1988 DAT gchem 60 and 56

[yahoogoogle]

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in which solvent should NaCl be most soluble?

1. CH3OH
2. C8H18
3. (C2H5)2O
4. CCl4
5. C6H6

Which one of the following .15M aqueous solutions has the lowest freezing point?

1. KCl
2. Al2(SO4)3
3. CH3OH
4. C2H5OH
5. NaOH

can anyone tell me what to study to prepare these types of questions

 

[joonkimdds]

Which one of the following .15M aqueous solutions has the lowest freezing point?

1. KCl
2. Al2(SO4)3
3. CH3OH
4. C2H5OH
5. NaOH

can anyone tell me what to study to prepare these types of questions

If i remember what i studied 2 days ago correctly, the freezing point varies
depends on the molarity of solute and how many particles are formed,
but all of them have the same molarity so I assum that it now only depends
on the number of particles that are formed.
I believe 2. Al2(SO4)3 is the correct answer becuz it has like...hm... 2 + 3 + 12 = 17 particles
so if i divide the freezing point by 17, that should give me the lowest value than the other choices.

I hope I am correct...PD3 explained it to be 2 days ago

 

[DentalKitty]

If i remember what i studied 2 days ago correctly, the freezing point varies
depends on the molarity of solute and how many particles are formed,
but all of them have the same molarity so I assum that it now only depends
on the number of particles that are formed.
I believe 2. Al2(SO4)3 is the correct answer becuz it has like...hm... 2 + 3 + 12 = 17 particles
so if i divide the freezing point by 17, that should give me the lowest value than the other choices.

I hope I am correct...PD3 explained it to be 2 days ago

You are close...freezing point depression has to do with molality and the number of particles in solution, NOT molarity. Also, Al2(SO4)3 is the correct answer, but it provides only 5 particles in solution, not 17. The SO4-2 part is a complex ion and the S and O remain bonded.

The answer to the first question is #1 because it is polar and the others are not. NaCl is ionic, therefore in solution its particles are charged and are attracted to other polar species.

 

[yahoogoogle]

You are close...freezing point depression has to do with molality and the number of particles in solution, NOT molarity. Also, Al2(SO4)3 is the correct answer, but it provides only 5 particles in solution, not 17. The SO4-2 part is a complex ion and the S and O remain bonded.

The answer to the first question is #1 because it is polar and the others are not. NaCl is ionic, therefore in solution its particles are charged and are attracted to other polar species.

for question 1, what about ether? it's polar too

 

[DentalKitty]

for question 1, what about ether? it's polar too

No, for the most part ethers are not polar. Especially not compared to an alcohol's O-H dipole. Ethers also do not H-bond like alcohols do...

 

[joonkimdds]

You are close...freezing point depression has to do with molality and the number of particles in solution, NOT molarity. Also, Al2(SO4)3 is the correct answer, but it provides only 5 particles in solution, not 17. The SO4-2 part is a complex ion and the S and O remain bonded.

Awwwww!!!!!!!!!!!!ouch... 🙁
by the way, if Al2(SO4)3 has 5 particles, shouldn't answer be
C2H5OH since it has 2+5+1+1 = 9?

 

[leeaf]

Awwwww!!!!!!!!!!!!ouch... 🙁
by the way, if Al2(SO4)3 has 5 particles, shouldn't answer be
C2H5OH since it has 2+5+1+1 = 9?

No C2H5H is just one particle

 

[joonkimdds]

No C2H5H is just one particle

I am sorry for being confused but could you tell me what is called 1 particle?

MX2 = 3 particles according to Kaplan blue book.
So I thought that C2H5OH is like M2X5N1 = 8 particles.

 

[leeaf]

I am sorry for being confused but could you tell me what is called 1 particle?

MX2 = 3 particles according to Kaplan blue book.
So I thought that C2H5OH is like M2X5N1 = 8 particles.

