6 replies to this topic

[AM4R]

Hi

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

Thanks

[Nathan]

my answer is this
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 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?

[AM4R]

i think i got something like that too im confuzzled!

[Nathan]

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

[AM4R]

how is the moles for the iodine and vitamin c the same?

[Nathan]

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

[AM4R]

ah right, thanks