P1 MCQ's preparation thread for chemistry ONLY!!!!

[mehranshaikh]

Can someone explain this to me? I have done all the calculation but answer is coming different.
Summer 2011, Variant 13, Q12.

My calculation
Al2O3 + 6HCl = 2AlCl3 + 3H20

As 1 mol of Al2O3 reacts with 6 mol of HCl, 0.02 mol react with 0.12 mol.
0.12/2= 0.06 dm3 of HCl 60 cm3. Whats wrong with this?

 

[Xeshan16]

12 M V12 28
12 M V11 Q.18
11 M V12 Q.29
10 M V12 Q.14, 25
09 M q.19, 20
08 M Q. 27, 30, 34
07 M Q. 34, 40
06 M Q.38
05 M Q.13, 32, 33
04 M Q.25

12 N V13 Q.27
10 N V11 Q.7, 34
09 N V12 Q.10, 12 ,27, 40
09 N V11 Q.13, 21, 31
08 N Q.7, 33
06 N Q.19
04 N Q.12. 23. 39

please help out. snowbrood

 

[itallion stallion]

Can someone explain this to me? I have done all the calculation but answer is coming different.
Summer 2011, Variant 13, Q12.

My calculation
Al2O3 + 6HCl = 2AlCl3 + 3H20

As 1 mol of Al2O3 reacts with 6 mol of HCl, 0.02 mol react with 0.12 mol.
0.12/2= 0.06 dm3 of HCl 60 cm3. Whats wrong with this?

View attachment 28684

From the start:
4Al+3O2 will give 2Al2O3
so the ratio of aluminium to aluminium oxide is 2:1.so Al2O3 moles will be 0.01moles.so the ratio of aluminium oxide to HCl is 1:6,so HCL moles will be 0.06moles.divide these moles by 2.00 and u will get the answer.hope u got it.

 

[itallion stallion]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_11.pdf plz help with q11.
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf q25(why not b)

 

[Rahma Abdelrahman]

Q18: It says that when it was heated without any air so no compound was added or removed from the Ammonium Cynate. Only the addition of CNO will balance both sides Ammonium will be Nh4. The compound was NH4CNO
Q20: I have no idea about that.
Q22: Hydration and oxidation all occur at the double bond but after polymerisation the double bond will be gone so they can not occur. Hydrolysis with Chloro ethene is neucholophillic substitution so it occurs ar Chlorine not double bond and can happen in Polymer.
Q40: In Polymer atleast 2 molecules join together to form them. 1 mol of monomer producing 1 mol of polymer is impossible as monomer has to join, so it will reduce. Any other number is possible so the last 2 are correct. Answer C.

Thanks ... but didn't get first one
Aries_95 Thanks. .. Any idea about Q 20? And in June 2010 P11 Q35? Please :/

 

[magnesium]

http://freeexampapers.com/#A Level/Physics/CIE/2002 Nov Q36..........

 

[Rahma Abdelrahman]

http://freeexampapers.com/#A Level/Physics/CIE/2002 Nov Q36..........

This is a chemistry thread.. plz post this in the physics thread!

 

[Obsidian Fl1ght]

B and C can be easily eliminated. The answer is D or A? And can you please tell me the year and number of this question?

My apologies. My "elimination" I wrote incorrectly up there.
Yeah it IS either A or D.
Answer is D - but why? That is the main question.
Yeah it's from the "early 90's".
J93 to be precise.
Thanx.

 

[snowbrood]

Can someone explain this to me? I have done all the calculation but answer is coming different.
Summer 2011, Variant 13, Q12.

My calculation
Al2O3 + 6HCl = 2AlCl3 + 3H20

As 1 mol of Al2O3 reacts with 6 mol of HCl, 0.02 mol react with 0.12 mol.
0.12/2= 0.06 dm3 of HCl 60 cm3. Whats wrong with this?

View attachment 28684

are u from bss hyderabad

 

[taimoor.08]

question no21 may june 2002
question no 2 may june 2002
question no 8 may june 20025
question no 3 ,8,9,10 oct nov 2002
question no 3,20,28 oct nov 2003
may june 2003 question no 8,20,32
question no 40 oct nov 2003
question no 5 may june 2004
quesyion no 5,8,18,28,35,39 may june 2004
oct nov 2005 queston no 2, 5,23,28,31
oct nov 2004 question 20
may june 2005 quetion 11,18
mayjune 06 q10,30
oct/nov questions 06 4,9,11,21
may june 07 questions 5,26,34,40
oct nov 07 questons 33
winter 08 2,8,30
winter 09 21,28,31
summer 10 qp 11 question no 4, 27
winter 10 questions 7 ,8

 

[mehranshaikh]

My apologies. My "elimination" I wrote incorrectly up there.
Yeah it IS either A or D.
Answer is D - but why? That is the main question.
Yeah it's from the "early 90's".
J93 to be precise.
Thanx.

