OCR B (Salters) F332 Tuesday 4th June 2013 Exam Discussion (Now Closed)

Heyy Guys does anyone have the Jan. 2013 paper and mark scheme available?

Would help alot; thaanks! :colondollar:

Reply 142

Original post by Branny101

Heyy Guys does anyone have the Jan. 2013 paper and mark scheme available?

Would help alot; thaanks! :colondollar:

I've put the mark scheme up, but you'll have to find the paper somewhere else :smile:

Reply 143

Hmmm thanks - where abouts is it?

And I need the paper more than anything :frown:

Reply 144

Original post by Branny101

Hmmm thanks - where abouts is it?

And I need the paper more than anything :frown:

Somewhere on this thread or the pre-release one. I'll try and get hold of the paper :smile:

Reply 145

nukethemaly

11

Original post by Branny101

Hmmm thanks - where abouts is it?

And I need the paper more than anything :frown:

2013_F332_January (1).pdf242.0KB

here!

2013_Markscheme_January_F332.pdf197.9KB

2013_F332_January_advance_notice.pdf234.7KB

Reply 146

Original post by nukethemaly

here!

THANKS MATE :ahee:

Reply 147

Original post by nukethemaly

here!

Thanks a lot! :biggrin:

Reply 148

Heyy could someone explain Jan. 2013, Q 4 c i) I don't understand why the mark scheme has drawn the isomer like that, isn't it just the methyl group that swaps?

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Reply 149

Original post by Branny101

Heyy could someone explain Jan. 2013, Q 4 c i) I don't understand why the mark scheme has drawn the isomer like that, isn't it just the methyl group that swaps?

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It's the C=O on the C=C on the left that 'swaps' so it on the other side of the carbon double bond.

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Reply 150

Original post by super121

It's the C=O on the C=C on the left that 'swaps' so it on the other side of the carbon double bond.

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Omg thanks but how would you know to swap that as apposed to the methyl groups?

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Reply 151

Original post by Branny101

Omg thanks but how would you know to swap that as apposed to the methyl groups?

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The molecule has E/Z isomers because of the carbon double bond on the right. It can't be the one on the left because there isn't 2 different groups on each carbon of the carbon double bond. You can swap the methyl groups because they are bonded to different carbon double bonds.
It may help to draw out the full structural formula to understand it better.

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Reply 152

nicolaa

Hi guys, I was just wondering if you'd be able to clarify which molecule shapes mean that the molecule will not have a dipole even though it's bonds are polar...tetrahedral and????

Thank you! :smile:

Reply 153

Original post by nicolaa

Hi guys, I was just wondering if you'd be able to clarify which molecule shapes mean that the molecule will not have a dipole even though it's bonds are polar...tetrahedral and????

Thank you! :smile:

It's not necessarily the shape of the molecule, even if you have a tetrahedral molecule, you still can have a polar molecule. For example, trichlorofluromethane is polar:

Because fluorine is more electronegative than chlorine, the charges will be unbalanced across the molecule and will not cancel out. Thus, it's important you know how electronegativity values can effect the overall charge of a shape.

If there's a symmetrical molecule, like tetrachloromethane, then that is non-polar. This is because all the negative charge can be cancelled down, as there is an equal electronegative difference between carbon and the chlorine atoms around the molecule.

In this exam, they'll mainly focus on more simple molecule shapes, such as tetrahedral, bent, triangle planers so it'll be obvious if the charges can get canceled down or not. Try some past paper questions, it has been asked quite a few times. Hope that helped! :smile:

Reply 154

blondie24

2

I'm confused on Q11 part C in Chemical Ideas problems for 1.5
I know the answers my question is WHY find out step c? What is its purpose? why is it necessary?

"The concentration of an acid solution can be found by titrating the acid solution with a solution of an alkali of a known concentration. In such a titration it was found that 19.00cm³ of 0.100mol dm^-3 sodium hydroxide were necessary to react with 25.00cm³ of a hydrochloric acid solution
NaOH(aq) +HCl(aq) --> NaCl(aq) + H₂O(l)

a) How many moles of NaOH are there in 19.00cm³ of 0.100mol dm^-3 solution ? Answer = 0.0019
b) How many moles of HCl are , therefore , in 25.00cm³ of the acid solution? Answer = 0.0019
c) How many moles of HCl are, therefore, in 1000cm³ of the acid solution? Answer = 0.0076
d) What is the concentration of the HCl solution? Answer = 0.0076"

Reply 155

Original post by blondie24

x

Sometimes there'll be a dilution in the process. For example Jan 2012, Q1e,v.

So you'll have to work out the dilution factor, which in the chemical ideas question will be 1000/25. Which is 40, so 40 times the initial moles you worked out of 0.0019, equals: 0.076. Check your answer again, I've checked the official answers, and it's 0.076 mol.

This is an important concept that is especially developed as you go into F334, where the calculations are far trickier and horrible!

Hope that helped :smile:

Reply 156

ashxx

10

hi guys, do we have to know how to do the practical? in my revision guide it has things like 'preperation of halogenoalkanes'???

Reply 157

blondie24

2

Original post by abzy1234

Sometimes there'll be a dilution in the process. For example Jan 2012, Q1e,v.

So you'll have to work out the dilution factor, which in the chemical ideas question will be 1000/25. Which is 40, so 40 times the initial moles you worked out of 0.0019, equals: 0.076. Check your answer again, I've checked the official answers, and it's 0.076 mol.

This is an important concept that is especially developed as you go into F334, where the calculations are far trickier and horrible!

Hope that helped :smile:

so you find out the moles in 1000cm3 because its the dilution factor :s-smilie:

Reply 158

Original post by blondie24

so you find out the moles in 1000cm3 because its the dilution factor :s-smilie:

Yes. Have you done some titration practicals?

Think about it. If I have an original solution of 1000cm³, it won't be wise for me to use all of it for my titrations. So I'll have to scale the volume down, and we do this by diluting it to a lower value of around 20 to 30 cm³ in most cases. This will obviously change the number of moles, so after finishing this modified practical, I'll have to scale it back up to reflect my original volume.

Practice the calculations, it'll help a lot :biggrin:

Reply 159