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

[Rabb94]

on 10 paper 11
9,10,12,35

 

[1597.2217]

Q#2, 29, 30 & 38
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf

Q2-
I have explained this yesterday, will find the post and copy here.

Q29-
Alcohol + Conc. H2SO4 --> Alkene
It must be either (B) or (C).
Dilute NaOH would react with any present Alcohol to purify the gas so it is (C).

Q30-
It cannot be (A) because another C atom is added.
Hydrolysis of Ketone gives Alcohol. So it should be (D).

Q38-
1 Step X involves a nucleophilic substitution. [True because Br- is replaced with OH-]
2 Hot aqueous sodium hydroxide is the reagent in step X [True again.]
3 Hot aqueous sodium chloride is the reagent in step Y. [Not true. Alcohol + Acid gives us haloganoalkane. HCl in this case]

So answer should be (B).

 

[1597.2217]

on 10 paper 11
9,10,12,35

Q9-
E=mcT [Where T is difference in the temperature]
E = 200 x 4.2 x (66-18)
E = 40320

It is mentioned that on 45% of energy is gained by water. Therefore, 1.6g x 0.45 = 0.72g.
To find the energy in 1g of fuel:

0.72g ---> 40320
1 ---> x

x = 40320/0.62 = 56,000

Answer is (B)

Q10-
I'm sorry, this question came about 2 or 3 times earlier today but I cannot solve it. I'm very weak in Kc questions.

Q12-
It is mentioned that mass of Hydrogen and Oxygen is same. 1 Mole of O is 16g and 1 Mole of H is 1g. Therefore 1 O atoms should be balanced by 16 H atoms. Answer is (C).

Q35-
1- Cl has initial charge of 0. After the reaction, 1 Cl atom has charge -1 and second atom has charge +1 (you can complete the equation ClO- will be formed]. True.
2- Cl has initial charge of 0. After the reaction, 1 Cl atom has charge of +7 and second atom has charge of -1 (you can complete the equation, 2 Cl2 is formed and 1 Cl-). True.
3- N has initial charge of 4. After the reaction, 1 N atom has charge of +5 and second N atom has charge of +3 (HNO2 is formed).

All are true, so answer is A.

 

[1597.2217]

Q#2, 29, 30 & 38
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf

Here you go, solution for Q2.

First residue: CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2
Second residue: CH3(CH2)7CH=CH(CH2)7CO2
Original branch: CH3(CH2)3CH=CHCH=CHCH=CH(CH2)7CO2

First, I omitted out the unchanged part of molecules resulting in:

First residue: (CH2)4...CHCH2CH...
Second residue: (CH2)7...

For first residue: [Looking at the breaking of double bonds]
One CH2 molecule broke up and moved to change (CH2)3 to (CH2)4. Hence this part is justified.


=CHCH=CHCH= (Break the middle double bond by adding 1 hydrogen atom on each carbon). [Total 2 Hydrogen atoms.]
This results in: =CHCH2CH2CH=
Take one CH2 away since it was already added to (CH2)3
Result: =CHCH2CH=

For second residue: [The end part remains same, so we look at breaking the double bonds again.]
-CH=CHCH=CHCH= (Break the first two double bonds by adding 2 Hydrogen atoms on each carbon.) [Total 4 Hydrogen atoms].
-CH2CH2CH2CH2CH= (Take three CH2 molecules to original (CH2)3 making it (CH2)7 again justifying this part.)

Therefore the first residue required 1 mole of Hydrogen and second required 2 moles. However, the second residue is formed with 2 branches hence it multiply the moles by 2, giving us 1 Mole + 2 Moles + 2 Moles = 5.

5 Moles of Hydrogen is required, which is (B).

 

[HubbaBubba]

Guys I'm a bit worried about Chem MCQ :/ When I solve past papers, sometimes I get around 30-35 which isn't bad, but in the next paper I'd get something as low as 24! :/ I revise constantly, but this still happens, does anybody have any tips?

 

[Lyfroker]

Here you go, solution for Q2.

