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Chapter 5: Solution Chemistry

Re: Chapter 5: Solution Chemistry

Postby metalwerew4603 » Tue Aug 04, 2015 6:17 am

Hi,
In the text regarding Ksp the statement is made
1.1*10^-12=[2x]^2[x]
solving for x, x=6.5x10^-5

Im totally frustrated as to how this is attained, could you put up a walk through for this
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Tue Aug 04, 2015 10:28 am

"In the text regarding Ksp the statement is made
1.1*10^-12=[2x]^2[x]
solving for x, x=6.5x10^-5"

I'll write it out step by step but with practice, most of the steps below should be done 'in your head'.

1.1*10^-12 = [2x]^2[x] = (4x^2)(x) = 4x^3

The first step to isolate x must be to divide both sides by 4:

x^3 = (1.1/4)*10^-12

I need to take the cube root of both sides but 1.1/4 doesn't mean anything to me. So I estimate that as 1/4 = 0.25 (of course it should be a higher number, but 0.25 is a fair, quick estimate), which still can't be cubed in my head, but . . .

x^3 = (1.1/4)*10^-12 = 250 * 10^-15 becomes easy

The cube root of 250 must be a number in between 6 and 7 (just a quick estimate in your head: 6 times 6 is 36 which we approx as 40 then times 6 = 240; 7 times 7 is 49 which we approx as 50 times 7 is 350).

Cube root of 10^-15 is the same as multiplying the exponent by 1/3 thus = 10^-5.

Thus

x^3 = 250 * 10^-15 = approx 6.5 * 10^-5

You would be expected to be able to do a calculation like this on the real exam in under a minute. Typically, because it is multiple choice, there would only be one answer between 6 and 7 * 10^-5. In the unlikely case where there would be more than one answer between those numbers, you would need to minimize the estimates and the calculation would take more time.

Because of the importance of math in Physics and Gen Chem, if you have the new edition of the book, I definitely suggest that you review the MCAT Math chapters and complete the chapter review practice questions (otherwise, that content is available in GAMSAT University; exponents are reviewed in Math Chapter 1 and 3).
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Wed Aug 05, 2015 2:54 am

If you want to memorize from a list, then there is a list in CHM 5.2.

However, it is rare to find a GAMSAT question that is solely dependent on memorization. A better strategy would be to read through the list in CHM 5.2 a couple of times, but more importantly, as you do practice questions, it is critical to focus on the molecules that appear most commonly, take notes and review those notes regularly.
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Re: Chapter 5: Solution Chemistry

Postby ElleRT » Fri Aug 21, 2015 8:07 pm

Hello,
In the working out of concentration of ions with solubility product.

I understand how ksp becomes 4X^3 where we then need to derive X.
But I'm unsure how it obtains 6.5 x 10^-5.

Do we need to know how to square root by 3 in the exam?

Thanks
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Sat Aug 22, 2015 6:50 am

This is definitely the type of mathematical reasoning that you are expected to be able to do for the GAMSAT. This is not about memorization, nor precision, but rather to work towards a reasonable estimate.

When dealing with a cube root, the first step is to adjust the number that you are evaluating so that the exponent is a multiple of 3. You don't want 10^-14 nor 10^-16. So the number must be adjusted to be multiplied by 10 to the exponent of a multiple of 3.

x^3 = 250 * 10^-15

Mathematically, a square root is the same as taking the exponent to the 1/2. So a cube root is the same as taking the exponent to the 1/3. So to isolate x in the equation above, you must take the cube root of both sides which makes (x^3)^1/3 become x^1 which is x. Now you can also see why the exponent of 10 must be a multiple of 3. So the cube root of 10^-15 must be 10^-5.

So now we need to figure out what is the cube root of 250. Start with easy numbers: 10 cubed is 1000, so the cube root of 1000 is 10. I know that 10 is too high because we want 250. So I try 5 x 5 x 5 (because it is easy) which is 125 but that's too low because I need 250. I try 7 x 7 x 7 and I estimate that as 50 x 7 (for speed) which is 350 (too high) but now I know the answer is between 6 and 7. Normally, the GAMSAT will not go further than that, meaning there will only be one answer between 6 and 7 times 10^-5. In a worse case, you just multiply an answer choice 3 times to be more precise.

GAMSAT Math section 1.5 in the GS book reviews exponents and roots.
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Re: Chapter 5: Solution Chemistry

Postby laura.cp897117 » Sat Oct 31, 2015 5:45 pm

I'm confused of when to use the van't Hoff factor in calculations of Boiling point and Freezing point calculations.

Normally I use the equation deltaTb = i . Kb . m

For question 7:
---
7) If the boiling point elevation constant, Kb, is 0.52 °C/m for water, what is the boiling point elevation of a mixture comprised of 0.1 mole of C2H6O2 and 0.1 mole of CH3OH in 18 g of H2O?

