Chemistry‎ > ‎

Answer page - Chem.

posted Sep 23, 2015, 1:02 PM by Upali Salpadoru   [ updated Sep 11, 2019, 1:53 AM ]
  The topics are in alphabetical order.
       A
  • Acids and Bases

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. 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.
 
 110 
 B C D  C  A
Q.1.0
Q.2.0

    i.    O 2-,   ii.   Cl-,   iii.  (SO4) 2-  iv. (CO3) 2-,  v. (PO4) 3-

H2O.jpgHCl.jpg


H2CO3.jpg
H2SO4.jpg

. H3PO4.jpg

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
  •  Dissolving.
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 Nto  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 6126 + 6O 2

       Respiration is: …..C 6H126 + 6O 2 ----->   6CO 2 + 6H 2O.

                     Glucose + Oxygen Carbon Dioxide + Water


                         
 (i)  Photosynthesis... 6CO 2 + 6H2O ----->  C 6126 + 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 =  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          = 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)SO =  2NH3   +   H2SO4                                           

                   Molecular Masses           132  =   …34…..  + ………..  

          
                             34g of Ammonia is from  =…132 g of (NH4)SO

                                     1g  of Ammonia is from = …132÷34g. of (NH4)SO

                                                                                                         =   3.88g

                     Therefore 8.5 g of Ammonia is from …3.88x 8.5g. of (NH4)SO

                                                     200g of fertilizer has  = ………33g. of  (NH4)SO                  

                       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 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 

  •  Dissolving.
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

Comments