WAEC GCE 2019 - CHEMISTRY PRACTICAL ANSWER

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ANSWERS:

(1a)
VA=11.30cm³
CA=5.3gldm³
VB=25.0cm³
CB=Ygldm³
HCl(aq)+NaOH(aq)----->NaCl(aq)+H²O(i)
For acid NA; NB =1;1

(1ai)
Conc of A in moldm-³
=conc of A in gldm-³/molar mass in glmol
=5.3/36.5
HCl=>1+35.5=36.5glmole
=0.145moldm-³

CAVA/CBVB=NA/NB
0.145*11.30/CB*25.0=1/1
1.6385=25CB
CB=1.6385/25
CB=0.06554moldm³

(1aii)
Conc of B in moldm-³=
Conc of A in gldm-³/molar mass in g/mol

0.0655=Yg/dm^3/40
Y=2.62g/dm^3
Molar mass of Y(NaOH)=23+16+1=40g/mol
The value of Y is 2.62g/dm^-3

(1aiii)
HCl+NaOH-->NaCl+H2O
1 mole of HCl gives 1 mole of NaCl
1.6385 mole of HCl would give 1.6385 mole of NaCl
n=CV
=0.145*11.30
=1.6385mole
No of mole =Mass/Molar mass
1.6385=mass/58.5
Molar mass of NaCl=58.5g/mol
Massof NaCl=58.5*1.6385
=95.85

(1b)
(i) Methyl orange
(ii) Orange






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(2ai)
TEST: E + H2O(stirred)
INFERENCE: E is a mixture of salts

(2aii)
TEST: Residue
INFERENCE: Gas evolved is CO2 from HCO3- or CO3²-

(2aiii)
TEST: residue + HNO3 + H2O
INFERENCE : CO2 gas is evolved from HCO3-or CO3²-

(2aiv)
OBSERVATION: Blue gelatinous ppt soluble in drops. Soluble in excess to give deep blue ppt.
INFERENCE: Cu2+ present
Cu2+ confirmed

(2bi)
INFERENCE : Zn²+, pb²+, Al³+ present
Zn²+, pb²+ present

(2bii)
OBSERVATION: White gelatinous ppt soluble in excess Nh3 solution
INFERENCE: Zn²+, pb²+, Al³+
Zn²+ confirmed

(2biii)
TEST: Portion of filtrate of Ba(NO3)2 + HCL
INFERENCE: SO4²- or CO3²- present.
SO4²- confirmed.



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(3a)
Sucrose + Benedict solution+heat => no colour change
Inference: non reducing sugar present.

Glucose+Benedict solution+heat=> brick red colour formed.
Inference: Reducing sugar like glucose present

(3bi)
(i) potassium iodide
(ii) throsulphate

(3bii)
Starch indicator

(3biii)
(i)before end point - dark color
(ii)at the end point - colourless

(3c)
(i) Soda lime is preferred because it does not attack rubber
(ii) Because soda lime is not deliquescent





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