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# NABTEB 2019 CHEMISTRY PRACTICAL ANSWER

(1a)
A = 20.0gdm^-3 NaHCO4
B = 5.60gdm^-3 KOH

Equation of reaction: 2NaHSO4(aq) + KOH(aq) = NaSO4(aq) + K2SO4(aq) + 2H2O(l)

Final burette reading (cm^3): 20.45, 18.93, 20.02

Initial burette reading (cm^3): 0.00, 0.00, 1.00

Volume of A used (cm^3) : 20.45, 18.93, 19.02

Average volume of Acid used =18.93+19.02/2
=37.95/2
Average volume of Acid used =18.98cm^3

(1bi)
Concentration of moldm^-3
Molarity = Mass concentration /molar mass
Molarity=5.60/56
Molarity= 0.10moldm^-3

(1bii)
Concentration in A moldm^-3
Cava/CbVb =na/nb
Ca=CbVbna / Vanb

=0.100*200*2/18.98*1
Ca = 5/18.98
Ca = 0.263moldm^-3
Concentration of A moldm^-3 is 0.263moldm^-3

(1biii)
Percentage of NaHSO4 in NaHSO4
NaOH(aq) + H2SO4 (aq) —> NaHSO4(aq) + H2O(l)

Molar mass of NaHSO4 = 104gmol^-1
% of sodium tetraoxosulphate(vi) =103/104*100/1
= 99.038
Percentage of sodium tetraoxosulphate(vi)= 99.04%
=============================

NOTE: DRAW TABLE HERE, TEST comes first, OBSERVATION follows, then INFERENCE

(2a)
Test: Sample C + 10cm of distilled water

Observation: Dissolves to form a colourless solution

Inference: C is a soluble salt of Na^+

(2aii)
Test: Sample C solution + litmus paper

Observation: It turns most blue litmus paper red

Inference: Acid gas present in salt sample C

(2aiii)
Test: 2cm^3 portion of sample + Bacl(aq) + Excess HCL(aq) + Warm

Observation: White precipitate formed which is soluble in excess dilute HCL(aq). A colourless gas with an irritating odour evolved which turns acidified K2Cr2O7 solution from orange to green

Inference; SO4^2- and SO3^2- is present while SO3^2- is confirmed. Also SO2 gas confirmed from SO3^2-

(2b)
Test: 2cm^3 portion of sample D + HCl(aq)

Observation; A reddish-brown precipitate is formed

Inference; Fe^3+ present

(2bi)
Test: 2cm^3 portion of sample D + HCL(aq) + 5cm^3 of solution C + Heat gently to boil then allowed to cool under water

Observation; White precipitate is formed which is insoluble in excess HCL. The precipitate turn brown at the upper part of the tube

Inference; Fe^2+ present, SO3^2- hence Fe^2+ oxidized to Fe^3+

(2bii)
Test; 2cm^3 portion of the coded solution in (2bi) + NH3(aq) + drops + excess

Observation: White precipitate formed, insoluble in excess NH3 solution

Inference; Fe^3+ present

(2biii)
Test; Sample in 2bi in test tube for 5minutes

Observation: Precipitate turned brown

Inference; Fe^3+ Confirmed
=============================

(3ai)
Substance of E + iodine solution gives a blue black colouration , hence E is a starch solution. E is boiled with HCL(aq) hence starch is hydrolysed, so visible reaction sample is Glucose or sucrose ie F=Glucose , G is zymase which convert Glucose to ethanol and CO2, hence H is carbon dioxide gas (CO2). Ethanol treated with propanoic acid ( C3H7COOH ) to obtain C3H7COOC2H5 (propylpropanoate) and water.
(3aii)
C3H7COOH(aq) + C2H5OH(aq) —-> C3H7COOC2H5(aq) (propylpropanoate) + H2O(l) (water)
(a) The concentration of propanoic acid used
(b) The functional group of the reactant as -COOH and -OH
(3aiii)
C3H7COOH(aq) + C2H5OH(aq) —-> C3H7COOC2H5(aq) + H2O(l)
(3b)
A gas jar is prepared containing water for upward collection of gas or downward collection
============================
COMPLETED
============================

Posted by on 14th May 2019.

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