First residue: CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2
Second residue: CH3(CH2)7CH=CH(CH2)7CO2
Original branch: CH3(CH2)3CH=CHCH=CHCH=CH(CH2)7CO2

First, I omitted out the unchanged part of molecules resulting in:

First residue: (CH2)4...CHCH2CH...
Second residue: (CH2)7...

For first residue: [Looking at the breaking of double bonds]
One CH2 molecule broke up and moved to change (CH2)3 to (CH2)4. Hence this part is justified.


=CHCH=CHCH= (Break the middle double bond by adding 1 hydrogen atom on each carbon). [Total 2 Hydrogen atoms.]
This results in: =CHCH2CH2CH=
Take one CH2 away since it was already added to (CH2)3
Result: =CHCH2CH=

For second residue: [The end part remains same, so we look at breaking the double bonds again.]
-CH=CHCH=CHCH= (Break the first two double bonds by adding 2 Hydrogen atoms on each carbon.) [Total 4 Hydrogen atoms].
-CH2CH2CH2CH2CH= (Take three CH2 molecules to original (CH2)3 making it (CH2)7 again justifying this part.)

Therefore the first residue required 1 mole of Hydrogen and second required 2 moles. However, the second residue is formed with 2 branches hence it multiply the moles by 2, giving us 1 Mole + 2 Moles + 2 Moles = 5.

5 Moles of Hydrogen is required, which is (B).

Thnx a lot

 

[Obsidian Fl1ght]

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.

Thanks loads!

 

[1597.2217]

Guys I'm a bit worried about Chem MCQ :/ When I solve past papers, sometimes I get around 30-35 which isn't bad, but in the next paper I'd get something as low as 24! :/ I revise constantly, but this still happens, does anybody have any tips?

30-35 is pretty good being in AS. Last time I got 14. If you read the examiner's report, the average usually lies around 22-26.

 

[Obsidian Fl1ght]

A question:
(I'll just summarize it

3 equations:

Cl2 + 2H20 + SO2 ---> 2HCl + H2SO4
Cl2 + H2S ----> 2HCl
SO2 + 2H2S ---> 2H2O + 3S

What's the correct order of strength of the 3 reacting gases as REDUCING agents?

Strongest Weakest
A Cl2 H2S SO2
B Chlorine SO2 H2S
C Hydrogen Sulfide SO2 Cl2
D Sulfur dioxide H2S Cl2


The answer is C.
Anyone who can explain why??
Thank u in advance!

 

[Lyfroker]

30-35 is pretty good being in AS. Last time I got 14. If you read the examiner's report, the average usually lies around 22-26.

But the minimum marks to get A is always around 35 :/

 

[1597.2217]

But the minimum marks to get A is always around 35 :/

Well try hard and pray to Allah that we get A, In sha Allah

 

[samiaaaaaaaaaaaaahhh]

can someone please explain m/june 2012 v12 q5 ,q14, q17, q18, q22, q29, q30
please asao

 

[Obsidian Fl1ght]

can someone please explain m/june 2012 v12 q5 ,q14, q17, q18, q22, q29, q30
please asao

q5: Iodine
Solid iodine (stable state) exists as lattice of simple molecules.
That is:
I2 molecules with Vander Waal's forces in between.
So: CVB WITHIN I2 molecules and Vander Waal's forces BETWEEN the molecules

 

[Obsidian Fl1ght]

can someone please explain m/june 2012 v12 q5 ,q14, q17, q18, q22, q29, q30
please asao

q14:
Thermal decomposition is ENDOTHERMIC.
Hydration (H20 is added see?), is EXOTHERMIC.

q17:
Generally, Grp (II) reaction with water increases down the group.

Mg reacts SLOWLY with cold water, so yeah - bubbles of H2 will form slowly.
But it reacts vigorously with steam.
Just go over the Grp (II) reactions once.

q18:
Atomic radii DECREASES across the period. (Nuclear charge incr. but shielding effect and dist. between valence shell nd nucleus remains roughly constant etc)
So Na will have gr8er radius than S.

