Chemistry Paper 1 (Practical) ,May/June 2011  
Questions:       1 2 3 4 5 6   Main
General Comments

Question 4

D is a solution of 0.100 mol dm-3 HNO3.
E is a solution containing 13.6 g of Na2CO3 yH2O perdm-3

Put D into the burette and titrate it against 20.0 cm3 or 25.0 cm3 portions of E using methyl organge as indicator.  Repeat the titration to obtain consistent titres.  Tabulate your readings and calculate the average volume of D used.  The equation for the reaction involved in the titration is:
2HNO3(aq)  + Na2CO3.yH2O(aq)→Na2CO3(aq)+ CO2(g)+ (y+1) H2O(1)

(i)From your results and the information provided above, calculate the:
concentration of E in mol dm-3;
concentration of E in g dm-3;
value of y in Na2CO3.yH2O.

[H = 1.00, C = 12.0, O = 16.0; Na = 23.0]



In part (a) candidates correctly tabulated their burette readings and calculated the average titre value.  Some candidates however manipulated their readings to agree with those of their teachers.

In part (b) (i) and (ii), most candidates that lost marks in this section did so for either using wrong units or not expressing their answers to three significant figures.  In section (iii), many candidates correctly calculated the molar mass of Na2CO3 and obtained the mass of anhydrous salt.  In the calculation of the value of y however, many candidates failed to express their final answer to the nearest whole number while others muddled up the substitution .  Performance was on the average.

The expected answers are:

(b)(i)    CDVD  =  2  mole ratio
pic            CEVE        1

CE  =          CDVD  making CE subj
      =   X moldm-3

(b(i)Amount of D used   =   0.100  x VD  = d mol

From the balanced equation of the reaction
2 moles of D  =    1 mole of E
D mol of D         =   d mole of E
i.e 20/25 cm3 of E contains d/2 mol
1000 cm3 of contained d x 1000 mol
                                    2 x 20/25
                        =  e mol
Conc. of E       =  e  moldm-3

  1. Molar mass of Na2CO3  =  106 gmol-

Mass of anhydrous Na2CO3 = 106 x X moldm-3
                                            =  Wg dm-3

  1. Mass of water = 13.6 - W

                        =  Z

Mass of Na2CO3  =  Molar mass of Na2CO3
Mass of water           (y) Molar mass of water

                   = 106
                   Z       18y

                  y   =  106  x Z
                           W x 18
                       =  m  (to the nearest whole number)


  1. (b)(iii)  Molar mass of Na2CO3.yH2O =   Concentration in gdm-3 Concentration in moldm-3
    =  - 13.6
    =  Q gmol-  say
            Thus (2 x 23)  + 12 + (3 x 16) + y(2 + 16) = Q
                                    106  +  18y          =  Q
                                     y                         =  Q – 106
                                                    =  s (to the nearest whole number)


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