Author: nashton
Company: RM plc

F321 module 1 Practice 1:

1.	(i)	Molar mass of CaCO_3 = 100.1 g mol^–1 (1)
2.68/100.1 = 0.0268/0.027 (1)	2
(ii)	0.0268 mol × 24,000 = 643 cm^3 (1)	1
(iii)	moles HNO_3 = 2 × 0.0268
= 0.0536 /0.054 mol (1)
(i.e. answer to (i) × 2)
	volume of HNO_3 = 0.0536 × 1000/2.50 = 21.4 cm^3 (1)	2
[5]

2.	Molar mass of anhydrous calcium nitrate = 164.1 g mol^–1 (1)
Ratio Ca(NO_3)_2 : H_2O = 69.50/164.1 : 30.50/18
or 0.4235 : 1.694 or 1 : 4 (1)
Formula = Ca(NO_3)_2•4H_2O (1)
[3]

3.	(i)	because Ca has changed from 0 to +2 (1)
and H has changed from +1 to 0 (1)	2
(ii)	Calcium reacts with water producing
hydrogen/H_2/calcium/hydroxide/Ca(OH)_2 (1) (i.e. one product)
Ca(s) + H_2O(l) → Ca(OH)_2(aq) + H_2(g) (1) (i.e. full equation)
Equation would subsume both two marks	2
[4]

4.	(a)	(atoms of) same element/same atomic number….. with
	different numbers of neutrons/different masses 	1
 
(b)		3

isotope

percentage
composition

number of

| | protons | neutrons | electrons
^85Rb | 71 to 73 | 37 | 48 | 37
^87Rb | 27 to 29 | 37 | 50 | 37

mark

must add
up to 100

ie 1 mark for each atomic structure; 1 for % compositions.

= 85.6  2nd mark for significant figures	2
71/29: 85.58 = 85.6
73/27: 85.54 = 85.5
 
(c)	carbon-12/^12C 	1
[7]

5.	(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]

6.	(a)	(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
 
(b)	Evidence that the oxidation number of Mn has reduced
and one of the oxidation numbers correct (ie MnO_2: ox no
of Mn = +4 or MnCl_2: ox no of Mn = +2 
The other oxidation number of Mn is correct,
ie in MnO_2: ox no of Mn = +4
or in MnCl_2: ox no of Mn = +2 	2
[5]

7.	(i)	mass = 0.0500 × 23.0 = 1.15 g 	1
(ii)	moles H_2 = 0.0250 
volume H_2 = 0.0250 × 24 = 0.600 dm^3 	2
ecf from calculated moles H_2
0.0500 mol in 50.0 cm^3	1
(iii)	concentration = 0.0500 × 20 = 1.00 mol dm^−3 
[4]

8.	(i)	2Na + O_2 → Na_2O_2 	1
(ii)	Na_2O_2 + 2H_2O → H_2O_2 + 2NaOH 	1
correct covalent bonds shown 
(iii)	electron count (14) for rest of molecule correct 	2
[4]

9.	(i)	loss (of electrons) 	1
(ii)	Ba 
0 → (+)2  (accept 2+)	2
[3]

10.	(i)	atoms of the same element with different numbers of
neutrons/different masses (1)	1
(ii)	^79Br 35 protons, 44 neutrons, 35 electrons (1)
^81Br 35 protons, 46 neutrons, 35 electrons (1)	2
(iii)	(1s^2)2s^22p^63s^23p^63d^104s^24p^5 (1)	1
[4]

The King's CE School