Atomic crystal are metals and those have very high boiling points. This has boiling point of just over 100 so it has to be Molecular crystal.

 

[!>@/^~^\@<!]

From the start:
4Al+3O2 will give 2Al2O3
so the ratio of aluminium to aluminium oxide is 2:1.so Al2O3 moles will be 0.01moles.so the ratio of aluminium oxide to HCl is 1:6,so HCL moles will be 0.06moles.divide these moles by 2.00 and u will get the answer.hope u got it.

Can u plz help me with these MCQ's of May/June 2002
2, 4, 5, 6, 7, 8 and 23.
I'll b really greatful.

 

[!>@/^~^\@<!]

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..

Can u plz help me with these MCQ's of May/June 2002
2, 4, 5, 6, 7, 8 and 23.
I'll b really greatful.

 

[Mairaxo]

b

Bro will acidic KMnO4 react with alkene?

u mean sis nope it does not

 

[Lyfroker]

q# 3, 9, 10, 23,28 & 32
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w04_qp_1.pdf

 

[Manobilly]

May June. 2005 q1

 

[Lyfroker]

q#3, 9, 18, 37 & 39
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

 

[simply adorable]

u mean sis nope it does not

It does react its the oxidation of alkene by KMno4 !

 

[1597.2217]

q#3, 9, 18, 37 & 39
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Q3:

n=50/2x(23+3x14) [2x because there are 2 moles NaN3.]
n= 5/13

3n=V/24 [3x because there are 3 moles of N2.]
3x5/13 x 24 =V
27.69 = V

The answer is (C) which is 27.7


Q9:

One molecule of HCl dissipates one H+ ion. Now looking at the options:
(A) Ethanoic Acid [It is weak acid and cannot dissipate the H+ ion completely.]
(B) Nitric Acid [The formula is HN)3. Each molecule can dissipate one H+ ion, same as HCl.]
(C) Sodium Hydroxide [It doesn't even dissipate H+ ion!!.]
(D) Sulphuric Acid [H2SO4 dissipates 2 H+ ions per molecule, twice that of HCl.]

So the answer is (B),


Q18: [I couldn't do it. Will try again soon.]

Q37:

1 All bond angles are approximately 120o. [This is true. I hope you know why.]
2 It will undergo electrophilic addition reactions. [Yes they undergo electrophlic addition because of presence of double bonds.]
3 It will undergo nucleophilic addition reactions. [Yes they undergo nucleophlic addition because of presence of lone pairs on Oxygen.]

So answer is (A).

Q39:

First of all, alkanes are not soluble. Arene rings are alkanes which aren't soluble. This gives direct answer of option 1 being correct. Option 1 is a Aldehyde which are oxidized to Carboxylic acids, which are quite soluble! So answer is (D).
Note: Option 2 is also an aldehyde and form a carboxylic acid but with poor solubility due to presence of arene ring. Option 3 is a Ketone and cannot be easily oxidized further.

(If I'm wrong somewhere, please correct me. I got my answers this way which are all correct.)

 

[1597.2217]

May June. 2005 q1

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s05_qp_1.pdf

We can form a equation with given information.
Let the hydrocarbon be X. We know hydrocarbons burn to give CO2 and H2O ONLY.
It is also given the Oxygen is in excess, but we can find the volume of Oxygen used.

So here we go. Volume of Oxygen used = Initial oxygen volume - Final oxygen volume
Which is 70 - 20 = 50cm^3

Using this information, we can get:

X + O2 ---> CO2 + H2O

Now we have to balance it. We have volume of X, O2 and CO2.

10X + 50 O2 ---> 30 CO2 + __ H2O [To make calculations easy, divide the number of moles by 10.]

X + 5O2 ---> 3 CO2 + __H2O [We can find the moles of H2O by balancing the Oxygen atoms on both sides. There are 10 Oxygen atoms in the reactants, and 6 in CO2. Therefore 10-6 = 4. There should be 4 oxygen atoms in H20.]

X + 5O2 ---> 3 CO2 + 5 H2O [Now calculate the number of carbon and hydrogen atoms on the product side.]

This gives us 3 C atoms and 8 H atoms. Looking at the options provided, the answer is (C) C3H8.

Good luck!