HKCEE Chemistry Volumetric Analysis & Calculations based on Equations (MC) Produced by William Tsang ©William’s production 2002

HKCEE 1990 (8) Which of the following contains the largest number of ATOMS at room temperature and pressure? (relative atomic mass: H = 1.0, N = 14.0, Cl = 35.5; molar volume of gas at room temperature and pressure = 24 dm3) 2 mol of ammonia gas 3 mol of nitrogen gas 7g of hydrogen gas 90 dm3 of HCl gas A: No. of moles of atoms in 2 mol NH3 = 4 X 2 = 8 B: No. of moles of atoms in 3 mol N2 = 3 X 2 = 6 C: No. of moles of atoms in 7 g H2 = 7 / 2 = 3.5 D: No. of moles of atoms in 90 dm3 of HCl gas = (90 / 24) X 2 = 7.5 The answer is A.

HKCEE 1990 (11) 2 NaHCO3 + H2SO4  Na2SO4 + 2CO2 + 2H2O From equation, What volume of 0.5M sulphuric acid is required to liberate 4.8 dm3 of carbon dioxide at room temperature and pressure from excess solid sodium hydrogencarbonate? (molar volume of gas at room temperature and pressure = 24 dm3) 0.2 dm3 0.4 dm3 2.0 dm3 4.0 dm3 2 NaHCO3 + H2SO4  Na2SO4 + 2CO2 + 2H2O From equation, No. of moles of H2SO4 = No. of moles of CO2 / 2 (0.5)(volume) = (4.8) / (24) / (2) Volume = 0.2 dm3 The answer is A.

HKCEE 1990 (31) 16.1 g of a hydrated sulphate was heated to constant mass. After cooling to room temperature, the residual anhydrous metal sulphate weighed 7.1 g. How many moles of water of crystallization are there in one mole of the hydrated metal sulphate? (Relative molecular masses: anhydrous metal sulphate = 142.0; water = 18.0) A. 4 B. 5 C. 7 D. 10 % by mass of water in crystal = (9 / 16.1) X 100% = 55.9 % Relative molecular mass of crystal X 44.1% = 142 Relative molecular mass of crystal = 322 Total relative molecular mass of water of crystallization = 322 – 142 = 180 Mass of 1 mol of crystal = 322 g Mass of water in 1 mol of crystal = 180 g No. of moles of water of crystallization in 1 mol of crystal = 180 / 18 = 10 1 mol of crystal contains 10 mol of water of crystallization  The answer is D.

HKCEE 1991 (3) Solid X undergoes complete thermal dissociation according to the following equation: X(s)  Y(g) + Z (s) On heating 4.90 g of X, 1.40 dm3 of gas Y and 2.30 g of solid Z are obtained at room temperature and pressure. What is the relative molecular mass of Y? A. 32.0 B. 39.4 C. 44.6 D. 84.0 No. of moles of Y formed = 1.4 / 24 = 0.0583 Mass of Y formed = 4.9 – 2.3 = 2.6 g Molar mass of Y = mass/ no. of moles Molar mass of Y = 2.6 / 0.0583 Molar mass of Y = 44.6 g Relative molecular mass of Y = 44.6 The Answer is C.

HKCEE 1991 (11) 2.60g of a metal X combine with 1.20g of oxygen to form an oxide in which the oxidation number of X is +3. What is the relative atomic mass of X? [Given relative atomic mass of O = 16] A. 11.6 B. 34.7 C. 52.0 D. 104 4X + 3O2  2X2O3 From equation, no. of mol of X / 2 = no. of mol of O2 /3 (2.6) / (relative atomic mass) / 4 = (1.2) / (32) / 3 Relative atomic mass of X = 52 The answer is C.

HKCEE 1991 (17) 22g of calcium carbonate are allowed to react with 200cm3 of 0.5M hydrochloric acid until no further reaction occurs. What is the mass of calcium carbonate left behind? (relative atomic masses: C= 12.0, O = 16.0, Ca = 40.0) 2g 5g 12g 17g CaCO3 + 2HCl  CaCl2 + H2O + CO2 From equation, No. of moles of CaCO3 = No. of moles of HCl / 2 Mass of CaCO3 / 100 = (0.2)(0.5)(0.5) Mass of CaCO3 used = 5 g Mass of calcium carbonate left = 22 – 5 = 17g The answer is D.

HKCEE 1992 (29) After 50cm3 of 0.6M H2SO4 have completely neutralized 100cm3 of 0.6M NaOH, the concentration of the resulting sodium sulphate solution is 0.2M 0.3M 0.6M 1.2M No. of moles of sulphate ions in 50cm3 of 0.6M H2SO4 = (0.05)(0.6) = 0.03 No. of moles of sodium ions in 100cm3 of 0.6M NaOH = (0.6)(0.1) = 0.06 0.03 mol sodium sulphate can be formed Con. = (0.03) / (0.15) = 0.2 The answer is A.

HKCEE 1993 (13) 0.12 g of sodium metal is added to a large volume of water. When the reaction is completed, the resulting solution is treated with 0.2M hydrochloric acid. What is the volume of the acid required, to the nearest cm3, for complete neutralization? (relative atomic mass: Na= 23) 13 26 39 52 No. of moles of Na = 0.12 / 23 = 0.005217 No. of moles of NaOH formed = 0.005217 NaOH + HCl  NaCl + H2O From equation, No of mol of HCl = no of mol of NaOH (0.2) (Volume) = 0.005217 Volume = 26 cm3 The answer is B.

