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Guys post all questions regarding chemistry MCQ's... I'll try to explain as much as i can... But all help would be appreciated....
So start posting chemistry questions
So start posting chemistry questions
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P1 MCQ's preparation thread for chemistry ONLY!!!!
- Thread starter nightrider1993
- Start date
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Good job !
So we'll start from June 2002 and we'll end to June 2012 with all its variants.
The work solutions will be posted here.
How much papers we'll do per day ?
- Messages
- 98
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- 127
- Points
- 43
Just post the questions and we'll do as many as we can
- Messages
- 389
- Reaction score
- 202
- Points
- 53
June 2002
- Messages
- 98
- Reaction score
- 127
- Points
- 43
I wont be able to explain each and everyone. I could help out in the one's your having problems with
- Messages
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- 81
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I hope it would be better it we could solve each past papers and post the solutions for each questions possible out here on the yearly basis as some of the folks did during their preparation for Physics during this May/June session. It would benefit a lot of people out there. I know its a lot of time consuming but its just a suggestion from me, don't take it in the wrong sense. I am also ready to contribute as much as I can and not only me but I guess other people will also join us and put their effort in this novel work.
- Messages
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- Reaction score
- 202
- Points
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- Messages
- 98
- Reaction score
- 127
- Points
- 43
So are there any doubts regarding the P1 chemistry paper?
- Messages
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- 81
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- 38
So, let's get started with May/June 2002.Agreed !
I'll post for P5 also..
- Messages
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- Reaction score
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- 53
P1 :: May/June 2003 'Solved Explanations'
============================
Q1. A
Step 1 : Under room conditions 1 mol of oxygen occupies 24dm^3
x mol of oxygen occupies 0.500 dm^3
Amount of oxygen = (0.500*1)/24 = 0.0208 mol.
Step 2 : There are 6.02 * 10^23 molecules in 1 mol of oxygen
There are 'y' molecules in 0.0208 mol of oxygen
Number of oxygen molecules = [0.0208* (6.020* 10^23)] / 1 = 1.25 * 10^22
Q2. B
Step 1 : Count the hydrogen atoms in the side-chains.
Upper side-chain and lower side-chain = 2*(3+6+6+14) = 14
Middle side-chain = 3+6+6+14 = 29
Step 2: Count the hydrogen atoms in each residue
First residue = 3+8+6+14 = 31
Second residue *2 (because two-side chains are converted) = 2*(3+14+2+14) = 66
Step 3 : Calculate how many hydrogen atoms had to be added to each side-chain in order to form the residues.
Upper & lower side-chain residues have 66-58 extra hydrogen = 8 extra hygrogens
Middle side-chain residue has 31-29 extra hydrogens = 2 extra hydrogens
Step 4 : We know that 10 hydrogen atoms have been added to the original molecule. Each molecule contains two hydrogen atoms,so there are 10/2 = 5
hydrogen molecule, = 5 moles of hydrogen.
Q3. D
Number of neutrons = relative atomic mass - atomic number (protons)
Neutrons in an atom of S = 32 - 16 = 16
Neutrons in an atom of P = 31 - 15 = 16
3115P
Q4 . C
Arsenic and Zirconium have a small number of protons compared to Tellerium and Iodine. Their number of protons is not big enough to be consistent with the high 7th ionization energy.
You're left with either Tellerium or Iodine. You can see the ionization energies of iodine in the Data Booklet, and they're not consistent with the ionization energies above, therefore the answer is Tellerium
Q5. D
Magnesium Oxide has a a giant ionic lattice.
