TRANSITION TEST MS

1.	(i)	(atoms of) same element/same atomic number/number of protons with
different numbers of neutrons/diff masses	1
(ii)	
	proton	neutron	electron
relative mass 1	1	/	negligible
relative charge	+1	0	−1	
i.e. 1 mark for each correct row
for electron, accept 1/1500 – 1/2000
for charges, accept +; 0; –
		2
[3]
 
2.	(a) (i)	153Eu has (2) more neutrons
	OR
153Eu has 90 neutrons AND 151Eu has 88 neutrons 
ALLOW There are a different number of neutrons
IGNORE Correct references to protons / electrons
DO NOT ALLOW Incorrect references to protons / electrons
1
   (ii)	(It has the) same number of protons AND electrons
OR
Both have 63 protons and 63 electrons 
ALLOW Same number of protons AND same electron configuration
DO NOT ALLOW ‘Same number of protons’ without reference to electrons (and vice versa)
1
	(b)	Mass of the isotope compared to 1/12th
OR
mass of the atom compared to 1/12th 
	(the mass of a) carbon-12 OR 12C (atom) 
IGNORE Reference to average OR weighted mean
(i.e. correct definition of relative atomic mass will score both marks)
ALLOW mass of a mole of the isotope/atom with 1/12th the mass of a mole OR 12 g of carbon-12 for two marks.
ALLOW 2 marks for:
‘Mass of the isotope OR mass of the atom compared to 12C atom given a mass of 12.0’
i.e. ‘given a mass of 12’ OR C12 is 12 communicates the same idea as 1/12th.’
ALLOW 12C OR C12
ALLOW 2 marks for:

i.e. fraction is equivalent to ‘compared to’
ALLOW 1 mark for a mix of mass of atom and mass of mole of atoms, i.e. ‘mass of the isotope/mass of an atom compared with 1/12th the mass of a mole OR 12 g of carbon-12.’
DO NOT ALLOW mass of ‘ions’ OR mass of element
2
(b)	(151 × 47.77) + (153 × 52.23)
100
	OR
	72.1327 + 79.9119
OR
	152.0446 (calculator value) 
	Ar = 152.04 
ALLOW Correct answer for two marks
ALLOW One mark for ECF from transcription error in first sum provided final answer is to 2 decimal points and is to between 151 and 153 and is a correct calculation of the transcription
2
[6]

3.	(a)		2

1, 2 or 3 p orbitals are OK
(b)	d orbital 2 
p sub-level 6 
3rd energy level 18 	3
 (i)	

(ii)		2
2s and 2p labels  Ignore any superscripted numbers.
8 electrons in correct levels with arrows correctly shown 
[7]

4.	(i)	79Br 35 protons, 44 neutrons, 35 electrons (1)
81Br 35 protons, 46 neutrons, 35 electrons (1)	2
(ii)	(1s2)2s22p63s23p63d104s24p5 (1)	1
[3]

5.	(i)	mass of Ni = 2.0g 
moles of Ni = 2.0/58.7 mol = 0.0341/0.034 mol 
(1 mark would typically result from no use of 25% ® 0.136 mol)	2
2nd mark is for the mass of Ni divided by 58.7
(ii)	number of atoms of Ni = 6.02 ´ 1023 ´ 0.0341	1
= 2.05 ´ 1022 / 2.1 ´ 1022 atoms 
Can be rounded down to 2.1 or 2.0 or 2 (if 2.0)
From 8 g, ans = 8.18/8.2 ´ 1022
(and other consequential responses)
[3]
 
6.	(i)	0.438/87.6 = 5.00 ´ 10−3 / 0.00500 mol 	1
(ii)	0.00500 ´ 24.0 = 0.120 dm3  (accept 120 cm3)	1
(iii)	0.00500 ´ 1000/200 = 0.0250 mol dm−3 	1
[3]
 
7.	(i)	Molar mass CaO = 56.1 (g mol–1)  (anywhere)	2
moles CaO =  = = 0.0267/0.027  calc: 0.0267379
Allow 56 which gives 0.0268
(ii)	moles HNO3 = 2 ´ 0.0267
= 0.0534 or 0.0535 /0.053 mol 
(i.e. answer to (i) x 2)
volume of HNO3 =  = 21.4 cm3 	2
calc from value above = 21.3903743
If 0.053 mol, answer is 21 cm3 but accept 21.2 cm3
If 0.054 mol, answer is 22 cm3 but accept 21.6 cm3
[4]

 8.	(i)	mass = 0.0500 ´ 23.0 = 1.15 g 	1
(ii)	moles H2 = 0.0250 
volume H2 = 0.0250 ´ 24 = 0.600 dm3 	2
ecf from calculated moles H2
0.0500 mol in 50.0 cm3	1
(iii)	concentration = 0.0500 ´ 20 = 1.00 mol dm−3 
[4]