4. (a) Describe the following terms:(i)empirical formula (ii)molecular formula(3 marks) (b) Terephthalic, a carboxylic acid on analysis was found to contain carbon 57.79% by mass and hydrogen 3.62% by mass. The molar mass of the acid is 166g/mol Calculate the empirical and molecular formulae of the acid. (H=1.00,C=12.00,O=16.00) (3 marks) 5. (a) A sample of K_(2)CO_(3) weighing 27600 mg is dissolved in water and the solution is made up to 250 ml . (R.A.M of K=39.001,C=12.001,O=16.0) (i)Calculate the molarity and normality of K_(2)CO_(3) solution. (3 marks) (ii) Calculate the concentration of CO_(3)^2- ions in ppm (1 mark) (b) What is the final concentration of 50.0 ml of a 2 M fructose solution is diluted to a volume of 250 ml. (2 marks) 6. (i) Explain why first ionization energy of Magnesium is greater than that of Aluminium (R.A.M.M.Mg=12,Al=13) (1 marks) (ii) Which two of the following elements would you expect to show the greatest similarity in chemical and physical properties: Ca,F,B,He,Mg,P ? Explain your answer. (2 marks) 7. Draw Lewis (dot - cross) structures of the following: PF_(3),NH_(2),SO_(3)^2-,NO_(3)^-,CO_(3)^2- and NH_(4)^+ (Atomic numbers: C=6,N=7,O=8,P=15,F=9, S=16) (4marks) SECTION B 8. (a) (i) State and explain five postulate of Bohr's theory. (3 mark) (ii)Discuss the success of Bohr's theory. (3 mark) (ii) Give three shortcomings of Bohr 's atom. (2 mark) (b) (i) Define each of the following terms:(i) Parts per million (ppm) (ii) molarity (iii) parts per billion (ppb) (3 marks) (c) (i)A 0.5M Sample of sulphuric acid solution was analysed by taking 200 ml portion and adding to it 100 ml of 0.4MNaOH_((aq)) After the reaction excess OH_((aq)) ions remained in the solution. This excess base required 26.4 ml of a 0.25M HNO_(3) solution for complete

4. (a) <br />(i) Empirical formula: The empirical formula of a compound represents the simplest whole-number ratio of the elements present in the compound. It does not provide information about the actual number of atoms in a molecule, but rather the relative proportions of the elements.<br />(ii) Molecular formula: The molecular formula of a compound represents the actual number of atoms of each element present in a molecule of the compound. It is a multiple of the empirical formula and provides more specific information about the composition of the compound.<br /><br />(b) <br />To calculate the empirical formula of terephthalic acid, we need to determine the moles of each element present in the compound based on the given percentages by mass and molar masses.<br /><br />Given:<br />Carbon (C): 57.79% by mass<br />Hydrogen (H): 3.62% by mass<br />Oxygen (O): 100% - 57.79% - 3.62% = 38.59% by mass<br /><br />Molar masses:<br />C = 12.00 g/mol<br />H = 1.00 g/mol<br />O = 16.00 g/mol<br /><br />Moles of each element:<br />C: 57.79 g / 12.00 g/mol = 4.82 mol<br />H: 3.62 g / 1.00 g/mol = 3.62 mol<br />O: 38.59 g / 16.00 g/mol = 2.41 mol<br /><br />To find the empirical formula, we need to divide each mole value by the smallest mole value (in this case, O) and round to the nearest whole number.<br /><br />C: 4.82 mol / 2.41 mol = 2<br />H: 3.62 mol / 2.41 mol = 1.5 (rounded to 2)<br />O: 2.41 mol / 2.41 mol = 1<br /><br />Therefore, the empirical formula of terephthalic acid is C2H2O2.<br /><br />To calculate the molecular formula, we need to determine the molar mass of the empirical formula and compare it to the given molar mass of the acid.<br /><br />Molar mass of C2H2O2 = (2 × 12.00 g/mol) + (2 × 1.00 g/mol) + (2 × 16.00 g/mol) = 74.00 g/mol<br /><br />Since the given molar mass of the acid is 166 g/mol, we can determine the molecular formula by dividing the molar mass of the acid by the molar mass of the empirical formula.<br /><br />166 g/mol / 74.00 g/mol = 2.24 (rounded to 2)<br /><br />Therefore, the molecular formula of terephthalic acid is (C2H2O2)2 or C4H4O4.<br /><br />5. (a) <br />(i) To calculate the molarity and normality of the K2CO3 solution, we need to determine the number of moles of K2CO3 dissolved in the solution.<br /><br />Given:<br />Mass of K2CO3 = 27600 mg = 27.6 g<br />Volume of the solution = 250 mL = 0.250 L<br />Molar mass of K2CO3 = (2 × 39.001 g/mol) + (12.001 g/mol) + (3 × 16.00 g/mol) = 138.21 g/mol<br /><br />Moles of K2CO3 = 27.6 g / 138.21 g/mol = 0.199 mol<br /><br />Molarity (M) = Moles of solute / Volume of solution in liters<br />Molarity = 0.199 mol / 0.250 L = 0.796 M<br /><br />Normality (N) = Molarity × Number of equivalents of solute<br />Normality = 0.796 M × 2 = 1.592 N<br /><br />(ii) To calculate the concentration of CO3^2- ions in ppm, we need to determine the mass of CO3^2- ions present in the solution.<br /><br />Moles of CO3^2- = Moles of K2CO3 = 0.199 mol<br /><br />Mass of CO3^2- = Moles of CO3^2- × Molar mass of CO3^2- = 0.199 mol × 60.01 g/mol = 11.99 g<br /><br />Concentration in ppm = (Mass of solute / Volume of solution in liters) × 10^6<br />Concentration = (11.99 g / 0.250 L) × 10^6 = 47960 ppm<br /><br />6. (i) The first ionization energy of magnesium is greater than that of aluminum because magnesium has a more stable electron configuration. Magnesium has a filled 3s orbital, which is more stable than the partially filled 3p orbital in aluminum. This increased stability requires more energy to remove an