Sodium has a metallic structure
Silicon (IV) Oxide has a giant covalent structure
The only chemical with a simple molecular lattice is sulphur
Q6. C
Ideal gases have negligible size, no intermolecular forces, and are always in the gaseous state. When temperature is low and pressure is high, the intermolecular forces and size become relevant, and there are phase changes, which would make the equation less accurate in calculating the relative molecular mass because the equation is made for ideal gases, not real gases... so the answer is a low pressure and high temperature
Q7. D The enthalpy of formation of PCl5 is 330 x 5 = -1650 kJ mol-1 (bond formation is exothermic)
The enthalpy of formation of PCl3 is 330 x 3 = - 990 kJmol-1
The enthalpy of formation of Cl2 is 240 x 1 = 240 kJmol-1
In order to be sure that your calculations are correct, you should draw a Hess's diagram*
ΔHfO + (-1650) = -990 + (-240)
ΔHfO = +420 kJ mol-1
Q8. A) In the first reaction:
Oxidation number of Nitrogen before reaction... x + 3 = 0... x = -3
Oxidation number of Nitrogen after reaction... x -2 = 0... x = +2
Q9. C) Aluminium oxide dissolves in cryolite and so the cryolite acts as a flux, reducing the melting point required to extract the aluminium from its ores
Q10. C) Step 1:
H2(g) + I2(g)
2HI(g)
Initial moles: 0.20 0.15 0
Equilibrium moles: (0.20 - x) (0.15 - x) 0.26
Step 2: The ratio of hydrogen to hydrogen iodide is 1:2 and from iodine to hydrogen iodide is 1:2. This means that both hydrogen and iodine gave half the number of moles of hydrogen iodide that were formed.
The number of moles (x) that each, hydrogen and iodine gave = 0.26 / 2 = 0.13 mol
Moles of hydrogen = 0.20 - 0.13 = 0.07 mol
Moles of iodine = 0.15 - 0.13 = 0.02 mol
Step 3:
Kc = [products] / [reactants]
= (0.26)2 / (0.07 x 0.02)
Q11. A) Ammonium ethanoate is not completely ionised in water and it's not acidic in aqueous solution. Water is more polar than ammonia, so the only choice left is that ammonia is a stronger base than water
Q12. The Boltzmann distribution shows that by increasing temperature, the number of molecules with a certain minimum energy increases (more than the activation energy) *Diagram will be uploaded soon
Q13. C) 6Na2O + P4O10 → 4Na3PO4
Na4PO4 + H2O → Na+ + PO-34
The result is slightly basic i.e. approximately neutral
Q14. Search for the lewis structure for AlCl3
There are only 6 electrons being shared between aluminium and chlorine atoms. According to the octet rule, the valence shell should be filled with 8 electrons, which means that aluminium has an incomplete octet of electrons. This is why when the chlorine atom (from RCl) is added onto the molecule, so as to fulfill the octet rule by forming a dative bond with the aluminium atom, thus leaving the carbon atom as a carbocation.
Q15. D) The thermal decomposition of metal nitrates is as follows:
Metal nitrate [heat] → Metal oxide + Nitrogen dioxide + Oxygen
2Mg(NO3)2 [heat] → 2MgO + 4NO2 + O2
Q16. D) A) AgNO3 + Cl- → AgCl + NO3- and then AgCl + 2NH3 → Ag[(NH3)2]+ + Cl- Oxidation number = -1
B) Cl- + H2SO4 → 2HCl + HSO4- Oxidation number = -1
C) 2NaOH + Cl2 → NaCl + NaClO + H2O Oxidation number = -1 and +1 (disproportionation)
D) 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O Oxidation number = -1 and +5
Q17. C) Ammonia and gaseous water molecules are attracted by hydrogen bonding, so A and D are incorrect.
Both hydrogen chloride and hydrogen iodide are attracted by Van der Waal's forces. the H-Cl bond is more polar than the H-I bond because Cl is more electronegative than I, sohydrogen iodide decomposes more readily into its elements (and form an equilibrium... H2 + I2
2HI)
Q18. C) Ammonium compounds are ionic salts (soluble in water). When heated with a base, they give off ammonia (Ammonium compound + base → salt + water + ammonia). The only base in the multiple choice is limewater.
(NH4)2SO4 + Ca(OH)2 → CaSO4 + 2H2O + 2NH3
Q19. D) The only compound that would end up forming an acid is NO. If not removed catalytically, NO combines with oxygen in the air forming NO2 which then reacts with water forming nitric acid (HNO3)
Q20. D) 7 isomers
*Diagram will be uploaded soon.
