In this section, we shall look at a few past year MCQ questions from top JCs.
In the A levels Chemistry, the MCQ paper (paper 1) is a 40question paper, in which students are given only 60 minutes to work out the solutions. That works out to 1.5 min / question.
A common problem amongst A level students is trying to complete the paper on time.
Hence, the challenge in paper 1 is not only to solve the question correctly, but to do so in a very quick manner.
Here we present 2 ways to do this question:
RI 2006, Paper 1:
Methane was burned in an incorrectly adjusted burner. The methane was converted into a mixture of carbon dioxide and carbon monoxide in the ratio of 98:2, together with water vapour.
What will be the volume of oxygen consumed when y dm^{3} of methane is burned?

B. (2y0.01y)dm^{3}  

D. (y0.01y)dm^{3} 
Solutions:
There are 2 ways to solve this question. One of the methods is presented below:
Method 1:
Balance the overall equation for the combustion of CH_{4} to produce CO_{2} and CO in 98:2 ratio:
100CH_{4} + 199O_{2} → 98CO_{2} + 2CO + 200H_{2}O
Reacted: y dm^{3}
Volume of O_{2} reacted = 

x199 = 1.99 y dm^{3} 
Answer is B (2y – 0.01y = 1.99 y)
Method 2:
Break up the overall equation into 2 equations:
CH_{4} + O_{2} → CO_{2} + H_{2}O ………(1)
CH_{4} + O_{2} → CO + H_{2}O ………(2)
Since CO_{2}:CO = 98:2,
0.98y dm^{3} of CH_{4} undergoes reaction (1) and 0.02 y dm^{3} CH_{4} undergoes reaction (2)
The total volume of O_{2} needed can then be calculated and the answer is still B.
RI, 2004, JC1BT1
A 10 cm^{3} sample of a gaseous hydrocarbon measured at room temperature and pressure (r.t.p) was reacted with 70 cm^{3} of oxygen.
The resulting gas mixture collected was found to have a volume of 60 cm^{3} (also measured at r.t.p). The gas mixture was completely absorbed by aqueous potassium hydroxide.
Calculate the empirical formula of the hydrocarbon
A. C_{2}H_{3}  B. C_{3}H_{2}  C. C_{4}H_{6}  D. C_{6}H_{4} 
Solutions
Let the formula of the unknown hydrocarbon be C_{x}H_{y}
The gas absorbed by potassium hydroxide is CO_{2}.
Equation:
C_{x}H_{y} + ( 

)O_{2} → xCO_{2} + 

H_{2}O 
Or:
C_{x}H_{y} + ( 

+ x)O_{2} → xCO_{2} + 

H_{2}O 
Reacted (cm^{3}):  10  70  60 
Mole ratio of  CH_{4}  :  CO_{2}  
1  :  X  
10  :  60 
X = 6
Mole ratio of  CH_{4}  :  CO_{2}  
1  : 


10  :  70 
⇒ 

+ X = 7 
⇒ y = 4(76) = 4
Thus the empirical formula of the hydrocarbon is C_{6}H_{4}. Answer is D.
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