BRAINBOX SPM EXAM formulae PAKAR ANALISIS Penulis: MARZUQI MOHD SALLEH Form 4 1. Learn formulae instantly through fast notes 2. Understand formulae through step-bystep problem-solving examples 3. Apply formulae in the practices Bilingual 100% Edisi 2024
2 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 KANDUNGAN CONTENTS Unit 1 Jirim dan Struktur Atom Matter and the Atomic Structure 5 - 15 1.1 Struktur Atom Atomic Structure 6 - 7 1.2 Isotop dan Penggunaannya Isotopes and Its Uses 8 Latihan 1.1 & 1.2 Exercise 1.1 & 1.2 9 - 10 Soalan Kertas 1 Paper 1 Questions 11 – 13 Soalan I-Think I-Think questions 14 - 15 Unit 2 Konsep Mol, Formula dan Persamaan Kimia The mole Concept, Chemical Formula and Equation 16 - 34 2.1 Jisim Atom Relatif, JAR Relative Atomic Mass, RAM 16 2.2 Konsep Mol Mole Concept 17 2.3 Bilangan Mol dan Jisim Bahan Number of Moles and Mass of Substances 18 2.4 Bilangan Mol dan Isi Padu Gas Number of Moles and Volume of Gases 19 – 20 2.5 Hubung kai tantara bilangan mol dengan bilangan zarah, jisim dan isi padu gas Relationship between the number of moles, number of particles, mass and volume of gases 21 – 22 2.6 Formula Empirik dan Formula Molekul Empirical Formula and Molecular Formula 22 – 23 Latihan 2.1, 2.2, 2.3, 2.4, 2.5 & 2.6 Exercise 2.1, 2.2, 2.3, 2.4, 2.5 & 2.6 24 – 29 Soalan Kertas 1 Paper 1 Questions 30 – 32 Soalan I-Think I-Think questions 33 - 34 Unit 3 Asid, Bes dan Garam Acid, Base and Salt 35 - 62 3.1 Nilai pH pH value 35 – 37 3.2 Kepekatan Larutan Akueus Concentration of Aqueous Solution 37 – 40 3.3 Larutan Piawai Standard Solution 41 3.4 Peneutralan Neutralisation 42 – 43
3 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Latihan 1.1 & 1.2 Exercise 1.1 & 1.2 43 - 51 Soalan Kertas 1 Paper 1 Questions 52 – 54 Soalan I-Think I-Think questions 55 – 62 Unit 4 Kadar Tindak Balas Rate of Reaction 63 - 74 4.1 Kadar Tindak Balas Purata Average Rate of Reaction 63 – 64 4.2 Kadar Tindak Balas pada Masa Tertentu Instantaneous Rate of Reaction 65 – 66 Latihan 1.1 & 1.2 Exercise 1.1 & 1.2 67 – 69 Soalan Kertas 1 Paper 1 Questions 70 – 72 Soalan I-Think I-Think questions 73 - 74 Unit 5 PRAKTIS SPM SPM PRACTICE 75 - 82 JAWAPAN LENGKAP SUGGESTION ANSWER 83 - 103 BRAINBOX CHEMISTRY-Calculation Practices dilutis oleh Cikgu Marzuqi. Setiap latihan berdasarkan soalan SPM mengikut standard KSSM terkini. Setiap aras soalan Latihan dan Kemahiran Berfikir Aras Tinggi, (KBAT) juga dimasukkan ke dalam latihan ini. Modul ini memang sesuai dijadikan sebagai soalan latihan persediaan bagi calon SPM 2024/2025. Semoga dapat membantu calon dalam menguasai kehendak soalan SPM sebenar. ©semua hak cipta terpelihara. Tiada mana-mana bahagian daripada penerbit ini boleh diterbitkan semula atau disimpan dalam bentuk yang boleh diperoleh semua atau disiarkan dalam sebarang bentuk dengan apa cara, elektronik, mekanikal, fotokopi, rakaman atau sebaliknya tanpa mendapat izin daripada penulis bahan ini.
4 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 UNIT 1 Jirim dan Struktur Atom Matter and the Atomic Structure Example 1 Atom magnesium mempunyai 12 proton dan 12 neutron. Apakah nombor proton dan nombor nucleon atom magnesium? Magnesium atom has 12 protons and 12 neutrons. What are the proton number and nucleon number of magnesium atom? Solution Nombor proton = bilangan proton Proton number = number of protons = 12 Nombor nucleon = nombor proton + bilangan neutron Nucleon number = proton + number of neutrons = 12 + 12 = 24 Example 2 Nombor nukleon atom karbon ialah 12. Atom karbon mempunyai 6 proton. Berapakah bilangan elektron dan neutron bagi atom karbon? The nucleon number of a carbon atom is 12. A carbon atom has 6 protons. How many neutrons and electrons are there in a carbon atom? Solution Bilangan elektron = bilangan proton Number of electrons = number of protons = 6 Bilangan neutron = bilangan nukleon – bilangan proton Number of neutrons = nucleon number – number of protons = 12 - 6 = 6 Formula Nombor, nukleon = bilangan proton + bilangan neutron Atau Nombor nucleon = nombor proton + bilangan neutron Nucleon number = number of protons + number of neutrons Or Nucleon number = proton number + number of neutrons 1.1 : Struktur Atom / Atomic Structure
