Which aqeuous solution has a freezing point closest to that of 0.30M C12H22O11?
A) 0.075M AlCl3
B) 0.15M CuCl2
C) 0.30M NaCl
D) 0.60M C6H12O6
I know the answer but I don't know the reason or rationale behind it. Please explain.
I suggest you review Raoult's Law at
http://answers.google.com/answers/threadview?id=547572
The freezing point lowering depends upon the solution molality and the number of ions the solute breaks up into, if any. The sugar molecule C12H22O11 does not dissociate. AgCl3 breaks up into 4 ions, so there will be the same number of ions in solution per unit volume as there are molecules of sugar, 0.30 moles per liter. So my guess is the answer is (A)
right.
delta T = i*Kf * m
where i is the number of ions (particles in solution), K is the same constant for all, and m is molality.
Therefore, we simply multiply i*m to see which = 0.3
AlCl3 = 0.075 x 4 = ??
CuCl2 = 0.15 x 3 = ??
NaCl = 0.30 x 2 = ??
C6H12O6 = 0.6 x 1 = ??
A) 0.075M AlCl3
Let me clown around and break it down for you!
We're looking for the aqeuous solution with the freezing point closest to that of 0.30M C12H22O11.
Now, using Raoult's Law, the freezing point lowering depends on the solution molality and the number of ions the solute breaks up into.
So, let's calculate the "Clownification Factor" (CF) for each solution. CF is just i*m, where i is the number of ions and m is the molality.
For AlCl3, we have CF = 0.075 x 4 = 0.3 bumper cars.
For CuCl2, we have CF = 0.15 x 3 = 0.45 funny clowns.
For NaCl, we have CF = 0.30 x 2 = 0.6 whoopee cushions.
And for C6H12O6, we have CF = 0.6 x 1 = 0.6 banana peels.
So, which solution has a CF closest to 0.3? The answer is (C) 0.30M NaCl with a CF of 0.6 whoopee cushions!
Congratulations! You've made it through the clownification of freezing point depression!
Let's calculate the products of i * m for each option:
A) 0.075M AlCl3: i = 4 (since AlCl3 dissociates into 4 ions) and m = 0.075
i * m = 4 * 0.075 = 0.3
B) 0.15M CuCl2: i = 3 (since CuCl2 dissociates into 3 ions) and m = 0.15
i * m = 3 * 0.15 = 0.45
C) 0.30M NaCl: i = 2 (since NaCl dissociates into 2 ions) and m = 0.30
i * m = 2 * 0.30 = 0.6
D) 0.60M C6H12O6: i = 1 (since C6H12O6 does not dissociate) and m = 0.60
i * m = 1 * 0.60 = 0.60
Comparing the values, i * m for option A is equal to 0.3, which is closest to the molality (m) of the C12H22O11 solution. Therefore, the correct answer is A) 0.075M AlCl3.
To determine which aqueous solution has a freezing point closest to that of 0.30M C12H22O11, we need to consider the concept of freezing point depression and Raoult's Law.
Freezing point depression occurs when a solute is added to a solvent, lowering the freezing point of the solution compared to that of the pure solvent. The degree of freezing point depression depends on the molality (moles of solute per kilogram of solvent) and the number of particles (ions or molecules) the solute breaks into when dissolved in the solvent.
Raoult's Law states that the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent in the solution. In the case of freezing point depression, the same principle applies, but for the freezing point instead of vapor pressure.
In this scenario, we have four options: A) 0.075M AlCl3, B) 0.15M CuCl2, C) 0.30M NaCl, and D) 0.60M C6H12O6.
To determine the number of particles (i) each solute breaks into, we need to look at its formula.
- AlCl3 dissociates into one Al3+ ion and three Cl- ions, totaling four particles.
- CuCl2 dissociates into one Cu2+ ion and two Cl- ions, totaling three particles.
- NaCl dissociates into one Na+ ion and one Cl- ion, totaling two particles.
- C6H12O6 does not dissociate, so it remains as one particle.
Now we can calculate i * m for each choice to see which one yields a value closest to 0.3:
A) 0.075M AlCl3: 4 * 0.075 = 0.3
B) 0.15M CuCl2: 3 * 0.15 = 0.45
C) 0.30M NaCl: 2 * 0.30 = 0.6
D) 0.60M C6H12O6: 1 * 0.60 = 0.6
As we can see, the option that gives us a value closest to 0.3 is choice A) 0.075M AlCl3, with a value of 0.3. Therefore, the aqueous solution of AlCl3 has a freezing point closest to that of 0.30M C12H22O11.