qVault error? Test 3

[ballyhoo]

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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: [Correct] NaCl
B: [Yours] CaCl2
C: KCl
D: NaI
E: KI

Somehow, they got that 10 g of NaCl would be 0.25 M of NaCl? I'm wondering how they got that ... according to their website, 75% of students put my answer, B, so I'm very confused! What am I missing?

Thank you!!

 

[ballyhoo]

bummmp? 😀

 

[Mr Dental]

I guess they meant to put NaOH instead of NaCl. If used NaOH, then its about 0.25 mol, and we use the molality of 0.25m. Since the Van Hoff factor is 2 for NaOH, it will yield 0.5m. In case of CaCl2, 10 grams is equal to 0.1 mol, and with van hoff=0.3m

**we use molality instead of molarity for freezing pt depression.

 

[mario0923]

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: [Correct] NaCl
B: [Yours] CaCl2
C: KCl
D: NaI
E: KI

Somehow, they got that 10 g of NaCl would be 0.25 M of NaCl? I'm wondering how they got that ... according to their website, 75% of students put my answer, B, so I'm very confused! What am I missing?

Thank you!!

I got that question wrong too
dT=imKf
where i is the number of ions and m is molality (I don't know why Qvaults uses molarity)
so NaCl is 58g/mol whereas CaCl2 is 110g/mol
so NaCl will have 0.17mol which equals to 0.17m and 0.34m of ions
and CaCl2 will have 0.09mol which equals to 0.09m and 0.27m of ions

since NaCl has greater molality of ions than CaCl2 does, NaCl will have greater change of freezing point

so basically if you have same number of ions formed and same mass of solute, the solute that has the smallest molecular weight will have greater effect on colligative properties (since lower molecular weight is bigger moles, and bigger moles bigger molality)
but seriously I didn't even think to compare NaCl and CaCl2...hope this is not going to be on real DAT

 

[ballyhoo]

Oh, thanks so much! I get it .. but yes I seriously hope this isn't going to be on the DAT haha - or maybe I'll just try this problem (and similar problems) over and over again until I get it, in which case I'd be fine with it!

Just to make sure I'm getting things right, when you go from NaCl having 0.17 mol to 0.17 m, you are dividing the moles by 1 kg because 1L = 1kg, correct?

Thanks again!

 

[mario0923]

Oh, thanks so much! I get it .. but yes I seriously hope this isn't going to be on the DAT haha - or maybe I'll just try this problem (and similar problems) over and over again until I get it, in which case I'd be fine with it!

Just to make sure I'm getting things right, when you go from NaCl having 0.17 mol to 0.17 m, you are dividing the moles by 1 kg because 1L = 1kg, correct?

Thanks again!

yeah since 1L of water is same as 1kg of water and molality m=(mols of solute)/(kg of solvent)