Last saved by: Luke

MOLES, EQUATIONS AND ACIDS TEST MS

1.	(i)	0.0268 OR 0.027 OR 0.02675 mol 
NO OTHER ACCEPTABLE ANSWER
1
(ii)	1.61 × 10^22
ALLOW 1.6 × 10^22 up to calculator value
ALLOW
ECF answer to (i) × 6.02 × 10^23
ALLOW any value for N_A in the range:
6.0 × 10^23 – 6.1 × 10^23
1
[2]
 
2.	(i)	0.438/87.6 = 5.00 × 10^−3 / 0.00500 mol 	1
(ii)	0.00500 × 24.0 = 0.120 dm^3  (accept 120 cm^3)	1
(iii)	0.00500 × 1000/200 = 0.0250 mol dm^−3 	1
[3]

3.	(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 HNO_3 = 2 × 0.0267
= 0.0534 or 0.0535 /0.053 mol 
(i.e. answer to (i) x 2)
volume of HNO_3 =  = 21.4 cm^3 	2
calc from value above = 21.3903743
If 0.053 mol, answer is 21 cm^3 but accept 21.2 cm^3
If 0.054 mol, answer is 22 cm^3 but accept 21.6 cm^3
[4]

4.		Calculates correctly:
Mol of Sr(NO_3)_2 =  
ALLOW 0.025
	Calculates correctly:
Mol of gas = 5/2 × 0.0250 = 0.0625 
ALLOW ECF for first answer × 2.5 as calculator value or correct rounding to 2 significant figures or more but ignore trailing zeroes
	Calculates correctly:
Volume of gas = 24.0 × 0.0625 = 1.50 dm^3 
ALLOW ECF for second answer × 24(.0) as calculator value or correct rounding to 2 significant figures or more but ignore trailing zeroes
DO NOT ALLOW ECF of first answer × 24(.0) (which gives 0.6(0) dm^3) as this has not measured the volume of any gas, simply 0.0250 mol of solid Sr(NO_3)_2 converted into a gas
i.e. This answer would give one mark
ALLOW 1.5 dm^3
ALLOW ECF producing correct volume of NO_2 only
i.e. 1.2(0) dm^3 would give two marks
OR
ALLOW ECF producing correct volume of O_2 only
i.e. 0.3(0) dm^3 would give two marks
3
[3]

5.		(i)	12 × 50/1000 = 0.600 mol 	1
(ii)	4 mol HCl → 1 mol Cl_2 / moles Cl_2 = 0.15 mol 
	vol of Cl_2 = 0.15 × 24 = 3.60 dm^3 	2
2nd mark is consequential on molar ratio given 
[3]

 6.	(i)	Mass Sb_2S_3 in stibnite = 5% of 500 kg = 25.0 kg 
Moles Sb_2S_3 =  / 73.5/ 73.529 /73.53/ 74 mol 
		(calculator value: 73.52941176)
If 5% is not used, 1471 mol; ecf for 2nd mark
		(calculator value: 1470.588235)
If 5% is used 2nd, 73.6 mol: OK for both marks	2
(ii)	moles Sb = 2 × 73.5 mol 
mass Sb = 2 × 73.5 × 122 g = 17.9 kg 
If the 2 isn’t used, answer = 73.5 × 122 = 8.95 
ecf ans from (i) x 2
ecf ans above x 2
OR
% Sb = 244/340 = 71.7% 
mass Sb = 25.0 × 71.7/100 = 17.9 kg  (ecf as above)	2
[4]

7.	(a)	(i)	Calculate correctly 
OR 0.00220 mol 
ALLOW 0.0022 OR 2.2 × 10^–3 mol
1
(ii)	Calculates correctly  = 1.10 × 10^–3 mol
OR 0.00110 mol 
ALLOW 0.0011 OR 1.1 × 10^–3 mol
ALLOW ECF for answer (i)/2 as calculator value or correct rounding to 2 significant figures or more but ignore trailing zeroes
1
(iii)	
OR 6.25 × 10^–2 mol dm^–3 
ALLOW 0.063 OR 6.3 × 10^–2 mol dm^–3
ALLOW ECF for answer (ii) × 1000/17.60
OR
ECF from (i) for answer (i)/2 × 1000/17.60 as calculator value or correct rounding to 2 significant figures or more but ignore trailing zeroes
1

(b)	(i)	(The number of) Water(s) of crystallisation 
IGNORE hydrated OR hydrous
1
(ii)	142.1 
ALLOW 142
ALLOW M_r expressed as a sum
ALLOW ECF from incorrect M_r and x is calculated correctly
	x =  = 10 
ALLOW ECF values of x from nearest whole number to calculator value
ALLOW 2 marks if final answer is 10 without any working
2
[6]

