Saturday, April 21, 2012


Assignment: Percent, ppm and ppb
1) Calculate the concentration of salt in a solution of water in percent if 45 grams is dissolved in 1,200 ml of water.

Solution:
            1200ml contain 45 g 
            1 ml contain 45/1200
            100 ml contain 45 x100/1200

2) Calculate the mass of solute in a 10% salt solution if the mass of the solvent is 350 grams.

Solution :
            Suppose the mass of solute is  "X"
            then the mas of solution is  equal 350 + X 
            we Know that 
            % of solution = mass of solute ÷mass of solution  x 100   
                          10 %= X ÷( 350+X) x 100
                     10 ÷ 100 =  X ÷(350 +X)
                              0.1=X÷(350 +X)
                                       X  = 0.1(350 +X)
                                        X= 35.0 +0.1X
                              X-0.1X = 35.0
                                  0.9X = 35.0
                                        X =35.0 ÷0.9
                                         X= 38.88  
3) Calculate the mass of solvent in a 6 ppm solution of a drug if the mass of the solute is 0.050 milligrams.  
  
Solution:
             we know that 
                   ppm     =mass of sloute ÷ mass of solvent x 106
                      6ppm      = 0.050 mg ÷mass of solvent x106
         mass of solvent= 6ppm ÷106 x 0.050mg 

4) What is the concentration in ppm of selenium if 1.3 milligrams is found in 2,500 kg of soil?

Solution:  
              ppm = 1.3 mg  ÷  2.5 x109  mg       x 106
              5.2 x10-2   ppm
5) What is the concentration in ppb of PCB's in a chemical spill, if their is 0.060 mg in 4,600 Kg of soil?



6) Calculate the mass of solute of hydrogen peroxide in a 35 % solution if 450 ml of solution is being used.

Solution: 
        100 ml solution contain = 35 g of hydrogen peroxide
             1ml solution contain = 35 ÷ 100
            450 ml solution contain = 0.35 x 450 ml 
7) Calculate the mass of solute PCB's in a 65 Kg person, if the concentration is 4 PPM?

Solution :
                         ppm = mass of sloute ÷ mass of solvent   x  106
                       4 ppm = mass  of solute  ÷  65 kg  x  106
      mass of sloute  = 4  ÷ (65 kg  x  106)

8) What mass of nickel is in a 2.4 Kg sample of propanol if the 
concentration is 20 ppb?
Solution : 
                

9) Calculate the mass of solvent that would contain 3.0 mg of a drug if the concentration was 3.5%.

Solution:
10) Calculate the concentration in ppm if 8 grams of CaCl2 is dissolved in 250 ml of water.

Thursday, April 19, 2012

Re -Sit Exam. of Chemistry

Re -Sit exam. of Chemistry on 21st of April 2012(  Saturday ) at 9:00 am to 12:00

Monday, April 2, 2012

Solution Of MCQs of XI year AKUEB First Shift


CLASS XI 
Chemistry Paper I
1ST SHIFT
Max. Marks: 30                                                                                       Time: 40 minutes

 1.  Which of the following  contains the largest  mass of chlorine ?
    A. 10.0 g of Cl2
    B. 50.0g of KClO3    
    C. 0.10   mol of NaCl
    D. 10.1g of NaCl
Sloution :
122.5 g of KClO3 contain 35.5 g of Chlorine
50g   of     KClO3 contain 35.5 ÷122.5 x 50
50g   of     KClO3 contain =14.49 g of Chlorine

2. The percentage of nitrogen in urea (NH2)2CO) is
    A. 23.3%
    B. 31.3%
    C. 38.0 %  
    D. 46.7%
Solution :
 % of Nitrogen =28g  ÷60 x100
                         =46.7

3.  Experiments with the cathode ray tube have shown
    A. that all nuclei contain protons                                       
    B. that all forms of matter contain electrons
    C. that all positive rays were actually protons    
    D. that of alpha  particles heavier than proton
Reason:
 The cathode ray does not depend  upon the material fill in the discharged tube

4. The particle with the smallest mass is the
    A. Alpha particle
    B. Proton
    C. Neutron  
    D. Beta particle
 
5. The reason why the atomic mass of chlorine is 35.453 rather than almost exactly 35 is because
    A. Every chlorine atom contain 17 proton
    B. all chlorine atoms have identical properties
    C. there are at least two naturally occurring isotops of chlorine                  
    D. every chlorine atom has a mass of 35.45 amu   

