Neco 2021 June/July Chemistry Practical Answers

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Neco 2021 June/July Chemistry Practical Answers
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    ""NECO JUNE/JULY 2021 CHEMISTRY PRACTICAL""

    The Chemistry Practical Comprises 3 Questions from Volumetric or Titrimetric Analysis And General Questions. You are to Attempt all Questions.

    Make sure you use your school Average Titration Values in other to get Accurate Solutions with your Lab. Attendant Report.





    NUMBER 1 STEP 3 VOLUME OF THE A USED IS THE LAST POINT, AND EACH SCHOOL HAS HIS LAST POINT ,SO ASK YOUR CHEMISTRY TEACHER FOR THE LAST POINT AND THEN CALCULATE YOUR AVERAGE VOLUME OF ACID USED




    (1a)
    Tabulate

    Volume of pipette used=25cm3.
    Indicator used= Methyl orange

    Under Titration, Rough, 1st, 2nd, 3rd,

    Final burette reading; |25.00| |23.40| |23.50| 23.60|

    Initial burette reading; |0.00| |0.00| |0.00| |0.00|

    Volume of A used; |25.00| |23.40| |23.50| |23.60|

    (1ai)
    Average volume of Acid used=23.40+23.50+23.60/3 =70.5/3=23.50cm3

    (1aii)
    This is because it is a reaction between a weak acid and strong base

    (1aiii)
    A funnel should be used while transferring the acid into the burette.

    (1bi)
    Conc of A in mol|dm³ =?
    Conc of A in g|dm³ = 12.75g|dm³
    Molar mass of A, NaHSO⁴ = 23+1+32+64
    =120g|dm³

    From.
    Conc(mol/dm³) = conc(g/dm³)/molar mass

    Conc =12.75/120
    Conc =0.106mol/dm³

    (1bii)
    Con of B in mol/dm³ =?
    Ca =0.106, Cb =? Va =23.50cm³, Vb =25.00cm³

    na=1 , nb=1
    From; CaVa/CbVb=na/nb
    Cb=CaVanb/Vbna
    Cb=0.106*23*50*1/25.00*1
    Cb=2.5/25
    Cb=0.1mol/dm³

    (1biii)
    Conc of B in g/dm³=?
    Molar mass of B , NaOH=23+16+1=40g/mol
    From; conc(g/dm³)=conc(mol/dm³)*molar mass
    =0.1*40
    =4.0g/dm³

    (1biv)
    Mass of salt formed =?
    Recall mole of NaHSO⁴=?
    From; mole =con(mol/dm³)*vol(cm³)/1000

    Mole=0.106*23.50/1000
    Mole=0.0025mol of NaHSO⁴

    By proportion 1mol of NaHSO⁴ produce 1mol of Na²SO⁴ 0.0025mol of NaHSO⁴ will produce X mol of Na²SO⁴ X=0.0025*1=0.0025mol of NaSO⁴
    Molar mass of Na²SO⁴=(23*2) + 32 + (16*4)
    =46+32+64
    =142g/mol

    Recall; mole =mole/molar mass
    Mass =mole * molar mass
    Mass =0.0025*142
    Mass =0.36g of Na²SO⁴
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    (2a)
    TEST
    C + 5cm³ of distilled water

    OBSERVATION
    It dissolve completely

    INTERFERENCE
    C is a soluble salt

    (2bi)
    TEST
    Solution + NaOH + Heat

    OBSERVATION
    Effervescence occurs in which a colourless gas with pungent smell and turns red litmus paper to blue is given off.

    INFERENCES
    NH³ gas from NH⁴+ is present

    (2bii)
    TEST
    Stirring rod of HCL + gas given off

    OBSERVATION
    a gas which gives a pop sound is given off

    INFERENCES
    H² is present

    (2ci)
    TEST
    Solution C + drops of BaCL²

    OBSERVATION
    a white precipitate is formed

    INFERENCES
    CO²^-3, SO²^-4, SO²^-3, May be present.

    (2cii)
    TEST
    Solution in C(i) + dil HCL

    OBSERVATION
    the white precipitate dissolve is dil HCL and effervescence occurs in which a colourless and odourless gas which turns blue litmus red and lime water milky is given off

    INTERFERENCE
    CO² gas from CO²^3- confirmed
    ====================================

    (3a)
    (i)Ca²+
    (ii)Pb²+

    (3bi)
    concentrated acid can be defined as an acid formed when a large quantity of an acid dissolve in a small or little volume of water.

    (3bii)
    A strong acid is defined as a type of acid that ionize completely in a solution

    (3c)
    Activated charcoal is used as an adsorbent material

    (3d)
    (i)Reddish - Brown
    (ii) I - red, II - brown.


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    Completed!.

    Note: Remember this, use your school end point to calculate number 1 question. Thanks





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