is 3.
d. HC2H3O2: This compound does not dissociate into ions in water, so the van't Hoff factor (i) for HC2H3O2 is 1.
e. C2H5OH: This compound does not dissociate into ions in water, so the van't Hoff factor (i) for C2H5OH is also 1.
Comparing the van't Hoff factors for each substance, we can see that Ca(NO3)2 and CuSO4 both have a factor of 3, NaC2H3O2 has a factor of 2, and HC2H3O2 and C2H5OH both have a factor of 1.
Given that the molality is the same for all substances, the substance with the highest van't Hoff factor (i) will have the lowest freezing temperature. Therefore, the substance with the lowest freezing temperature would be Ca(NO3)2, as it has a van't Hoff factor of 3, the highest among the given substances.