Which of the following aqueous solution of equal molality has the lowest freezing point?

a. H2SO3
b. NaCl
c. NaOH
d. Ca(NO3)2

How do you answer this without looking at their i values from textbooks? I know that
delta(Fpt) = m * K,f * i
and that i is approximately equal to the number of ions dissociated, but in the choices, letters A and D have the same number of ions? Please help.

Van't Hoff factor can be associated to the property of the substance (if it's strong electrolyte, weak electrolyte, etc.), and can be approximated by counting the number of ions dissociated in water.

a. i ~ 1 (weak acid, therefore weak electrolyte)
b. i = 2 (strong electrolyte)
c. i = 2 (strong base, therefore strong electrolyte)
d. i = 3 (strong electrolyte)

Therefore, the greatest delta freezing point that you would get is using the van't Hoff factor of Ca(NO3)2.

Hope this helps~ :)

To determine the aqueous solution with the lowest freezing point, we need to compare their van't Hoff factors (i values) as well as their molality (m) values. However, since we can't access specific i values for H2SO3 and Ca(NO3)2, we can make an assumption based on their molecular formulas.

From the molecular formulas:
a. H2SO3 - No apparent ions present
b. NaCl - Na+ and Cl- ions (i = 2)
c. NaOH - Na+ and OH- ions (i = 2)
d. Ca(NO3)2 - Ca2+ and NO3- ions (i = 3)

Based on this assumption, we can consider NaCl, NaOH, and Ca(NO3)2 as strong electrolytes, meaning they will dissociate completely in water, while H2SO3 is a weak electrolyte and will only partially dissociate.

Since H2SO3 is a weak electrolyte, it will have the lowest van't Hoff factor (i = 1) among the choices. Therefore, the aqueous solution of H2SO3 will have the lowest freezing point when compared to the other options.

In summary, option (a) H2SO3 will have the lowest freezing point among the given choices.

To determine which of the given aqueous solutions of equal molality has the lowest freezing point, we can use the concept of van't Hoff factor (i). The van't Hoff factor represents the extent of dissociation of a solute into ions when it is dissolved in a solvent.

As you correctly mentioned, the formula for calculating the change in freezing point (ΔTf) is given by ΔTf = m * Kf * i, where m is the molality, Kf is the cryoscopic constant, and i is the van't Hoff factor.

To compare the freezing point depression, we need to determine the van't Hoff factor (i) for each compound, even if the number of ions appears to be the same. However, it is important to note that not all compounds dissociate completely in water. Some weak electrolytes might partially dissociate, whereas strong electrolytes will dissociate almost completely.

To find the van't Hoff factor for each compound in the given options without relying on textbook values, we can use some general guidelines:

1. For covalent molecular compounds like H2SO3 (sulfurous acid), they exist as molecules in solution rather than ions since they do not fully dissociate. In this case, the van't Hoff factor (i) can be assumed to be 1.

2. For ionic compounds like NaCl (sodium chloride), they dissociate completely in water, meaning they fully dissociate into their constituent ions. Hence, the van't Hoff factor (i) for NaCl is 2 since it forms two ions, Na+ and Cl-.

3. Similarly, for Ca(NO3)2 (calcium nitrate), it dissociates completely into three ions: Ca2+, 2NO3-, making the van't Hoff factor (i) equal to 3.

4. Lastly, NaOH (sodium hydroxide) also dissociates completely, producing one Na+ ion and one OH- ion. Therefore, the van't Hoff factor (i) for NaOH is 2.

Comparing the van't Hoff factors for the given options, we have:
a. H2SO3: i = 1
b. NaCl: i = 2
c. NaOH: i = 2
d. Ca(NO3)2: i = 3

Since Ca(NO3)2 has the highest van't Hoff factor (i = 3), it will cause the greatest freezing point depression. Therefore, option (d) Ca(NO3)2 would have the lowest freezing point among the given solutions.