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SophieL1996

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-qp/6CH02_01_que_20100607.pdf could someone explain how to do all the calculations? as I am really struggling! Thanks in advance :smile:

Reply 1

A learner

question 17:
you've got concentration and volume of NaOH so work out no. of moles
(you've been given concentration in cm3 so you have to convert it to dm3 by dividing by 1000)
no. mol = conc x vol so n= 0.1x(20/1000) n=0.002

H2SO4 + 2NaOH → Na2SO4 + 2H2O

moles of sulfuric used:moles of sodium hydroxide used is 1:2 from the eq.
from the ratio, twice as much NaOH is used up (in terms of moles) so just half the number of moles of NaOH (0.002)
moles of sulfuric acid used = 0.002/2 =0.001

vol.=no. mol/conc. 0.001/0.5 = 0.002 dm3 (convert to cm3 by x1000), 2 cm3

Reply 2

A learner

question 19:
c) mean titre is (27.95+28.05+28.00)/3 so answer is 28.00 cm3 (so when using formula, you've to divide by 1000 again to convert it to dm3)

d)(i) 0.1(the concentration of NaOH) x 28.00/1000 = 0.0028 mol

(ii) CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
ratio of moles of ethanoic acid used : moles of NaOH used is 1:1
so the answer is 0.0028 mol again

(iii) c (mol dm^-3) = n(mol)/v(dm^3) so 0.0028/(25/1000) = 0.112 mol dm^-3

(iv) if our previous answer was the concentration for 250 cm^3 solution (which was diluted) then 25 cm^3 (which was not diluted) will be 10x more concentrated, right? so answer is 1.12 mol dm^-3

(v)units is gdm^-3 which is the same as g/dm^3
to get g/dm^3 you have to multiply g/mol x mol/dm^3 (so the mol cancels out from top and bottom and you're left with g on top and dm^3 at the bottom so g/dm^3
60(g/mol) x 1.12(mol/dm^3) = 6.72 g/dm^3

f) add and subtract 0.05 from 27.95 to get 28.00 and 27.90 so your answer is X :wink: