The topics are in alphabetical order.
A
Q. 1.0 .
|
Acid
|
Acidic
|
Alali
|
Alkaline
|
Salts
|
Others
|
|
HCl
HNO3
H3PO4
H2S
|
Lime juice
Vinegar
Soda water
Gastric juice
|
NaOH
NH4OH
Slaked lime(aq)
|
Soap
Saliva
|
NaCl
CuSO4
Na2CO3
PbCl2
|
Water
Wax
Oil
|
2.1.Give formulae of salts formed by the acids marked in red
|
Ion
|
Salts
|
|
H+
|
HCl
|
HNO3
|
H2SO4
|
H2CO3
|
|
Cu2+
|
CuCl2
|
Cu(NO3)2
|
CuSO4
|
CuCO3
|
|
Hg2+
|
HgCl2
|
Hg(NO3)2
|
HgSO4
|
HgCO3
|
|
Ag+
|
AgCl
|
AgNO3
|
Ag2SO4
|
Ag2CO3
|
|
NH4+
|
NH4Cl
|
NH4NO3
|
(NH4)2SO4
|
(NH4)2CO3
|
|
Fe3+
|
FeCl3
|
Fe (NO3)3
|
Fe2SO4)3
|
Fe2(CO3)3
|
Which ion gives blue or bluish coloured salts. Cu+
3. Which ion give thermally unstable salts. NH4+
4. Name the following:-
a. A soluble carbonate (NH4)2CO3 b. c. Insoluble chloride. AgCl
Q.3.0
Complete these equations.
1. KOH HCl → H2O + KCl
2. H2SO4 + Ca(OH)2 → CaSO4 + 2H2O
3. .NH4OH +HNO3 → NH4NO3 + H2O
4, MgO + .HCl→ MgCl2 + H2O
5. NH4OH + H2CO3.→ (NH4)2CO3 + H2O
4.0
1. d. 2. c 3. a. 4. c. 5. a.
Activity Series
Q.1.0
A
|
Metal
|
Ion
|
With Oxygen
|
With water
|
|
Copper
|
Cu++
|
2Cu + O2 → 2CuO
|
Cu + H2O→ No reaction
|
|
Calcium
|
Ca++
|
2Ca + O2 →2CaO
|
Ca + 2H2O→ Ca(OH)2 +H2
|
|
Silver
|
Ag+
|
Ag + O2 →No reaction
|
Ag + H2O→No reaction
|
|
Sodium
|
Na+
|
4Na + O2 → 2Na2O
|
2Na+ 2H2O→2NaOH +H2
|
B . Na, Ca, Cu, Ag
Q. 2.0
C , A, D B
Q. 3.0
1-A, 2. B. 3. A. 4. - a, 5. d
Q. 4.0
- 1. Copper, 2. Potassium, 3.Aluminium, 4. Calcium, 5. Mercury, 6. Iron,
7. Lead, 8. Silver. 9 Zinc, 10. Sodium.
C, A D, B.
Atomic Mass Unit
|
Question
|
1
|
2
|
3
|
4
|
5
|
|
Answer
|
12.
|
2.
|
1.
|
9
|
2:1
|
|
Question
|
6
|
7
|
8
|
9
|
10
|
|
Answer
|
2x.
|
12g.
|
35.49amu
|
8 x 10 -7 g.
|
6:71
|
Method of working for no.8 and 9.
75.5 % atoms should have a mass of 75.5x 35 = 2642.5
24.5 % of atoms will have a massof 24.5x 37 = 906.5
Total = 3549
Average mass of an atom = 35.49 amu.
9.
Volume of breath in L = 10/1000 L = 0.01 L
50,000 L of air would have 4g of perfume.
1 L …………………. 4/50,000. g.
0.01 L …………………. 4/50,000 x 0.01 g
= 8 x 10 -7
Avagadro's number and Mole.