Reason:
            35.5 is the  average relative atomic mass naturallyally occurring isotopes of Chlorine  

6. The total number of valence electron in the BrO-3 :
    A. 25
    B. 26
    C. 27
    D. 24
Solution :
            Br =7e x1=7e
            O   = 6e x3=18e
            Charge=1e
Total=7e+18e=1e=26e

7. Of the pairs of element listed, which would form the most ionic bond?

    A. B, N
    B. H, Cl      
    C. K , Cl                   
    D. C ,O                          
Reason:
            Due to High electron Affinity of halogen and llow ionization of alkali metal

8. The shape of the phosphate (PO4-3) ion
    A.  Angular
    B. Triangular   
    C. Tetrahedral   
    D. Square planer

9. London dispersion forces are also called:
    A. Ion dipole forces
    B. Dipole- induced dipole forces 
    C. Dipole dipole forces
    D. Instantaneous dipole induced dipole forces

10. Which crystal system which has,  a≠ b≠ c and abg ≠ 900
    A. Monoclinic
    B. Triclinic   
    C. Trigonal 
    D. Hexagonal

11. The moleclues of which of the following gases have the greater average molecular speed 
at 300K? 
    A.  Argon
    B.  All have the same average molecules speed at 300K
    C. Fluorine
    D. Neon
Reason:
          According to KMT of gas at same temperature of different gases have same
 Kinetic energy and kinetic energy depends the velocity of gas molecules and the
mass of gas . Due to this larger the mass  lesser the  velocity of gas and smaller the
mass of gas the larger of the velocity of molecule.
      

12. At a temperature of about 4250C ,  H2 (g) + I2(g)  ↔ 2HI(g)
the equilibrium mixture contain  [H2] =0.10  ,[I2]= 0.10 and [HI]= 0.70. If suddenly 0.050 mol/dm3 of H2, 0.050 mol/dm3 of I2 0.350 mol/dm3 of HI were added , the result would be that:

A. more HI would from
            B.  the equilibrium would be shifted to the right
            C.  More H2  and I2 would from 
D. No net Chemical change would occur
Solution:
Initial state the equilibrium constant(Kc) is =(0.70)2 ÷ (0.10x0.10)
                                                                          =49
Final state the equilibrium constant  (Kc)     =0.70+0.350)2 ÷ {(0.10+0.05)x(0.10+0.050)}
                                                                        =49

13 In atomic crystals, which type of bonding is present?
    A. Ionic bonding
    B. Covalent bonding
    C. Metallic bonding
    D. Vander Waal forces 

14. A real gas most closely approaches the behavior of an ideal gas under condition of :    
      A.  High pressure and low temp.
      B. Low pressure and high temp.
      C. Low pressure and low temp. 
      D. High pressure and high temp.

15. At normal boiling point of substance, the vapour pressure of the substance :                
      A. is equal to 760cm of Hg
      B. differs for different substance
      C.  does not have a definite value vapour pressure at boiling
      D. is equal to 760 torr

16. Which of the following reaction has tendency to go to completion?
            A.  H2 (g) + O2(g)  ↔ 2H2O(g)     K= 1.7x 1027
B.  N2 + O2  ↔ 2NO               K= 5.0 x 10-31
C.  2HF  ↔ + F2 + H2             K= 1.0x 10-13
D.  2NOCl   ↔ 2NO + Cl2     K= 1.7x 10-4

             
     
17. Three acids HX, HY & HZ have the following Ka values
            i.   Ka of HX = 1 x 10-5
            ii.  Ka of  HY = 2 x 10-5
            iii. Ka of HZ    =1 x 10-6
      What is the correct increasing order of acidic strength?
            A. HX< HY<, HZ
            B. HZ< HX< HY
            C. HZ< HY< HX
            D. HY< HX< HZ

     
18. Which of the following  solutions has the smallest [H3O+]?
            A. 0.10M HCl
            B. 0.10M H2SO4
            C. 0.10M HNO3
            D. 0.10M NH4Cl
             