============================
Q1. A
Step 1 : Under room conditions 1 mol of oxygen occupies 24dm^3
x mol of oxygen occupies 0.500 dm^3
Amount of oxygen = (0.500*1)/24 = 0.0208 mol.
Step 2 : There are 6.02 * 10^23 molecules in 1 mol of oxygen
There are 'y' molecules in 0.0208 mol of oxygen
Number of oxygen molecules = [0.0208* (6.020* 10^23)] / 1 = 1.25 * 10^22
Q2. B
Step 1 : Count the hydrogen atoms in the side-chains.
Upper side-chain and lower side-chain = 2*(3+6+6+14) = 14
Middle side-chain = 3+6+6+14 = 29
Step 2: Count the hydrogen atoms in each residue
First residue = 3+8+6+14 = 31
Second residue *2 (because two-side chains are converted) = 2*(3+14+2+14) = 66
Step 3 : Calculate how many hydrogen atoms had to be added to each side-chain in order to form the residues.
Upper & lower side-chain residues have 66-58 extra hydrogen = 8 extra hygrogens
Middle side-chain residue has 31-29 extra hydrogens = 2 extra hydrogens
Step 4 : We know that 10 hydrogen atoms have been added to the original molecule. Each molecule contains two hydrogen atoms,so there are 10/2 = 5
hydrogen molecule, = 5 moles of hydrogen.
Q3. D
Number of neutrons = relative atomic mass - atomic number (protons)
Neutrons in an atom of S = 32 - 16 = 16
Neutrons in an atom of P = 31 - 15 = 16
3115P
Q4 . C
Arsenic and Zirconium have a small number of protons compared to Tellerium and Iodine. Their number of protons is not big enough to be consistent with the high 7th ionization energy.
You're left with either Tellerium or Iodine. You can see the ionization energies of iodine in the Data Booklet, and they're not consistent with the ionization energies above, therefore the answer is Tellerium
Q5. D
Magnesium Oxide has a a giant ionic lattice.
Sodium has a metallic structure
Silicon (IV) Oxide has a giant covalent structure
The only chemical with a simple molecular lattice is sulphur
Q6. C
Ideal gases have negligible size, no intermolecular forces, and are always in the gaseous state. When temperature is low and pressure is high, the intermolecular forces and size become relevant, and there are phase changes, which would make the equation less accurate in calculating the relative molecular mass because the equation is made for ideal gases, not real gases... so the answer is a low pressure and high temperature
Q7. D The enthalpy of formation of PCl5 is 330 x 5 = -1650 kJ mol-1 (bond formation is exothermic)
The enthalpy of formation of PCl3 is 330 x 3 = - 990 kJmol-1
The enthalpy of formation of Cl2 is 240 x 1 = 240 kJmol-1
In order to be sure that your calculations are correct, you should draw a Hess's diagram*
ΔHf
ΔHf
Q8. A) In the first reaction:
Oxidation number of Nitrogen before reaction... x + 3 = 0... x = -3
Oxidation number of Nitrogen after reaction... x -2 = 0... x = +2
Q9. C) Aluminium oxide dissolves in cryolite and so the cryolite acts as a flux, reducing the melting point required to extract the aluminium from its ores
Q10. C) Step 1:
H2(g) + I2(g)
Initial moles: 0.20 0.15 0
Equilibrium moles: (0.20 - x) (0.15 - x) 0.26
Step 2: The ratio of hydrogen to hydrogen iodide is 1:2 and from iodine to hydrogen iodide is 1:2. This means that both hydrogen and iodine gave half the number of moles of hydrogen iodide that were formed.
