Chemistry Past Paper PDF - M.E.S Indian School 2024-2025

M.E.S INDIAN SCHOOL, DOHA -QATAR NOTES 2024-2025 Section: BOYS’ & GIRLS’ Date : 18-05-2024 Class &Div.: CLASS XI (all divisions) Subject: CHEMISTRY Lesson: CH: 1 SOME BASI...

M.E.S INDIAN SCHOOL, DOHA -QATAR NOTES 2024-2025 Section: BOYS’ & GIRLS’ Date : 18-05-2024 Class &Div.: CLASS XI (all divisions) Subject: CHEMISTRY Lesson: CH: 1 SOME BASIC CONCEPTS OF CHEMISTRY Xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx LAWS OF CHEMICAL COMBINATIONS 1. Law of Conservation of Mass : It states that matter/ mass can neither be created nor destroyed C+ O2 → CO2 12 g + 2x16g → 12+2x16 g 44g → 44g 2. Law of Definite Proportions/ composition. It states that a given compound always contains same elements in the same proportion by mass H2O CO2 2:16 12: 2x16 1:8 3: 8 3. Law of Multiple Proportions: It states that if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the small whole number ratio. H + O Hydrogen+ Oxygen → Water 2g 16g 18g Hydrogen + Oxygen → Hydrogen peroxide H2O H2O2 2g 32g 34g Hence 16:32 = 1:2, a simple whole number ratio. F 061, Rev 01,dtd 10th March 2020 1  Masses of oxygen (i.e. 16 g and 32 g) which combine with a fixed mass of hydrogen (2g) bear a s imple ratio..ie 16:32 or 1: 2. 4. Gay Lussac’s Law of Gaseous Volumes : It states that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure 2H2 + O2 → 2H2O 2V + 1V → 2V H2 and O2 combine in the simple ratio of 2:1. 5. Avogadro Law : It states that equal volumes of gases should contain equal number of molecules at same temperature and pressure. 6. Relative Atomic Mass It is the mass of the atom relative to 1/12th the mass of one carbon – 12 atom. 7. Atomic mass unit / Unified mass (u) : One atomic mass unit is defined as a mass exactly equal to 1/12th the mass of one carbon – 12 atom 8. Molecular Mass : Molecular mass is the sum of atomic masses of the elements present in a molecule Molecular mass of methane, (CH4) = 12 + 4 x1 = 16u Molecular mass of water (H2O) = 2 x1 +16 = 18 u 9. Formula Mass : Formula mass is the sum of atomic masses of the elements present in a formula unit of a compound Formula mass of NaCl = 23.0 u + 35.5 u = 58.5 u Mole Concept molar mass mass 1 mole [GAM/GMM] 6.023 x1023 particle Volume molar volume (22.4L at STP) F 061, Rev 01,dtd 10th March 2020 2 Mole: One mole is the amount of a substance that contains 6.022x1023 particles (OR ) One mole is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of the 12C isotope. 6.022x1023 is known as ‘Avogadro constant’, denoted by NA 1 mol of hydrogen atoms = 6.022×1023 atoms 1 mol of water molecules = 6.022×1023 water molecules 10. Molar mass : The mass of one mole of a substance in grams is called its molar mass.. Molar mass in grams = atomic/molecular/formula mass in grams. Molar mass of water = 18 g mol-1 which contain 6.022×1023 water molecules 1. Calculate the molecular mass of (i) H2O (ii)CO2 (iii) CH4 2. Calculate the number of atoms in the following? a) 52 moles of He Number of atoms = mole x 6.022x1023 = 52 x 6.022x1023 = 313.144 x 1023 atoms b)52u of He 4 u mass of He = one gram atom 52 u mass of He= gram atoms c) 52 g of He Moles of He = moles 3 th F 061, Rev 01,dtd 10 March 2020  Number of atoms = mole x 6.022x1023 = 13 x 6.022x1023 =78.286 x 1023 11. Molar volume : Volume occupied by I mole of gas Molar volume of a gas at STP is 22.4L which contain 6.022×1023 molecules. a) Calculate volume occupied by 2 moles of oxygen gas? Volume of gas = mole x 22.4 L = 2 X 22. 4 L = 44.8 L b) Calculate the moles present in 5.6 L of He? Mole Try: Which gas occupies more volume a} 11 g CO2 b) 2 moles of O2 ? 12. Percentage composition It is the percentage of a particular element present in a compound Mass % of an element i) Calculate the percentage of carbon , hydrogen and oxygen in ethanol( C2H50H) ? Molar mass of ethanol is = 2x12 + 6x1+ 1x16) = 46g/mole Mass % of an element Mass% of an carbon Mass % of an Hydrogen Mass % of oxygen (ii)Calculate the mass percent of different elements present in sodium sulphate (Na 2 SO4). Molar mass of Na2SO4= (23x2)+(32x1) +(16x4)= 142g/mol 4 th F 061, Rev 01,dtd 10 March 2020  Try : Calculate the percentage of hydrogen and oxygen in H 2O ( H=1g/mole, O=16g/mole 13. Empirical formula Empirical formula represents the simplest whole number ratio of atoms present in a compound. 14. Molecular formula Molecular formula shows the exact number of atoms present in a molecule ofa compound Molecular Formula Emp.Formula Glucose C6H12O6 CH2O Ethene C2H4 CH2 Ethane C2H6 CH3 Q. A compound contains 4.07% Hydrogen, 24.27% Carbon, and 71.65% of Chlorine. Its molar mass Is 98.96g. Determine its empirical and molecular formula. Sol: Empirical formula CH2Cl 5 th F 061, Rev 01,dtd 10 March 2020 Try :.Determine molecular formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass. Stoichiometry ⮚ Stoichiometry is the quantitative relationship between the reactants and products in a balanced chemical equation. It deals with the calculation of masses (or volumes) of the reactants and the products involved in a chemical reaction CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) 1 mole of CH4 + 2 moles of O2 → 1 mole of CO2 +2 moles of H2O 1 molecule of CH4 + 2 molecules of O2 → one molecule of CO2+ 2 molecules of H2O 22.4 L of CH4 + 2x 22.4 L of O2 → 22.4 L of CO2 + 2x 22.4 L of H2O 16 g of CH4 +2x32 g of O2 → 44 g of CO2 + 2x18 g of H2O. 1. Calculate the amount of water (g) produced by the combustion of 4 g of methane? CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) 1 mole of CH4 + 2 moles of O2 → 1 mole of CO2 +2 moles of H2O 1 x 16 g of CH4 forms 2 x18 g of H2O. 4 g of CH4 forms 2. How much copper can be obtained from 100 g of copper sulphate (CuS0 4) ? 