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Waec GCE 2019 Chemistry Solutions & Answer Available Here

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2019 Waec Gce Chemistry Practical Answer.
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(1a)
VA=11.30cm^3
CA=5.3g/dm^3
VB=25.0cm^3
CB=Yg/dm^3
HCL(aq)+NaOH(aq)–>NaCl(aq)+H2O(l)

(i)
For acid nA:nB=1:1
Conc. of A in mol/dm^3=Conc of A in g/dm^3/molar mass in g/mol
HCl=1+35.5=36.5g/mol
=5.3/36.5
=0.145mol/dm^3
CAVA/CBVB=nA/nB
(0.145*11.30)/(CB*25.0)=1/1
1.6385=25CB
CB=1.6385/25
CB=0.06554mol/dm^3

(ii)
Conc of B in mol/dm^3=Conc of A in g/dm^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

(iii)
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)
-Methyl orange
-Yellow

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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.

======================

(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)
Soda lime is preferred to sodium hydroxide in the preparation of methane because soda lime is not deliquescent and sodalime does not attack glass.
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