Some Gchem Q's

[prsndwg]

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Someone help me out here:

I chose B and it was WRONG!?!

Wouldn't it endothermic? but that's not that right answer 😕

 

[caloriefree]

For this question, I think the answer should be A (assuming they are asking for the molality of the electrolytes.)

B is equivalent to 0.5M of NaCl because 0.5M of NaCl dissociates into Na+ and Cl- ions to create 1m solution.

C is incorrect because mole fraction of NaCl is more than 0.01. (1000g of water = 55.55 mol, thus, 1 mol of NaCl / 55.55 + 1 mol = mole fraction = 0.017)

D is incorrect because molar mass of NaCl is 58.5g/mol. So 55g is less than 1 mol.

CaF2 <--> Ca++ and 2F-

Due to common ion effect and Le Chatelier's principle, B should be the right answer, since we are adding F- to the right side, the equilibrium should shift to the left. Since F- ions are being added, it can't be determined whether the conc. of F- will decrease/increase/stay the same compared to the original conc., I think, that depends on the amount of NaF added.

 

[hausee]

second one choose D because Q is talk about volume percentage.

 

[caloriefree]

Edit: deleted. didn't realize the % for the second question was for volume.

I'm not a chemistry major, so someone correct me if my reasoning is wrong. Since the vapor pressure of the mixed solution is less than the vapor pressure of either one in its pure form, the two solvent particles must exert stronger attractive forces on each other compared to the attraction between the same molecules in pure form. The stronger attraction means the stability of the system is increased. And what happens when a system gets stabilized? The internal energy decreases to a lower level, and the extra energy is given off as heat, thus the process is exothermic. (Think of formation of new bonds as an analogy: new bonds formed --> more stability --> energy given off)

I could be wrong.. please let me know what the solution says.

 

[UCB05]

caloriefree already explained why C and D are definitely wrong and I'm assuming his math is correct for C. Remember when you had to make solutions in chem lab? Molarity is mols solute/L solution whereas molality is mols solute/Kg solvent. Also, a solute tends to occupy a little bit of volume.
1m NaCl = (1mol NaCl)/(1 kg H2O) which is equivalent to 1mol/1L
1M NaCl = (1mol NaCl)/(slightly-less-than 1L H2O).

Since you have to use slightly less H2O to make the 1M solution, it is more concentrated than 1m solution. Answer should be A.

Take a look at A and D. A basically says 79%/21% composition by mols, D says 79%/21% by mass. Since the two gases have different molar masses, either A or D has to be wrong if the other is right. You don't even have to look at B and C. % composition of air is conventionally given in volume, which can be directly related to mols (PV=nRT). That means A is true. D is false.

caloriefree explained teh last two questions quite well

 

[myheart01]

I chose B and it was WRONG!?!

The answer is D
21% of Oxyen = 0.21 mol
79% of Nitrogen = 0.79 mol

A. Correct

B. Correct
N2=14+14=28g/mol
28g/mol*o.79mol = 22.12g

C. Correct
Pp = Xp*Pt Pp=partial pressure, Xp=mole fraction, Pt= total pressue
600=0.79/(0.21+0.79)*Pt
Pt=760 torr
Pt=Pn + Po
760=600+Po
Po=160

D. Incorrect
O2=32g/mol
32g/mol*X =21g
X=0.656 mol of O2
0.21mol of O2/o.79 mol of N2 = 0.656 mol of O2/ X mol of N2
X mol of N2 = 2.47 mol
N2 = 28 g/mol * 2.47 mol = 69.16 g

 

[prsndwg]

Thank for the info and you are all very helpful. I am still confused about the last Q with CaF2. The answer key shows Ca decreases. Can someone explain that please.

 

[loveoforganic]

Calcium fluoride is a sparingly soluble salt. Sodium fluoride is a fairly soluble salt. The sodium ions present in solution comment is sort of irrelevant as far as I can see, a distractor. Anyway, you add a bunch of fluoride ions to your solution which causes the calcium to precitipate out as calcium fluoride.