6 th F 061, Rev 01,dtd 10 March 2020  Molar mass of CuSO4 = 63.5 + 32 + 4 × 16 =63.5 + 32 + 64 = 159.5 amu or u 159.5 g of CuSO4 contains copper = 63.5 g 100 g of CuSO4 contains copper = 63.5159.5×100 = 39.81 g Try: 1. Three moles of ethane (C2H6), calculate the following : (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane. 2. How many moles of methane are required to produce 22 g CO2 (g) after combustion? ⮚ Limiting reagent [ present in lesser amount] The reactant that is completely consumed in a reaction and limits the amount of product formed. ⮚ Excess reagent : The reagent that is present in excess ⮚ In a reaction A + B2 → AB2. Identify the limiting reagent, if any, in the following reaction mixtures. (i) 300 atoms of A + 200 molecules A + B2 → AB2. One atom of A + 1 Molecule of B2 300 atoms required 300 molecules but have only 200, hence B is the limiting reagent. (ii) 2 mole A + 3 mole B A + B2 → AB2. One mole of A + 1 mole of B 2 mole of A + 2 moles of B but have only 3 moles of B, hence A is limiting reagent. ⮚ 3g of H2 reacts with 29g of O2 to form water. Which is the limiting reagent? Calculate the amount of water formed and also calculate the amount of unreacted substance? 2H2 (g) + O2 (g) → 2 H2O (g) 2 moles of H2 reacts with 1 mole of O2 7 th F 061, Rev 01,dtd 10 March 2020  2 x 2 g of H2 reacts with 1 x 32g of O2 3g of H2 reacts with of O2. 4 To find the amount of water 2 moles of H2 gives 2 moles of H2O. 4g of H2 = 2 x 18g of H2O 3 g of H2gives 3x36/4 =27g H2O Try : a) N2 (g) + 3H2 (g) →2NH3 (g) (i) Calculate the mass of NH 3 produced if 2 × 103 g N2 reacts with 1×103 g of H2 (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass? b) How much MgS can be obtained from 2g of Mg and 2g of S by the reaction, Mg + S→ MgS? Which is the limiting reagent? Calculate the amount of the reactants which remain unreacted. Reactions in Solutions The concentration of a solution can be expressed in any of the following ways. (1) Mass per cent (2) Mole fraction (3) Molarity (4) Molality 8 th F 061, Rev 01,dtd 10 March 2020 Mass per cent Mass per cent = Mass of solute x 100 Mass of solution 1. A solution is prepared by adding 2 g of a substance to 18 g of water. Cal. the mass % of the solute? Mole Fraction (X) It is the ratio of no. of moles of a particular component to the total number of moles ofthe solution. Mole fraction of A Mole fraction of B = no: of moles of B = nB No: of moles of solution nA + nB Also XA + XB = 1 Molarity [Unit-mol/L or Molar (M)] It is defined as the number of moles of the solute in 1 litre of the solution. Molarity= 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 = mass of solute x1000 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 Litre molar mass of solute x volume of solution 9 th F 061, Rev 01,dtd 10 March 2020  Molarity Equation Suppose if we want to dilute a solution of particular concentration, then the concentration of this dilute solution can be calculated using M1 V1 = M2 V2 where M and V are molarity and volume resp. 4. Molality[Unit-mol/Kg or molal (m)]: It is defined as the number of moles of solute present in 1 kg of solvent. Molality = mass of solute x 1000 molar mass of solute x mass of solvent in gram Q. How does molarity and molality varies with temperature? Molarity depends upon temperature because volume of a solution is temperature dependent. Molality does not change with temperature since mass does not depend on temperature. Stock solution The solution of higher concentration is also known as stock solution. Questions: a) Cal. the molarity of NaOH in the solution prepared by dissolving its 4 g in water to form 250 mL of the solution? F 061, Rev 01,dtd 10th March 2020 10  b) Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1. Volume of solution= 500ml Molarity of solution = 0.375M c) How are 0.50 mol Na2C03 and 0.50 M Na2C03 different? Answer: Molar mass of Na2C03= 2 x 23 + 12 + 3 x 16 = 106g mol-1 0.50 mol Na2C03 means 0.50 x 105 g = 53 g 0. 50 M Na2C03 means 0.50 mol, i.e., 53 g Na2C03 are present in 1 litre of the solution. F 061, Rev 01,dtd 10th March 2020 11  d)Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL- 1 and the mass per cent of nitric acid in it being 69%. e) What is the concentration of sugar (C12H22O11) in mol L -1 if its 20 g are dissolved in enough water to make a final volume up to 2 L? Answer: F 061, Rev 01,dtd 10th March 2020 12