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Monday, 11 June 2012
NECO 2012 (Internal): Chemistry Paritical
Note that wat you are going to see right now is not the 2012 NECO question but most likely question. This is been set with the specimen given to schools.
But i'm sure 85% of wat you about to see is comming out.
1. A is 0.050 moldm-3 tetraoxosulpate(VI)acid. B is a solution containing 1.0g of impure sodium hydroxide pellets per 250cm3 of solution. a. Put A into burette and titrate with 20 or 25cm3 portions of B using methyl orange or screened methyl orange as indictor. Tabulate your burette readings and calculate the average volume of A used. b. From your results and the information provided, calculate the: i. concentration of solution B in moldm3 ¬ii. concentration of solution B in gdm3 iii. percentage purity of the sodiumhydroxide pellets. The equation for the reaction is: 2NaOHaq + H2SO4 aq H2SO4aq + 2H2Ol (H = 1, O = 16, Na = 23) c. Give one possible source of the impurity in the sodium hydroxide pellets.
ANSWER 1. Indicator used: methylorange Volume of pipette: 25.00cm3 a. Burette reading: TITRATIONS Rough 1st 2nd 3rd Final Burette cm3 25.00 24.60 25.60 26.50 Initial Burette cm3 0.00 0.00 1.00 2.00 Volume of acid used cm3 25.00 24.60 24.60 24.50 Average volume of acid used = (24.60 + 25.60 + 26.50)cm3 ÷ 3 = 73.70cm3 = 24.57cm3 bi. Using (CAVA) ÷ (CBVB) = nA ÷ nB CB = (CBVB nB) ÷ VB nA CB = (0.050 moldm-3 x 24.57 cm3 x 2 cm3) ÷ (25.00 cm3 x 1) = 2.457 moldm-3 ÷ 25.00 = 0.098 moldm-3 Thus, the conc. Of solution is B in moldm-3 is 0.098 moldmC ii. Molar mass of Na OH = (23 + 16 + 1)gmol-1 = 40 gmol-1 therefore concentration of solution B in gdm-3 = conc. of solution B in moldm-3 x molar mass of NaOH in gmol-1 = 0.098 moldm-3 x 40 gmol-1 = 3.92 gdm-3 iii. 250 cm3 of solution B contain 1.0g of impure NaOH pellets. Therefore 1000cm3 of solution B will contain (1.0 x 1000) ÷ 250 of impure NaOH(s) = 4.0g of impure NaOH(2) Thus, the percentage purity of the NaOH pellets = {conc. Of solution B in gdm-3 / conc. Of impure NaOH pellets (in gdm-3) } x 100/1 = (3. 92 gdm-3/4.0gdm-3) x (100/1) = 98% c. Presence of sodium chloride
2. X NH4Cl + CuCO3 is a mixture of two simple salts. Carry out the following tests on them. Record your observation and inferences. Identify any gas(es) given off. (filtrate = NH4Claq and Residue = CuCO3.)
a. TEST Put all of X in a boiling tube and acid about 5cm of distilled water. Shake thoroughly and then filter. Keep both the filtrate and residue. OBSERVATION A false solution formed. The solution is slightly bluish. INFERENCE X is a mixture of soluble and insoluble salts.
b. Divide the filterate into two portions: i. TEST To the first portion, add sodium hydroxide solution, in drop until it is in excess. OBSERVATION No visible reaction is observed. INFERENCE Na+, K+ or NH4+ present.
ii TEST Warm the mixture in (b)(i) gently. OBSERVATION An effervescence of gas with choking smell observed. The gas turns moist red litmus paper blue INFERENCE The gas is ammonia (NH3) Therefore NH4present iii. TEST To the 2nd portion, add a few drops of dilute trioxomtrate(v) acid and then a few drops of silver trioxonitrate (v) solution OBSERVATION White precipitate formed on adding AgNO3 (aq) INFERENCE Cl – present
c TEST Put the residue in a test tube and add about 3cm3 of dilute HCl. Warm gently and indentify any gas evolved. Keep the mixture. OBSERVATION A colourless and odourless gas is evolved. The gas turns moist blue litmus paper faint red. INFERENCE The gas is CO2. Therefore CO2-3 present.
d TEST d. Divide the clear solution from (c) above into two portion. i. To the 1st portion, add NaOH solution in drops and then in excess OBSERVATION Blue gelatinous precipate which is insoluble in excess NaOH solution is observed. INFERENCE CU2+ present
ii TEST To the 2nd portion, add aqueous ammonia in drops and then in excess OBSERVATION Blue precipitate which is soluble in excess NH3(aq) is formed. INFERENCE CU2+ present
Question
ReplyDelete1. A is 0.050 moldm-3 tetraoxosulpate(VI)acid. B is a solution containing 1.0g of impure sodium hydroxide pellets per 250cm3 of solution.
a. Put A into burette and titrate with 20 or 25cm3 portions of B using methyl orange or screened methyl orange as indictor. Tabulate your burette readings and calculate the average volume of A used.
b. From your results and the information provided, calculate the:
i. concentration of solution B in moldm3
¬ii. concentration of solution B in gdm3
iii. percentage purity of the sodiumhydroxide pellets.