A) 5.2 X 102 °C
B) 5.2 X 101 °C
C) 5.2 °C
D) 52 °C

Your Answer: B
Correct Answer: C

Explanation: ∆Tb= Kbm= (0.52)(0.2/0.018)= (0.52)(10); therefore, ∆Tb= 5.2 °C

--

I understand that the calculation of molality (0.2/0.018) takes into account that there are 2 ions not one. But if the formula is deltaTb = i . Kb . m, why do we do we not have a value for i?
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Sun Nov 01, 2015 11:10 am

laura.cp897117 wrote:I understand that the calculation of molality (0.2/0.018) takes into account that there are 2 ions not one. But if the formula is deltaTb = i . Kb . m, why do we do we not have a value for i?


The key is understanding the following: Be sure that you have accounted for all particles (i.e. ions or molecules).

So you could have done it this way:

deltaTb = i . Kb . m

There are 2 particles so i = 2

So: deltaTb = i . Kb . m = 2(0.52)(0.1/0.018)

which arrives at the same answer.

Yet a 3rd way: Calculate the effect of lowering the freezing point for each molecule separately then add the effects which would also get the same answer. Many ways to arrive at the same answer, the important thing is to understand why: Take into account the effect of all particles (= 'colligative properties').
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A question about 13) in Q & A

Postby li.zhang96890 » Mon Nov 02, 2015 3:23 pm

13) A protein solution with concentration 80 g/L has an osmotic pressure of 0.0205 atm at 27oC. What is the approximate molecular weight of the protein? (The gas constant is 0.082 L-atm/K-mol.)

A) 96,000
B) 192,000
C) 48,000
D) 1,500

Result: Your answer is wrong.
Your choice: N/A Correct choice: A

Explanation: n/V = Π/RT
= 0.0205 / (0.082 x 300)
= 0.00083 M
Now, 1 L contains 80 g of protein.
Therefore, mw = 80 g / 0.00083 mol
= 96,000 g/mol
(Note: Π is osmotic pressure. Also, ensure the units are consistent, especially concerning R.)

In the answer,the equation n/V = Π/RT is used. How do you get this equation. I found this equation is similar to PV = nRT, but PV = nRT is used in terms of ideal gas. But in this question, it is the osmotic pressure of the protein solution not the pressure of a gas. So could you explain why you use the equation n/V = Π/RT to solve this question?
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Tue Nov 03, 2015 9:15 am

Yes, the equations for osmolarity and pressure for an ideal gas are very similar (i.e. they have the same structure). In fact, if you know PV = nRT then you know both equations.

The equation Π = MRT where M is the molarity in moles per liter (i.e. n/V) is discussed in the book in section CHM 5.1.3. This is why this question is among the Chapter 5 chapter review questions.
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Thu Dec 03, 2015 1:58 pm

1. It's not just that the reactant is solid, but more precisely, it is a sparingly soluble solid. This means that very, very little of the original solid dissolves and thus the Ksp is always a very small number (i.e. a very tiny fraction).
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Re: Chapter 5: Solution Chemistry

Postby Neda » Mon May 30, 2016 9:34 am

Hello,
Could you please explain about the molarity of H2SO4 on page CHM-68 that how this value (=1.2) is calculated?
Regards
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Mon May 30, 2016 1:32 pm

Neda wrote:Hello,
Could you please explain about the molarity of H2SO4 on page CHM-68 that how this value (=1.2) is calculated?
Regards


The value 1.2 M in CHM 5.3.1 is not calculated. It is a "given", which is to say, "If the concentration of H2SO4 is provided as 1.2 M, what would be the normality of that solution?"
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Re: Chapter 5: Solution Chemistry

Postby AnitaK » Wed Jan 31, 2018 7:35 pm

please explain question 6)
more specifically how do you know that Fm= 3.72

I thought that this was the freezing point of the sucrose soln. NOT the change in freezing point??
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Thu Feb 01, 2018 9:14 pm

AnitaK wrote:please explain question 6)
more specifically how do you know that Fm= 3.72

I thought that this was the freezing point of the sucrose soln. NOT the change in freezing point??


Correct, -3.72 oC (that's negative 3.72 degrees Celcius) is the actual freezing point of the sucrose solution. But water freezes at 0 degrees and so the change in freezing point is 3.72.
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Re: Chapter 5: Solution Chemistry

Postby zimengy17106 » Sun Feb 28, 2021 4:21 pm

Hi,

I have questions on Question 19 and Question 29 and would like your help please.

In Question 19, "A solution of 50 gram of KCl at 85 degree was slowly cooled to 60 degree without precipitate". But the volume/weight of solution is unknown. The online solutions choose D, but it is based on the assumption that it is 50g KCl in 100g water. However, if it is 50g KCl in 119g water (which equals 42g KCl in 100g water), then at 85 degree it is unsaturated and at 60 degree it is just saturated without any precipitate. In this way, Choice A is also correct?

In Question 29, I thought choice C is also correct? Because when A and B contribute equally, which is at the joint point of the 2 green lines, on the x axis it looks like XA=0.3 XB=0.7, so component B is in greater concentration than A. Why is C wrong?

Many thanks
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