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

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Presentation on theme: "HKCEE Chemistry Volumetric Analysis & Calculations based on Equations (MC) Produced by William Tsang ©Williams production 2002."— Presentation transcript:

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

2 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 dm 3 ) A.2 mol of ammonia gas B.3 mol of nitrogen gas C.7g of hydrogen gas D.90 dm 3 of HCl gas A: No. of moles of atoms in 2 mol NH 3 = 4 X 2 = 8 B: No. of moles of atoms in 3 mol N 2 = 3 X 2 = 6 C: No. of moles of atoms in 7 g H 2 = 7 / 2 = 3.5 D: No. of moles of atoms in 90 dm 3 of HCl gas = (90 / 24) X 2 = 7.5 The answer is A.

3 HKCEE 1990 (11) What volume of 0.5M sulphuric acid is required to liberate 4.8 dm 3 of carbon dioxide at room temperature and pressure from excess solid sodium hydrogencarbonate? ( molar volume of gas at room temperature and pressure = 24 dm 3 ) A.0.2 dm 3 B.0.4 dm 3 C.2.0 dm 3 D.4.0 dm 3 2 NaHCO 3 + H 2 SO 4 Na 2 SO 4 + 2CO 2 + 2H 2 O From equation, No. of moles of H 2 SO 4 = No. of moles of CO 2 / 2 (0.5)(volume) = (4.8) / (24) / (2) Volume = 0.2 dm 3 The answer is A.

4 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.

5 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 dm 3 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 B C D No. of moles of Y formed = 1.4 / 24 = 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 / Molar mass of Y = 44.6 g Relative molecular mass of Y = 44.6 The Answer is C.

6 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 B C D X + 3O 2 2X 2 O 3 From equation, no. of mol of X / 2 = no. of mol of O 2 /3 (2.6) / (relative atomic mass) / 4 = (1.2) / (32) / 3 Relative atomic mass of X = 52 The answer is C.

7 HKCEE 1991 (17) 22g of calcium carbonate are allowed to react with 200cm 3 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) A.2g B.5g C.12g D.17g CaCO 3 + 2HCl CaCl 2 + H 2 O + CO 2 From equation, No. of moles of CaCO 3 = No. of moles of HCl / 2 Mass of CaCO 3 / 100 = (0.2)(0.5)(0.5) Mass of CaCO 3 used = 5 g Mass of calcium carbonate left = 22 – 5 = 17g The answer is D.

8 HKCEE 1992 (29) After 50cm 3 of 0.6M H 2 SO 4 have completely neutralized 100cm 3 of 0.6M NaOH, the concentration of the resulting sodium sulphate solution is A.0.2M B.0.3M C.0.6M D.1.2M No. of moles of sulphate ions in 50cm 3 of 0.6M H 2 SO 4 = (0.05)(0.6) = 0.03 No. of moles of sodium ions in 100cm 3 of 0.6M NaOH = (0.6)(0.1) = mol sodium sulphate can be formed Con. = (0.03) / (0.15) = 0.2 The answer is A.

9 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 cm 3, for complete neutralization? (relative atomic mass: Na= 23) A.13 B.26 C.39 D.52 No. of moles of Na = 0.12 / 23 = No. of moles of NaOH formed = NaOH + HCl NaCl + H 2 O From equation, No of mol of HCl = no of mol of NaOH (0.2) (Volume) = Volume = 26 cm 3 The answer is B.

10 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 cm 3 with distilled water cm 3 of this solution required 20.0 cm 3 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) A.23.0 B.24.0 C.39.0 D.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

11 HKCEE 1995 (31) Refer to the following chemical equation: Fe 2 O CO 2 Fe + 3CO 2 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 dm 3 ) A.16 dm 3 B.36 dm 3 C.72 dm 3 D.144 dm 3 From the equation, No. of moles of Fe / 2 = no. of moles of CO 2 / / 56 / 2 = volume / 24 / 3 Volume = 144 dm 3 The answer is D.

12 HKCEE 1997 (14) The formula of a metal carbonate is X 2 CO cm 3 of a solution containing 0.69g of the carbonate requires 50 cm 3 of 0.20M hydrochloric acid for complete reaction. What is the atomic mass of metal X? (relative atomic masses: C= 12, O=16) A.19.0 B.23.0 C.39.0 D.78.0 X 2 CO HCl 2XCl + H 2 O + CO 2 From equation, No. of moles of X 2 CO 3 = 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.

13 HKCEE 1998 (10) The formula for hydrated iron(II) sulphate is FeSO 4.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) A.5 B.6 C.7 D.8 18 x / ( x) = x = 7 Answer is C

14 HKCEE 1998 (16) The formula of a solid dibasic acid is H 2 X. 2.88g of the acid is dissolved in some distilled water and the solution is then diluted to cm 3 with distilled water cm 3 of the diluted solution requires 16.0 cm 3 of 0.40 M sodium hydroxide solution for complete neutralization. What is the molar mass of H 2 X? A.22.5g B.45.0g C.90.0g D.180.0g (25/1000) (M) / 2 = (16/1000) (0.4) Molarity = M 2.88 / molar mass = (0.128)(0.25) Molar mass = 90g The answer is C

15 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) A.17.5M B.18.3M C.18.7M D.19.8M Overall Concentration = 1830 / 98 = 18.67M Acid Concentration = x 94% = 17.5M The answer is A.

16 HKCEE 2001 (27) Suppose the Avogadro number is L. How many atoms does 600cm 3 of oxygen at room temperature and pressure contain? (Molar volume of gas at room temperature and pressure = 24dm 3 ) A.1/40 L B.1/20 L C.25 L D.50 L No. of moles of oxygen = 600 / = mol No. of moles of oxygen atoms = X 2 = 0.05 = 1/ 20 L The answer is B.

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

18 HKCEE 2002 (16) 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 cm 3 of X and 60 cm 3 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.) A.40 cm 3 B.60 cm 3 C.80 cm 3 D.100 cm 3 By mole ratio, X : Y : Z = 1:3:2 Volume ratio = 1:3:2 20 cm 3 of X react with 60 cm 3 of Y to give 40 cm 3 of Z 20 cm 3 of X will be left Resultant gaseous mixture = = 60 cm 3 The answer is B.


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