The number of moles (x) that each, hydrogen and iodine gave = 0.26 / 2 = 0.13 mol
Moles of hydrogen = 0.20 - 0.13 = 0.07 mol
Moles of iodine = 0.15 - 0.13 = 0.02 mol
Step 3:
Kc = [products] / [reactants]
= (0.26)2 / (0.07 x 0.02)
Q11. A) Ammonium ethanoate is not completely ionised in water and it's not acidic in aqueous solution. Water is more polar than ammonia, so the only choice left is that ammonia is a stronger base than water
Q12. The Boltzmann distribution shows that by increasing temperature, the number of molecules with a certain minimum energy increases (more than the activation energy) *Diagram will be uploaded soon
Q13. C) 6Na2O + P4O10 → 4Na3PO4
Na4PO4 + H2O → Na+ + PO-34
The result is slightly basic i.e. approximately neutral
Q14. Search for the lewis structure for AlCl3
There are only 6 electrons being shared between aluminium and chlorine atoms. According to the octet rule, the valence shell should be filled with 8 electrons, which means that aluminium has an incomplete octet of electrons. This is why when the chlorine atom (from RCl) is added onto the molecule, so as to fulfill the octet rule by forming a dative bond with the aluminium atom, thus leaving the carbon atom as a carbocation.
Q15. D) The thermal decomposition of metal nitrates is as follows:
Metal nitrate [heat] → Metal oxide + Nitrogen dioxide + Oxygen
2Mg(NO3)2 [heat] → 2MgO + 4NO2 + O2
Q16. D) A) AgNO3 + Cl- → AgCl + NO3- and then AgCl + 2NH3 → Ag[(NH3)2]+ + Cl- Oxidation number = -1
B) Cl- + H2SO4 → 2HCl + HSO4- Oxidation number = -1
C) 2NaOH + Cl2 → NaCl + NaClO + H2O Oxidation number = -1 and +1 (disproportionation)
D) 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O Oxidation number = -1 and +5
Q17. C) Ammonia and gaseous water molecules are attracted by hydrogen bonding, so A and D are incorrect.
Both hydrogen chloride and hydrogen iodide are attracted by Van der Waal's forces. the H-Cl bond is more polar than the H-I bond because Cl is more electronegative than I, sohydrogen iodide decomposes more readily into its elements (and form an equilibrium... H2 + I2
Q18. C) Ammonium compounds are ionic salts (soluble in water). When heated with a base, they give off ammonia (Ammonium compound + base → salt + water + ammonia). The only base in the multiple choice is limewater.
(NH4)2SO4 + Ca(OH)2 → CaSO4 + 2H2O + 2NH3
Q19. D) The only compound that would end up forming an acid is NO. If not removed catalytically, NO combines with oxygen in the air forming NO2 which then reacts with water forming nitric acid (HNO3)
Q20. D) 7 isomers
*Diagram will be uploaded soon.
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Q21. B) Citric acid (from top to bottom): The first carbon is bonded to 2 hydrogens, so it's not chiral. The second carbon is bonded to the top and bottom chains which are the same, so it's not chiral. The third carbon is bonded to 2 hydrogens, so it's not chiral.
Isocitric acid (from top to bottom): The first C is bonded to 2 hydrogens, so it's not chiral. The second carbon is bonded to the top chain, an H, a CO2H, and the bottom chain, so it is chiral. The third carbon is bonded to the upper part of the molecule, an OH, an H, and a CO2H, so it is chiral.
Conclusion: Citric acid has 0 chiral centres and isocitric acid has 2 chiral centres
Q22. D) P is an alkene and a primary alcohol. Alkenes are oxidized by hot concentrated KMnO4 to carboxylic acids with two carboxylic acid groups. Primary alcohols are oxidized to a carboxylic acid. This means that the products of this oxidation are CH3CH2CO2H (from primary alcohol oxidation) and HO2CCH2CO2H (from alkene oxidation)
Q23. B) Write down some equations for the complete combustion of simple alkanes so as to determine how the ratio of carbon atoms to moles of oxygen changes as there are more carbon atoms in the alkane.