5 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 2 Nombor nukleon atom karbon ialah 12. Atom karbon mempunyai 6 proton. Berapakah bilangan elektron dan neutron bagi atom karbon? The nucleon number of a carbon atom is 12. A carbon atom has 6 protons. How many neutrons and electrons are there in a carbon atom? Solution Bilangan elektron = bilangan proton Number of electrons = number of protons = 6 Bilangan neutron = nombor nukleon – bilangan proton Number of neutrons = nucleon number – number of protons = 12 - 6 = 6 Example 3 Jadual menunjukkan perbandingan bilangan proton, neutron dan elektron apabila atom oksigen menerima elektron untuk membentuk ion oksida. Apakah perubahan pada bilangan proton, neutron dan elektron? Table shows the comparison among the number of protons, neutrons and electrons when an oxygen atom accepts an electron to form an oxide ion. What are the changes in the number of protons, neutron and electrons? Jenis zarah Type of particle Atom oksigen, O Oxygen atom, O Ion oksida, O2- Oxide ion, O2- Bilangan proton Number of protons 8 8 Bilangan neutron Number of neutrons 8 8 Bilangan elektron Number of electrons 8 10 Solution Bilangan proton bagi atom oksigen = bilangan proton bagi ion oksida Number of protons in an oxygen atom = Number of protons in an oxide ion = 8 Bilangan neutron bagi atom oksigen = bilangan neutron bagi ion oksida Number of neutrons in an oxygen atom = Number of neutrons in an oxide ion = 8 Bilangan elektron bagi atom oksigen = 8 Number of electrons in an oxygen atom Bilangan elektron bagi ion oksida = 8 (-2) = 10 Number of an electron in an oxide ion
6 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 4 Fosforus terdiri daripada dua isotop, P dan P. Kelimpahan semula jadi P ialah 99% dan P ialah 1%. Hitungkan jisim atom relative fosforus. Phosphorus consists of two isotopes, P and P. The natural abundance of P is 99% and P is 1%. Calculate the relative atomic mass of phosphorus. Solution Jisim atom relative fosforus = (% isotop P X jisim P)+(% isotop P x jisim P) 100 Relative atomic mass of phosphorus = (% isotope P X mass P)+(% isotope P x mass P) 100 = (99 31) + (1 32) 100 = 31.01 Exercise 1.1 & 1.2 1 Tentukan bilangan elektron dan neutron dalam jadual di bawah. Determine the number of electrons and neutrons in the table below. Zarah Particle N Be Bilangan elektron Number of electron Bilangan neutron Number of neutron 31 15 31 15 32 15 32 15 31 15 31 15 32 15 32 15 32 15 32 15 31 15 31 15 31 15 9 4 32 15 1.2 : Isotop dan Penggunaannya / Isotopes and Its Uses Formula Jisim atom relative = ∑(% ) Relative atomic mass = ∑(% ) 14 7 31 15 32 15
7 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 UNIT 2 Konsep Mol, Formula dan Persamaan Kimia The mole Concept, Chemical Formula and Equation Example 1 Hitungkan jisim molekul relatif bagi ammonia, NH3. [Jisim atom relatif: H = 1, N = 14] Calculate the relative molecule mass of ammonia, NH3. [Relative atomic mass: H = 1, N = 14] Solution JMR bagi ammonia, NH3 = 1(JAR bagi N) + 3(JAR bagi H) RMM of ammonia, NH3 = 1(RAM of N) + 3(RAM of H) = 14 + 1 = 15 Example 2 Hitungkan jisim formula relatif bagi ferum(III) sulfat, Fe2(SO4)3. [Jisim atom relatif: O = 16, S = 32, Fe = 56] Calculate the relative formula mass of iron(III) sulphate, Fe2(SO4)3. [Relative atomic mass: O = 16, S = 32, Fe = 56] Solution JFR bagi ferum(III) sulfat, Fe2(SO4)3 = 2(JAR bagi Fe) + 3[JAR bagi S + 4(JAR bagi O)] RFM of iron(III) sulphate, Fe2(SO4)3 = 2(RAM of Fe) + 3[RAM of S + 4(RAM of O)] = 2(56) + 3[32 + 4(16)] = 400 Formula Jisim atom relatif sesuatu unsur, JAR = − Relative atomic mass of an element, RAM= − 2.1 : Jisim Atom Relatif, JAR / Relative Atomic Mass, RAM Formula Jisim molekul relatif sesuatu unsur, JMR, = − Relative molecular mass of element, RMM = −
8 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 3 Berapakah bilangan atom yang terdapat dalam 0.01 mol zink, Zn? How many atoms are there in 0.01 mol of zinc, Zn? Solution Bilangan atom zink, Zn = 0.01 mol x 6.02 x 1023 mol-1 Number of zinc atoms, Zn = 6.02 x 1021 atom Example 4 Satu sampel natrium klorida, NaCl mengandungi 2.408 x 1024 unit NaCl. Hitungkan bilangan mol natrium klorida, NaCl yang terdapat di dalam sampel itu. A sample of sodium chloride, NaCl contains 2.408 x 1024 units. Calculate the number of moles of sodium chloride, NaCl found in the sample. Solution Bilangan mol natrium klorida, NaCl = 2.408 1024 6.02 1023 Number of moles of sodium chloride, NaCl = 4 mol Example 5 Sebuah balang gas berisi 5 mol gas hidrogen, H2. A gas jar is filled with 5 mol of hydrogen gas, H2. (a) Berapakah bilangan molekul hidrogen di dalam balang gas itu? How many molecules of hydrogen are there in the gas jar? (b) Berapakah bilangan atom hidrogen di dalam balang gas itu? How many atoms of hydrogen are there in the gas jar? Solution (a) Bilangan molekul hidrogen, H2 = 5 mol x 6.02 x 1023 mol-1 = 3.01 x 1024 molekul (b) Setiap molekul hidrogen, H2 terdiri daripada 2 atom hidrogen, H Jadi, bilangan atom hidrogen, H = Bilangan molekul H2 x 2 = 3.01 x 1024 x 2 = 6.02 x 1024 atom 2.2 : Konsep Mol / Mole Concept Formula Bilangan mol, n = Number of moles, n = Note: NA = 6.02 x 1023 mol-1
