Author: nashton
Company: RM plc

F321 module 1 Practice 4:

1.	(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]

2.	(i)	dative covalent, bonded pair comes from same atom/	1
electron pair is donated from one atom/
both electrons are from the same atom 
(ii)	Ca(NO_3)_2 → CaO + 2NO_2 + ½O_2 
or double equation with 2/2/4/1	1
[2]

3.	(i)	203.3 g mol^−1 	1
Accept 203
(ii)	white precipitate / goes white 	1
(iii)	Ag^+(aq) + Cl^−(aq) → AgCl(s)
equation 
state symbols 	2
AgCl dissolves in NH_3(aq) 
(iv)	AgBr dissolves in conc NH_3(aq)/
partially soluble in NH_3(aq) 
AgI insoluble in NH_3(aq) 	3
[7]

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

5.	
isotope | protons | neutrons | electrons
^12C | 6 | 6 | 6 |
^13C | 6 | 7 | 6 |
[2]

6.	(i)	mass spectrometry 	1
(ii)	mass of an isotope compared with carbon-12 
1/12th of mass of carbon-12/on a scale where carbon-12 is
12 		2
	mass of 1 mole of the isotope/mass of 1 mole of carbon-12
is equivalent to the first mark
	“mass of the isotope that contains the same number of
atoms as are in 1 mole of carbon-12” → 1 mark (mark
lost because of mass units)
(iii)	12 × 95/100 + 13 × 5/100 OR 12.05 
= 12.1 (mark for significant figures) 
(12.1 scores both marks)	2
[5]

7.	(i)	moles CO_2 = 1000 /44 mol = 22.7 mol 
volume CO_2 in 2000 = 22.7 × 24 = 545 dm^3 
(ii)	reduction = 545 × 60/100 = 327 dm^3 
[3]

8.	(a)	……Ca(s) + …..2  HCl(aq) …….CaCl_2(aq) + .H_2(g). 	2
(g) not required for H_2
 
(b)	In Ca, oxidation state = 0  and	2
In CaCl_2, oxidation state = +2 
Oxidation number increases from Ca to CaCl_2
[4]

9.	(i)	moles HCl = 2.0 × 50/1000 = 0.10 	1
(ii)	moles Ca = ½ × moles HCl = 0.050 
	mass Ca = 40.1 × 0.050 = 2.00 g / 2.005 g 	2
(accept 40 × 0.050 = 2.0 g)
(mass Ca of 4.0 g would score 1 mark as ‘ecf’ as molar ratio
has not been identified)
(iii)	Ca has reacted with water 
Ca + 2H_2O → Ca(OH)_2 + H_2 
state symbols not required
	1st mark for H_2	3
2nd mark is for the rest of the balanced equation
[6]

10.	(a)	(i)	atoms of same element/same atomic number….. with
different numbers of neutrons/different masses 	1
(ii)	isotope	protons	neutrons	electrons	2
	^46Ti		22	24	22 
	^47Ti		22	25	22 
 
(b)	 
     = 47.7 	2
[5]

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