Question

Based on the freezing point depression data for the two substances, what conclusion is MOST accurate?

Answer 1

I think I could have been more helpful if you had mentioned what the other substance in the problem was. But, I did some research and hopefully it can help you figure out the right answer anyway.  


Lets compare: NaCl in water vs. $CaCl_{2}$ in water.
 This question is centered trying to see how much you know about colligative properties such as Vapor Pressure, Boiling Point, Freezing Point Depression, and Osmotic pressure(Osmotic pressure isn't mentioned in this problem. So, I am goin to ignore that last property.) 

When you compare Colligative properties of compounds it is all about the concentration of ions in solution:

For example:
NaCl can only break into two ions in solution with water (Na+) and (Cl-). 

$CaCl_{2}$ can break into three ions in solution( 2Cl-) and (Ca2+). Because this compound can break into more ions than NaCl when added to water, it is more effective at Freezing point Depression.
It will have lower freezing point depression and lower vapor pressure, but it will have a HIGHER boiling point.

A)Unable to determine without more info.
B) unable to determine without more info.
C) FALSE- we know this is not true because when you combine a metal(Na) with a non-metal(Cl) you form an IONIC compound, not a covalent compound(two or more non-metals combined).
D) FALSE- I am going to guess that this is false because NaCl can only produce two particles in solution and that is the lowest number of particles that can be produced in solution from a compound. For example, where as, $CaCl_{2}$ Produces three particles when put into solution with water.  (2 Cl- and an Ca) 

Answer 2

The answer is D I took the test and that’s the answer