9 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 6 Berapakah jisim bagi 0.7 mol berilium, Be? What is the mass of 0.7 mol of beryllium, Be? [Jisim atom relatif: Be = 9] [Relative atomic mass: Be = 9] Solution Jisim molar berilium, Be = 9 mol-1 Molar mass of beryllium, Be Jisim berilium, Be = 0.7 mol x 9 g mol-1 Mass of beryllium, Be = 6.3 g Example 7 Berapakah bilangan mol molekul yang terdapat di dalam 4.4 g gas karbon dioksida, CO2? How many moles of molecules are found in 4.4 g of carbon dioxide gas, CO2? [Jisim atom relatif: C = 12, O = 16] [Relative atomic mass: C = 12, O = 16] Solution Jisim molekul relative karbon dioksida, CO2 = 12 + 2(16) Relative molecular mass of carbon dioxide, CO2 = 44 Jadi, jisim molar karbon dioksida, CO2 = 44 g mol-1 Thus, the molar mass of carbon dioxide, CO2 Bilangan mol molekul karbon diosid, CO2 = 4.4 g 44 g mol−1 Number of moles of carbon dioxide molecules, CO2 = 0.1 mol Formula Bilangan mol, n = () ( −) Number of moles, n = () ( −) 2.3 : Bilangan Mol dan Jisim Bahan / Number of Moles and Mass of Substances
10 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 8 Hitungkan isi padu 1.5 mol gas oksigen, O2 dalam dm3 pada STP. Calculate the volume of 1.5 mol of oxygen gas, O2 in dm3 at STP. [Isi padu molar = 22.4 dm3 mol-1 pada STP] [Molar volume = 22.4 dm3 mol-1 at STP] Solution Isi padu gas oksigen, O2 = Bilangan mol x Isi padu molar pada STP Volume of oxygen gas, O2 = Number of moles x Molar volume at STP = 1.5 mol x 22.4 dm3 mol-1 = 33.6 dm3 Example 9 Berapakah isi padu 0.2 mol gas klorin, Cl2 dalam cm3 pada keadaan bilik? What is the volume of 0.2 mol chlorine gas, Cl2 in cm3 at room conditions? [Isi padu molar = 24 dm3 mol-1 pada keadaan bilik] [Molar volume = 24 dm3 mol-1 at room conditions] Solution Isi padu gas klorin, Cl2 = Bilangan mol x Isi padu molar pada keadaan bilik Volume of chlorine gas, Cl2 = Number of moles x Molar volume at room conditions = 0.2 mol x 24 dm3 mol-1 = 4.8 dm3 = 4.8 x 1000 cm3 = 4800 cm3 Formula Bilangan mol, n = Number of moles, n = 2.4 : Bilangan Mol dan Isi Padu Gas / Number of Moles and Volume of Gases
11 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 10 Berapakah bilangan mol gas nitrogen diosida, NO2 yang mempunyai isi padu 500 cm3 pada keadaan bilik? How many moles of nitrogen dioxide gas, NO2 has the volume of 500 cm3 at room conditions? [Isi padu molar = 24 dm3 mol-1 pada keadaan bilik] [Molar volume = 24 dm3 mol-1 at room conditions] Solution Isi padu gas nitrogen, NO2 = 500 cm3 Volume of nitrogen gas, NO2 = 500 1000 dm3 = 0.5 dm3 Bilangan mol gas nitrogen, NO2 = Isi padu gas Isi padu molar pada keadaan bilik Number of moles of nitrogen gas, NO2= Volume of gas Molar volume at room conditions = 0.5 3 24 3−1 = 0.02 mol
12 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 11 Berapakah isi padu bagi 32 g gas bromin, Br2 pada keadaan bilik? What is the volume of 32 g of bromine gas, Br2 at room conditiond? [Jisim atom relative: Br = 80 ; Isi padu molar = 24 dm3 mol-1 pada keadaan bilik] [Relative atomic mass: Br = 80 ; Molar volume = 24 dm3 mol-1 at room conditions] Solution JMR bagi gas bromin, Br2 = 80(2) RMM of bromine gas, Br2 = 160 Jadi, jisim molar bagi gas bromin, Br2 = 160 g mol-1 Thus, the molar mass of bromine gas, Br2 Bilangan mol gas bromin, Br2 = Jisim Jisim molar Number of moles of bromine gas, Br2 = Mass Malar mass = 32 g 160 g mol−1 = 0.2 mol ÷ NA Number of moles Volume of gas Formula Formulae 2.5 : Hubung kai tantara bilangan mol dengan bilangan zarah, jisim dan isi padu gas Relationship between the number of moles, number of particles, mass and volume of gases Bilangan mol Bilangan zarah Isi padu gas Jisim (g) Number of particles ÷ Molar mass X Molar mass X NA X Molar volume ÷ Molar volume Mass (g) X NA ÷ NA X Jisim molar ÷ Jisim molar X Isi padu molar ÷ Isi padu molar
