Edexcel Official Chemistry Paper1:Core Inorganic and Physical Chemistry - 27th of May

Original post by loveire&song

Sorry for not being clear - on the exam paper, I'm pretty sure? I know they wrote it as "ClF3" but I also think, somewhere in the paragraph, they did write "Chlorine triflouride". I remember thinking about it by name, and I'm not great at chemistry, so I'd only have done that if I'd read the name somewhere :wink:

Okay i hope so, i assummed too as this compound is commmon and we have talked about it in class before :smile:

Reply 203

Original post by sc67

Has anyone made or started making an unofficial mark scheme for this paper ?

Remembering the questions would be the hard bit, we could try and make a list?

Reply 204

Original post by richpanda

Remembering the questions would be the hard bit, we could try and make a list?

Question 1 as I remember but I could be wrong!

Spoiler

Reply 205

Original post by Cryptokyo

Question 1 as I remember but I could be wrong!

Spoiler

yep I agree with all of this, the next one was the dodgy experiment one with the maths and the volume of hydrogen?

Reply 206

For the magnesium reaction one.

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(edited 7 years ago)

Reply 207

Original post by Cryptokyo

For the magnesium reaction one.

Spoiler

I can't remember getting that for the moles of H2, but I got 40.22 for the volume and so must be correct! I rounded my answer for the moles but kept the un rounded answer for my volume calculation, lucky I did that! Otherwise I would have got 40.3! I think the next question was bout the relative mass?

Reply 208

Original post by Cryptokyo

Question 1 as I remember but I could be wrong!

Spoiler

Yes I would agree. We can probably miss out some of the easily forgettable 1 markers, as most people probably get them correct.

Reply 209

Promethean_M

6

Original post by richpanda

It was still stupid of them though, but I also did it as chlorine trifluoride. Wasn't the question asking about the bond polarities? What did you put? I said that the bonds would be polar due to electronegativity difference, and that Cl would be delta positive as it is less electronegative. I also quoted the Pauling values from the data book.

That's all I did too. It said bonds, not the overall molecule

Reply 210

Original post by clairebear101

I can't remember getting that for the moles of H2, but I got 40.22 for the volume and so must be correct! I rounded my answer for the moles but kept the un rounded answer for my volume calculation, lucky I did that! Otherwise I would have got 40.3! I think the next question was bout the relative mass?

For the relative atomic mass question. There was 5.000g of lithium of which 0.460g was Lithium-6 and the rest is Lithium-7.
Molar mass of Li-6: 6.015
Molar mass of Li-7: 7.016

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(edited 7 years ago)

Reply 211

Does anyone know the acid used in the titration? I will post a solution to the titration once I know the acids and the concentrations.

(edited 7 years ago)

Reply 212

Original post by Cryptokyo

For the relative atomic mass question. There was 5.000g of lithium of which 0.460g was Lithium-6 and the rest is Lithium-7.
Molar mass of Li-6: 6.015
Molar mass of Li-7: 7.016

Spoiler

well i didnt do that methid but i think i got that answer or something like that, defo was 6.9 something!

Reply 213

Promethean_M

6

Original post by Cryptokyo

Does anyone know the acid used in the titration? I will post a solution to the titration once I know the acids and the concentrations.

It was nitric acid. Mean titre was 19.95 cm3. NaOH was in burette of conc 0.08moldm-3? 25cm3 aliquots from larger 250 cm3, make up from 10cm3 pure. Thats all I remember

Reply 214

Original post by clairebear101

well i didnt do that methid but i think i got that answer or something like that, defo was 6.9 something!

Perhaps this way? But I am not sure if this is correct as it gives a different answer.

\frac{0.460\times 6.015 + 4.540\times 7.016}{5}=6.923

Reply 215

Original post by SolomonP

It was nitric acid. Mean titre was 19.95 cm3. NaOH was in burette of conc 0.08moldm-3? 25cm3 aliquots from larger 250 cm3, make up from 10cm3 pure. Thats all I remember

Cheers, sounds about right! You are correct just checked.

(edited 7 years ago)

Reply 216

Original post by Cryptokyo

Perhaps this way? But I am not sure if this is correct as it gives a different answer.

\frac{0.460\times 6.015 + 4.540\times 7.016}{5}=6.923

it was 6.9 something. don't fuss over a few decimal places.

Reply 217

For titration:

Average titre of 19.95cm^3 of 0.08 moldm^3 NaOH, so 0.08 x 0.01995 = 0.001596mol of acid in 25cm^3.

Therefore 0.01956 mol of acid in 250cm^3. This is the diluted bit of acid. They added 10cm^3 of pure acid, so concentration is 0.01956/0.01 = 1.596 moldm^3.

Multiply this by the molar mass of nitric acid (63) to get 100.548 gdm^3.

Therefore the acid is suitable.

Reply 218

My solution to the titration with help of SolomonP. The question was that 10.00cm^3 of nitric acid was made into 250cm^3 using distilled water. It was then titrated using 0.08moldm^-3 NaOH in 25cm^3 aliquots.

Spoiler

(edited 7 years ago)

Reply 219