Chemistry www.tiwariacademy.in (Chapter 1)(Some Basic Concepts of Chemistry)(Class 11) Easy Question 1: State the number of significant figures in (i) 0.436 and (ii) 802.42 Answer 1: (i) 0.436 has three significant figures (ii) 802.42 has five significant figures. Question 2: What do you mean by Mole fraction? Answer 2: Mole fraction is the ratio of number of moles of one component to the total number of moles (solute & solvents) present in the solution. It is expressed as ‘x’. Question 3: Calculate the molecular mass of H2SO4. Answer 3: Molecular Mass of H2SO4 = 2 × 1 + 32 + 4 × 16 = 2 + 32 + 64 = 98u Question 4: What is the SI unit of mass? How is it de – fined? Answer 4: The S.I. unit of mass is Kilogram. Mass of a substance is the amount of matter present in it. Question 5: What is limiting reagent? Answer 5: The reactant which gets consumed first or limits the amount of product formed is known as limiting reagent. www.tiwariacademy.com A free web support in education Chemistry www.tiwariacademy.in (Chapter 1)(Some Basic Concepts of Chemistry)(Class 11) Average Question 1: What is the value of Avogadro number? Answer 1: 6.022 × 1023 Question 2: What is the law called which deals with the ratios of the volumes of the gaseous reactants and products of the volumes of the gaseous reactants and products? Answer 2: Gay Lussac’s law of gaseous volumes. Question 3: How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? Answer 3: Molar mass of Na2CO3 = 2 × 23 + 12 × 3 × 16 = 106 g/mol. 0.50 mol Na2CO3 means 0.50 × 106 = 53 g. 0.50 mol Na2CO3 means 0.50 mol i.e. 53 g of Na2CO3 are present in 1 L of the solution. Question 4: What is the percentage of carbon, hydrogen and oxygen in ethanol? Answer 4: Molecular formula of ethanol is C2H5OH Molar mass of ethanol is = 2 × 12 + 5 × 1 + 16 + 1 = 24 + 5 + 17 = 46 g Mass percent of carbon = 24 46 Mass percent of hydrogen = Mass percent of oxygen = 16 46 × 100 = 52.17% 6 46 × 100 = 13.1% × 100 = 34.7% Question 5: Classify the following as pure substances or mixture. (a) ethyl alcohol (b) oxygen (c) carbon (d) distilled water(e) blood (f) steel Answer 5: Pure substance – ethyl alcohol, oxygen, carbon, distilled water mixture – blood, steel www.tiwariacademy.com A free web support in education Chemistry www.tiwariacademy.in (Chapter 1)(Some Basic Concepts of Chemistry)(Class 11) Difficult Question 1: What is the difference between 160 cm and 160.0 cm? Answer 1: 160 has three significant figures while 160.0 has four significant figures. Hence160.0 represents greater accuracy. Question 2: In the combustion of methane in air, what is the limiting reagent & why? Answer 2: Methane is the limiting reagent because the other reactant is oxygen of air which is always present in excess. Thus the amounts of carbon dioxide & water formed will depend upon the amount of methane burnt. Question 3: Which aqueous solution has higher concentration, 1 molar or 1 molal solution of the same solute? Give reason. Answer 3: 1 molar aqueous solution has higher concentration than 1 molal solution. A molar solution contains one mole of solute in one litre of solution while one molal solution contains one mole of solute in 1000 g of solvent. If density of water is 1, then one mole of solute is present in 1000 mL of water in 1 molal solution while one mole of solute is present in less than 1000mL of water in 1 molar solution (1000 mL sol = amount of solute + amount of solvent). Thus 1 molar solution is more concentrated. Question 4: Calculate the molarity of water if its density is 1000 kg/m3. Answer 4: Molarity of water means the number of moles of water in 1 litre of water 1L of water = 1000 cm3 = 1000 g (1000 kg/m3 = 1g/cm3) 1000 g of water = 1000 18 = 55.56 moles Molarity = 55.56 M www.tiwariacademy.com A free web support in education