Q.1.0
Complete the blank spaces in this table
|
Formula
|
Molar mass.g/mol.
|
No. of molecules in a mole.
|
Mass of 1 molecule in grams.
|
No. of atoms in a molecule.
|
Mass of an atom in grams.
|
|
Oxygen O2
|
32g.
|
AN
|
32/AN
|
2
|
16/AN
|
|
Ozone. O3
|
48 g.
|
AN
|
48/AN
|
3
|
16/AN
|
|
Iodine I2
|
254g
|
AN
|
254/ AN
|
2
|
127AN/
|
|
White phosphorus P4
|
124g
|
AN
|
124/AN
|
4
|
31/AN
|
|
Sulphur S6
|
192.6g
|
AN
|
192.6/AN
|
6
|
32/AN
|
|
Bucky ball
B60
|
720 g
|
AN
|
720/AN
|
60
|
12/AN
|
Q.2.0
|
Q.No.
|
i.
|
ii.
|
iii.
|
iv.
|
v.
|
vi.
|
vii.
|
viii.
|
ix.
|
x.
|
|
Answer
|
128
|
2.
|
2NA
|
37
|
2
|
2NA
|
0.01
|
0.01NA
|
1
|
1
|
Q. 3.0
-
1 mol. 2. 0.25 3. 26.5g. 4. 100m.l.
Q.4.0
4.1 110.1g. 4.2. AN 4.3. 2AN
Q. 5.0
5.1 607 5.2 394g. 5.3 2AN 5.4 6AN
5.5 Au2 Cl6 ----> 2Au+++ + 6 Cl-
B
C
Chemical bonds.
Multiple choice questions.
| 1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
| B |
C |
D |
B |
D |
B |
C |
A |
A |
B |
Q.1.0
Q.2.0
i. O 2-, ii. Cl-, iii. (SO4) 2- iv. (CO3) 2-, v. (PO4) 3-
. 
Q.3.0
I. [Na+] + [OH-] <-------> NaOH
Ii. [Mg2+] + 2[OH-] ←-----> Mg(OH)2
Iii. [Al3+] + 3[OH-] ←----> Al(OH)3
Iv. [Pb2+] + 2[OH-] ←-----> Pb(OH)2
V, [Ag +] + [OH-] ←-----> AgOH.
Q. 4.0
D
|
1.1 Dissolving (Physical)1.2 Mixing and partial dissolving. 1.3.May mix but does not dissolve. (Immiscible).1.4 Chemical dissolving. 1.5 Physical as well as chemical dissolving.
A- Partly soluble. B - Soluble, C- Soluble
1….A, 2…..B, 3…..C, 4……D. 5………C
4.1
Find the solubility at 10°C for A ----5---- and B ---10----- . Find the solubility at 50°C for A --25-----and B---25------.
4.2…
4.2 What mass of solute A is necessary to saturate 100g of water at 30°C? 10g What mass of B is necessary to saturate 50g of water at 10°C ? 5g
5.1
5.2 Solute 2. Solvent 3 Solution (mixture) 4 Dissolving
|
20
15
25
20
20
|
E
Empirical formulae
Answers
1. Mass ratio of Hg:O is 46:4
Atom ratio is 46/201 : 4/16 = 0.23 : 0.25
This is approximately 1 :1 ( Leaving for experimental errors)
Therefore the formula must be HgO.
2. Mass ratio of H:O is 20: 320
Number rato is 20/1 : 320/16
= 20 : 20
The formula must be HO (There is no compound like this. What we get is called the empirical formula)
The real compound is H2O2. Hydrogen peroxide.
3. Mass ratio Cu: O = 1.6 : 0.2
Number ratio = 1.6/64 : 0.2/16 = .025 : 0.0125
= 2:1
Formula Cu2O.
4. Percentage of Hydrogen is 100 – (54.05+43.24) = 2.71%
Mass ratio Ca:O:H = 54.05:43.24:2.71
Number ratio = 54.05/40.1 : 43.24/16: 2.71/1
= 1.35 : 2.7 : 2.71
= 1: 2:2
Formula has to be CaO2H2
The real formula is Ca(OH)2 Calcium hydroxide.
5. Mass ratio is Pb:O = 2.07: 0.32
Number Ratio = 2.07/207 : 0.32/16 = 0.01 : 0.02
= 1:2
Formula is PbO2 Lead dioxide
6. Mass ratio C: H:O= 40 : 6.7 : 53.3
Atom ratio is = 40/ 12 : 6.7/1 : 53.3/16
= 3.33 : 6.7: 3.33
= 1:2:1
Empirical formula is CH2O.
