Hi

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

Thanks

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

### Nathan

Posted 23 May 2006 - 05:50 PM

my answer is this

n

1:1 ratio, so 1.07x10

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

1.07x10

n

_{iodine}=0.005 x 0.0214 = 1.07x10^{-4}1:1 ratio, so 1.07x10

^{-4}iodine means there is 1.07x10^{-4}vitamin Cmass 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?__### Nathan

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

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