AS-LEVEL AP1 MS
 
1.      (a)     Proton        mass = 1                                           charge = +1
Electron     mass  1/1800 Or  5.6 × 10–4            charge = –1
(Do not accept +1 for proton mass or ‘g’ units)
2
(b)     (i)      13
1
(ii)     Si
1
Mass number = 28 and atomic number = 14
(Do not accept 28.1 or 28.0 or ‘Silicon’)
5
(c)     Mean (average) mass of an atom / all the isotopes
1/12th mass of atom of 12C 
Or   Mass of 1 mole of atoms of an element (1)
       1/12th mass of 1 mole of 12C (1)
Or   Average mass of an atom / all the isotopes (1)
       relative to the mass of a 12C atom taken as exactly 12 / 12.000 (1)
(Penalise ‘weight’ once only) (Ignore ‘average’ mass of 12C)
(Do not allow ‘mass of average atom’)
2
(d)     Ar = (24 × 0.735) + (25 × 0.101) + (26 × 0.164) 1 = 24.4 1
(mark M2 conseq on transcription error or incorrect addition of %)
(e)     Mr = highest m/z value 1
(NOT ‘highest/largest/right-hand’ peak)
3
[10]
 
2.	(a)     5s2 4d10 5p4           /           4d10 5s2 5p4
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p4
or 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p4
Allow any order but must finish with 5p4
1

(b)     (i)       or 
 
M1 for top line
1
127.8
M2 for correct denominator
1
127.8 with no working shown scores 3 marks
1
Or

1
Mark for 100 dependent on top line correct
1
127.8
1
(ii)     Other isotopes present / some isotopes absent / different abundances of isotopes
1
(c)     Te+ + e(−)  Te
Ignore state symbols
Allow Te2+ + 2e(−)  Te
1
(d)     128
Only
1
Most abundant ion (QoL − superlative)
M2 dependent on correct M1
1
(e)     2+ ion formed / 2 electrons removed
Due to 128Te2+ = 2 marks
1
From 128 (Te)
Mark independently
1
(f)     Same
If not same CE = 0 / 2
1
(Each isotope has the) same number of protons / same nuclear charge and same number of electrons / electronic configuration
Ignore more neutrons in 130Te
1
[12]

3.

4.      (a)     (i)      0.013;
1
(ii)     0.0065;
Answer to (i) ÷ 2
1
(iii)      = 84.3;
Allow 0.548 ÷ answer to (ii)
Allow 84.1 – 84.4
1
(b)     84.3 – 60 = 24.3;
1 mark for -60
1
          Mg;
          If 147.6 used the answer is 87.6 (1)
And this is Sr (1)
Allow consequential metal from their calculated A
Answer has got to be a metal to score M2
1
[5]

 5.    (a)     0.943 g water (M1)
If Mr of NiSO4 wrong, can allow M1 and M3 from
method 1 i.e. max 2
NiSO4                       H2O
  (M2)            (M3)
(8.68 × 10−3            0.052)
1                               6           or x = 6   (M4)
Allow Mr = 155
Allow other methods e.g.
Mr (NiSO4) = 58.7 + 32.1 + 64.0 = 154.8
n(NiSO4) =  = 0.008682 mol   (M1)
Mr (NiSO4.xH2O) =  = (263.4) (M2)
so 18x = 263.4 − 154.8 = (108.6)   (M3)
so x =  = 6   (M4)
 
If using alternative method and Mr of NiSO4 wrong, allow ecf to score M2 and M3 only i.e. max 2
4
(b)     re-heat
Heat to constant mass = 2 marks
1
check that mass is unchanged
M2 dependent on M1
Allow as alternative:
M1: record an IR spectrum
M2: peak between 3230 and 3550 (cm−1)
1
[6]

6.	(a)     The value of the titre would be higher (than the true value.)
1
(b)     It should have no effect.
1
The first titration can be ignored / subsequent titrations would be accurate
Allow references to the first titration being a ‘rough’ or ‘trial’ value.
1
[3]

7. (a)

(b)

(c)

(d)

(e)

	

(f)

(g)

(h)

(i)
	
---

	
(j)	

       

8.	(a)     M1   550 ×  = 579 g would be 100% mass
Allow alternative methods.
There are 4 process marks:
1
M2   So  = 8.91 moles NaN3
or
M1    = 8.46 moles NaN3 (this is 95%)
M2   So 100% would be 8.46 ×  = 8.91 moles NaN3
1: mass ÷ 65
2: mass or moles × 100 / 95 or × 1.05
3: moles NaN3 × 2
4: moles NaNH2 × 39
1
Then M3 Moles NaNH2 = 8.91 × 2 = ( 17.8(2) moles)
1
M4   mass NaNH2 = 17.8(2) × 39
1
M5   693 or 694 or 695 (g)
If 693, 694 or 695 seen to 3 sig figs award 5 marks
1
(b)     M1   308 K and 150 000 Pa
1
M2   n =  or 
1
M3   = 4.4(0) or 4.395   moles N2
Allow only this answer but allow to more than 3 sig figs
1
M4   Moles NaN3 = 4.395 ×  (= 2.93)
M4 is for M3 × 
1
M5   Mass NaN3 = (2.93) × 65
M5 is for moles M4 × 65
1
M6 = 191 g
Allow 190 to 191 g allow answers to 2 sig figs or more
1

(c)     (i)      150 / 65 = 2.31 moles NaN3 or 2.31 moles nitrous acid
1
Conc = 2.31 × 
 
M2 is for M1 × 1000 / 500
1
4.6(1) or 4.6(2) (mol dm−3)
Only this answer
1
(ii)     3HNO2  HNO3 + 2NO + H2O
Can allow multiples
1
 [15]

9.	ZnCO3 → ZnO + CO2
Ignore state symbols.
If equation incorrect, allow one mark only for correct atom economy method.
1
Percentage atom economy =
Mark consequentially for incorrect formula mass(es)
1
× 100 = 64.9
Accept answer to at least 2 significant figures
1
[3]

10.	D
[1]
11.	D
[1]
12.	C
[1]
13.	D
[1]