CH4 + 2O2 → CO2 + 2H2O
2C2H6 + 7O2 → 4CO2 + 6H2O
C3H8 + 5O2 → 3CO2 + 4H2O
The ratio of carbon atoms to moles of oxygen in each equation respectively is:
1:2, 2:7, 3:5
The number of carbon atoms is directly proportional to the number of moles of oxygen, so the answer has to be B
Q24. B) When an alkene reacts with cold liquid bromine, dibromoalkene is formed. Since the compound above is made up of two alkenes, then four bromine atoms are added on to the molecule. This reaction is an electrophilic addition reaction. Each bromine atom adds on to a carbon engaged in the double bond. The answer is 1,3,4,6 - tetrabromocyclohexane
Q25. D) Reaction D is a nucleophilic addition (CN- is the nucleophile)
Q26. A) When a halogenoalkanes reacts with KCN in ethanol, a nucleophilic substitution reaction takes place, in which the CN- ions substitute the halogen atoms (in this case the bromine atoms from 1,4-dibromobutane)
Q27. C) The weakest bond is the C-Cl bond. The C-H bond is very strong because it is an ionic bond. The C-F bond is strong because fluorine is very electronegative. The C-Cl bond however, is covalent and Cl is not as electronegative as Fluorine, which means that it is a relatively weak bond (relative to the other bonds in the molecule). This is why in the reaction, the most susceptible atom to leave the molecule is chlorine, and so the radical in option "C" is formed
Q28. D) When an alcohol reacts with sodium an alkoxide ion is formed (O-Na+). When a carboxylic acid reacts with sodium, since it's an acid, a salt is formed (COONa). The only compound that forms one mole of hydrogen is the last one.
CH3CH(OH)CO2H + 2Na → CH2CH(ONa)COONa + H2
Q29. C) An alcohol and concentrated sulphuric acid under reflux will produce an alkene which can be purified by dilute sodium hydroxide (base hydrolysis of esters).
Q30. D) When propanone reacts with hydrogen cyanide, nucleophilic addition takes place and a hydroxynitrile is formed.
CH3COCH3 + HCN → CH3C(OH)CH3CN
2-hydroxybutanenitrile is hydrolysed under acidic conditions to Butanoleic acid
CH3C(OH)CH3CN + H2O + H+ → (CH3)2C(OH)CO2H + NH4
Q31. C) Silicon tetrachloride doesn't have co-ordinate bonding because it follows the octet rule, sharing all of its valence electrons with the chlorine atoms.
Both silicon and chlorine are non-metals, so, it has covalent bonding.
There are instantaneous dipole-induced dipole forces between the molecules (Van der Waals forces)
Q32. A) The right-hand side of the structure is polar and since water is a dipole, it is attracted to water.
The alkyl chain is non-polar and attracted to other alkyl chains by Van der Waals forces. Since oil is of a similar character to this alkyl chain, the alkyl chain is soluble in oil droplets.
In alkanes, each carbon atom forms a tetrahedral structure (due to sp3 hybridisation), so the C-C-C bond angles are tetrahedral
Q33. A) The reaction is endothermic, which means that diamond has more energy than graphite. The enthalpy change of atomisation is the enthalpy change when one mole of gaseous atoms is formed from one mole of the element in the standard state. Since diamond has more energy than graphite, it requires a smaller enthalpy change of atomisation.
Since the enthalpy change of atomisation is smaller in diamonds, it means that the C-C bonds in diamond are weaker than in graphite because it requires less energy to change into the gaseous state.
Since diamond has more energy than graphite, there is a higher energy requirement to break the C-C bond to form new C=O bonds (in carbon dioxide) in combustion.
Q34. C) The electronegativity difference decreases between the elements (3.05, 2.13, 1.43, 0.65)
All of the compounds fulfill the octet rule and are isolelectronic.
The compounds become increasingly covalent (starting from ionic)
Q35. B) Sulphur dioxide is a reducing agent which prevents oxidation.
Since it's an anti-oxidant, it prevents alcohols from oxidizing to carboxylic acids (prevents sour-tasting acids).
It does smell and is toxic in large quantities.
Q36. C) Iodide ions are strong reducing agents and so they reduce the sulphuric acid, first to sulphur dioxide, then to sulphur and then hydrogen sulphide. Barely any hydrogen iodide is formed because it's displaced by the sulphuric acid.
