2017 Nabteb Chemistry Practical Answer – May/june Expo

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*2017 NABTEB CHEMISTRY PRACTICAL SOLUTION BY MR.GALANT*

(No1a)
Volume of Acid used is 25cm3
Indicator used =Methyl orange

Table of observation
========================
Burette reading: Rough, 1st, 2nd, 3rd
========================
Final reading(cm):24.50, 23.50, 23.40, 23.60
========================
Initial reading(cm):0.00, 0.00,0.00, 0.00
=========================
Volume of Acid used:24.50, 23.50, 23.40, 23.60
==========================

Volume of Acid used

= 1st +2nd +3rd
––––––––––
3

= 23.50 +23.40+23.60
–––––––––––––
3

= 70.3
–––– =23.50cm3
3
+++++++++++++++++++++++++
(1bi)
Concentration of B in mold – 3

CAVA NA
–––– = –––
CBVB NB

CA= 0.03moldm-3
VA=23.50moldm-3
NA=2
VB=25cm3
NB=1
CB=?

CB = CA VA NB
––––––
VB NA

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CB = 0.03 × 23.50 × 1
––––––––––
25×2

= 0.075
–––– = 0.0141
50

Concentration of solution B in moldm-3 = 0.0141moldm-3
++±++++++++++++++++++++++
(1bii)
Milan mass of NaCl
= 23 + 35.5 =58.5g/mol
Amount of NaCo3 in 1dm3

= 0.0141 × 1
= 0.0141mol

From the equation
NaCo3 + 2HCL –––> 2NaCl +H2O + CO2

1mole of Na2CO3 produces
1mole of Nacl

0.0141mole of Na2CO3 liberate
2 × 0.0141mole of NaCl
=1.6497g
=1.65g
+++++++++++++++++++++++++
(1biii)

From the equation
1mole of Na2Co3 produce 1mole of Co2 at S.T.P

1mole of NaCo3 produce 22.4dm3 of CO2 at S. T. P

Therefore 0.0141mol of Na2Co3 will produce
22.4 × 0.0141

Volume of CO2 = 0.316dm3
(1biv)
Molar mass of H2X = (2×1) + X
= (2+X) g/mol.
+++++++++++++++++++++++
(No2)

(2a)
*TEST* : C + water, Mixed thoroughly and filtrate

*OBSERVATION* : partially soluble in water with colourless filtrate and white residue

*INFERENCE*: C is a mixture of soluble and insoluble salt
===========================
(2bi)
*TEST*:Filtrate +NaCl(aq)

*OBSERVATION*:white precipitate is formed
The precipitate soluble in excess NaOH(aq)

*INFERENCE*:Pb2+, Zn2+, or Al2+ is present
==========================
(2bii)
*TEST*:Filtrate + NH3(aq) in drop, then in excess

*OBSERVATION*:white precipitate is formed
The precipitate is insoluble in excess NH3(aq)

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*INFERENCE*: Pb2+ or Al2+ present

==========================
(2biii)
*TEST*:Filtrate + KI(aq) + warm and allow to cool

*OBSERVATION*:Yellow precipitate is formed
The precipitate reappears after cooling

*INFERENCE*:Pb2+ confirmed

===========================
(2c)
*TEST*:Residue + Hcl(aq)

*OBSERVATION*:The residue dissolves, librating a colourless gas which turn lime water milky

*INFERENCE*:Gass is CO2 from CO3^2+ is present
=========================
(2d)
*TEST*:Residue + NH3(aq) in drops, then in excess

*OBSERVATION*:white gelatinous precipitate is formed
The precipitate is. in excess NH3(aq)

*INFERENCE*:Zn^2+ Confirmed
==========================
(3ai)
NaCl4 and NaNo3 dissolves readily in cold water.
+++++++++++++++++++++++
(3aii)
CuCo3 leave a black residue on heating
+++++++++++++++++++++++
(3bi)
E = CuO
+++++++++++++++++++++++
(3bii)
Blue precipitate is formed, the precipitate dissolves in aqueous ammonia to give a blue deep solution
+++++++++++++++++++++++
(3biii)
G is water (H2O)
+++++++++++++++++++++++
(3biv)
A dark brown precipitate is formed at the button of the test tube, which shows that copper has been displaced by zinc for its solution
+++++++++++++++++++++++

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