This could be anyone of these:
Formaldehyde CH2O, Acetic acid C2H4O2 Or Glucose C6H12O6
F
G
H
හදුනාගන්න.
1 Na සෝඩියමි, 2.Mg මැග්නීසියමි, 3.Al අැලුමීනියමි, 4. C කාබන්, 5. S ගෙනදගම්, 6.Au රත්තරන්.
I
J
K
L
M
Mass of Molecules and atoms.
Part 2 Answers
1.
| Protons |
Electrons |
Nucleons |
Atomic no. |
Mass no. |
| 17 |
17 |
37 |
17 |
3 |
10
2.
| Laft |
Right |
|
Left |
Right |
| 6C12 ...........2 |
12Mg24-. ...........1 |
|
6C14-- a. 1 |
1H1 b. 14 |
| 16S32 c. 1 |
8O16-- d. 2 |
|
26Fe56 e. 1 |
14Si28 f. 2 |
| 7N14 g. 2 |
2He4 h. 7 |
|
24Cr52 i. 4 |
8O16......... .j. 13 |
20
3.
| O3 |
HCl |
NO2 |
H2SO4 |
Ca(OH)2 |
NH4Cl |
C2H6 |
H3PO4 |
Al2(SO4)3 |
CuSO4.5H2O |
| 48 |
36.5 |
46 |
98 |
74 |
53.5 |
30 |
98 |
162 |
250 |
2x10=20
.4. If the number of oxygen atoms in 16 g. of oxygen is ‘x’ find the number of atoms in 2 g. of oxygen.
16 g of O2 contains x atoms. 1g of O2 contains x/16 atoms. 2g of O2 contains x' x2÷16 atoms.......... =x/8 atoms.
5. Pure chlorine has two isotopes. Cl-35 and Cl- 37. If Cl-35 makes 75.5% while the balance is
Cl-37. What is the average Atomic mass of chlorine?
35x75.5 =2642.5
37x 24.5 =906.5 Total = 35.49 mass of 100 atoms.
Relative Molecular mass of Chlorine = 35.5.
6. Air consists mainly of four parts of Nitrogen N2 to one of Oxygen O2. What is the average mass of an air molecule?
Answer:-
Mass of 4 Nitrogen molecules= 4 x 28 = 112
Mass of an Oxygen molecule = 32
Mass of 5 air molecules= 112 = 32 = 144
Mass of 1 air molecules= 144/5 =28.8
30
5. Mass from Formulae
1. Find the amounts of the elements formed by the electrolysis of water.
| Question |
Answer |
| Mass of Hydrogen from 18 g*. of H2O. *molar mass of water. Mass of Hydrogen from 1 g. of water Mass of Hydrogen from 100g. of water Mass of Oxygen from 100g of water. *This method is possible as there is no other element in water. |
= 2g = 2/18 g =2x100/18= 11.11g. = 100-11.11=88.8 |
2. Find the mass of Oxygen in 27g of HgO.
|
Mass of Oxygen in 201+ 16= 217g of HgO
Mass of Oxygen in 1g of HgO.
Mass of Oxygen in 27g. of HgO.
|
16 g.
16/ 217 g.
16x27/ 217g =1.99g.
|
3.Find the mass of silver that may be obtained from 100g of silver nitrate.
|
Molar mass of AgNO3
Mass of silver from 138g of AgNO3
Mass of silver from 1g of AgNO3
Mass of silver from 100g of AgNO3
|
108 +14+16=138g/mol.
108g.
108/138
108x100/ 138g
= 78.26 g.
|
4. Find the mass of Carbon dioxide that can be obtained by heating 100g of baking powder.NaHCO3
|
Molar mass of Lime stone NaHCO3 .
Mass of CO2 in 72g of the chemical
In 100g
|
23+1+48 = 72g/mol
60g.
60x100/ 72 =83.33g.
|
5. A sample of lime stone was found to contain 90% of CaCO3. Find the mass of CaO in 1 tonne of lime stone.
|
Pure CaCO3. In 100 tonnes of sample
200 tonnes ……………………………..
CaO in 100 tonnes of pure CaCO3 has
Therefore 180 tonnes of CaCO3.
