Mole Concept Exercise # 3

MOLE CONCEPT Exercise # 3 1. Part # I [Matrix Match Type Questions] Column I Column II (A) Zn(s) + 2HCl(aq)  ZnCl2(s) + H2(g) above reaction is carried out by taking 2 moles each of Zn and HCl (p) 50% of excess reagent left (B) AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(g) above reaction is carried out by taking 170 g AgNO3 and 18.25 g HCl (Ag = 108) (q) 22.4 L of gas at STP is liberated (C) CaCO3(s)  CaO(s) + CO2(g) 100 g CaCO3 is decomposed] (r) 1 moles of solid (product) obtained. (D) 2KClO3(s)  2KCl(s) + 3O2(g) 2/3 moles of KClO3 decomposed (s) HCl is the limiting reagent (p) Column - II One mole of compound contains 4NA atoms of Hydrogen. 2. (A) Column - I A gaseous organic compound containing C = 52.17%, H = 13.04% & O = 34.78% (by weight) having molar mass 46 g/mol. (B) 0.3 g of an organic compound containing C, H and O on combustion yields 0.44 g of CO2 and 0.18 g of H2O, with two O atoms per molecule. (q) The empirical formula of the compound is same as its molecule formula (C) A hydrocarbon containing C = 42.857% and H = 57.143% (by mole) containing 3C atoms per molecule. A hydrocarbon containing 10.5 g carbon per gram of hydrogen having vapour density 46. (r) Combustion products of one mole of compound contains larger number of moles of CO2 than that of H2O. CO2 gas produced by the combustion of 0.25 mole of compound occupies a volume of 11.2 L at NTP. (D) 3. (s) (B) (C) (D) Column-I 100 ml of 0.2 M AlCl3 solution + 400 ml of 0.1 M HCl solution 50 ml of 0.4 M KCl + 50 ml H2O 30 ml of 0.2 M K2SO4 + 70 ml H2O 200 ml 24.5% (w/v) H2SO4 (A) Column-I Molarity (p) (B) Molality (q) MA × n A n A M A + n BM B (C) Mole fraction (r) Independent of temperature (D) Mass % (s) XA X B M B × 1000 (A) 4. (p) Column-II Total concentration of cation(s) = 0.12 M (q) (r) (s) [SO42–] = 0.06 M [SO42–] = 2.5 M [Cl¯] = 0.2 M Column-II Dependent on temperature x 100 Where MA , MB are molar masses, nA , nB are no of moles & XA , XB are mole fractions of solute and solvent respectively. 5. Column-I Column-II (A) Law of conservation of mass (p) CH4 has carbon and hydrogen in 3 : 1 mass ratio. (B) Law of multiple proportion (q) 10 mL N2 combines with 30 mL of H 2 to form 20 mL of NH3 (C) Law of definite proportion (r) S and O2 combine to form SO2 and SO3 (D) Law of reciprocal proportion (s) In H2S and SO2 mass ratio of H and O w.r.t. sulphur is 1 : 16, hence in H2O, mass ratio of H and O is 1 : 8. (E) 6. Gay Lussac's Law (t) Column-I 4.2 g MgCO3 gives 2.0 g residue on heating. Column-II (mass of product) (A) 2H2 + O2 2H2O 1g 1g (p) 1.028 g (B) 3H2 + N2 2NH3 1g 1g (q) 1.333 g (C) H2 + Cl2 2HCl 1g 1g (r) 1.125 g (D) 2H2 + C  CH4 1g 1g (s) 1.214 g Part # II [Comprehension Type Questions] Comprehension # 1 According to the Avogadro’s law, equal number of moles of gases occupy the same volume at identical condition of temperature and pressure. Even if we have a mixture of non-reacting gases then Avogadro’s law is still obeyed by assuming mixture as a new gas. Now let us assume air to consist of 80% by volume of Nitrogen (N2) and 20% by volume of oxygen (O2). If air is taken at STP then its 1 mol would occupy 22.4 L. 1 mol of air would contain 0.8 mol of N2 and 0.2 mol of O2 hence the mole fractions of N2 and O2 are given by XN2  0.8 , X O2  0.2 1. 2. 3. Volume occupied by air at NTP containing exactly 11.2 gm of Nitrogen : (A) 22.4 L (B) 8.96 L (C) 11.2 L If air is treated as a solution of O2 and N2 then % W/W of oxygen is : 10 200 700 (A) (B) (C) 9 9 9 (D) 2.24 L (D) 350 9 Density of air at NTP is : (A) 1 g/L (B) 9 g/L 7 (C) 2 g/L 7 (D) can’t be determined MOLE CONCEPT Comprehension # 2 A chemist decided to determine the molecular formula of an unknown compound. He collects following informations (I) Compounds contains 2 : 1 'H' to 'O' atoms(number of atoms). (II) Compounds has 40% C by mass (III) Approximate molecular mass of the compound is 178 g (IV) Compound contains C, H and O only. 1. 