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Chapter 5: Solution Chemistry

Soloution Chem Chapter Review Question

Postby maddiewall2699 » Fri Feb 11, 2011 9:53 am

7) If the boiling point elevation constant, Kb, is 0.52 °C/m for water, what is the boiling point of a mixture comprised of 0.1 mole of C2H6O2 and 0.1 mole of C2H6 in 18 g of H20?

This question given in the online chapter review seems to require the equation for boiling pt elevation to be memorised. There was no mention of needing to memorise such equations in the GS general chem chapter of solution chem.
I have come across several such problems that require memorisation of knowledge not specified in the chapter index. What should I do?
Please help,
Thanks so much :(
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Postby goldstanda3269 » Sat Feb 12, 2011 11:42 pm

Stick to the chapter guides since they have been formulated based on past GAMSAT testing patterns. This means that ACER would normally provide the equation for boiling point elevation or freezing point depression. However, sometimes they provide equations without fully identifying the variables or by using non standard letters to represent the variables.

Nonetheless, consider the chapter review questions more like an open book exam with no time limits in order to get more familiar with the most important aspects of what you are studying. Though you may not have to memorize that equation, you should have a sense of the relationships.

The 4 ACER booklets +/- the 5 full length GS tests will be your practicing ground for the March 26 exam with respect to timing, etc.
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Postby domhnall.h7396 » Tue Mar 01, 2011 8:52 pm

Hi,
I am confused as to the answer to this question.
If the equation for boiling point elevation is iBm, do we need the number of particles for this problem and if so, what is the number of particles?
Thank you!
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Postby goldstanda3269 » Thu Mar 03, 2011 3:12 am

i = the number of particles

If you have, for example, a solution with NaCl which dissociates into 2 particles, then i = 2. If the molecules do not dissociate (as is the case in this problem) then i = 1.

The bottom line with "colligative properties" (which includes boiling point elevation and freezing point depression) is that they depend on the number of particles present and not the types of particles.
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Flashcards Bpt Q

Postby shehzad6090 » Mon Jan 23, 2012 11:52 pm

Hi :)
In the general chemistry flashcards, card number 7, the second question is
- A certain molal concentration of a sugar in H2O gives a boiling point of 102.4 C. the same molal conc. of NaCl in H2O gives a bp of?
- The answer on the other side says that, because NaCl has double the number of particles and this is a concept of colligative properties, the boiling point will change twice as much. (the delta T will be 2 times larger since i in the equation iBm is 2). Then it says, "thus the bp would be 104.8C".

My question is, that because the change in boiling point should be twice as much as before, shouldn't 102.4 x 2 = 204.8 C, not 104.8 C?
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Postby goldstanda3269 » Tue Jan 24, 2012 12:47 am

The question is testing an important area of chemistry called colligative properties which depend on the number, not the type, of particles (atoms or molecules). When a solute is added to a solvent, one gets boiling point elevation and freezing point depression. The math therefore does not give you the boiling point or freezing point but rather by how much they have changed. So, in the problem you mentioned, it is the difference that doubled and then you must add this to the boiling point of the solvent (100 degrees for water).
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Postby shehzad6090 » Tue Jan 24, 2012 6:10 pm

thankyou. that is rather helpful.
:D
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boiling point elevation and freezing point depression

Postby JWC01 » Wed Jul 25, 2012 1:12 am

For boiling point elevation the forumla delta TB = KB' XB = KBm is given.

I undertand what all of these mean except for KB?

I have the same problem for KF in freezing point depression.

In the DVD the forumla iBm is given for boiling point elevation.

Are i and KB the same, measuring number of particles?

Your help please.
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Postby goldstanda3269 » Wed Jul 25, 2012 5:49 am

Anytime you see the letter K or k in chemistry, you should assume that it is a constant.

KB is the Boiling point constant which can be symbolized many different ways including just the letter B (example: iBm).

The equation TB = KB' XB = KBm is a very common formulation found in most textbooks; however, it does not include the term "i" which is the number of particles which must be kept in mind if you are using the latter equation.

NB: of course you need to know how to use the preceding equations; however, on the GAMSAT, they will be presented and explained prior to the need to use them.
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Postby JWC01 » Wed Aug 08, 2012 7:28 pm

In section 5.1.2 there is the statement 'if the mass concentration of a solute (in kg solute/ kg of solvent) is known and the molality is determined from the freezing point of the solution, the mass of 1 mole of a solute can be calculated.

Can you please explain this to me in a bit more detail?
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Postby goldstanda3269 » Sun Aug 12, 2012 4:52 am

Basically, if you consider:

TB = KBm

If you do an experiment and you know TB and m then you can calculate KB.

Well, the sentence is saying that if you know mass concentration of a solute (in kg solute/ kg of solvent) and the molality (moles solute/kg solvent) then you can calculate the mass of 1 mole of a solute (this is dimensional analysis):


(kg solute)/(kg solvent) x 1/(moles solute/kg solvent) = (kg solute)/(moles solute)
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Re: Chapter 5: Solution Chemistry

Postby suncoast » Tue Nov 05, 2013 6:22 pm

Question 4, Help.

Could someone give me a detailed explanation on how to work out the answer for question 4. e.g. the given ksp.


Thankyou.
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Fri Nov 08, 2013 5:35 am

First, please look carefully at the explanation and examples of Ksp in the book.

In the explanation, it shows the 2 key steps:

1) solubility product "sp" takes the products raised to the power of the coefficients (i.e. the number in front of that element) in the balanced equation, and the products are multiplied by each other.

balanced equation: MgF2 <--> Mg2 + 2F-
So Mg has a coefficient of 1 and F has a coefficient of 2. So both concentrations (always symbolized by square brackets) are raised by those respective coefficients:

Ksp = [Mg2 ][F-]2 = 10-8

Given


2) Now just replace Mg and F with their concentrations as given in the question, so:

Ksp = [10-4][10-4]2

The rest is just multiplication.

Ksp = [10-4][10-4]2 = 10-12 (it is better to see this in the actual explanation to not get confused with missing exponents)

Now repeat with the next equation which is even easier because the balanced equation produces only coefficients of 1.
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Re: Chapter 5: Solution Chemistry

Postby hluo » Wed Mar 05, 2014 10:44 am

Dear Goldstandard,

I have a question regarding question 7. Perhaps it may be a little too easy to figure out the correct answer since the other values given seem a little too great to be true.
Also I felt that perhaps the temperature raised should be 2.6 degrees Celsius rather than the 5.2 degrees Celsius given. This is because C2H6 which I presume is ethane is non-soluble in water due to it being non-polar and also that it cannot form hydrogen bond with water. If I did not understand wrong, colligative property only applies to solutes is that right?

Thank you,
Hao
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Re: Chapter 5: Solution Chemistry

Postby goldstanda3269 » Wed Mar 05, 2014 6:14 pm

You are correct, ethane is immiscible with water (does not mix) and therefore the bp change would be as you mentioned. We updated the question to replace ethane with methanol which would dissolve in the water due to the H bonding/polarity and thus would contribute to the boiling point elevation. Good of you to notice.
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