8.	(i)	M(MgSO_4) = 120.4 OR 120 (g mol^–1) 
	mol MgSO_4 =  = 0.0125 mol 
ALLOW 0.013 up to calculator value of 0.012541528 correctly rounded (from M = 120.4 g mol^–1)
ALLOW 0.013 up to calculator value of 0.012583333 correctly rounded (from M = 120 g mol^–1)
ALLOW ecf from incorrect M i.e. 1.51 ÷ M
2
 (ii)	 = 0.0872(2) (mol) 
ALLOW 0.09 up to calculator value of 0.08722222
1
 (iii)	× = 7 
ALLOW ecf i.e. answer to (ii) ÷ answer to (i)
ALLOW correctly calculated answer from 1 significant figure up to calculator value, ie, × does not have to be a whole number. Likely response = 6.95 
1
[4]

9.	(i)	Simplest (whole number) ratio of atoms/moles/elements 	1
(ii)	ratio Rb : Ag : I = 7.42/85.5 : 37.48/108 : 55.10/127
		or 0.0868 : 0.347 : 0.434
			or 1 : 4 : 5 
			= RbAg_4I_5 	2
[3]

10.	(i)	moles Ti = 1.44/47.9 = 0.0301 mol/0.03 mol	1
(accept use of answer from (b))
(ii)	mass of Cl = 5.70-1.44 = 4.26 g 
moles Cl = 4.26/35.5 = 0.120 mol 	2
5.70/35.5 = 0.161 mol gets 1 mark
(iii)	Ti:Cl = 0.0301 : 0.12 = 1:4.
Empirical formula = TiCl_4 
0.0301 : 0.161 mol gives TiCl_5 for 1 mark	1
(iv)	Ti + 2Cl_2 → TiCl_4 	1
(ecf possible from (iii)
covalent 
[5]

11.	(i)	a proton donor 	1
(ii)	MgO + 2HCl → MgCl_2 + H_2O 	1
[2]

 12.	(i)	CaCO_3 (s) + 2HCl(aq) → CaCl_2(aq) + CO_2(g) +H_2O (l)
CaO(s) + 2HCl(aq) → CaCl_2(aq) + H_2O (l)
each balanced equation 1 mark (2)
all state symbols (1)	3
(ii)	CaCO_3 fizzes/ gas given off/ gas evolved / carbon dioxide evolved (1)	1
[4]

13.	(i)	The H^+ ion in an (nitric) acid has been replaced by a metal ion
OR by a Ca^2+ ion 
DO NOT ALLOW it has been produced by the reaction of an acid and a base as this is stated in the question.
IGNORE references to replacement by NH_4^+ ions or positive ions.
ALLOW H OR Hydrogen for H^+;
DO NOT ALLOW Hydrogen atoms
ALLOW Ca OR Calcium for Ca^2+.
DO NOT ALLOW Calcium atoms
ALLOW ‘metal’ for ‘metal ion
1
(ii)	2HNO_3(aq) + Ca(OH)_2(aq) → Ca(NO_3) _2 (aq)+ 2H_2O(l)
Formulae 
Balance AND states 
ALLOW multiples
ALLOW (aq) OR (s) for Ca(OH) _2
2

(iii)	Accepts a proton OR accepts H^+ 
ALLOW H^+ + OH^–→ H_2O
ALLOW OH^– reacts with H^+ OR OH^– takes H^+
ALLOW OH^– ‘attracts’ H^+ if ‘to form water’ is seen
DO NOT ALLOW OH^– neutralises H^+ (‘neutralises’ is in the question)
1
[4]

14.	(i)	2NH_3 + H_2SO_4 → (NH_4)_2SO_4 
ALLOW 2NH_4OH + H_2SO_4 → (NH_4)_2SO_4 + 2H_2O
ALLOW NH_3 + H^+ → NH_4^+
ALLOW any correct multiple
IGNORE state symbols
1
(ii)	when the H^+ in an acid is replaced by a metal ion OR an
ammonium ion OR a + ion 
ALLOW H for H^+;
ALLOW ‘metal’ for ‘metal ion’
i.e.: H in an acid can be replaced by a metal
1
(iii)	accepts a proton OR accepts H^+ 
ALLOW donates a lone pair
ALLOW removes H^+
ALLOW forms OH^− ions
1
(iv)	132.1 
IGNORE units
NO OTHER ACCEPTABLE ANSWER
1
[4]

Paddington Academy