Therefore 200 tonnes of sample will yield
|
90 tonnes. CaCO3.
90x2=180 tonnes
40.1+16= 56.1 tonnes
56.1 x100/180
100.98 tonnes.
|
6 Agricultural fertilisers usually display a NPK value. These are related to the percentages of Nitrogen, Phosphorus and Potassium present. P actually gives the phosphorus pentoxide( P2O5) value. If a fertiliser has 30 .15 10 What will be mass of Phosphorus in a 10kg fertiliser bag.
|
Given Percentage of P2O5 in the 10 kg bag
Mass of P2O5 in bag.
Molar mass of P2O5
Mass of pure P in 142 kg of P2O5 will have
1.5 kg of P2O5 will give
|
15%
1.5kg.
(2x 31) + (5x16)=142
62 kg
62x1.5 /142 kg
=0.65 kg.
|
|
Mixtures
Q.1.0 Are these Elements, Mixtures or compounds.
|
A-Solids
|
B-Liquids
|
C-Gases
|
|
Name
1. Starch
2. Carbon
3. Concrete
4. Bronze
5. Iron
|
Answer
Compound
Element
Mixture
Mixture
Element
|
Name
1.H2O
2.Alcohol
3.Milk
4.Palm oil
5.Mercury
|
Answer
Compound Compound
Mixture
Mixture
Element
|
Name
1.Air
2.CO2
3.NH3
4.Oxygen
5.Argon.
|
Answer
Mixture
Compound
Compound
Element
Element
|
15 Marks.
Q.2.0 Question 1 :- 1- D, 2- A, 3- C, 4- B . 5 - E.
Question 2:- 1- E, 2- D, 3- C, 4- B, 5- A, .
6. Mass from Equations. I.
Answers
1.
Equation: Zn + 2 HCl → ZnCl2 + H2
Relavent masses: 119 ……………… …….2
Mass of Zn for 2g of hydrogen……………..119g
For 1g of hydrogen………………………..119/2 = 59.5g. of Zinc.
2.
Equation: Mg + 2 H2O → Mg(OH)2 + H2
Formula masses: 24.3+36 …. 24.3 +2(16+1) + 2
Mass of Magnesium hydroxide from 24.3g ……..58.3 g
From 1g ……………58.3/24.3 g = 2.4g
Mass of hydrogen from 24.3g………………………2g
From 1g………………2/24.3 = 0.08 g
3.
2 C8 H18 +25 O2 → 16 CO2 + 18H2O
Formula masses 2(96 +18)….. 25x32 = 16x 44 ……..18x 18
228……+……800 = 704 ….+ … 324
Mass of carbon dioxide produced from 228 t octane 704 t
Mass of carbon dioxide from 10 t = 704x10 /228 = 30.9 t.
Mass of Oxygen used from 228 t of octane = 800 t
Mass of oxygen from 10 t of octane = 800x10/228 = 35.1 t
4.
(NH4)2SO4 → H2SO4 + 2 NH3
Molar masses:…(36) +32+64 ….= 2+32 +64……2( 14+3)
132 …= 98……….+ 34
34g of ammonia from………….132 ammonium sulphate.
8,5g ……. 132x 8.5 / 34
= 33g
200g of fertiliser has…………33g
100g of fertiliser has……….33/2 = 16.5%
5. Find (i) the mass of Oxygen produced during photosynthesis of 100 g of glucose
(ii) the amount of oxygen consumed for 100g of Glucose.
Photosynthesis... 6CO 2 + 6H2O -----> C 6H 12O 6 + 6O 2
Respiration is: …..C 6H12O 6 + 6O 2 -----> 6CO 2 + 6H 2O.
Glucose + Oxygen Carbon Dioxide + Water
(i) Photosynthesis... 6CO 2 + 6H2O -----> C 6H 12O 6 + 6O 2
Formula mass 264 + 108 = 180 + 192
Making 180g of glucose will produce 192g of oxygen
100g of glucose .......= 192x100/180 =106.7 g
(ii) Same answer...............106.7g.
7. Mass from equations II
1.
i. 63.6 g of Cu combine to form 95.7g of copper sulphide.
