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Waec Gce 2018 Chemistry Alternative To Practical Questions and Answers - Nov/Dec Expo - solutionfans.com
   

  • Waec Gce 2018 Chemistry Alternative To Practical Questions and Answers – Nov/Dec Expo - solutionfans.com
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    Chemistry-Practicals

    1ai)
    tabulate
    Burette reading 1cm^3|1|2|3|
    final reading|24.80|34.50|20.65|
    initial reading|0.00|10.50|6.55|
    volume of acid used 1cm^3|24.80|24.00|24.10|
    1aii)
    avaerage volume of acide used= VA=24.80+24.00+24.10)cm^3/n =3
    VA=72.9/3=24.30cm^3
    Hence average volume of A used = 24.30cm^3
    1bi)
    Concentration of A in moldm^-3
    A contains 0.79g of KMnO4 per 250cm³ of solution.
    Hence since 250cm³ = 0.79g
    100cm³ = Xg
    X = 1000×0.79/250 = 3.16g
    Hence Conc in g/dm³ of A = 3.16gdm^-³
    But molar conc = mass conc(gdm^-³)/molar mass(g/mol)
    Hence molar Conc.Of A= 3.16gdm^-3/molar mass ofA
    But molar mass of A = KMnO4
    = 39 + 55 + 4(16) = 39 + 55 + 64
    = 158g/mol
    Hence Conc in moldm^-³ of A = 3.16gdm^-3/158g/mol
    = 0.020moldm^-3
    1bii)
    To get concentration of B in moldm^-³, we use the relation
    CAVA/CBVB = na /nb
    Where,
    CA = 0.02moldm^-³
    VA = 24.05cm^-³
    na = 1
    CB = ?
    VB = 25.0cm³
    nb = 5
    0.02×24.05/CB×25 = 1/5
    CB = 5 × 0.02×24.05/25
    = 2.405/25
    =0.0962
    CB = 0.096moldm^-³
    1biii)
    (NH4)2 SO4-FeSO4, 6H2O——>250^2-4 +Fe^2+2NH^+4
    N(SO4^2-)/N(Fe^2+)=2/1===>N(SO4^2-)=2*N(Fe^+
    C(SO4^2-)=2*0.0962
    =0.192moldm^-3
    1biv)
    m(SO4^2-)=C*M*Vdm^3
    SO4^2-
    32+16*14=96gmol
    =
    =0.192*96*250/1000
    =4.608
    ===================================

    2)
    A is ZnSO4(aq)
    B is NaOH(aq)
    C is Pb(NO3)2 (aq)
    D is HCL(aq)

    -Reasons-
    A- since it has two no observable reaction and one white ppt
    B- Reasons
    Since it forms a white precipitate soluble with NaOH
    C – Reason’s
    Since it has no reaction and forms a white gelatinous precipitate soluble in excess
    D- Reasons
    – since it gives three white precipitate
    ==============≠====================

    3ai)
    Sodium is stored under paraffin oil to prevent its oxidation by atmospheric gases
    3aii)
    desiccator lid must be greased with a thin layer of grease, to ensure an airtight seal.
    3aiii)
    Because it is hygroscopic
    3bi)
    Add an acid eg HCL and this gives a neutralization reaction and the indicator react with the acids. Heat the solution to obtain the NaCL(s)
    3bii)
    Dip a rod in aqueous Ammonia and dip in a beaker containing the acids, only HCl produces dense white fumes of ammonium chloride
    3ci)
    i)Beaker
    ii)glass rod
    iii)volumetric flask
    3cii)
    On adding the dilute HCL to Na3SO3 a gas with a chocking smell is allow
    ===================================
    Completed
    ========≠==========================



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    Master Solution September 13, 2018 Categories: WAEC 835


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