13 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Isi padu gas bromin, Br2 = Bilangan mol x Isi padu molar Number of bromine gas, Br2 = Number of moles x Molar volume = 0.2 mol x 24 dm3 mol-1 = 4.8 dm3 Jadi, 32 g gas bromin, Br2 menepati isi padu 4.8 dm3 pada keadaan bilik. Hence, 32 g dm3 of bromine gas, Br2 occupies a volume of 4.8 dm3 at room conditions. Example 12 Jadual menunjukkan komposisi sebatian X. Table shows the composition of compound X. Komposisi unsur dalam sebatian X (%) Composition of elements in compound X (%) Karbon, C Carbon, C Hidrogen, H Hydrogen, H Oksigen, O Oxygen, O 48.65 8.11 43.24 Tentukan formula empiric bagi sebatian X. Determine the empirical formula of compound X. [Jisim atom relatif: H = 1, C = 12, O = 16] [Relative atomic mass: H = 1, C = 12, O = 16] Formula Menentukan formula empirik Determining the empirical formulae Unsur Element X Y Jisim (g) Mass (g) Bilangan mol atom Number of moles of atoms Nisbah mol atom Mole ratio Nisbah mol atom paling ringkas Simplest mole ratio of atom Formula empirik ialah ……. Empirical formula is 2.6 : Formula Empirik dan Formula Molekul / Empirical Formula and Molecular Formula Formula molekul = (Formula empirik)n Molecular formula = (Empirical formula)n
14 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Solution Unsur Element Karbon Carbon Hydrogen Hydrogen Oksigen Oxygen Jisim (%) Mass (%) 48.65 8.11 43.24 Bilangan mol atom Number of moles of atoms 48.65 12 = 4.0542 8.11 1 = 8.1100 43.24 16 = 2.7025 Nisbah mol atom Mole ratio 4.0542 2.7025 = 1.5 8.1100 2.7025 = 3 2.7025 2.7025 = 1 Nisbah mol atom paling ringkas Simplest mole ratio of atom 1.5 x 2 = 3 3 x 2 = 6 1 x 2 = 2 Formula empirik ialah C3H6O2 Empirical formula is C3H6O2 Example 13 Satu sebatian mempunyai formula empirik CH2O. Jisim molekul relatifnya ialah 180. Apakah formula molekul sebatian itu? A compound has the empirical formula CH2O. Its relative molecular mass is 180. What is the molecular formula of the compound? [Jisim atom relatif: H = 1, C = 12, O = 16] [Relative atomic mass: H = 1, C = 12, O = 16] Solution (CH2O)n = 180 [12 + 2(1) + 16]n = 180 30n = 180 = 6 Formula molekul ialah C6H12O6 Molecular formula is C6H12O6
15 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 1 Rajah 1 menunjukkan graf bagi keputusan eksperimen pemanasan jisim serbuk zink nitrat, Zn(NO3)2 yang berbeza dalam oksigen membentuk zink oksida, oksigen dan nitrogen dioksida. Diagram 1 shows a graph of the results for the experiment of heating different masses of zinc nitrtae powder in oxygen to form zinc oxide, oxygen, nitrogen dioxide. Rajah 1 Diagram 1 (a) Berdasarkan Rajah 1, tentukan Based on Diagram 1, determine (i) jisim zink nitrat yang digunakan dalam eksperimen ini. the mass of zinc nitrate used in this experiment. [1 markah] [1 mark] (ii) bilangan mol zink nitrat. number of moles of zinc nitrate. [Jisim atom relatif: N = 14, O = 16, Zn = 65] [Relative atomic mass: N = 14, O = 16, Zn = 65] [1 markah] [1 mark] Jisim zink nitrat, Zn(NO3)2 (g) Mass of zinc nitrate, Zn(NO3)2 (g) Jisim nitrogen dioksida, NO2 (g) Mass of nitrogen dioxide, NO2 (g) 1 Soalan I-Think / I-Think questions 2 3 4 1.0 0.5
16 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 (iii) jisim nitrogen dioksida yang dihasilkan. the mass of nitrogen dioxide produced. [2 markah] [2 marks] (iv) isi padu oksigen yang dibebaskan the volume of oxygen released [Isi padu molar = 24 dm3 mol-1 pada keadaan bilik] [Molar volume = 24 dm3 mol-1 at room conditions] [3 markah] [3 marks] (b) Tulis persamaan kimia yang terlibat. Write the chemical equation involved. [2 markah] [2 marks] (c) Merujuk persamaan kimia di 1(b), nyatakan maklumat dari segi kualitatif dan kuantitatif. Referring to the chemical equation in 1(b), state information in term of qualitatively and quantitatively. [2 markah] [2 marks]
17 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 UNIT 3 Asid, Bes dan Garam Acid, Base and Salt Example 1 Hitungkan nilai pH bagi asid hidroklorik, HCl dengan kepekatan ion hydrogen, H+ 0.01 mol dm-3 . Calculate the pH value of hydrochloric acid, HCl with 0.01 mol dm-3 of hydrogen ion, H + . Solution pH = -log[0.01] = -(-2) = 2 Nilai pH asid hidroklorik, HCl = 2 pH value of hydrochloric acid, HCl = 2 Example 2 Tentukan kemolaran asid nitric, HNO3 dengan nilai pH 3.0. Determine the molarity of nitric acid, HNO3 with pH value of 3.0. Solution pH = -log[H+ ] 3.0 = -log[H+ ] Log[H+] = - 3.0 [H+] = 10-3 Kemolaran asid nitric, HNO3 = 0.001 mol dm-3 Molarity of nitric acid, HNO3 = 0.001 mol dm-3 Formula pH = - log [H+] 3.1 : Nilai pH / pH value
