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I need help on this question

I need help on part Bi I want to know why is it that you can't do it by working out moles and multiply by Mr why is that you you have to use concentration which 23.9gdm3 and multiply by volume

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Original post by Selekt1234
I need help on part Bi I want to know why is it that you can't do it by working out moles and multiply by Mr why is that you you have to use concentration which 23.9gdm3 and multiply by volume


Maybe it might help to actually tell us what 'part Bi' is... :facepalm:
Do you have the volume?
Reply 4
AAnother picture
Reply 5
The question is from the as paper1 last question edexcel chemistry
Reply 6
Original post by Selekt1234
The question is from the as paper1 last question edexcel chemistry


We're getting there... a year might help, don't you think?
Reply 7
i couldnt get the attachment to attahc but here it is
find the number of moles of NaOH used
find the number of moles of trichloroethanoic acid used in the average neutralisation volume.
find the number of moles of trichloroethanoic acid in 250cm3.
hence find the mass of trichloroethanoic used
hence find percentage purity
it is less than what the manufacturer claims
Reply 9
by doing that i get 5.31 g in 250cm3 but the answer is 5.795g
the question is from as paper 1 specimen paper
post the full question and your working
so i worked out the moles of NAOH 0.130x25/1000=0.00325 mol/dm3
Ratio is 1:1
therefore the moles of the acid is 0.00325 as well in 25cm2
moles in 250cm2 is 0.0325
the i multiplied thatby the mr which is 163.5 and i got 5.31375
% purity = 5.31375/6.2=85.7%
bu the answer is 96.4% and their mass is 5.975g
i want to know why is that when you do it the normal way you get the worng answer.
Their method is that they used vthe concentration which was the answer from the previous question to find the mass in 250cm3
they did
23.9 g in 10000cm3
and devide by 4 to get the mass in 250cm3 which gave them 5.975g
Here i have attached the concentratrion question
Original post by Selekt1234
so i worked out the moles of NAOH 0.130x25/1000=0.00325 mol/dm3
Ratio is 1:1
therefore the moles of the acid is 0.00325 as well in 25cm2
moles in 250cm2 is 0.0325
the i multiplied thatby the mr which is 163.5 and i got 5.31375
% purity = 5.31375/6.2=85.7%
bu the answer is 96.4% and their mass is 5.975g
i want to know why is that when you do it the normal way you get the worng answer.
Their method is that they used vthe concentration which was the answer from the previous question to find the mass in 250cm3
they did
23.9 g in 10000cm3
and devide by 4 to get the mass in 250cm3 which gave them 5.975g
Here i have attached the concentratrion question


% error in the procedure:

mass used = 100 x 0.05/6.2 = 0.81%
250 ml of solution = 100 x 0.16/250 = 0.064%
pipette aliquot = 100 x 0.15/25 = 0.60%
burette reading 100 x 0.1/20 = 0.5% (OK I had to guess the titre as it wasn't given)
------------------------------------------ total
% error total = 2.0 %

99.9 - 2.0 = 97.9
96.4 + 2.0 = 98.4

Both answers are the same within the accuracy limits of the determination method.
(edited 5 years ago)
Original post by charco
% error in the procedure:

mass used = 100 x 0.05/6.2 = 0.81%
250 ml of solution = 100 x 0.16/250 = 0.064%
pipette aliquot = 100 x 0.15/25 = 0.60%
burette reading 100 x 0.1/20 = 0.5% (OK I had to guess the titre as it wasn't given)
------------------------------------------ total
% error total = 2.0 %

99.9 - 2.0 = 97.9
96.4 + 2.0 = 98.4

Both answers are the same within the accuracy limits of the determination method.

i dont want to know % error i want % purity
Original post by Selekt1234
i dont want to know % error i want % purity


Maybe, but as you omitted the titration data I thought I'd give another perspective ... :smile:
Could you please tell me how to find out % purity please and why is it the way i didi it is wrong
Original post by Selekt1234
Could you please tell me how to find out % purity please and why is it the way i didi it is wrong


moles of acid is found in the titre volume not 25cm3
Original post by BobbJo
moles of acid is found in the titre volume not 25cm3

Fine I will do it this way but I still get it wrong
0.036516675×163.5=5.970
5.970/6.2×100=96.29%

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