NABTEB 2018 - CHEMISTRY PRACTICAL ANSWERS
Friday 11/05/2018
Chemistry Paper II(Practical)
9.00am - 11.00am
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COMPLETED
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(1a)
Volume of Pipette = 25.0cm
TABULATE
Burette reading |rough |1st |2nd|3rd|
Final burette reading (cm³) |24.90| 39.50|34.80|24.50
Initial Burette reading (cm³) |0.00|15.00|10.00|0.00
Volume of E used (cm³) |24.90|24.50|24.80|24.50
Average volume of acid used
= (24.50+24.80+24.50)cm/3
= 24.50cm³
= 24.50cm³
(1bi)
Concentration of E (CE) in moldm
= mole x 1000/volume
4.90g ==> 500cm³
xg ==> 1000cm³
x = 4.90x1000gdm³/500
=9.80gdm³
Molar mass (H2S04) = 1x2 + 32 + 16x4
= 2+ 32 + 64
= 98gmol
Concentration (mole dm³) = 9.80gdm³/98
= 0.100moldm³
(1bii) CEVE/CFVF = Ne/Nf
= 0.100x24.50/Cf x 25.0 =1/1
Cf = 0.100x24.50/25x1
Cf = 2.45/25 moldm³
= 0.0980moldm³
(1biii) Concentration of F in g/dm
Conc (gdm³) = molar mass x conc
Molar mass (Na3CO3) = 23x2 + 12 + 16 x3
=106gmol³
Conc (gdm³) = 106gmol x 0.0980
=10.39gdm³
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(2)
TABULATE:
Test | Observation| Inference
(2a)
TEST:
G + 8cm³ of distilled water
OBSERVATION:
Sample G dissolves partially in water colourless
INFERENCE:
G is a mixture of soluble and insoluble salt
(2bi)
TEST:
1st portion + HNO3(aq) + AgNO3(aq)
OBSERVATION:
A white precipitate is formed
INFERENCE:
Cl- present
(2bii)
TEST:
2nd portion + NaOH + heat
OBSERVATION:
A purgent irritating, choking gas is liberated which turns moist red litmus paper blue
INFERENCE:
Gas is NH3 from NH4+
(2c)
TEST:
Residue + H2SO4(aq)
OBSERVATION:
Residue dissolve completely on warming
INFERENCE:
Residue is soluble in acid
(2ci)
TEST:
1st portion + NaOH(aq) in drops than excess
OBSERVATION:
A pale blue precipitate is formed which is insoluble in excess
INFERENCE:
Cu2+ present
(2cii)
TEST:
2nd portion + NH3(aq) in drops, then excess
OBSERVATION:
A pale blue precipitate is formed which dissolved in excess to give deep blue solution
INFERENCE:
Cu2+ present
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(3a)
Effervescence of brownish gas which turns a moist litmus paper red and a black precipitate or residue of lead sulphide is formed.
Pb (NO3)(aq) + H2S(g) ==> Pbs(s)black + H2O(i) + NO(g)
(3bi)
-H2SO4 ==> strong acid
-CH3COOH ==> weak acid
(3bii)
H2SO4 is strong because it ionises completely in water while CH3COOH is weak acid because it ionise partially in water
(3c)
Heat the mixture, ammonium chloride sublime and it then condenses to solid NH4Cl
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COMPLETED
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