Since C2H5OH is volatile, it is considered as 1 particle
(Same for C6H12O6)


Hope it helps

 

[DentalKitty]

I am sorry for being confused but could you tell me what is called 1 particle?

MX2 = 3 particles according to Kaplan blue book.
So I thought that C2H5OH is like M2X5N1 = 8 particles.

Think about dissociation. Things like NaCl, or Mg(OH)2 dissociate into ions in water forming 2 and 3 particles respectively (Na+, Cl- and Mg2+, 2OH-). Things like sugars, volatile compunds like alcohols, or other soluble, but not ionic compounds do not dissociate into ions for this purpose. Therefore, they only count as one particle. It has nothing to do with the number of atoms in the chemical formula. When Kaplan refers to MX2, they mean an ionic compound that will dissociate into 3 ions (1-M, 2-X). An example is Mg(OH)2 like I mentioned above.

 

[Comlan]

Think about dissociation. Things like NaCl, or Mg(OH)2 dissociate into ions in water forming 2 and 3 particles respectively (Na+, Cl- and Mg2+, 2OH-). Things like sugars, volatile compunds like alcohols, or other soluble, but not ionic compounds do not dissociate into ions for this purpose. Therefore, they only count as one particle. It has nothing to do with the number of atoms in the chemical formula. When Kaplan refers to MX2, they mean an ionic compound that will dissociate into 3 ions (1-M, 2-X). An example is Mg(OH)2 like I mentioned above.

Even though the answer is correct, I don't think that it's due to the number of particles. From the freezing point depression formula, I believe that this answer is based on the molecular mass of the molecule.😕

 

[Bob Faraples]

Even though the answer is correct, I don't think that it's due to the number of particles. From the freezing point depression formula, I believe that this answer is based on the molecular mass of the molecule.😕

Freezing/boiling point depression/elevation are "colligative," which basically means that they depend on the number of particles of solute in the solvent. The formula for the change in freezing point is: change in temp = Kf * molality * the van't hoff factor "i"

Molality just depends on the number of moles of solute per kg solvent, so it is independant of the solute's mass. It does not matter what the solute is, it is only concerned with how many moles are in a kg of solvent.

The van't hoff factor "i" takes into consideration the fact that a solute might dissociate into ions (such as NaCl --> Na+ and Cl-), the number of particles floating around actually increases (1 NaCl molecule produced 2 particles). Since all we are really concerned about is the number of particles in the solvent (doesn't matter what kind of particle - a Na+ has the same effect as a Cl-), if we add a solute that dissociates we have really doubled (or tripled/quadrupled/etc - just depends on the solute) our molality (1 mol NaCl/kg --> 1mol Na+/kg and 1 mol Cl-/kg -- two moles really exist in the 1 kg of solvent) and so we have to multiply accordingly. If our solute dissociated into three particles, we multiple by i=3, if it dissociated into 4, we multiply by i=4.

Hopefully this helps - for further reference Wikipedia "freezing point," "colligative properties" and "van't hoff factor"

 

[DentalKitty]

Even though the answer is correct, I don't think that it's due to the number of particles. From the freezing point depression formula, I believe that this answer is based on the molecular mass of the molecule.😕

Nope, the question states they are all .15m which means the molar mass factor has been removed. Yahoogoogle, did the question actually say m vs. M like you typed? Usually FP depression has to do with molality vice molarity, and here they give you really no way (or need) to convert molarity to molality....

 

[Bob Faraples]

Yeah in this case you really don't need to convert molarity to molality - .15 moles of any of these compounds in 1 L of water isn't really going to rock the boat that much going from molarity to molality. You can just *pretend* it said molality and look for the compound that dissociates the most.

 

[jeezy]

where do you find these old DATs?

 

[jdmsamurai]

when you think of dissociation think of ionic bonds...ionic bonds dissociate completely, while polar colvalent bonds alittle...

for example ethanol can dissociate to H+ and CH3CH2O- , but small compare to an ionic compound as NaCl which is 2