Chemistry question help (1 Viewer)

[A.void.that.thinks]

Hi guys, if anyone could help me with the attached questions it would be greatly appreciated. The correct answer is highlighted, I just need help with the working out.

 

[jimmysmith560]

Are you sure the answer to question 19 is (C)? I am looking at the Baulkham Hills High School 2019 Chemistry trial exam (where this question appears) and it states the answer to this question is (D).

Now that I'm looking at a VCE variant of this question, i.e. same type of question (see below), it seems their answer is consistent with yours, despite the minor differences in the question:




This possibly means that the answer provided in the Baulkham Hills High School paper is incorrect.

Regardless I hope this helps for question 19!

 

[Vaibhav123456]

19) 1000 monomer units. Therefore 999 water molecules are eliminated. MM(polymer) = MM(1000 monomers) - MM(999 water molecules).

MM(polymer) = 1000(104) - 999(18) = 8.6 x 10^4 i.e option C.

 

[jazz519]

Are you sure the answer to question 19 is (C)? I am looking at the Baulkham Hills High School 2019 Chemistry trial exam (where this question appears) and it states the answer to this question is (D).

Now that I'm looking at a VCE variant of this question, i.e. same type of question (see below), it seems their answer is consistent with yours, despite the minor differences in the question:

View attachment 32302
View attachment 32304

This possibly means that the answer provided in the Baulkham Hills High School paper is incorrect.

Regardless I hope this helps for question 19!

If you look at the question here it says 500 monomer units. In this one the OP made it contains 1000 monomer units

 

[jazz519]

19) There are two types of polymer processes. These are the equations we use for the calculations for both:

Addition polymerisation: n monomer --> polymer

Condensation polymerisation: n monomer --> polymer + (n-1) H2O

All you have to do is select the correct equation and then substitute molar masses in and change the --> to an equals sign.

Here it is a condensation polymerisation process because 4-hydroxybutanoic acid reacts through OH and COOH to eliminate water molecules.

It cannot be an addition polymerisation process because these require the monomer to contain a carbon double bond (C=C). For example, propene, ethene, chloroethene are all things that would undergo addition polymerisation since they have a C=C.


Applying this to the question:
n monomer --> polymer + (n-1) H2O
as it is condensation polymerisation

n = 1000
monomer = 1 + 16 + 3(12 + 2x1) + 12 + 2x16 + 1 = 104 g/mol

1000 x (104) = polymer + (1000-1) (2x1 + 16)
104000 = polymer + 17982
86018 = polymer
polymer = 8.6 x 10^4 g/mol (2 sig figs)

 

[jazz519]

20) This is a back titration question. These are usually in the short answer because the calculation is long. Important thing to remember in questions like this is that you need to do it in a step-wise process. Don't think too much about the final answer and just calculate quantities you can see directly. It may not always be possible unless you are very good at calculation questions that you will see the full steps pathway to the answer straightaway and so if you attempt to find some quantities first, this can then potentially present the pathway to you as you go through the question.

Two reactions occurring:
MgO(s) + 2HCl(aq) --> MgCl2(aq) + H2O(l)
HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)

With back-titrations you always start at the bottom as it's easier to work through the answer this way:
v(NaOH) = 19.7 mL = 0.0197 L
C(NaOH) = 0.200 mol/L

Whenever you have a conc and volume, find the moles
n(NaOH) = cv = (0.200)(0.0197) = 0.00394 moles

They said this reacts with the excess acid so do a molar ratio for it. See how in the working I am clearly labelling which type of HCl it is. This is to avoid confusion with the different HCl moles we will find shortly.
n(HCl) excess = n(NaOH) (1:1 molar ratio)
n(HCl) excess = 0.00394 moles

v(HCl) = 100 mL = 0.1 L
C(HCl) = 2.00 mol/L
n(HCl) initial = cv = (2.00)(0.1) = 0.2 moles

Once you have initial and excess moles of a certain reagent, you can find the amount that reacted with the main compound.
n(HCl) reacted with MgO = initial - excess
n(HCl) reacted with MgO = 0.2 - 0.00394 = 0.19606 moles

Do molar ratio with MgO, using equation 1:
n(MgO) in 100 mL aliquot = 1/2 n(HCl)
n(MgO) in 100 mL aliquot = 1/2 (0.19606)
n(MgO) in 100 mL aliquot = 0.09803 moles

This is only the moles in the 100 mL aliquot, we had a solution that was 1.00 L prior, so we need to multiply by 10 to account for only taking one-tenth of the volume:
n(MgO) in 1 L = 0.09803 x 10 = 0.9803 moles

This is the sample as the 500 mL as it's just a dilution so we can ignore this number.
m(MgO) = n x MM = (0.9803) (24.31 + 16) = 39.515....g

%m(MgO) = m(MgO) / m(tablet) x 100
%m(MgO) = 39.515... / 40.6 x 100
%m(MgO) = 97.3%

Personally I would say this is an unsuitable question to put in the multiple choice. It's like the equivalent of doing a 5-6 marker in the short answer, which is way too much work for a multiple choice but that's the working above for how to get to the answer. Once you understand the method properly you could probably reduce some of the lines to make it faster to calculate for the multiple choice

 

[A.void.that.thinks]

Wow! Thank you so much, this really helped!

 

[jimmysmith560]

If you look at the question here it says 500 monomer units. In this one the OP made it contains 1000 monomer units

despite the minor differences in the question

 

[CM_Tutor]

Are you sure the answer to question 19 is (C)? I am looking at the Baulkham Hills High School 2019 Chemistry trial exam (where this question appears) and it states the answer to this question is (D).

Now that I'm looking at a VCE variant of this question, i.e. same type of question (see below), it seems their answer is consistent with yours, despite the minor differences in the question:

View attachment 32302
View attachment 32304

This possibly means that the answer provided in the Baulkham Hills High School paper is incorrect.

Regardless I hope this helps for question 19!

I am looking at the copy of the 2019 BHHS trial that I have, and @jimmysmith560 is correct that it has the question with a 1000 monomer units, and gives the answer as D. Jimmy is also correct that the VCE version he shows, with 500 units, is answered correctly. I concur with @jazz519 that the correct answer is unquestionably C, without any shadow of a doubt. Thus, the only available conclusion is that the BHHS answer is wrong.

Interestingly, the version of the paper that I has already has two other MCQ answers changed (question 8, where B has been changed to A, and question 14, where C has been changed to B). Both of these corrections are right, and in both cases the incorrect answer given originally arises from a common student mistake - in other words, if I was writing the exam, each of these wrong answers would be ones I would include as distractors, but there is no way that even a half-competent chemistry teacher would make these mistakes if they were going through a paper writing solutions. And, I just did a quick check, and 19 is wrong as well - the answers give C but it's actually B. Right below the MCQ is a sample answer for the graph for 21(a) and there is no way I could give that more than 1 mark out of 2 as there are two significant mistakes in it.

I don't understand how a BHHS paper could have 4 MCQ answers wrong, plus a badly drawn equilibrium sketch... though looking at the cut-and-paste solutions that I have, I wonder if the source for the answers is actually BHHS.

 

[jazz519]

despite the minor differences in the question

Sorry my bad I missed that