Back
to top
Error in AQA A-Level Chemistry CGP revision guide
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
I think I’ve found an error in a math calculation in the AQA A-Level Chemistry CGP guide... can someone check it - I can’t post a pic so I’ll type the whole question in, alternatively it is page 127 topic transition metals titrations with transition metals.
Example: 27.5cm3 of 0.0200 mol dm-3 aqueous potassium manganate(VII) reacted with 25.0cm3 if acidified sodium ethanedioate solution. Calculate the concentration of C2O4- ions in the solution.
2MnO4- + 16H+ + 5C2O4- —> 2Mn^2+ + 8H2O + 10CO2
So above is the question, and then the working out is shown in the question, but at the end the answer is shown to be 0.552mol dm-3 but in my own working out I got 0.0552 and even when I typed in the working out from the actual page into my calculator I got 0.0552 - not 0.552. Can someone good at math check this for me and get back to me, please. Would be very appreciated. Thank you!
Example: 27.5cm3 of 0.0200 mol dm-3 aqueous potassium manganate(VII) reacted with 25.0cm3 if acidified sodium ethanedioate solution. Calculate the concentration of C2O4- ions in the solution.
2MnO4- + 16H+ + 5C2O4- —> 2Mn^2+ + 8H2O + 10CO2
So above is the question, and then the working out is shown in the question, but at the end the answer is shown to be 0.552mol dm-3 but in my own working out I got 0.0552 and even when I typed in the working out from the actual page into my calculator I got 0.0552 - not 0.552. Can someone good at math check this for me and get back to me, please. Would be very appreciated. Thank you!
0
reply
Report
#2
Start by converting the volumes to dm^3:
27.5 * 10^-3 = 0.0275
25 * 10^-3 = 0.025
The total volume is 0.0525
Then find the moles of manganate using the volume and concentration
Moles = concentration * volume
0.0275*0.02 = 0.00055 mol manganate
Next calculate the moles of ethanedioate needed to react with the manganate
0.00055 * 5/2 = 0.001375
And find the concentration of ethanedioate
Concentration = moles/volume
= 0.001375/0.025
= 0.055 mol/dm^3
I think you’re right, but that’s not surprising. Those books are full of mistakes, especially for subjects like chemistry. If you have good notes, don’t use CGP. And make sure you look at the specification for your exam to see exactly what you have to know, sometimes those books will miss something or tell you that you don’t need to learn an equation that you actually do need to know; people in my class at A level learnt that the hard way
27.5 * 10^-3 = 0.0275
25 * 10^-3 = 0.025
The total volume is 0.0525
Then find the moles of manganate using the volume and concentration
Moles = concentration * volume
0.0275*0.02 = 0.00055 mol manganate
Next calculate the moles of ethanedioate needed to react with the manganate
0.00055 * 5/2 = 0.001375
And find the concentration of ethanedioate
Concentration = moles/volume
= 0.001375/0.025
= 0.055 mol/dm^3
I think you’re right, but that’s not surprising. Those books are full of mistakes, especially for subjects like chemistry. If you have good notes, don’t use CGP. And make sure you look at the specification for your exam to see exactly what you have to know, sometimes those books will miss something or tell you that you don’t need to learn an equation that you actually do need to know; people in my class at A level learnt that the hard way
0
reply
Thank you, Charlie. I don’t really have notes for chemistry, would you recommend the exam board specific textbook? In addition to the CGP books?
(Original post by Charlie133)
Start by converting the volumes to dm^3:
27.5 * 10^-3 = 0.0275
25 * 10^-3 = 0.025
The total volume is 0.0525
Then find the moles of manganate using the volume and concentration
Moles = concentration * volume
0.0275*0.02 = 0.00055 mol manganate
Next calculate the moles of ethanedioate needed to react with the manganate
0.00055 * 5/2 = 0.001375
And find the concentration of ethanedioate
Concentration = moles/volume
= 0.001375/0.025
= 0.055 mol/dm^3
I think you’re right, but that’s not surprising. Those books are full of mistakes, especially for subjects like chemistry. If you have good notes, don’t use CGP. And make sure you look at the specification for your exam to see exactly what you have to know, sometimes those books will miss something or tell you that you don’t need to learn an equation that you actually do need to know; people in my class at A level learnt that the hard way
Start by converting the volumes to dm^3:
27.5 * 10^-3 = 0.0275
25 * 10^-3 = 0.025
The total volume is 0.0525
Then find the moles of manganate using the volume and concentration
Moles = concentration * volume
0.0275*0.02 = 0.00055 mol manganate
Next calculate the moles of ethanedioate needed to react with the manganate
0.00055 * 5/2 = 0.001375
And find the concentration of ethanedioate
Concentration = moles/volume
= 0.001375/0.025
= 0.055 mol/dm^3
I think you’re right, but that’s not surprising. Those books are full of mistakes, especially for subjects like chemistry. If you have good notes, don’t use CGP. And make sure you look at the specification for your exam to see exactly what you have to know, sometimes those books will miss something or tell you that you don’t need to learn an equation that you actually do need to know; people in my class at A level learnt that the hard way
0
reply
Report
#4
The correct answer is 0.0550 moldm^-3. (not sure where they or you get the 2 on the end).
Text books and revision guides are riddled with errors and omissions.
I have done a fair bit of checking for publications and am appalled at the errors I find in the authors' work.
It is possible this was simplified and in a previous version of the question there was a x10 dilution which would explain the book answer but it all likelyhood it is just poor error checking by CGP.
Text books and revision guides are riddled with errors and omissions.
I have done a fair bit of checking for publications and am appalled at the errors I find in the authors' work.
It is possible this was simplified and in a previous version of the question there was a x10 dilution which would explain the book answer but it all likelyhood it is just poor error checking by CGP.
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
Today on TSR
-
Ofqual consultationTeachers to use 'mini-exams' and other work to decide grades
- Is it too late to do well in my A-levels?
- How are you feeling today?
- 2021 wellbeing blog launch
- Student tips on lockdown mental health