 

[myheart01]

CaF2 <--> Ca++ and 2F-

Due to common ion effect and Le Chatelier's principle, B should be the right answer, since we are adding F- to the right side, the equilibrium should shift to the left. Since F- ions are being added, it can't be determined whether the conc. of F- will decrease/increase/stay the same compared to the original conc., I think, that depends on the amount of NaF added.[/QUOTE]
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Yes, that is right! Answer is B.

Solubility of CaF2 (Ksp) :
CaF2 <--> Ca++ and 2F-
In this reversible reaction, Ksp is equal state at 4*10^-11. It means this net reaction in this reaction hardly moves to the right or left.

If CaF2 is added in the solution, this net reaction in the direction shifts to the right.
If Ca++ or 2F-is added in the solution, this net reaction in the direction shifts to the left.

When NaF is added to the solution, it becomes ions, Na+ and F- (NaF --> Na+ and F-, because it is not equal state).
Therefore, F- would be increased, then the net reaction in the direction shifts to the right. Because of this reaction,
A. Na+ will increase
B. Ca++ will decrease
C. F- will increase
D. doesnt make sense.

 

[prsndwg]

CaF2 <--> Ca++ and 2F-

Due to common ion effect and Le Chatelier's principle, B should be the right answer, since we are adding F- to the right side, the equilibrium should shift to the left. Since F- ions are being added, it can't be determined whether the conc. of F- will decrease/increase/stay the same compared to the original conc., I think, that depends on the amount of NaF added.

-----------------------------------------------------
Yes, that is right! Answer is B.

Solubility of CaF2 (Ksp) :
CaF2 <--> Ca++ and 2F-
In this reversible reaction, Ksp is equal state at 4*10^-11. It means this net reaction in this reaction hardly moves to the right or left.
why is this? How can we tell from Ksp value?
If CaF2 is added in the solution, this net reaction in the direction shifts to the right.
If Ca++ or 2F-is added in the solution, this net reaction in the direction shifts to the left.

When NaF is added to the solution, it becomes ions, Na+ and F- (NaF --> Na+ and F-, because it is not equal state).
Therefore, F- would be increased, then the net reaction in the direction shifts to the right, you mean left? Because of this reaction,
A. Na+ will increase
B. Ca++ will decrease
C. F- will increase
D. doesnt make sense.[/QUOTE]


thank you

 

[myheart01]

-----------------------------------------------------
Yes, that is right! Answer is B.

Solubility of CaF2 (Ksp) :
CaF2 <--> Ca++ and 2F-
In this reversible reaction, Ksp is equal state at 4*10^-11. It means this net reaction in this reaction hardly moves to the right or left.
why is this? How can we tell from Ksp value?
If CaF2 is added in the solution, this net reaction in the direction shifts to the right.
If Ca++ or 2F-is added in the solution, this net reaction in the direction shifts to the left.

When NaF is added to the solution, it becomes ions, Na+ and F- (NaF --> Na+ and F-, because it is not equal state).
Therefore, F- would be increased, then the net reaction in the direction shifts to the right, you mean left? Because of this reaction,
A. Na+ will increase
B. Ca++ will decrease
C. F- will increase
D. doesnt make sense.

thank you[/QUOTE]
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Sorry, I made my mistake. Thank you for correction as the right word.
Ksp is the solubility constant and is not changeable at a specific temperature. Each compound has each specific Ksp.
For example, Ksp of BaCO3 is 2.6*10^-9, CaF2 is 1.5*10^-10, and PbI2 is 8.5*10^-9 at 25 oC. We should find these values at the index of Ksp.

The reason why we should find these values is to know concentrations (M) of Ions at the equal state (Ksp).

For example, if we know CaF2 <--> Ca++ and 2F- ,ksp=4*10^-11, and the concentraton of Ca++ = 2*10^-13, then we can know the concentration of 2F- like this.
CaF2 = (Ca++)*(2F-)^2
4*10^-11 = (2*10^-13)*(2F)^2
2F-=14.14
Therefore, when Ksp is 4*10^-11 as a equal state, the concentraton of Ca++ = 2*10^-13 and the concentration of 2F-=14.14. It means this net reaction in the direction hardly moves to the right or left at the ion concentrations.