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Chemistry Practical Questions ====================== 1) Burette Readings (Initial And Final) Must Be Given To Two Decimal Place Volume Of Pipette Used Must Also Be Recorded.All Calculations Must Be Experimental Procedure Is Required.All Calculations Must Be Done In Your Answers Book. A Is A Solution Containing 4.50g Of An Acid H2C204 Per Dm^3 B Contains 3.6g Of KMnO4 Per Dm^3 Of Solution A) Put A Into The Burette And Titrate It Against 20.0cm^3 Or 25.00Cm^3 Portions Of B Using Methyl Orange As Indicator. Repeat The Tritration To Obtain Consistent Titres. TABULATE Your Burette Readings And Calculate The Average Volume Of Acid A Used B) From Your Result And The Information Provided Above Calculate. (1) TABULATE: Volume Of Pipette Burette Reading(Cm3)|Rough Titre(Cm3)|1st Titre(Cm3)|2nd Titre(Cm3) Final Reading(Cm3) |20.00|19.60|19.60 Initial Reading(Cm3)|0.00 |0.00 |0.00 Volume Of Acid Used |20.00|19.60|19.60 Average Titre=(19.60+19.60)/2 Cm3 =39.20/2 Cm3 =19.60cm3 Equation=>2MnO4^-+5C2O4^2-+16H^+->2NA^2++8H2O+10CO2 (1bi) Conc Of B In Mol/Dm3 CaVa/CbVb=Na/Nb (0.05*19.6)/(Cb*25.0)=5/2 5*Cb*25=2*0.05*19.6 Cb=(2*0.05*19.6)/(5*25)=1.96 Cb=0.0157mol/Dm3 Cb=0.016mol/Dm3 (1bii) Conc In B In G/Dm3 =0.0157*158 =2.4806g/Dm3 =2.481g/Dm3 ========================== 2) I) TEST A = C + distilled H20 OBSERVATION = Partly soluble colorless filtrate white residue INFRENCE = C is mixture of soluble & insoluble salts II) TEST B = Filtrate + AgNo3 , | HNO3 , | excess NH3 OBSERVATION = White PPT | PPT Insoluble | PPT dissolve INFRENCE = Cl^- , SO3^2-, CO3^2- present | Cl- presnt | Cl- confirmed III) TEST = Residue + HNO3 OBSERVATION = Effervescence colorless,odorless gas, turns lime water milky. INFRENCE = CO2(g) , CO3^2- is present Iv) TEST = Filtrate + NaOH in drop OBSERVATION = White Gelatinous PPT INFRENCE = IV) (ii) TEST = Filtrate + NaOH in Excess OBSERVATION = PPT dissolve INFRENCE = Zn2+ , Pb2+ present IV) (iii) TEST = Filtrate + NH3 in drop OBSERVATION = White Gelatinous PPT INFRENCE = IV) (iv) TEST = Filtrate + NH3 in Excess OBSERVATION = PPT Dissolve INFRENCE = Zn2+ confirmed (3a) The differences in Boiling point of a liquid is (I) they are used to separate the mixture of color (II) with close different boiling point (3b) (i) To make the accurate result of the solution (II) To avoid the error of decimal 3c) C1=2.5moldm³ V2=500cm³ Ç2=0.2moldm³ V1=? C1V1=C2V2 V1=C2V2/C1 V1=0.2*500/2.5 V1=40.0cm³ Chem-Pract-Answers ============ 3a) I) They are used to seperate mixture of colour II) With colour different boiling point 3bi) To make accurate result of solution 3bii) To avoid error of decimal 3c) C1 = 2.5mol/dm3 V2 = 500cm C2 = 0.2mol/dm3 V1 = ? C1V1 = C2V2 V1 = C2V2/C1 V1 = 0.2 x 500 / 2.5 V1 = 40.0cm3 ============ ( NOTE THAT of this ^ MEANS Raise to power / means ÷ divide * means multiplication sign. + means plus = means equal to) ============