18 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 3 Hitungkan nilai pOH bagi larutan kalium hidroksida, KOH yang mengandungi kepekatan ion hidroksida, OH0.03 mol dm-3 . Calculate the pOH value of potassium hydroxide solution, KOH with 0.03 mol dm-3 of hydroxide ion, OH - . Solution Diberikan bahawa kepekatan ion hidroksida, OH- = 0.03 mol dm-3 Given concentration of hydroxide ion, OH- = 0.03 mol dm-3 pH = -log[0.03] = -(-1.523) = 1.5 Nilai pOH larutan kalium hidroksida, KOH = 1.5 pOH value of potassium hydroxide solution, KOH = 1.5 Example 4 Hitungkan nilai pH bagi larutan natrium hidroksida, NaOH yang mengandungi kepekatan ion hidroksida, OH0.02 mol dm-3 . Calculate the pH value ofsodium hydroxide solution, NaOH with 0.02 mol dm-3 of hydroxide ion, OH- . Solution Diberikan bahawa kepekatan ion hidroksida, OH- = 0.02 mol dm-3 Given concentration of hydroxide ion, OH- = 0.02 mol dm-3 pOH = -log[0.02] = -(-1.699) = 1.7 Nilai pOH larutan natrium hidroksida, NaOH = 1.7 pOH value of sodium hydroxide solution, NaOH = 1.7 pH larutan natrium hidroksida, NaOH = 14 – 1.7 pH value of sodium hydroxide solution, NaOH = 12.3 Formula pOH = - log [OH-]
19 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 5 Tentukan kemolaran larutan litium hidroksida, LiOH dengan nilai pH 11. Determine the molarity of lithium hydroxide solution, LiOH with pH value 11. Solution pH + pOH = 14.0 11 + pOH = 14.0 = 3.0 pOH = - log[OH- ] 3.0 = -log[OH- ] Log[OH- ] = -3.0 [OH- ] = 10-3 = 0.001 mol dm-3 Kemolaran larutan litium hidroksida, LiOH = 0.001 mol dm-3 Molarity of lithium hydroxide solution, LiOH = 0.001 mol dm-3 Formula Kepekatan (g dm-3 ) () () Concentration (g dm-3 ) () () Kepekatan (mol dm-3 ) () () Concentration (mol dm-3 ) () () 3.2 : Kepekatan Larutan Akueus / Concentration of Aqueous Solution
20 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 6 Hitungkan kepekatan, dalam unit g dm-3 , bagi setiap larutan yang terhasil. Calculate the concentration in g dm-3 , for each solution produced. a) 6 g pepejal zink sulfat, ZnSO4 dilarutkan di dalam air untuk menghasilkan 60 dm3 larutan. 6 g of solid zinc sulphate, ZnSO4 is dissolved in water to produce 60 dm3 solution. b) 13.75 g pepejal natrium hidroksida, NaOH dilarutkan di dalam air untuk menghasilkan 550 cm3 larutan. 13.75 g of sodium hydroxide pellets, NaOH is dissolved in water to produce 550 cm3 solution. Solution a) Kepekatan larutan zink sulfat, ZnSO4, = Jisim zat terlarut (g) Isi padu larutan (dm3) Concentration of zinc sulphate, ZnSO4, = Mass od solute (g) Volume of solution (dm3) = 6 g 60 dm3 = 0.1 g dm-3 b) Kepekatan larutan natrium hidroksida, NaOH Concentration of sodium hydroxide, NaOH = Jisim zat terlarut (g) Isi padu larutan (dm3) = Mass od solute (g) Volume of solution (dm3) = 13.75 (g) 0.55 dm3 = 25 g dm-3 Example 7 Hitungkan kemolaran bagi setiap larutan yang terhasil. Calculate the concentration in g dm-3 , for each solution produced. a) 10 mol pepejal kalsium karbonat, CaCO3 dilarutkan di dalam air untuk menghasilkan 25 dm3 larutan. 10 mol of solid calcium carbonate, CaCO4 is dissolved in water to produce 25 dm3 of solution. b) 0.3 mol pepejal zink klorida, ZnCl2 dilarutkan di dalam 250 cm3 air suling. 0.3 mol of solid zinc chloride, ZnCl2 is dissolved in 250 cm3 of distilled water.
21 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Solution a) Kemolaran larutan kalsium karbonat, CaCO3, = Bilangan mol terlarut (mol) Isi padu larutan (dm3) Molarity of calcium carbonate, CaCO3, = Number of moles of solute (mol) Volume of solution (dm3) = 10 mol 25 dm3 = 0.4 mol dm-3 b) Kemolaran larutan zink klorida, ZnCl2, = Bilangan mol terlarut (mol) Isi padu larutan (dm3) Molarity of zinc chloride, ZnCl2, = Number of moles of solute (mol) Volume of solution (dm3) = 0.3 (mol) 0.25 dm3 = 1.2 mol dm-3 Example 8 Apakah kepekatan, dalam unit g dm-3 , bagi asid hidroklorik, HCl berkemolaran 0.1 mol dm-3 ? What is the concentration of hydrochloric acid, HCl with a molarity of 0.1 mol dm-3 in unit g dm-3 ? [Jisim atom relatif: H = 1, Cl = 35.5] [Relative atomic mass: H = 1, Cl = 35.5] Solution Kepekatan = Kemolaran x Jisim molar HCl Concentration = Molarity x Molar mass HCl = 0.1 mol dm-3 x [1 + 35.5] g mol-1 = 0.1 mol dm-3 x 36.5 g mol-1 = 3.65 g dm-3 Example 9 Tukarkan kepekatan larutan natrium hidroksida, NaOH 8.0 g dm-3 kepada kemolaran, mol dm-3 . Concert the concentration of 8.0 g dm-3 sodium hydroxide, NaOH to molarity, mol dm-3 . [Jisim atom relatif: H = 1, O = 16. Na = 23] [Relative atomic mass: H = 1, O = 16. Na = 23] Solution Kemolaran = Kepekatan Jisim molar NaOH Molarity = = 8.0 g dm3 23+16+1 g mol−1 = 0.2 mol dm-3
