2HKCEE 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 gas3 mol of nitrogen gas7g of hydrogen gas90 dm3 of HCl gasA: No. of moles of atoms in 2 mol NH3 = 4 X 2 = 8B: No. of moles of atoms in 3 mol N2 = 3 X 2 = 6C: No. of moles of atoms in 7 g H2 = 7 / 2 = 3.5D: No. of moles of atoms in 90 dm3 of HCl gas = (90 / 24) X 2 = 7.5The answer is A.
3HKCEE 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 dm30.4 dm32.0 dm34.0 dm32 NaHCO3 + H2SO4 Na2SO4 + 2CO2 + 2H2OFrom equation,No. of moles of H2SO4 =No. of moles of CO2 / 2(0.5)(volume) = (4.8) / (24) / (2)Volume = 0.2 dm3The answer is A.
4HKCEE 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. 4B. 5C. 7D. 10% by mass of water in crystal= (9 / 16.1) X 100% = 55.9 %Relative molecular mass of crystalX 44.1% = 142Relative molecular mass of crystal = 322Total relative molecular mass of water of crystallization = 322 – 142 = 180Mass of 1 mol of crystal = 322 gMass of water in 1 mol of crystal = 180 gNo. of moles of water of crystallization in 1 mol of crystal = 180 / 18 = 101 mol of crystal contains 10 mol of water of crystallization The answer is D.
5HKCEE 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. 32.0B. 39.4C. 44.6D. 84.0No. of moles of Y formed= 1.4 / 24 =Mass of Y formed= 4.9 – 2.3 = 2.6 gMolar mass of Y = mass/ no. of molesMolar mass of Y = 2.6 /Molar mass of Y = 44.6 gRelative molecular mass of Y = 44.6The Answer is C.
6HKCEE 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. 11.6B. 34.7C. 52.0D. 1044X + 3O2 2X2O3From equation,no. of mol of X / 2 =no. of mol of O2 /3(2.6) / (relative atomic mass) / 4= (1.2) / (32) / 3Relative atomic mass of X = 52The answer is C.
7HKCEE 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)2g5g12g17gCaCO3 + 2HCl CaCl2 + H2O + CO2From equation,No. of moles of CaCO3= No. of moles of HCl / 2Mass of CaCO3 / 100 = (0.2)(0.5)(0.5)Mass of CaCO3 used = 5 gMass of calcium carbonate left= 22 – 5 = 17gThe answer is D.
8HKCEE 1992 (29)After 50cm3 of 0.6M H2SO4 have completely neutralized 100cm3 of 0.6M NaOH, the concentration of the resulting sodium sulphate solution is0.2M0.3M0.6M1.2MNo. of moles of sulphate ions in 50cm3 of 0.6M H2SO4= (0.05)(0.6) = 0.03No. of moles of sodium ions in 100cm3 of 0.6M NaOH= (0.6)(0.1) = 0.060.03 mol sodium sulphate can be formedCon. = (0.03) / (0.15) = 0.2The answer is A.
9HKCEE 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)13263952No. of moles of Na= 0.12 / 23 =No. of moles of NaOH formed=NaOH + HCl NaCl + H2OFrom equation,No of mol of HCl = no of mol of NaOH(0.2) (Volume) =Volume = 26 cm3The answer is B.
10HKCEE 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 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)23.024.039.040.0(25/1000)(M) = (20/1000)(0.1)Molarity of XOH = 0.08MNo. of moles of XOH= (molarity) (volume)= (0.08) (0.25)= 0.02Relative atomic mass of XOH= mass / no.of moles= 1.12 / 0.02 = 56Relative atomic mass of X= 56 – 1 – 16= 39 The Answer is C
11HKCEE 1995 (31) Refer to the following chemical equation: Fe2O3 + 3 CO 2 Fe + 3CO2What 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 dm336 dm372 dm3144 dm3From the equation,No. of moles of Fe / 2= no. of moles of CO2 / 3224 / 56 / 2 = volume / 24 / Volume = 144 dm3 The answer is D.
12HKCEE 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)19.023.039.078.0X2CO3 + 2 HCl 2XCl + H2O + CO2From equation,No. of moles of X2CO3= no. of moles of HCl / 2(0.69)/ (2x + 60) = (0.05)(0.2) / 2x = 39Relative atomic mass of X is 39The answer is C.
13HKCEE 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)567818 x / ( x) =x = 7 Answer is C
14HKCEE 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 cm3 with distilled water 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?22.5g45.0g90.0g180.0g(25/1000) (M) / 2 = (16/1000) (0.4)Molarity = M2.88 / molar mass = (0.128)(0.25)Molar mass = 90g The answer is C
15HKCEE 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)17.5M18.3M18.7M19.8MOverall Concentration= 1830 / 98 = 18.67MAcid Concentration= x 94%= 17.5MThe answer is A.
16HKCEE 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 L1/20 L25 L50 LNo. of moles of oxygen= 600 / 24000= molNo. of moles of oxygen atoms= X 2= 0.05= 1/ 20 LThe answer is B.
17HKCEE 2002 (3)An oxide of element X has the formula X2O g of this oxide contains 5.4 g of X. What is the relative atomic mass of X? (Relative atomic mass: O= 16.0)12.018.027.036.05.4 / mass / 2 = 4.8 / 16 / 3mass = 27Relative atomic mass of x = 27The answer is C.
18HKCEE 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 cm360 cm380 cm3100 cm3By mole ratio, X : Y : Z= 1:3:2Volume ratio = 1:3:220 cm3 of Xreact with 60 cm3 of Yto give 40 cm3 of Z20 cm3 of X will be leftResultant gaseous mixture= = 60 cm3 The answer is B.