2009 National Qualifying Exam – Chemistry · OLYMPIADS (Ô OH (iii) 2009 National Qualifying Exam...
Transcript of 2009 National Qualifying Exam – Chemistry · OLYMPIADS (Ô OH (iii) 2009 National Qualifying Exam...
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2009 National Qualifying Exam – Chemistry
Solutions mw of G(OH)a = X + 17.016 a
No. of mmoles G(OH)a from 1000 mg G = 1000/392.2 = 2.550 mmoles G
Mass of G(OH)a = 229 mg (from (c))
∴ mw of G(OH)a = 229/2.550 = 89.8
∴ aw of G = 89.8 − 17.016 a .
a can be 1,2 or 3 (not 4 as this would mean no NH4+)
[a, aw of G]: [1, 72.8] [2, 55.8] [3, 38.8]
55.8 corresponds to Fe. ∴ a = 2, the formula is Fe2+
(h) MnO4− + 5 Fe2+ + 8 H+ � Mn2+ + 5 Fe3+ + 4 H2O
(i) For total cation charge of +4, x must be 2
mw of salt = 392.2 of which Fe(NH4)2(SO4)2 contributes 55.8 + 2 × 17.06 + 2 × 96.06 = 282.0
∴ water contributes 392.2 − 282.0 = 110.2
∴ z = 110.2/18.016 = 6.11. Nearest whole number is 6.
(j) Fe(OH)2
(k) Fe(NH4)2(SO4)2.6 H2O
(l) Fe(OH)2 is rapidly oxidised in air to Fe(OH)3.
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