2. 3. What is the % by mass of oxygen in the compound (A) 53.33% (B) 88.88% (C) 33.33% (D) None of these What is the empirical formula of the compound (A) CH3O (B) CH2O (C) C2H2O (D) CH3O2 Which of the following could be molecular formula of compound (A) C6H6O6 (B) C6H12O6 (C) C6H14O12 (D) C6H14O6 Comprehension # 3 In chemistry, oxidation and reduction are taken as two mutually exclusive events. For example, if life is oxidation then death is taken as reduction, taking off a flight is oxidation then standing would be reduction and so many other. In brief it is used as redox in chemical science. There are so many conceptual facts regarding redox such as adding oxygen or oxygenation, removing hydrogen or dehydrogenation, removing electron or dielectronation are fixed for oxidation and their corresponding antonyms would be reduction processes. Simple way of judging whether a monatomic species has under gone oxidation or reduction is to note if the charge number of species has changed. It is possible to assign to an atom in polyatomic species an operative charge number called their oxidation number or state. (O. N. or O. S.). There is no standard symbol for this quantity so we say it is . An O. N. is assigned to an element in a compound by assuming that it is present as ion with a characteristic charge for instance oxygen is present as O(-II) and fluorine as F(–I) and some time it may be hypothetical also. For example For ZnO Zn O  H For NH3 H N –2 +2 +1 H  +1 –3 +1 In continuation to our study, species promoting oxidation are named as oxidant and those promoting reduction are termed as reductant. At the same time their equivalent weights is the ratio of their molecular weight and change is O. N. () involving one molecule/formula unit of the reactant i.e., molecular weight divided by number of electrons lost or gained by one molecule/formula during their respective action. Based on the above discussion answer the following objective question having one best answer. 1. Which corresponds to oxidation action (A) = 0 (B) = 0 2. A compound contain P(II), Q(V) R(-II). The possible formula of the compound is (A) PQR2 (B) Q2(PR3)2 (C) P3[QR4]2 (D) P3(Q4R)2 3. A compound has number of carbon, number of hydrogen and number of oxygen their equation of finding oxidation number (x) of carbon will be  2 (A) 3 + 4x2 + = 0 (B) x + – 2 = 0 (C) x + – =0 (D) none of these 3 x (C) > 0 (D)  < 0 Comprehension # 4 The concentrations of solutions can be expressed in number of ways; viz : mass fraction of solute (or mass percent), Molar concentration (Molarity) and Molal concentration (molality). These terms are known as concentration terms and also they are related with each other i.e. knowing one concentration term for the solution, we can find other concentration terms also. The definition of different concentration terms are given below : Molarity : It is number of moles of solute present in one litre of the solution. Molality : It is the number of moles of solute present in one kg of the solvent moles of solute Mole Fraction = moles of solute + moles of solvent If molality of the solution is given as 'a' then mole fraction of the solute can be calculated by Mole Fraction = a 1000 a+ M solvent ; = a × M solvent (a × M solvent + 1000) where a = molality and Msolvent = Molar mass of solvent We can change : Mole fraction  Molality  Molarity 1. 