64g ………………. = 95.7 x64/ 63.6 = 96.3 g of CuS.
ii. Sulphur is in excess.
iii. 64g of copper will need = 32.1 x64/ 63.6 = 32.3 sulphur
Mass of sulphur left = 40 – 32.3 =7.7 g.
2.
2. i. B.
ii. Use more oxygen or reduce the amount of Carbon.
3.
3. Equation: 3C2H2 + 3O2 ---à CO2 +CO + 4C +3H2O
3x 26 + 3x32 = 44 + 28 + 48 +3x 18
78 + 96 = 44 + 28 + 48 + 54
1kg gives 1 96/78 = 44/78 +28/78+ 48/78+ 54/78
1.23 = 0.56 + 0.36 + 0.62 + 0.69
4.
Equation: Cu + 4HNO3 ----à Cu(NO3)2 +2NO2 + 2H2O
Formula mass: 63.6 + 4x63 = 63.6 + 2x62 + 2x 46 + 2x 18
63.6 + 252 = 187.6 + 92 + 36
1 + 252/63.6 = 187.6/63.6+ 92/63.6 +36/63.6
1 + 3.96 = 2.95 + 1.45 + 0.57
For 10 g 39.6g = 29.5g + 14.5g + 5.7g.
Answer
Combining ratio for Zn: S = 119 : 32
This is Zn / S = 3.72
|
Mass of Zn
|
(i) 10
|
(ii) 20
|
(iii) 30
|
(iv) 40
|
(v) 50
|
|
Mass of S.
|
5
|
8
|
10
|
11
|
30
|
|
Mass of Zn / S
|
2.0
|
2.5
|
3.0
|
3.64
|
1.67
|
5.1 The ratio nearest to the combining ratio is no. (iii).
5.2 According to this ratio most of the elements will combine with very little wastage.
6. 6.1 Find the mass of the precipitate.
Equation: KCl + AgNO3 ----> AgCl ↓ + KNO3
Formula mass: 39.1+35.5 + 108 +14 +48 = 108 +35.5 + 39.1 + 14 +48
74.6 + 170 = 143.5 + 101.1
Using 1.7g AgNO3 0.75 + 1.7 = 1.44 + 1.01
Formula mass: 74.6 + 170 = 143.5 + 101.1
Using 11g KCl 11 + 25.1 ? = 21.16 + 14.9
Which is the correct answer?
To make 21.16g of precipitate there must be 25.1g of AgNO3.
There isn’t that much. So the answer is 1.44g. of precipitate.
6.2 What substances would be present in the filtered solution.
There will be KNO3 newly formed and the remaining KCl.
2. Molar Mass
Find the Molar mass of the following compounds by adding the atomic masses.
|
Name
|
Formula
|
Atomic massx No. of atoms
|
Molecular
mass in
amu.
|
Molar mass
M
|
|
Example 1. Hydrogen (di atomic)
|
H2
|
H + H
= 1.01 + 1.01
|
2.02
|
2.02 g/mol
|
|
Example 2.
Ammonium sulphate.
|
(NH4)2 SO4
|
2( 14 +4.04) + 32 + (4x16)
= 36.08 + 32 + 64 = 132.08
|
132.08
|
132.08 g/mol
|
|
1. 1.Oxygen
(di atomic)
|
O2
|
O + O
=16.0 =16.0
|
32
|
32.g/mol
|
|
2. 2.Hydrogen peroxide
|
H2O2
|
2H + 2O
=2.02 + 32
|
34.02
|
34.02g/mol
|
|
3. 3.Potassium sulphate.
|
K2SO4
|
2K +S + 4O
= 2x39.1 + 32.1 4x16 = 174.2
|
174.2
|
174.2 g/mol
|
|
4. 4.Phosphorous pentoxide
|
P2O5
|
P + P + 5x O
= 2x31 + 5x16 = 142
|
142
|
142 g/mol
|
|
5. 5.Acetic acid
|
CH3COOH
|
= 12+ 3x1.01 +12 +16+16 +1.01
= 12 + 3.03 +12 +32 +1.01 =60.04
|
60.04
|
60.04 g/mol
|
|
6. 6. Glucose
|
C6H12O6
|
6x12 + 12x1.01 +6x16
=72 + 12.12 + 96 = 180.12
|
180.12
|
180.12 g/mol
|
|
7. 7.Sodium carbonate decahydrate.
|
Na2CO3.
