100% Correct NABTEB 2015 CHEMISTRY
PRACTICAL ANSWER
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CHEMISTRY PRACTICAL ANSWERS
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3a.
i. kipps apparatus: it is used in intermittant supply
to gasses.
ii. it is used in cooling or condensing vapour into
liquid during distillation defenselessness
iii. hoffmanis voltameter: it is used oin
determination of the chemical composition of water
by volume.
3b.
i.add three or four drops of concentrated
trioxonitrate (v) acid of egg-white solution the
formation of an intense yellow colour indicate the
presence of protein.
ii. crystallization
iii. when water is added to din hydrous copper (ii)
trioxosulphate (vi) to the colour will change from
white to blue.
2a. Tabulate Test, Observatiob, Inference
i.
UNDER TEST PUT:
G + 5cm3 of distilled water
G solution + litmuss paper
UNDER OBSERVATION PUT:
G is soluble in water
No effect in litmus paper
UNDER INFERENCE PUT:
A soluble salt is present
The soft in neutral to litmus paper.
ii.
UNDER TEST PUT:
G solution + oe or two drop of iondine
UNDER OBSERVATION PUT:
The soltuion turns blue black
UNDER INFERENCE PUT:
Glucose is present
iii.
UNDER TEST PUT:
G solution + a few drop of feliling solution and
warn
UNDER OBSERVATION PUT:
A brick red pricipitate is formed
UNDER INFERENCE PUT:
Glucose confirm.
2b. Tabulate Test, Observatiob, Inference
i.
UNDER TEST PUT:
H + 5cm3 of dilute tetraoxosulphate acide and heat
UNDER OBSERVATION PUT:
a blue gelate haay precipitate is formed
UNDER INFERENCE PUT:
CU2+ is present
ii.
UNDER TEST PUT:
2cm3 b (i) + NH(A) in drop then in excess
UNDER OBSERVATION PUT:
a deep blue precipitate is formed and it dissolved in
excess
UNDER INFERENCE PUT:
CU2+ confirm
1a.
Volume of pipette used = 25cm3
Tabulate: —- , 1st , 2nd , 3rd
UNDER —- PUT:
Burette Reading (cm3)
Initial burette reading (cm3)
Volume of acide used
UNDER 1ST PUT:
23.50
00.00
23.50
UNDER 2nd PUT:
47.10
23.50
23.60
NDER 3rd Put:
23.40
00.00
23.40
Average volume of E used
= 23.50 + 23.60 + 23.40 / 3
= 23.50cm3
HNO3(aq) + NaOH(aq) –>NaNO3(aq) + H20(aq)
1b. F contains 2.00g NaOH(aq) in 5.00cm3
500cm3 of F = 100cm3
2.00g of F = X
500X = 2 x 1000
X = 2000/5
X = 4gldm3 of F
conc of F in gdm3 = 4gdm3
molar concentration = mass concentration / molar
mass
molar mass = NaOH = 23 + 16 + 1
= 40gldm
= 4gldm3 / 40gldm
molar conc of F = 0.1moldm3
Using CAVA / CBVB = nA / nB
CA x 23.50 / 0.1 x 25 = 1/1 ->ca x 23.5 = 2.5
CA = 2.5/23.50 ->CA = 0.1064moldm3
Concentration of E = 0.1064moldm3 recall that
mass concentrate = mass/volume(dm3)
form he balanced chemical equation
1 mole of F = 1 mole of NaNO3
500dm3 of F = 500dm3 of NaNO3
->volume of NaNo3 = 500dm3
also 1 mole of F = 1 mope of NaNO3
0.1 moldm3 of F = 0.1 moldm3 NaNO3
->Number of mole of NaNO3 = volume x
concentration
= 500 x 0.1
= 50 mole
Molar Mass NaNO3 = 23 + 14 + 3(16)
= 85gmol
mass of NaNO3 = number of mole x molar mass
= 50 x 85
= 4250g
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