60 gm of solution containing 40% by mass of NaCl are mixed with 100 gm of a solution containing 15% by mass NaCl. Determine the mass percent of sodium chloride in the final solution. (A) 24.4% 2. (C) 48.8% (D) 19.68% (C) 24.4 m (D) none What is the molality of the above solution. (A) 4.4 m 3. (B) 78% (B) 5.5 m What is the molarity of solution if density of solution is 1.6 gm/ml (A) 5.5 M (B) 6.67 M (C) 2.59 M (D) none Comprehension # 5 Calcium lactate is used in the food and beverage industries. It has also been used medicinally for treatment of various allergies, for treatment of muscular leg cramps, and as an antidote for a variety of poisons, including lead, arsenicals and carbon tetrachloride. A 0.8274 g sample of anhydrous calcium lactate is found by analysis to contain 0.2732 g of C, 0.0382 g H, 0.1520 g Ca and 0.3640 g O. Each mole of calcium lactate is found to contain one mole of calcium ions. Calcium lactate can be crystallised from water as pentahydrate salt. 1. Simplest formula of the calcium lactate is : (A) CaO6C6H10 2. (C) CaO2C3H3 (D) CaO2C3H5 (C) 218 g mol–1 (D) 113 g mol–1 Formula weight of calcium lactate is : (A) 129 g mol–1 3. (B) CaO3C3H5 (B) 111 g mol–1 How many grams of calcium lactate pentahydrate would be recovered from 1 g of anhydrous salt : (A) 1.41 g (B) 1.00 g (C) 1.27 g (D) 1.51 g MOLE CONCEPT Comprehension # 6 Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride (KCl) and potassium sulphate (K2SO4). Potash production is often reported as the potassium oxide (K2O) equivalent or the amount of K2O that could be made from a given mineral. KCl costs Rs. 50 per kg. 1. What is the cost of K per mole of the KCl sample? (A) Rs. 13.42 mol–1 2. (C) Rs. 1.00 mol–1 (D) Rs. 2.00 mol–1 For what price must K2SO4 be sold in order to supply the same amount of potassium as in KCl ? (A) Rs. 58.40 kg–1 3. (B) Rs. 3.73 mol–1 (B) Rs. 50.00 kg–1 (C) Rs. 42.82 kg–1 (D) Rs. 25.00 kg–1 What mass (in kg) of K2O contains the same number of moles of K atoms as 1.00 kg KCl? (A) 0.158 kg (B) 0.315 kg (C) 1.262 kg (D) 0.631 kg Comprehension # 7 342 g of 20% by mass of Ba(OH)2 solution (sp. gr. 0.57) is reacted with 200 mL of 2 M HNO3 according to given balanced reaction : Ba(OH)2 + 2HNO3  Ba(NO3)2 + 2H2O 1. The nature of the final solution is : (A) acidic 2. (B) neutral (C) basic (D) can't say Find the molarity of the ion in resulting solution by which nature of the above solution is identified, is (A) 0.5 M (B) 0.8 M (C) 0.4 M (D) 1 M Comprehension # 8 NaBr, used to produce AgBr for use in photography can be self prepared as follows : Fe + Br2 FeBr2 .....(i) FeBr2 + Br2 Fe3Br8 .....(ii) (not balanced) Fe3Br8 + Na2CO3 NaBr + CO2 + Fe3O4 .....(iii) (not balanced) How much Fe in kg is consumed to produce 2.06 × 103 kg NaBr 1. Mass of iron required to produce 2.06 × 103 kg NaBr (A) 420 g 2. (B) 420 kg (C) 4.2 × 105 kg (D) 4.2 × 108 g If the yield of (ii) is 60% & (iii) reaction is 70% then mass of iron required to produce 2.06 × 103 kg NaBr. (A) 105 kg 3. .....(iv) (B) 105 g (C) 103 kg (D) none If yield of (iii) reaction is 90% then mole of CO2 formed when 2.06 × 103 kg NaBr is formed. (A) 20 (B) 10 (C) 40 (D) none Comprehension # 9 A monobasic acid of weight 15.5 g is heated with excess of oxygen & evolved gases when passed through KOH solution increased its weight by 22 g and when passed through anhydrous CaCl2, increased its weight by 13.5 g. When the same mass of this organic acid is reacted with excess of silver nitrate solution form 41.75 g silver salt of the acid which on ignition gave the residue of weight 27 g. 1. The molecular formula of the organic acid is. (A) C2H6 2. (B) C2H5O2 (C) C2H6O2 (D) C2H4O The molar masses of the acid & its silver salt respectively are: (A) 60, 168 (B) 167, 60 (C) 60, 167 (D) 168, 60 EXERCISE - 3 : PART # I 1. A (p, q, r, s), B (p, s), C (q, r), D (q) 2. A (q, s), B (p, s), C (p, q, r), D (q, r) 3. A (p, s), B (s), C (p, q), D (r) 4. A (p), B (r, s), C (r), D (r, q) 5. A (t), B (r), C (p), D (s), (E) (q) 6. A (r), B s), C (p), D (q) PART # II Comprehension # 1 : 1. C 2. B 3. B Comprehension # 2 : 1. A 2. B 3. B Comprehension # 3 : 1. C 2. C 3. B Comprehension # 4 : 1. A 2. B 3. B Comprehension # 5 : 1. A 2. C 3. A Comprehension # 6 : 1. B 2. C 3. D Comprehension # 7 : 1. C 2. A Comprehension # 8 : 1. B 2. C 3. B Comprehension # 9 : 1. C 2. A