# 2005 - Written Paper - 15(b)(ii)

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### #1AM4R

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Posted 23 May 2006 - 05:38 PM

Hi

Can someone take me through how you do Q.15 b iii

Thanks
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### #2Nathan

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Posted 23 May 2006 - 05:50 PM

niodine=0.005 x 0.0214 = 1.07x10-4

1:1 ratio, so 1.07x10-4 iodine means there is 1.07x10-4 vitamin C

mass of 1 mole = 12 x 6 + 8 x 1 + 6 x 16 =72+8+96=180g

1.07x10-4 mole= 1.07x10-4 x 180 = 0.01926g

but unfortunately it's wrong...any ideas why?

### #3AM4R

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Posted 23 May 2006 - 05:53 PM

i think i got something like that too im confuzzled!
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### #4Nathan

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Posted 23 May 2006 - 06:01 PM

i dunno if this'll help you any, but have a look at PPA3 here: http://www.hsn.uk.net/resources/HSN14310

edit:
from 2005 marking solution

(iii) moles I2 = 21.4/1000 x 0.005
= 0.000107 (1.07 x 10-4)

moles vitamin C in 500 cm3 = 1.07 x 10-3

relative formula mass = 176
mass = 1.07 x 10-3 x 176
= 0.188 (or 0.2) g

so basically all got wrong was the RFM and forgot to multiply by ten

### #5AM4R

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Posted 23 May 2006 - 06:17 PM

how is the moles for the iodine and vitamin c the same?
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### #6Nathan

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Posted 23 May 2006 - 06:19 PM

if you look at the equation, it's a 1 to 1 ratio

### #7AM4R

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Posted 23 May 2006 - 06:23 PM

ah right, thanks
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