BUT:
In ions:
Cations (Na+) are smaller than THEIR corresponding atoms.
So Na+ is smaller than Na.
Anions (S2-) are larger than THEIR corresponding atoms.
So S2- is bigger than S.
And anions in general are larger than cations. (Same period).
So:
Na > S
Na+ < S2-

Btw, just refer to the Data Booklet if u get confused.

 

[Obsidian Fl1ght]

can someone please explain m/june 2012 v12 q5 ,q14, q17, q18, q22, q29, q30
please asao

q22

It's either A or C.
Look at C first.
If you add ethanolic KCN to CH2BrCH2CH3, you'll get: CH3CH2CH2CN...
Now, adding HCl will hydrolyse CN to COOH (carobxylic grp).
So you get an acid.
But do you get Propanoic Acid?

No.
U don't.
Remmbr that whenever we add do the above, we add an extra CARBON to the chain.
SO the acid formed will be - NOT propanoic acid. Ie CH3CH2CH2CO2H (C3H5CO2H = Butanoic Acid)

So we cross out C.

A: With first reagant u get alcohol, CH3CH2CHOH (Propanol)
U oxidise it. (It's a Primary alcohol so u get acid. (With reflux - but we'll assume they refluxed it.)
U get - Propanoic Acid.
CH3CH2CO2H

So A's the winner.

 

[Obsidian Fl1ght]

can someone please explain m/june 2012 v12 q5 ,q14, q17, q18, q22, q29, q30
please asao

q29:
Let's see what we have.
An ester and a ketone.
Let's quickly look at the obviously NOT right choices:

A
HCl - HCl will hydrolyse (though partially) the ester.

D NaBH4 is a reducing agent which will reduce the ketone.

B and C:
B - KCN: Consider what KCN gives - CN- ions. These are nucleophiles. Nucleophiles take part in nucleophilic reactions.
Ester doesn't take part in any such reaction.
But ketones and aldehydes do.
Nuclophilic Addition reactions. So B is not possible.

That leaves us with C.

Alternatively, u know that C (Na) reacts with acids or .... alcohols.
We do not see any carboxylic grps or -OH grps or acids in the ketone and ester.
So it'll be C.

Hope this helps.

 

[Obsidian Fl1ght]

can someone please explain m/june 2012 v12 q5 ,q14, q17, q18, q22, q29, q30
please asao

q30:

Since the cmpd will NOT decolorise bromine water, it means it doesn't have a double CVB. It cannot be an alkene.
Eliminate C - Ethene.

It decolourises dil. KMNO4, which means it can be oxidised by KMNO4.
We know that all alcohols (except tertiary) are oxidised by KMNO4 and K2Cr2O7.
So the answer is B.

(Butane and ethanoic acid are not oxidised by KMNO4)

 

[queen of the legend]

A sample of ethyl propanoate is hydrolysed by heating under reflux with aqueous sodium
hydroxide. The two organic products of the hydrolysis are separated, purified and weighed.

M/J 2012 varinat 11 question 22 plss

 

[Obsidian Fl1ght]

A sample of ethyl propanoate is hydrolysed by heating under reflux with aqueous sodium
hydroxide. The two organic products of the hydrolysis are separated, purified and weighed.

M/J 2012 varinat 11 question 22 plss

You'll get Alcohol and Salt with this hydrolysis.​

​

Ethanol and Sodium Propanoate if I'm not mistaken.​

C2H5OH and (CH3CH2CH2CO2)-(Na)+ ​

Mr (C2H5OH): 24+6+16 = 46​

Mr (CH3CH2CH2CO2)-(Na)+) = 48+7+23 + 32 = 110​

Total mass: 46+66= 146​

​

% mass alchohol: 46/146 * 100 = 31.5%​

% mass salt: 110/146* 100 = 75.3%​

​

Well, my answers roughly match with A. So there u go.​

Hope I was of some help.​

​

 

[Obsidian Fl1ght]

s10_qp11:
Q22:

What will react differently with the two isomeric alcohols, (CH3)
3
CCH2
OH and
(CH3)
2
CHCH2
CH2
OH?
A acidified aqueous potassium manganate(VII)
B concentrated sulfuric acid
C phosphorus pentachloride
D sodium

Answer's B. WHY?