10H2O
|
2x23 + 12 + 3x16 + 10x18.02
46 +12 +48 +180.2 =
|
286.2
|
286.2 g/mol
|
|
8. 8.Potash Alum.
|
KAl(SO4)2.
|
K + Al + 2 ( S + 4xO)
= 39.1 +27 + 2 ( 32.1 + 64)=
|
258.3
|
258.3 g/mol
|
9. Using the Avogadro’s number as NA find the real mass of one molecule of chlorine.
10. Find the real mass of 1 atom of normal chlorine.
9. Using the Avogadro’s number as NA find the real mass of one molecule of chlorine.
10. Find the real mass of 1 atom of normal chlorine.
N
O
P
Q
R
S
Q 3. Find the mass of Magnesium required to get 1 g of Hydrogen from hot water.
Working:- Equation Mg + 2H2O = Mg(OH)2 + H2
Molar masses …24…. +.. …...= …42.. + …2…
Answer 24÷2…………. = 12g
Q 3. The percentage of baking powder in a sample was 30%. If 50g of this was heated what will be the mass loss due to the escaping of water and Carbon dioxide.
Working:- Mass of pure sodium bicarbonate in baking powder =Image not availableThe original image was not present in the supplied archive.
X 50 g
=15 g.
Equation 2NaHCO3 = Na2CO3 + H2O + CO2
Molar masses …2X84 = _____ + …18…. + …44……
168g of sodium bicarbonate will give = 18+44……g. of gases.
1g of sodium carbonate will give = 62÷168 g.
15g will give = Image not availableThe original image was not present in the supplied archive.
62÷168) x15
Answer:- Therefore loss in mass will be = 5.54g.
Q 4. Barium chloride was dissolved in 500ml of water. Mass of BaCl2 was 5.2 g. When excess of dilute sulphuric acid was added to it a white precipitate resulted. Find the mass of the precipitate.
Working :-
Equation: BaCl2 + Na2SO4 = BaSO4 + 2 NaCl
molecular masses 137+71+ …… = …233…… + ………
…208.g of BaCl2 gives = 233g. of BaSO4
1g of BaCl2 gives = 233÷ 208 = 1.12g
5.2 g of BaCl2 = 1.12 x 5.2 = 5.83g
Answer:- Mass of precipitate = 5.83g.
Q 5. The only ammonium compound in fertilizer sample was Ammonium sulphate. 200g of the sample yielded 8.5g of Ammonia. Find the percentage of Ammonium sulphate in the fertilizer.
Working:-
Equation:……… (NH4)2 SO4 = 2NH3 + H2SO4
Molecular Masses 132 = …34….. + ………..
34g of Ammonia is from =…132 g of (NH4)2 SO4
1g of Ammonia is from = …132÷34g. of (NH4)2 SO4
= 3.88g
Therefore 8.5 g of Ammonia is from …3.88x 8.5g. of (NH4)2 SO4
200g of fertilizer has = ………33g. of (NH4)2 SO4
Therefore percentage of fertilizer is = (33÷200) x100
Answer = 16.5%
T
12-Titrations
Q 1.0
i. Na2CO3 = 106g/mol.
ii. 1 L of solution has 1 mol
0.25 L ………………1x0.25 mol
iii. 1 mol Na2CO3 neutralizes = 2 mol HCl
0.25 mol………………….= 2x 0.25 = 0.5 mol
iv. 0. 201 L had … = .5 molof acid
1L should have 0.5 / 0.201 = 2.48 mol
Strength of acid = 2.48 M.
2.0
i.=58.3g/mol
ii. Equation:- Mg(OH)2 + 2 HCl -----à MgCl2 + H2O
Molar Ratio 1 : 2
1000mL of juice will have = 0.1 mol.
20 mL ……………… = 0.1x20 / 1000 = 0.002 mol
This will need …………..= 0.002/2 mol of Mg(OH)2
Mass of 1 mol of Mg(OH)2 = 58
Mass of 0.001mol…………….= 58.3 x0.001 = 0.06g
.3.0 Molar mass of calcium phosphate.