22 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 10 Hitungkan bilangan mol natrium hidoksida, NaOH yang terkandung di dalam 20 dm3 larutan natrium hidroksida, NaOH 0.1 mol dm-3 . Calculate the number of moles of sodium hydroxide, NaOH found in 20 dm3 of 0.1 mol dm-3 sodium hydroxide solution, NaOH. Solution = MV = 0.1 mol dm-3 x 20 dm3 = 0.2 mol NaOH Example 11 Sebuah bikar mengandungi 100 cm3 larutan zink sulfat, ZnSO4 0.5 mol dm-3 . Berapakah bilangan mol zink sulfat, ZnSO4 di dalam bikar itu? A beaker contains 100 cm3 of 0.5 mol dm-3 zinc sulphate solution, ZnSO4. How many moles of zinc sulphate, ZnSO4 is in the beaker? Solution Bilangan mol, n = MV 1000 Number of moles, n = MV 1000 = 0.5 mol −3 x 100 3 1000 = 0.05 mol ZnSO4
23 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 12 50 cm3 asid hidroklorik, HCl 3.0 mol dm-3 yang dicairkan kepada x mol dm-3 apabila 25 cm3 air suling ditambahkan. Hitung nilai x. 50 cm3 of 3.0 mol dm-3 hydrochloric acid, HCl that is diluted to x mol dm-3 when 25 cm3 distilled water is added. Calculate the value of x? Solution M1 = 3.0 mol dm-3 ; V1 = 50 cm3 M2 = x mol dm-3 ; V2 = (50 + 25) = 75 cm3 3.0 mol dm-3 X 50 cm3 = x mol dm-3 X 75 cm3 x mol dm-3 = 3.0 mol dm−3 X 50 cm3 75 3 = 2.0 mol dm-3 Then, x = 2.0 Example 13 Tentukan isi padu asid nitrik, HNO3 3.0 mol dm-3 yang perlu dipipetkan ke dalam sebuah kelalang volumetrik 250 cm3 untuk menghasilkan asid nitric, HNO3 0.5 mol dm-3 . 50 cm3 of 3.0 mol dm-3 hydrochloric acid, HCl that is diluted to x mol dm-3 when 25 cm3 distilled water is added. Calculate the value of x. Solution M1 = 4.0 mol dm-3 ; V1 = …… cm3 M2 = 0.5 mol dm-3 ; V2 = 250 cm3 4.0 mol dm-3 X V1 = 0.5 mol dm-3 X 250 cm3 V1 = 0.5 mol dm−3 X 250 cm3 4 3 = 31.25 mol dm-3 3.3 : Larutan Piawai / Standard Solution Formula Bilangan mol zat terlarut sebelum pencairan = bilangan mol zat terlarut selepas pencairan Number of moles of solute before dilution = Number of moles of solute after dilution n1 = n2 =
24 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 14 25 cm3 larutan kalium hidroksida, KOH 0.2 mol dm-3 dineutralkan dengan asid sulfurik, H2SO4 0.2 mol dm-3. Hitungkan isi padu asid sulfurik, H2SO4 yang diperlukan dalam tindak balas peneutralan ini. 25 cm3 of 0.2 mol dm-3 potassium hydroxide solution, KOH is neutralized with 0.2 mol dm-3 sulphuric acid, H2SO4. Calculate the volume of sulphuric acid, H2SO4 needed for this neutralization reaction. [Jisim atom relatif: O = 16, Pb = 207] [Relative atomic mass: O = 16, Pb = 207] Solution 2KOH + H2SO4 K2SO4 + 2H2O a = 2 mol ; b = 1 mol Ma = 0.2 mol dm-3 ; Va = 25 cm3 Mb = 0.2 mol dm-3 ; Vb = ……. cm3 0.2 mol dm−3 X 25 cm3 0.2 = 2 1 Vb = 12.5 cm3 Example 15 6.69 g plumbum(II) oksida diperlukan untuk melengkapkan peneutralan 50 cm3 asid hidroklorik, HCl. Hitungkan kepekatan asid dalam mol dm-3 . 6.69 g of lead(II) oxide, PbO is needed to complete the neutralization of 50 cm3 of hydrochloric acid, HCl. Calculate the concentration of the acid in mol dm-3 . Solution PbO + 2HCl PbCl2 + H2O Jisim molekul relatif plumbum(II) oksida, PbO = 207 + (16) Relative molecular mass of lead(II) oxide, PbO = 233 Jadi, jisim molar plumbum(II) oksida, PbO = 223 g mol-1 Thus, the molar mass of lead(II) oxide, PbO Bilangan mol molekul plumbum(II) oksida, PbO = 6.69 g 223 g mol−1 Number of moles of lead(II) oxide, PbO = 0.03 mol Formula = 3.4 : Peneutralan / Neutralisation
25 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 1 mol PbO : 2 mol HCl 1 mole of PbO : 2 moles of HCl 0.03 mol PbO : 0.06 mol HCl 0.03 moles of PbO : 0.06 moles of HCl Bilangan mol, n = MV 1000 Number of moles, n = MV 1000 0.06 mol = M x 50 3 1000 = 1.2 mol HCl
26 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 1 Rajah 1 menunjukkan nilai pH bagi satu larutan. Diagram 1 shows the pH value of a solution. Rajah 1 Diagram 1 Apakah kemolaran larutan itu? What is molarity of the solution? A 0.005 mol dm-3 B 0.010 mol dm-3 C 0.020 mol dm-3 D 0.200 mol dm-3 2 Rajah 2 menunjukkan nilai pH larutan kalium hidroksida. Diagram 2 shows the pH value of potassium hydroxide solution. Apakah nilai X? What is the value of X? A 0.3 B 1.5 C 8.1 D 13.7 3 Berapakah isi padu air suling yang perlu ditambah kepada 5.7 g magnesium klorida untuk menghasilkan larutan dengan kepekatan 0.3 mol dm-3 ? What is the volume of distilled water that needed to be added to 5.7 g of magnesium chloride to obtain a solution with concentration of 0.3 mol dm-3 ? [Jisim atom relatif: Mg = 24, Cl = 35.5] [Relative atomic mass: Mg = 24, Cl = 35.5] A 100 cm3 B 200 cm3 C 250 cm3 D 300 cm3 4 Rajah 2 menunjukkan susunan radas bagi satu tindak balas peneutralan. Diagram 2 shows the apparatus set-up for a neutralisation reaction. Rajah 2 Diagram 2 Apakah kepekatan asid sulfurik? What is the concentration of the sulphuric acid? A 0.10 mol dm-3 B 0.15 mol dm-3 C 0.20 mol dm3 D 0.40 mol dm-3 Buret Burette 25.0 cm3 asid sulfurik 25.0 cm3 of 50.0 cm sulphuric acid 3 larutan natrium jidroksida dengan nilai pH 13 50.0 cm3 of potassium hydroxide with pH value 13 Nilai pH pH value Meter pH pH meter Larutan kalium hidroksida 0.5 mol dm-3 0.5 mol dm-3 of potassium hydroxide solution Paper 1 Questions