Iodide ions are reducing agents, and become oxidised to iodine.
The majority of the products of the reaction are sulphur compounds (as explained above)
Q37. B) A chiral centre is an atom bonded to four different groups.
An optical isomer (geometric isomer) occurs when there's a chiral centre.
Chiral carbon atoms DO NOT need to have structural isomers.
Q38. B) Step X is a nucleophilic substitution because the reagent is hot aqueous sodium hydroxide (OH- being the nucleophile).
A chloroalkane cannot be formed by reacting sodium chloride with alcohol. It can be done with phosphorus (III) chloride or phosphorus (V) chloride.
Q39. C) Only an aldehyde forms a brick-red precipitate with Fehling's solution. Aldehydes are formed by the oxidation (in acidified dichromate) of primary alcohols. The two primary alcohols are CH3CH2CH2OH and CH3OH
Q40. B) It only has one chiral carbon (The one in the middle).
It has a carboxylic acid group, so it can be esterified by ethanol. It has an OH group, so it can be esterified by ethanoic acid.
The molecule contains tertiary and primary alcohols, not secondary.
P.S :- Others please post work solutions for all P1 yearly like this..
Isocitric acid (from top to bottom): The first C is bonded to 2 hydrogens, so it's not chiral. The second carbon is bonded to the top chain, an H, a CO2H, and the bottom chain, so it is chiral. The third carbon is bonded to the upper part of the molecule, an OH, an H, and a CO2H, so it is chiral.
Conclusion: Citric acid has 0 chiral centres and isocitric acid has 2 chiral centres
Q22. D) P is an alkene and a primary alcohol. Alkenes are oxidized by hot concentrated KMnO4 to carboxylic acids with two carboxylic acid groups. Primary alcohols are oxidized to a carboxylic acid. This means that the products of this oxidation are CH3CH2CO2H (from primary alcohol oxidation) and HO2CCH2CO2H (from alkene oxidation)
Q23. B) Write down some equations for the complete combustion of simple alkanes so as to determine how the ratio of carbon atoms to moles of oxygen changes as there are more carbon atoms in the alkane.
CH4 + 2O2 → CO2 + 2H2O
2C2H6 + 7O2 → 4CO2 + 6H2O
C3H8 + 5O2 → 3CO2 + 4H2O
The ratio of carbon atoms to moles of oxygen in each equation respectively is:
1:2, 2:7, 3:5
The number of carbon atoms is directly proportional to the number of moles of oxygen, so the answer has to be B
Q24. B) When an alkene reacts with cold liquid bromine, dibromoalkene is formed. Since the compound above is made up of two alkenes, then four bromine atoms are added on to the molecule. This reaction is an electrophilic addition reaction. Each bromine atom adds on to a carbon engaged in the double bond. The answer is 1,3,4,6 - tetrabromocyclohexane
Q25. D) Reaction D is a nucleophilic addition (CN- is the nucleophile)
Q26. A) When a halogenoalkanes reacts with KCN in ethanol, a nucleophilic substitution reaction takes place, in which the CN- ions substitute the halogen atoms (in this case the bromine atoms from 1,4-dibromobutane)
Q27. C) The weakest bond is the C-Cl bond. The C-H bond is very strong because it is an ionic bond. The C-F bond is strong because fluorine is very electronegative. The C-Cl bond however, is covalent and Cl is not as electronegative as Fluorine, which means that it is a relatively weak bond (relative to the other bonds in the molecule). This is why in the reaction, the most susceptible atom to leave the molecule is chlorine, and so the radical in option "C" is formed
Q28. D) When an alcohol reacts with sodium an alkoxide ion is formed (O-Na+). When a carboxylic acid reacts with sodium, since it's an acid, a salt is formed (COONa). The only compound that forms one mole of hydrogen is the last one.