The equation for the reaction is:
2NaOHaq + H2SO4 aq H2SO4aq + 2H2Ol
(H = 1, O = 16, Na = 23)
c. Give one possible source of the impurity in the sodium hydroxide pellets.
ANSWER
ReplyDelete1.
Indicator used: methylorange
Volume of pipette: 25.00cm3
a.
Burette reading:
TITRATIONS
Rough 1st 2nd 3rd
Final Burette cm3 25.00 24.60 25.60 26.50
Initial Burette cm3 0.00 0.00 1.00 2.00
Volume of acid used cm3 25.00 24.60 24.60 24.50
Average volume of acid used = (24.60 + 25.60 + 26.50)cm3 ÷ 3 = 73.70cm3 = 24.57cm3
bi. Using
(CAVA) ÷ (CBVB) = nA ÷ nB
CB = (CBVB nB) ÷ VB nA
CB = (0.050 moldm-3 x 24.57 cm3 x 2 cm3) ÷ (25.00 cm3 x 1) = 2.457 moldm-3 ÷ 25.00 = 0.098 moldm-3
Thus, the conc. Of solution is B in moldm-3 is 0.098 moldmC
ii. Molar mass of Na OH = (23 + 16 + 1)gmol-1 = 40 gmol-1
therefore concentration of solution B in gdm-3 = conc. of solution B in moldm-3 x molar mass of NaOH in gmol-1
= 0.098 moldm-3 x 40 gmol-1
= 3.92 gdm-3
iii. 250 cm3 of solution B contain 1.0g of impure NaOH pellets.
Therefore 1000cm3 of solution B will contain (1.0 x 1000) ÷ 250 of impure NaOH(s) = 4.0g of impure NaOH(2)
Thus, the percentage purity of the NaOH pellets
= {conc. Of solution B in gdm-3 / conc. Of impure NaOH pellets (in gdm-3) } x 100/1
= (3. 92 gdm-3/4.0gdm-3) x (100/1) = 98%
c. Presence of sodium chloride
Be careful there are some subscript and superscript
Delete2. X NH4Cl + CuCO3 is a mixture of two simple salts. Carry out the following tests on them. Record your observation and inferences. Identify any gas(es) given off.
ReplyDelete(filtrate = NH4Claq and Residue = CuCO3.)
a.
TEST
Put all of X in a boiling tube and acid about 5cm of distilled water. Shake thoroughly and then filter. Keep both the filtrate and residue.
OBSERVATION
A false solution formed. The solution is slightly bluish.
INFERENCE
X is a mixture of soluble and insoluble salts.
b. Divide the filterate into two portions:
i.
TEST
To the first portion, add sodium hydroxide solution, in drop until it is in excess.
OBSERVATION
No visible reaction is observed.
INFERENCE
Na+, K+ or NH4+ present.
ii
TEST
Warm the mixture in (b)(i) gently.
OBSERVATION
An effervescence of gas with choking smell observed. The gas turns moist red litmus paper blue
INFERENCE
The gas is ammonia (NH3)
Therefore NH4present
iii.
TEST
To the 2nd portion, add a few drops of dilute trioxomtrate(v) acid and then a few drops of silver trioxonitrate (v) solution
OBSERVATION
White precipitate formed on adding AgNO3 (aq)
INFERENCE
Cl – present
c
TEST
Put the residue in a test tube and add about 3cm3 of dilute HCl. Warm gently and indentify any gas evolved. Keep the mixture.
OBSERVATION
A colourless and odourless gas is evolved. The gas turns moist blue litmus paper faint red.
INFERENCE
The gas is CO2. Therefore CO2-3 present.
d
TEST
d. Divide the clear solution from (c) above into two portion.
i. To the 1st portion, add NaOH solution in drops and then in excess
OBSERVATION
Blue gelatinous precipate which is insoluble in excess NaOH solution is observed.
INFERENCE
CU2+ present
ii
TEST
To the 2nd portion, add aqueous ammonia in drops and then in excess
OBSERVATION
Blue precipitate which is soluble in excess NH3(aq) is formed.
INFERENCE
CU2+ present
Be careful there are some subscript and superscript
DeleteOnce again this is meant to be the likely question.
ReplyDeleteThis is the same that was asked for in 2002 NECO.
Please get back to me if you need other assistance.
Expect oderz by tomorrow.
I'm sry, others will be added by tomorrow.
ReplyDeleteI'm happy to have you online
ReplyDelete