MARK SCHEME
UJIAN PENGESANAN 1 KIMIA TING. 4 (2018)
SECTION A / BAHAGIAN A
1. C
6. C
11. C
16. A
21. C
2. B
7. D
12. A
17. C
22. D
3. D
8. D
13. B
18. D
23. A
4. A
9. D
14. A
19. C
24. B
5. B
10. A
15. D
20. C
25. D Total = 25 marks
SECTION B / BAHAGIAN B
Question 1 (a)
(b)(i)
Mark Scheme Able to record all the temperature readings accurately with one decimal place 70.0, 78.0, 80.0, 80.0, 80.0, 80.0, 82.0, 90.0
Marks 3
Able to record at least 6 temperature readings 70, 78, 80, 80, 80, 80, 82, 90
2
Able to record at least 4 temperature readings // Records the bottom meniscus 69, 77, 79, 79, 79, 81, 89
1
Correct label of axes + units Able to plot all 8 points transferred correctly Smooth curve Refer attachment
1 1 1 Total
(b)(ii) Able to mark and state the melting point correctly
3 3
1
MARK SCHEME
(c)
UJIAN PENGESANAN 1 KIMIA TING. 4 (2018)
Able to mark the value of the melting point
2
Able to mark any value of the curve or on the temperature axis
1
Able to state all the physical state of acetamide correctly
3
Time range (s) Julat masa (s) 0 – 60
Physical state Keadaan fizik Solid Pepejal 60 – 150 Solid and liquid Pepejal dan cecair 150 – 210 Liquid Cecair Able to state any two physical state of acetamide correctly
2
Able to state idea of physical state of acetamide correctly
1
(d)
Molecule Molekul
1
(e)
To ensure even / uniform heating // Acetamide is highly flammable Untuk memastikan pemanasan adalah sekata // Asetamida mudah terbakar
1
(f)
To ensure even / uniform heating Untuk memastikan pemanasan adalah sekata
1
Total
15
2
MARK SCHEME
UJIAN PENGESANAN 1 KIMIA TING. 4 (2018)
3
MARK SCHEME
UJIAN PENGESANAN 1 KIMIA TING. 4 (2018)
SECTION C / BAHAGIAN C
1
(a)(i)
n=
0.3 dm3 // 0.0125 mol 3 -1 24 dm mol
1
No. of atoms = 0.0125 mol × 6.02×1023 mol-1 × 3 // 2.26×1022 Total (a)(ii)
Molar mass = [12 + (2×16)] g mol-1 // 44 g mol-1 Mass = 0.0125 mol × 44 g mol-1
// 0.055 g Total
1 2 1 1 2
(b) Element Unsur Percentage (%) Peratus (%) Mass (g) Jisim (g) No. of moles (mol) Bil. mol (mol) Simplest mol ratio Nisbah mol teringkas
C
H
85.7
14.3
85.7
14.3
85.7 // 7.142 12 7.142 // 1 7.142
1 1
14.3 // 14.3 1 14.3 // 2 7.142
1
Empirical formula = CH2 Formula empirik
1
(CH2)n = 42 (12 + 2)n = 42 14n = 42 n=3
1
Molecular formula = (CH2)3 = C3H6 Total Total
1 6 10
END OF MARK SCHEME Prepared by:
Checked by:
Verified by:
......................................................... (CIK INTAN ZAREENA ZAHEDI)
......................................................... (PN. NURRUL ‘AINNI MD. TENI)
................................................. (PN. NORLIZA MD DERUS)
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