Another datQvault Error #2??????

[GoBlue24]

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This is also from the 3rd chem test of qvaultDAT. Is this also wrong? I thought the answer would be B due to a bigger i-factor (3) and a bigger molality (approx 11 mol/1kg). I don't know where the heck the explanation got .25M and .3M from.... can anyone help?

10 grams of the following solutes is added to 1L of pure water. Which of the five ensuing solutions would have the lowest freezing point?


A) NaCl

B) CaCl2

C) KCl

D) NaI

E) KI



Explanation:
The correct answer is A. The solution with the lowest freezing point will be the solution with the greatest number of ions. It is not necessary to calculate the molarity of each solution. The solutes with the smallest gram formula masses will have the greater molarity. Because NaCl yields 2 ions and CaCl2 yields three ions, you need to compare these solutes. 10g of NaCl is approximately 0.25 M, which yields 0.5 M for both ions 10g CaCl2 is approximately 0.1M which yields 0.3M for all the ions. Thus 10 g of NaCl will have the greater effect on the freezing point of the solution

 

[premac]

This is also from the 3rd chem test of qvaultDAT. Is this also wrong? I thought the answer would be B due to a bigger i-factor (3) and a bigger molality (approx 11 mol/1kg). I don't know where the heck the explanation got .25M and .3M from.... can anyone help?

10 grams of the following solutes is added to 1L of pure water. Which of the five ensuing solutions would have the lowest freezing point?


A) NaCl

B) CaCl2

C) KCl

D) NaI

E) KI



Explanation:
The correct answer is A. The solution with the lowest freezing point will be the solution with the greatest number of ions. It is not necessary to calculate the molarity of each solution. The solutes with the smallest gram formula masses will have the greater molarity. Because NaCl yields 2 ions and CaCl2 yields three ions, you need to compare these solutes. 10g of NaCl is approximately 0.25 M, which yields 0.5 M for both ions 10g CaCl2 is approximately 0.1M which yields 0.3M for all the ions. Thus 10 g of NaCl will have the greater effect on the freezing point of the solution

The answer is right, but the explanation doesn't seem right.
10g NaCl = 0.17 mol = 0.17M
10g CaCl2 = 0.09 mol = 0.09 M
NaCl -> i=2
CaCl2 -> i=3
Since the equation is -iKm and we don't know the value of K, we just compare by i and m.
NaCl = 0.17*2 = -0.34
CaCl2 = 0.09*3 = -0.27

Since NaCl has lower value, it would mean lower freezing point.
What the problem is trying to teach you is that you can't simply compare the freezing points by comparing "i". You need to take the whole equation into account, and calculate the molarity, too.

 

[GoBlue24]

Oh fml, i made an amateur mistake. I did 10g X (58g/1mol) instead of 10g X (1mole/58g) to get m (molality). Thanks for clearly that up!

The answer is right, but the explanation doesn't seem right.
10g NaCl = 0.17 mol = 0.17M
10g CaCl2 = 0.09 mol = 0.09 M
NaCl -> i=2
CaCl2 -> i=3
Since the equation is -iKm and we don't know the value of K, we just compare by i and m.
NaCl = 0.17*2 = -0.34
CaCl2 = 0.09*3 = -0.27

Since NaCl has lower value, it would mean lower freezing point.
What the problem is trying to teach you is that you can't simply compare the freezing points by comparing "i". You need to take the whole equation into account, and calculate the molarity, too.