I think your answer to the first problem is flawed because the problem states that the concn is the same which means the molarity is the same.
I think the second problem looks ok.
My answer & reason: solution A has a smaller molecular weight and therefore, a larger Molarity and higher osmotic pressure (pi=iMRT)
Potassium chloride is used in the kitchen as a salt substitute (in place of sodium chloride). Suppose you were planning to make ice cream but discovered too late that you were out of table salt for making the ice bath. If the recipe called for one pound of NaCl, how would it work to use one pound of KCl instead?
My answer & reason:
so the 1 lb of KCl produces less moles than 1lb of NaCl because the molecular weight of KCl is greater than the molecular weight of NaCl? therefore, the molarity for KCl is less than the molarity for NaCl. As a result, KCl has a slightly higher freezing point (due to dTb=ikm).
I think the second problem looks ok.
For the second question, you correctly pointed out that the molecular weight of KCl is greater than the molecular weight of NaCl. Consequently, if you used 1 pound of KCl instead of 1 pound of NaCl, there would be fewer moles of KCl compared to NaCl. This would lead to a lower molarity of KCl compared to NaCl. Regarding the freezing point, you mentioned the equation dTb=ikm, where "dTb" represents the change in boiling point elevation, "i" is the van't Hoff factor, "k" is the cryoscopic constant, and "m" is the molality of the solution. It's worth noting that freezing point depression and boiling point elevation calculations have different constants in their respective equations, and in this case, the boiling point elevation equation might not be directly relevant to the freezing point depression caused by the substitution of KCl for NaCl in the ice bath for making ice cream.