27 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 5 Persamaan kimia berikut menunjukkan tindak balas antara larutan kalium iodida dan larutan plumbum(II) nitrat. The following chemical equation shows the reaction between the solution of potassium iodide and the solution of lead(II) nitrate. Pb(NO3)2 + 2KI → PbI2 + 2KNO3 Hitungkan jisim maksimum mendakan yang terbentuk apabila larutan plumbum(II) nitrat berlebihan ditambah ke dalam 50 cm3 larutan kalium iodide 0.2 mol dm-3 . Calculate the maximum mass of precipitate formed when excess lead(II) nitrate solution is added to 50 cm3 of 0.2 mol dm-3 potassium iodide solution. [Jisim atom relatif: N = 14, K = 39, I = 127, Pb = 207] [Relative atomic mass: N = 14, K = 39, I = 127, Pb = 207] A 1.010 g B 2.020 g C 2.305 g D 4.610 g 6 Persamaan berikut menunjukkan proses penguraian kalium klorat(V). The following equation shows the decomposition process of potassium chlorate(V). 2KCIO3 → 2KCI + 3O2 Apakah jisim kalium klorat(V) yang diperlukan untuk menghasilkan 2.0 mol gas oksigen? What is the mass of potassium chlorate(V) needed to produce 2.0 mol of oxygen gas? [Jisim atom relatif: O = 16, Cl = 35.5, K = 39] [Relative atomic mass: O = 16, Cl = 35.5, K = 39] A 162.93 g B 245.00 g C 325.85 g D 490.00 g 7 Persamaan kimia di bawah mewakili tindak balas kimia bagi penyediaan garam terlarutkan, zink sulfat. The chemical equation below represents the chemical reaction for the preparation of the dissolved salt, zinc sulphate. Zn + H2SO4 → ZnSO4 + H2 3.5 g serbuk zink ditambah ke dalam 50 cm3 asid sulfuric 0.60 mol dm-3 . Berapakah jisim bagi serbuk zink yang tidak bertindak balas? 3.5 g of zinc powder is added to 50 cm3 of sulfuric acid 0.60 mol dm-3 . What is the mass of unreacted zinc powder? [Jisim atom relatif: Zn = 65] [Relative atomic mass: Zn = 65] A 0.43 g B 0.75 g C 1.55 g D 1.95 g 8 Persamaan berikut menunjukkan pemanasan zink karbonat. The following equation shows the heating of zinc carbonate. ZnCO3 → ZnO + CO2 Berapakah jisim bahan yang diperlukan sekiranya isi padu gas karbon dioksida yang terhasil adalah 350 cm3 ? What is the mass of reactant required if the volume of carbon dioxide gas produced is 350 cm3 ? [Jisim atom relatif: C = 12, O = 16, Zn = 65] [Relative atomic mass: C = 12, O = 16, Zn = 65] [Isi padu molar = 24 dm3 mol-1 pada keadaan bilik] [Molar volume = 24 dm3 mol-1 at room conditions] A 2.055 g B 1.825 g C 0.125 g D 0.0146 g
28 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 9 Rajah 4 menunjukkan dua tindak balas berbeza bagi plumbum(II) karbonat. Diagram 4 shows two different reactions of lead(II) carbonate. Rajah 1 Diagram 1 Berapakah jisim pepejal Z yang terbentuk apabila jisim plumbum(II) karbonat yang sama digunakan? What is the mass of solid Z formed when the same mass of lead(II) carbonate is used? [Jisim atom relatif: N = 14, O = 16, Pb = 207] [Relative atomic mass: N = 14, O = 16, Pb = 207] A 14.34 g B 13.38 g C 11.95 g D 11.15 g 10 Apakah jisim bagi magnesium yang diperlukan untuk bertindak balas dengan asid nitric berlebihan dan membebaskan 448 cm3 gas hydrogen pada suhu dan tekanan piawai? What is the mass of magnesium needed to react with excess nitric acid to produce 448 cm3 of hydrogen gas at standard temperature and pressure? [Isi padu molar = 22.4 dm3 mol-1 pada STP] [Molar volume = 22.4 dm3 mol-1 at STP] A 0.240 g B 0.448 g C 0.480 g D 0.960 g Plumbum(II) nitrat Lead(II) nitrate Pepejal Z + CO2 Solid Z 16.5 g garam X + CO2 + H2O 16.55 g salt X + Asid nitrik Nitric acid Pananskan Heated