HKCEE 1994 (9) A metal X forms a hydroxide XOH. 1.12g of XOH were dissolved in some distilled water and then made up to 250 cm3 with distilled water. 25.0 cm3 of this solution required 20.0 cm3 of 0.10 M hydrochloric acid for complete neutralization. What is the relative atomic mass of X? (relative atomic masses: H = 1.0, O = 16.0) 23.0 24.0 39.0 40.0 (25/1000)(M) = (20/1000)(0.1) Molarity of XOH = 0.08M No. of moles of XOH = (molarity) (volume) = (0.08) (0.25) = 0.02 Relative atomic mass of XOH = mass / no.of moles = 1.12 / 0.02 = 56 Relative atomic mass of X = 56 – 1 – 16 = 39  The Answer is C

HKCEE 1995 (31) Refer to the following chemical equation: Fe2O3 + 3 CO  2 Fe + 3CO2 What volume of carbon dioxide, measured at room temperature and pressure, is produced if 224g of iron are formed? (relative atomic mass: Fe = 56; molar volume of gas at room temperature and pressure = 24 dm3) 16 dm3 36 dm3 72 dm3 144 dm3 From the equation, No. of moles of Fe / 2 = no. of moles of CO2 / 3 224 / 56 / 2 = volume / 24 / 3 Volume = 144 dm3  The answer is D.

HKCEE 1997 (14) The formula of a metal carbonate is X2CO3. 100cm3 of a solution containing 0.69g of the carbonate requires 50 cm3 of 0.20M hydrochloric acid for complete reaction. What is the atomic mass of metal X? (relative atomic masses: C= 12, O=16) 19.0 23.0 39.0 78.0 X2CO3 + 2 HCl  2XCl + H2O + CO2 From equation, No. of moles of X2CO3 = no. of moles of HCl / 2 (0.69)/ (2x + 60) = (0.05)(0.2) / 2 x = 39 Relative atomic mass of X is 39 The answer is C.

HKCEE 1998 (10) 18 x / (152 + 18x) = 0.4527 x = 7  Answer is C The formula for hydrated iron(II) sulphate is FeSO4.xH2O. On strong heating, 20.1g of the sulphate produces 9.1g of water. What is the value of x? (relative atomic masses: H= 1.0, O= 16.0, S = 32.1 Fe= 56.0) 5 6 7 8 18 x / (152 + 18x) = 0.4527 x = 7  Answer is C

HKCEE 1998 (16) The formula of a solid dibasic acid is H2X. 2.88g of the acid is dissolved in some distilled water and the solution is then diluted to 250.0 cm3 with distilled water. 25.0 cm3 of the diluted solution requires 16.0 cm3 of 0.40 M sodium hydroxide solution for complete neutralization. What is the molar mass of H2X? 22.5g 45.0g 90.0g 180.0g (25/1000) (M) / 2 = (16/1000) (0.4) Molarity = 0.128 M 2.88 / molar mass = (0.128)(0.25) Molar mass = 90g  The answer is C

HKCEE 2000(20) A sample of concentrated sulphuric acid has a density of 1.83 gcm-3 and contains 94.0% of sulphuric acid by mass. What is the concentration (correct to 1 d.p.) of sulphuric acid in the sample? (relative atomic masses: H=1.0, O=16.0, S=32.1) 17.5M 18.3M 18.7M 19.8M Overall Concentration = 1830 / 98 = 18.67M Acid Concentration = 18.67 x 94% = 17.5M The answer is A.

HKCEE 2001 (27) Suppose the Avogadro number is L. How many atoms does 600cm3 of oxygen at room temperature and pressure contain? (Molar volume of gas at room temperature and pressure = 24dm3) 1/40 L 1/20 L 25 L 50 L No. of moles of oxygen = 600 / 24000 = 0.025 mol No. of moles of oxygen atoms = 0.025 X 2 = 0.05 = 1/ 20 L The answer is B.

HKCEE 2002 (3) An oxide of element X has the formula X2O3. 10.2 g of this oxide contains 5.4 g of X. What is the relative atomic mass of X? (Relative atomic mass: O= 16.0) 12.0 18.0 27.0 36.0 5.4 / mass / 2 = 4.8 / 16 / 3 mass = 27 Relative atomic mass of x = 27 The answer is C.

HKCEE 2002 (16) By mole ratio, X : Y : Z = 1:3:2 Volume ratio = 1:3:2 Gases X and Y react to give a gaseous product Z. The reaction can be represented by the equation: X(g) + 3Y (g)  2Z (g) In an experiment, 40 cm3 of X and 60 cm3 of Y are mixed and are allowed to react in a closed vessel. What is the volume of the resultant gaseous mixture? (All volumes are measured at room temperature and pressure.) 40 cm3 60 cm3 80 cm3 100 cm3 By mole ratio, X : Y : Z = 1:3:2 Volume ratio = 1:3:2 20 cm3 of X react with 60 cm3 of Y to give 40 cm3 of Z 20 cm3 of X will be left Resultant gaseous mixture = 20 + 40 = 60 cm3  The answer is B.