CH3CH(OH)CO2H + 2Na → CH2CH(ONa)COONa + H2
Q29. C) An alcohol and concentrated sulphuric acid under reflux will produce an alkene which can be purified by dilute sodium hydroxide (base hydrolysis of esters).
Q30. D) When propanone reacts with hydrogen cyanide, nucleophilic addition takes place and a hydroxynitrile is formed.
CH3COCH3 + HCN → CH3C(OH)CH3CN
2-hydroxybutanenitrile is hydrolysed under acidic conditions to Butanoleic acid
CH3C(OH)CH3CN + H2O + H+ → (CH3)2C(OH)CO2H + NH4
Q31. C) Silicon tetrachloride doesn't have co-ordinate bonding because it follows the octet rule, sharing all of its valence electrons with the chlorine atoms.
Both silicon and chlorine are non-metals, so, it has covalent bonding.
There are instantaneous dipole-induced dipole forces between the molecules (Van der Waals forces)
Q32. A) The right-hand side of the structure is polar and since water is a dipole, it is attracted to water.
The alkyl chain is non-polar and attracted to other alkyl chains by Van der Waals forces. Since oil is of a similar character to this alkyl chain, the alkyl chain is soluble in oil droplets.
In alkanes, each carbon atom forms a tetrahedral structure (due to sp3 hybridisation), so the C-C-C bond angles are tetrahedral
Q33. A) The reaction is endothermic, which means that diamond has more energy than graphite. The enthalpy change of atomisation is the enthalpy change when one mole of gaseous atoms is formed from one mole of the element in the standard state. Since diamond has more energy than graphite, it requires a smaller enthalpy change of atomisation.
Since the enthalpy change of atomisation is smaller in diamonds, it means that the C-C bonds in diamond are weaker than in graphite because it requires less energy to change into the gaseous state.
Since diamond has more energy than graphite, there is a higher energy requirement to break the C-C bond to form new C=O bonds (in carbon dioxide) in combustion.
Q34. C) The electronegativity difference decreases between the elements (3.05, 2.13, 1.43, 0.65)
All of the compounds fulfill the octet rule and are isolelectronic.
The compounds become increasingly covalent (starting from ionic)
Q35. B) Sulphur dioxide is a reducing agent which prevents oxidation.
Since it's an anti-oxidant, it prevents alcohols from oxidizing to carboxylic acids (prevents sour-tasting acids).
It does smell and is toxic in large quantities.
Q36. C) Iodide ions are strong reducing agents and so they reduce the sulphuric acid, first to sulphur dioxide, then to sulphur and then hydrogen sulphide. Barely any hydrogen iodide is formed because it's displaced by the sulphuric acid.
Iodide ions are reducing agents, and become oxidised to iodine.
The majority of the products of the reaction are sulphur compounds (as explained above)
Q37. B) A chiral centre is an atom bonded to four different groups.
An optical isomer (geometric isomer) occurs when there's a chiral centre.
Chiral carbon atoms DO NOT need to have structural isomers.
Q38. B) Step X is a nucleophilic substitution because the reagent is hot aqueous sodium hydroxide (OH- being the nucleophile).
A chloroalkane cannot be formed by reacting sodium chloride with alcohol. It can be done with phosphorus (III) chloride or phosphorus (V) chloride.
Q39. C) Only an aldehyde forms a brick-red precipitate with Fehling's solution. Aldehydes are formed by the oxidation (in acidified dichromate) of primary alcohols. The two primary alcohols are CH3CH2CH2OH and CH3OH
Q40. B) It only has one chiral carbon (The one in the middle).
It has a carboxylic acid group, so it can be esterified by ethanol. It has an OH group, so it can be esterified by ethanoic acid.
The molecule contains tertiary and primary alcohols, not secondary.
P.S :- Others please post work solutions for all P1 yearly like this..
- Messages
- 97
- Reaction score
- 23
- Points
- 8
Hey, guys! Good effort here! hope this can carry on and benefit those taking in november 2012...
I thought this thread was about P1? :0
I have posted my question regarding P1 in another thread, but I'll post it here too. this question bothers me a lot, I don't understand!