Ca3(PO4)2 = (3x40) + 2(31+64) = 120 + 2x 95 = 310 g/mol
25mL of acid has 0.067 mol
As the molar ratio of acid : salt = 2:1
0.067 can produce ½ of 0.067 moles.
This is …………..=310 x 0.034= 10.54g of calcium phosphate
U
9.- Using Equations.
Answers.
1.1 H2 : Cl2 = 1:1, 1.2 3:1, 1.3. 73 g.
2.1 C: O2: CO2 = 1:1:1.
2.2 0.5 mol.
2.3 0.5 mol.
2.4 mass = mol x molar mass
= 0.5 x 44 = 22 g.
2.5 Volume = 12.5 litres.
Answers
(5x7 = 35 points)
3.1
207 +2(14 + 48) = 331
Converting 33.1g to moles = 33.1/ 331 = 0.1 mol.
3.2
Converting 5.85g of NaCl to moles = 5.85/ 58.5 =0.1 mol
3.3
Equation: Pb(NO3)2 + 2 NaCl ------> PbCl2 + 2 NaNO3
Mole ratios 1 : 2
3.4
1 mol. Of PbCl2
3.5
Equation: Pb(NO3)2 + 2 NaCl ------> PbCl2 + 2 NaNO3
Mole ratios 1 2 1 2
Reacting moles 0.1 0.2 0.1 0.2
4.0 Answers
4..1 CaCO3 : CaO : CO2 = 1. : 1. : 1.
4.2 12 litres under given conditions = 1 mol
36 litres………………………….= 3 mol.
4.3 3 mol of CO2
4.4 Molar mass of CaCO3 100g/mol.
Mass of 3 mol………..300g.
4.5 Molar mass of CaO = 56 g/mol.
Mass of 3 moles … 3x56 = 168 g
5.0
5.1 The mass of iron that has reacted= 15 - 9.4 = 5.6g
5.2 Number of moles of iron reacted. 5.6/ 56 = 0.1 moles.
5.3 Number of copper moles produced. = 0.1 mol.
5.4 Mass of copper formed.= 0.1 x63.6 = 6.36g.
5.5 Number of copper sulphate moles reacted = 0.1 mol.
6.1 1000ml of solution has 3 mol.
100ml………………..3x100/1000 = 0.3 mol. NaOH.
6.2 2mol of OH- can be removed by 1 mol of CuSO4
0.3 mol………………………..= ½ x 0.3 = 0.15 mol. CuSO4
6.3 CuSO4 + 2 NaOH -----> Cu(OH)2 + Na2SO4
2 mol of NaOH forms………… 1mol of Cu(OH)2
0.3 mol………………………. 1/2x o.3 = 0.15 Cu(OH)2
6.4 Molar mass of Cu(OH)2 ……….63.6+34 = 97.6
0.15 mol……………………...97.6x 0.15= 14.64g
6.5 1 mol of Cu(OH)2 has 63.6g of copper.
0.15 mol……………….= 63.6x0.15 =9.54g.
7.1 Molar mass of Na2SO4 = 46+32 + 64 =142
Converting 1.42 to moles = 1.4 / 142 =0.01 mol of Na2SO4
7.2 1 mol Na2SO4 is from 2 mol of NaOH
0.01 …………………… 2x0.01 = 0.02 mol
500 ml solution has 0.02mol of NaOH
1000ml ………………=0.02x2 = 0.04 mol of NaOH.
V
W
X
Y
Z
| Q. No. |
Answer |
Marks |
| |
D
|
|
|
1.0
2.0
3.0
4.0
5.0
|
1.1 Dissolving (Physical)1.2 Mixing and partial dissolving. 1.3.May mix but does not dissolve. (Immiscible).1.4 Chemical dissolving. 1.5 Physical as well as chemical dissolving.
A- Partly soluble. B - Soluble, C- Soluble
1….A, 2…..B, 3…..C, 4……D. 5………
4.1….80-55 = 25g. 4.2…….(40 – 10)/2 = 15 g.
5.1 1.Mixture. 2.Residue 3. Filtering 4 Filtrate.
5.2 Solute 2. Solvent 3 Solution (mixture) 4 Dissolving
|
20
15
25
20
20
|