    
19. The term  “Kb for the ammonia” refers directly to 
A. NH3            +   H2O   ↔ NH+4         +  OH-
B. NH3                  +   H3O+ ↔ NH+4      +   H2O
C. NH+4 +   H2O ↔   NH3      +   H3O+
D. NH+4 +   OH- ↔   NH3      +   H2O

     
20. The compositions of different solutions of sugar in water are below
 i) 10g of sugar dissolved in 100g of water = “A” solution
ii) 20g of sugar dissolved in 100g of water = “B” solution
iii) 30g of sugar dissolved in 100g of water = “C” solution
            iv) 40g of sugar dissolved in 100g of water  =  “D”  solution
            which solution has highest boiling point  
             A.   “D” solution
             B.   “B” solution      
             C.   “A” solution
             D.   “C” solution

21.If solution  A has  a density  of 1.1g/mL . What is the molarity of the solute in this solution?  
A. 0.38  M
B. 0.50 M
            C. 0.45  M
 D. 0.10 M

22. What is osmotic pressure at 0.0 0C of  solution containing 1.0g/L of sucrose (sugar cane )
A. 0.066 atm
B. 0.11   atm
C. 0.24   atm
D. 0.29 atm
Solution:
     ЛV=ST
Where  
            Л    is the osmotic pressure  (atm )
          S    is solution constant      =0.0821atm.dm3.mol-1K-1
          T is the  temperature of solution (K)
          V is the volume of solution which contain one mole of solute dm3/mol
              
       Л=0.082 x 273  ÷ 0.342
            =0.066
23.  If the 150 mL of 12.0 M HCl is diluted with water to a total volume of 650mL, the molar concentration of H3O+ is:   

A.   2.77 M                          
  1. 12.0 M
  2. 3.56 M            
  3. 1.80 M
Solution:
     M2V2=M1V1
           M2=12.0 x150÷650
               =2.769

24. A positive catalyst  can:

A. diminish the enthalpy of reaction
B. shift the equilibrium of reaction
C. makes a reaction thermodynamically more feasible
D. diminish the activation of reaction 

25.    
Experiment No.

Initial concentrations
Initial Rate of reaction
Mol/dm3/hr
[A]
      [B]
01
0.10
0.02
10
02
0.10
0.03
15
03
0.2
0.02
20
 The rate law derived for the reaction from the above data is 

A. Rate law  =k [A][B]
B. Rate law  =k [A]2[B]
C. Rate law  =k [A] 2 [B]1/2
D. Rate law  =k [A]1/2[B]2
Solution:

           According  to Initial rate method 
For A reactant 
                     Rate 2 ÷ Rate1 =  (Concentration  2 ÷ Concentration 1)n
                         20 ÷10  = (0.2 ÷0.1)n
                              2  =(2) n
                            Therefore
                                        n =1
                               order of A is 1st
  For  B reactant

                       Rate 2 ÷ Rate1 =  (Concentration  2 ÷ Concentration 1)n
                                  15 ÷10  = (0.3 ÷0.2)n

                                       1.5  =(1.5) n
                            Therefore
                                   n =1
                           order of B  is also 1st
26. The quantity of heat evolved or absorbed during a chemical reaction is called __________. 


               A. Heat or Reaction  

B. Heat of Formation          
C. Heat of Combination 

D. Lattice Energy


27. When the bond being formed are more than those being broken in a chemical reaction, then the ∆H will be __________. 


               A. Positive  

B. Negative       
C .Zero 

D. All of above


28.  A voltaic is prepared by    two cell between Zn  and Cu electrode using ZnSO4 and CuSO4  and           
 salt bridge . The Standard reduction potentials are  Zn2+/Zn  of   0E= -0.76v,   Cu2+/Cu of  0E= +0.34v
The cell potential (0E)of the cell is :


 A.   - 1.1 v
 B .   + 1.1 v
 C.   - 4.2 v
 D.   Zero

29.  Sulphur has the lowest  Oxidation state  in __________. 



A.    SO2
B.   H2S
C.     H2SO4
D.    S


30. When we passed same of electric charge in electrolytic cells of  AgNO3  , Cu(NO3)2 and Al(NO3)3  in same time the mass deposited on cathode of Al(NO3)3  is 9g of Al. What is mass deposited of Ag in AgNO3
A.   108 g
B.   54 g
C.   36 g
D.   9 g