Cyanohydrins can be made from carbonyl compounds by generating CN–ions from HCN in the presence of a weak base.
In a similar reaction, –CH2CO2CH3 ions are generated from CH3CO2CH3 by strong bases. Which compound can be made from an aldehyde and CH3CO2CH3 in the presence of a strong base?
A CH3CH(OH)CO2CH3
B CH3CO2CH2CH(OH)CH3
C CH3CH2CH(OH)CH2CO2CH3
D (CH3)2C(OH)CH2CO2CH3
I understand how A is impossible, but what about B, C, and D? they only differ in the arrangement and number of carbon atoms, while the question does not specify what is the aldehyde.
an explanation is greatly appreciated.
if you wonder from which year is it, it is from ON2009.
I have posted my question regarding P1 in another thread, but I'll post it here too. this question bothers me a lot, I don't understand!
Cyanohydrins can be made from carbonyl compounds by generating CN–ions from HCN in the presence of a weak base.
In a similar reaction, –CH2CO2CH3 ions are generated from CH3CO2CH3 by strong bases. Which compound can be made from an aldehyde and CH3CO2CH3 in the presence of a strong base?
A CH3CH(OH)CO2CH3
B CH3CO2CH2CH(OH)CH3
C CH3CH2CH(OH)CH2CO2CH3
D (CH3)2C(OH)CH2CO2CH3
I understand how A is impossible, but what about B, C, and D? they only differ in the arrangement and number of carbon atoms, while the question does not specify what is the aldehyde.
an explanation is greatly appreciated.
if you wonder from which year is it, it is from ON2009.
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Answer is C. Since, –CH2CO2CH3 ions are produced in the presence of strong base, therefore A is not possible. And since a aldehyde should react with it, B and D are not possible since both are ketones and not aldehydes.
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Thanks @bamtek for those May June 2003 solutions ! I just did that past paper yesterday and scored 33/40 in the MCQs ! Had the Cis Trans Isomers Question wrong, hope you will upload the picture soon ! And i also hope that some other helpful-fella will upload similar answers to the next MCQ on the list ! I wish it goes serially as there in the xtremepapers so next would be May June 2004 MCQ ~
Anyways may i get the threshold for this may june 2003 mcq please !
Anyways may i get the threshold for this may june 2003 mcq please !
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could anyone help me with this ! Actually, i was goin on serially with the Chem MCQ as in xtremepapers ! Did MJ2003 ! This is from MJ2004 !
Q. 26, 27 , 33 , 35 ..
These are the ones i made mistake !
Q. 26, 27 , 33 , 35 ..
These are the ones i made mistake !
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bamteck I was supposed to solve out some of the MCQ's but due to my school's semester exam I was not able to do that. I am really willing to help but my Oct/Nov exams are knocking at my doors and I have lot more to practice and I'm not appearing for the Chemistry paper this season. So I'm really sorry for that. Hope you understood my situation.
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Please Please Please someone give me the link for May June 2003 and May june 2004 Paper 2 Thresholds ! PLEASE ASAP !
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Q.26: You have to consider that there's an intermediate so there's two humps. The front hump is always higher than the second hump. I'm not sure why but it's just a basic rule, I think. XD
Q.27: Hydrolysis means substituting Br with OH group. So if all the Br are substituted, they would give you the same diol, wouldn't they? It's a nucleophilic substition reaction. And only the second organic compound can form H bond due to presence of OH group.
Q.33: CAtalyst does not increase KE of molecules. They lower the activation e so that molecules with lower KE can react. Catalyst also increases the rate of reaction whether it's backward or forward. It doesn't affect enthalpy change, same amount of energy released or used by reaction whether catalyst is there or not. It just speeds up the reaction.
Q.35: For the first: CaO + SO2 = CaSO3 thats correct
second: SO2+O2 = 2SO3 due to excess air then CaO + SO3 = CaSO4
third: CaO + CO = CaCO3 (lazy to balance) this is not a likely reaction as CO is neutral (but CO2 is acidic!) My lecturer helped with this question