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JAMB DAY 3 FREE QUESTIONS AND ANSWERS FOR 13TH APRIL 2019
Most Applied Faculty in JAMB 2019: Medicine, Pharmacy & Health Sciences
JAMB - Prohibited Items Candidates Should Avoid Taking to UTME
JAMB Fixes 2019 Mock Exam March 23rd
i) TABULATE
burette reading/cm^3 ___ 1 ____ 2 _____ 3
final reading__________ 39.5 __ 29.80 __ 19.40
initial reading ______ 19.55 __ 10.50 ___ 0.0
volume of E used _____ 19.70 ___ 19.30 __ 19.40
ii) Na2CO3(aq)^2HNO3(aq) — 2NaNO3(aq) + CO2(g) +
H2O(c)
average volume of E used = 19.70+19.30+19.40/3
=19.47cm^3
=========
(2a)
Inference: SO3^2- suspected
(2b)
(i) Inference: insoluble salt
(ii) Inference: SO3^2- confirmed
(iii) Inference: Cu2 suspected Cu2 confirmed
(iv) Obseravation: -Blue gelatineous precipitate -
Precipitate dissolves
(2c)
(i) CuSO3
(ii) Copper
(ii) trioxosulphate (iv)
=====================
(3ai)
SO2(g)
(3aii)
A->To dry the gas
B->To detect the type of gas that was dried
(3aiii)
A->P4O10
B->KMno4/Blue litmus paper
(3bi)
-It is an acid that turns moist blue litmus paper
red
-It is colourless and poisonous gas with an
irritating smell
(3bII)
The irritating smell
(3bii)
Downward delivery
Chlorine gas (Cl2(g))
(3ci)
I-T,U,V
II-X,Y,Z
(3cii)
W
(3ciii)
I-U
II-V
++++++++++++++++++++++++++++++
(1ai)
CE=0.905mol/dm^3
VD=20.0cm^3
For the first titration
Burette reading|1|2|3
Final reading|39.25|29.80|19.40
Initial reading|19.55|10.50|0.00
Volume|19.70|19.30|19.40
(1aii)
Volume of E used=(2nd+3rd)titre/2
=(19.30+19.40)/2
=(38.70/2)cm^3
Volume of E used=19.35cm^3
Hence VE=19.35cm^3
(1bi)
From the information given,
nD=1, nE=2,
CD=?, CE=0.0905mol/dm^3
VD=20.0cm^3, VE=19.35cm^3
Using: CDVD/CEVE=nD/nE
CD=CEVEnD/VDnE
=(0.0905*19.35*1)/(20.0*2)
=1.751175/40
=0.04377
CD=0.0438mol/dm^3
(1bii)
Since 0.04377mol is in 1000cm^3
xmol will be 350cm^3
x=0.04377*35.0/1000
=1.5322/1000
=0.00153mol/dm^3 at 25C
Hence the solubility is 0.00153mol/dm^3
(1biii)
From the stoichiometry
1 mole of Na2CO3 produces 1 mole of CO2
0.00153mol of Na2CO3 produces 0.00153mole of
CO2
mass=molar mass*number of mole
molar mass Na2CO3=23*2+12+16*3
=46+12+48
=106g/mol
mass of CO2=12+16*2
=12+32
=44g/mol
106g Na2CO3 produces 44g of CO2
0.0153*106 will produce y
y=0.153*106*44/106
=0.153*44
y=6.732g of CO2
Hence mass of CO2 produced is 6.732g
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