How do the boiling points of 3.0 m C6H12O6 and 3.0 m CaCl2 compare?
A. The boiling point of 3.0 m C6H12O6 is higher.
B. The boiling point of 3.0 m CaCl2 is higher.
C. They are the same.
D. The difference cannot be determined from the information given.
I think the answer is D. Is that correct?
No.
delta T = i*Kb*m
Since Kb is the same for both, the differences are due to i*m and since m is the same the differences are due to i.
i for C6H12O6 = 1
i for CaCl2 = 3
Look up the van't Hoff factor.
To determine how the boiling points of 3.0 m C6H12O6 (glucose) and 3.0 m CaCl2 (calcium chloride) compare, we need to consider the colligative properties of solutions. Colligative properties rely on the number of solute particles (i.e., molecules or ions) present in the solution, rather than their identity.
The main colligative property related to boiling point is boiling point elevation. According to Raoult's law, the presence of a non-volatile solute in a solvent raises the boiling point of the solution compared to the pure solvent.
In this case, both glucose (C6H12O6) and calcium chloride (CaCl2) are soluble in water. Glucose is a non-electrolyte, meaning it does not dissociate into ions when dissolved in water. Calcium chloride, on the other hand, is an electrolyte and dissociates into Ca2+ and 2Cl− ions when dissolved in water.
Since glucose does not dissociate into ions, it will contribute fewer solute particles compared to calcium chloride at the same concentration. Therefore, 3.0 m CaCl2 will have a higher boiling point than 3.0 m C6H12O6.
Therefore, the correct answer is B. The boiling point of 3.0 m CaCl2 is higher.