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2013/2014 NABTEB Alternative To Chemistry Practical, Specimen, Real Questions And Answers

(1) tabulate
TABLE OF RESULT
Burette Readings (cm^3), Rough Titration, First Titration, second Titration, Third Titration

Final Burette Reading(cm^3),21.40,42.70,21.20,42.40

Initial Burette Reading(cm^3),0.00,21.40,0.00,21.20

Volume Of Acid Used, 21.40,21.30,21.20,21.20



(1a)
Average Volume of Acid Used(cm¿3) = (21.30+21.20+21.20)/(3)cm^3

Average Volume of acid Used= (63.70)/(3)cm^3

Average Titre  (VA)cm^3= 21.23cm^3

(i) Volume of pipette (solution B)=25.00cm^3

(ii) The Average Volume of A used= 21.23cm^3

(iii) Equation of Reaction: 2HX(aq)+Na2CO3(aq)--> Na2X(aq)+CO2(g)+H2O(L)

(1b)
(i)  Concentration of B in mol/dm^3

250cm^3 of solution B contained 2.65g of Na2CO3(aq)

:1000cm^3 of solution B will contain (1000cm^3)/(250cm^3)*2.65g of Na2Co3(aq)

mass concentration of Na2CO3= 10.6gdm^3 of Na2CO3(aq)

solution:
But Molar concentration of B in mol/dm^-3= molar conc of B gdm^-3/ Molar mass of B gmol^-1

: molar conc of B in mol/dm¿-3 = (10.6gdm^-3)/(106gmol^-1)

conc of B in mol/dm^-3 = 0.100mol/dm^3

But molar mass of B (Na2CO3)= [(23*2)+(12*1)+(16*3)]

Molar mass of B = 46+12+48 =106gmol^-1


(1bii)
Concentration of A in mol/dm^-3:.
from the equation above;
NA=2, Nb=1,

VA=21.23cm^3, VA= 25.00cm^3

CB= 0.100mol/dm^-3

Using  CAVA/CBVB=NA/NB
(CA * 21.23)/(0.100*25.00)= 2/1
cross multiply;
CA*21.23*1 = 2*0.100*25.00
CA= (2*0.100*25.00)/(21.23*1)
CA= 5.0/21.23= 0.236mol/dm^-3

(1biii)
Molar mass of the Acid, Hx:
Given that 1000cm3 of solution A contained 6.30g of the Acid
6.30gdm^-3 of solution A = Mass conc of A in gdm^-3
Using molar conc of A in Mol/dm^-3 = Mass conc of A in gdm^-3/ Molar mass of A in gmol^-1
:. Molar mass of Hx= Mass conc of HX in gdm^-3/ Molar conc of HX in mol/dm^-3
: molar mass od HX= 6.30gdm-3/ 0.236mol/dm^-3
:Molar  mass of Hx= 26.7gmol^-1

(2)TABULATE
(ai)
TEST:  C+ Water, and filter

OBSERVATION: part of sample dissolves, colourless  filterate and black residue.

INFERENCE: C is a mixture of soluble and insoluble salt

(aii)
TEST:   filtrate + Litmus
OBSERVATION: The filtrate is neutral to litmus
INFERENCE: the filtrate is neutral

(2b)
TEST: To about 2cm3 of the filterate + Bacl2[aq] + dil. HCl and warm or heat.

OBSERVATION:  The white precipitate dissolves on adding dilute HCL solution. Gas evolved which is colourless with pungent choking or sharp irritating smell which decolourises KNMO4. this gas can also decolourise k2Cr2O7

INFERENCE: SO3^(-2) or CO3^(-2) is present. Gas is SO2 for SO3^(-2)

(ci)TEST:  Residue + H2SO4 in excess

OBSERVATION: Residue dissolve to form a blue solution

INFERENCE: CU2+ is likely present

(2cii)
TEST: Residue solution from (ci) above + NH3 in drops; then in excess

OBSERVATION: Pale (light) blue precipitate or blue percipitate. Blue percipitate dissolved give a deep blue solution

INFERENCE: CU2+ is confirmed and present

(3ai) consider [liebog condenser].
(3aii) delivery tube.

(3bi) sugar charcoal is obtained in warming concentrated tetraoxosulphate [vi] acid with a cube of sugar. conc H2SO4 is used as a dehydrating agent, it removes elements of water in the sugar. It is a chemical process.
C12 H22 O11 (S) ----------->  12 C(s) + 11H2O(S)

(3bii)
 A deep red colour or precipitate is formed indicating the presence of protein or

 A white precipitate is observed which turns to brick red in warming .

(3biii) filter paper stained red indicating the presence of oil.

(3c) using aqueous ammoniacal solution of copper[1] chloride or silver[1] chloride

But-1-yne will react with ammoniacal solution of copper[1]chloride or silver[1] chloride while but-1-ene will not react.

FOR MORE ANSWERS VISIT THE BELOW LINK;
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Guy na comfirm answer be dat we dn giv fake tinx here