29 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 8 Rajah 8 menunjukkan satu carta alir bagi tindak balas yang berlaku apabila garam X dipanaskan. Diagram 8 shows a flow chart for the reactions occurred when salt X is heated. Rajah 8 Diagram 8 Pepejal Y yang terhasil bertukar daripada perang ke kuning setelah disejukkan dan gas Q yang terbebas berwarna perang. Solid Y produced turns from brown to yellow when cooled and the liberated gas Q is brown. (a) Tulis persamaan kimia yang terlibat bagi kedua-dua tindak balas tersebut. Write the chemical equations involved for both reactions. [4 markah] [4 marks] (b) Jika 50 cm3 larutan X 1.0 mol dm-3 digunakan. Hitungkan, If 50 cm3 of 1.0 mol dm-3 solution X is used. Calculate, (i) jisim garam X. mass of salt X [Jisim molar: garam X = 331 g mol-1 ] [Molar mass: salt X = 331 g mol-1 ] (ii) jisim pepejal Y mass of solid Y [Jisim molar: pepejal Y= 223 g mol-1 ] [Molar mass: solid Y = 223 g mol-1 ] (iii) jisim pepejal M mass of solid M [Jisim molar: pepejal M = 267 g mol-1 ] [Molar mass: solid M = 267 g mol-1 ] (iv) isi padu gas Q dan gas J. volume of gas Q and gas J [10 markah] [10 marks] Garam X Salt X Pepejal Y Solid Y Gas Q Gas Q Gas J Gas J + + Panaskan Heated Larutan X Solution X Pepejal M Solid M + NaNO3 + Air Water + larutan natrium karbonat Sodium carbonate solution Tindak balas II Reaction II Tindak balas I Reaction I Soalan I-Think / I-Think questions
30 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 UNIT 4 Kadar Tindak Balas Rate of Reaction Formula Kadar tindak balas purata dari t1 hingga t2 The average rate of reaction from t1 to t2 = Jumlah isi padu gas yang terkumpul dari t1 hingga t2 Tempoh masa yang diambil = 1 2 = V2−V1 t2−t1 cm3 s -1 = V t cm3 s -1 Formula Kadar tindak balas purata bagi keseluruhan tindak balas. The overall average rate of reaction. = Jumlah isi padu gas yang terkumpul Tempoh masa yang diambil = = V t cm3 s -1 Formula Kadar tindak balas purata dalam t1 saat yang pertama The average rate of reaction for first t1 seconds = Jumlah isi padu gas yang terkumpul dalam t1 saat yang pertama Tempoh masa yang diambil = ℎ 1 = V1−0 t1−0 cm3 s -1 = V t cm3 s -1 4.1 : Kadar Tindak Balas Purata / Average Rate of Reaction
31 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 1 Jadual 1 menunjukkan isi padu gas yang terkumpul apabila magnesium karbonat bertindak balas dengan asid sulfurik. Table 1 shows the volume of gas that accumulates when magnesium carbonate reacts with sulphuric acid. Masa (minit) Time (minute) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Isi padu gas (cm3 ) Volume of gas (cm3 ) 0.0 25.0 40.0 51.0 58.0 63.0 68.0 70.0 70.0 Jadual 1 Table 1 (a) Kira kadar tindak balas purata dalam dua minit pertama? Calculate the average rate of reaction in the first two minutes? (b) Kira kadar tindak balas purata dalam satu minit pertama? Calculate the average rate of reaction in the first one minute? (c) Tentukan kadar tindak balas bagi keseluruhan tindak balas. Determine the overall average rate of reaction. Solution (a) = Jumlah isi padu gas yang terkumpul dari t1 hingga t2 Tempoh masa yang diambil = 1 2 = 58.0−40.0 2.0−1.0 cm3 min-1 = 18 cm3 min-1 (b) = Jumlah isi padu gas yang terkumpul dalam t1 saat yang pertama Tempoh masa yang diambil = ℎ 1 = 40.0−0 1.0−0 cm3 min-1 = 40.0 cm3 min-1 (c) = Jumlah isi padu gas yang terkumpul Tempoh masa yang diambil = = 70.00 3.5 cm3 min-1 = 20 cm3 min-1
32 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 Example 2 Jadual 2 menunjukkan jumlah isi padu gas yang terkumpul pada sela masa yang tetap dalam suatu tindak balas. Table 2 shows the total volume of gas collected at regular interval in a reaction. Masa (s) Time (s) 0 30 60 90 120 150 180 210 Isi padu gas (cm3 ) Volume of gas (cm3 ) 0 2.0 3.7 5.2 6.5 7.3 8.0 8.0 Jadual 2 Table 2 (a) Berdasarkan Jadual 2, plotkan graf isi padu gas melawan masa. Based on Table 2, plot a graph of the volume of gas against time. (b) Tentukan kadar tindak balas bagi keseluruhan tindak balas. Determine the overall average rate of reaction. (c) Berdasarkan graf yang diplot, kira kadar tindak balas pada 60 saat. Based on the plotted graph, calculate the rate of reaction at 60 seconds. Solution (a) rujuk sebelah refer next (b) = Jumlah isi padu gas yang terkumpul Tempoh masa yang diambil = = 8.0 180 cm3 s -1 = 0.044 cm3 s -1 4.2 : Kadar Tindak Balas pada Masa Tertentu / Instantaneous Rate of Reaction Formula Kadar tindak balas pada masa t = Kecerunan tangen pada masa t Rate of reaction at time t = Gradient of the tangent at time t = ∆ ∆ = 2−1 2−1 = V t cm3 s -1
33 4541 4541 © BRAINBOX CHEMISTRY – CALCULATION PRACTICES FORM 4 (c) = 5.7−2.0 97.5−27 = 0.05 cm3 s -1 Isi padu gas (cm3 ) Volume of gas (cm3 ) Masa (s) Time(s) 0 30 60 3 3.0 4.0 90 3 120 3 5.0 150 180 210 6.0 1.0 2.0 7.0 8.0 9.0 x x x x x x x x (97.5, 2.0) (27, 2.0) (97.5, 5.7)