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 40-question 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 dm3 of methane is burned?
|
B. (2y-0.01y)dm3 | |||||
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D. (y-0.01y)dm3 |
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 CH4 to produce CO2 and CO in 98:2 ratio:
100CH4 + 199O2 → 98CO2 + 2CO + 200H2O
Reacted: y dm3
Volume of O2 reacted = |
|
x199 = 1.99 y dm3 |
Answer is B (2y – 0.01y = 1.99 y)
Method 2:
Break up the overall equation into 2 equations:
CH4 + O2 → CO2 + H2O ………(1)
CH4 + O2 → CO + H2O ………(2)
Since CO2:CO = 98:2,
0.98y dm3 of CH4 undergoes reaction (1) and 0.02 y dm3 CH4 undergoes reaction (2)
The total volume of O2 needed can then be calculated and the answer is still B.
RI, 2004, JC1BT1
A 10 cm3 sample of a gaseous hydrocarbon measured at room temperature and pressure (r.t.p) was reacted with 70 cm3 of oxygen.
The resulting gas mixture collected was found to have a volume of 60 cm3 (also measured at r.t.p). The gas mixture was completely absorbed by aqueous potassium hydroxide.
Calculate the empirical formula of the hydrocarbon
A. C2H3 | B. C3H2 | C. C4H6 | D. C6H4 |
Solutions
Let the formula of the unknown hydrocarbon be CxHy
The gas absorbed by potassium hydroxide is CO2.
Equation:
CxHy + ( |
|
)O2 → xCO2 + |
|
H2O |
Or:
CxHy + ( |
|
+ x)O2 → xCO2 + |
|
H2O |
Reacted (cm3): | 10 | 70 | 60 |
Mole ratio of | CH4 | : | CO2 | |
1 | : | X | ||
10 | : | 60 |
X = 6
Mole ratio of | CH4 | : | CO2 | |||||
1 | : |
|
||||||
10 | : | 70 |
⇒ |
|
+ X = 7 |
⇒ y = 4(7-6) = 4
Thus the empirical formula of the hydrocarbon is C6H4. Answer is D.
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