1.02 x 10. = 3 mol lit

Transcription

1 XII ISC CEMISTRY SPECIMEN QUESTION PAPER FOR BOARD EXAM:013 CEMISTRY PAPER 1 ANSWERED BY: S. NARAYANA IYER,M.Sc.M.Phil (TEORY) (Three ours) (Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.) Answer all questions in Part I and six questions from Part II, choosing two questions from Section A, two from Section B and two from Section C. All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer. The intended marks for questions or parts of questions are given in brackets [ ]. Balanced equations must be given wherever possible and diagrams where they are helpful. When solving numerical problems, all essential working must be shown. In working out problems use the following data: Gas constant R = cal deg -1 mol -1 = JK -1 mol -1 = dm3 atm K -1 mol -1 1 l atm = 1 dm 3 atm = J. 1 Faraday = Coulombs. Avogadro s number = PART I (0 Marks) Answer all questions. Question 1 (a) Fill in the blanks by choosing the appropriate word/words from those given in the brackets: (hydrolysis, reduction, oxidation, vacant, osmotic, above, benzoic acid, phenol, aniline, below, can, decreases, increases, cannot, crystal, ionization, rate, rate constant.) [5] (i) A catalyst start a reaction but it can increase the of the reaction. cannot, rate (ii) Electrons trapped in the sites of the lattice are called F-centres. vacant, crystal (iii) An aqueous solution of sugar boils o C and freezes o C. above, below (iv) Toluene on with alkaline potassium permanganate gives benzoic acid (v) The degree of of ammonium hydroxide on addition of ammonium chloride. ionization, decreases (b) Complete the following statements by selecting the correct alternative from the choices given:- [5] (i) For reaction N O 5 NO + O, the rate and rate constants are 1.0 x 10-4 mole litre -1 sec -1 and 3.4 x 10-5 sec -1 respectively. The concentration of N O 5 at that time will be: (1) 1.73 mol lit -1 () 3 mol lit -1 (3) 1.0 x 10-4 mol lit -1 (4) 3. x 10 5 mol lit -1 () 3 mol lit -1 (First order rate = K [N O 5 ] and x 10 So [N O 5 ] = = 3 mol lit x 10 (ii) Ethanoic acid dimerises in solution. Its molecular mass determined from its depression of freezing point of the solution will be: (1) Same as the theoretical value () alf its theoretical value (3) Double its theoretical value (4) One third of its theoretical value. (3) Double its theoretical value (iii) Magnesium displaces hydrogen from dilute acid solution because: (1) The oxidation potential of magnesium is less than that of hydrogen. -1 1

2 () The reduction potential of magnesium is less than that of hydrogen. (3) Both magnesium and hydrogen have same oxidation potential. (4) Both magnesium and hydrogen have same reduction potential. () The reduction potential of magnesium is less than that of hydrogen. (iv) In the series of reactions C 3 COO N 3 A heat B P O 5 C the product C is: (1) Acetyl chloride () Ammonium acetate (3) Acetic anhydride (4) Methyl cyanide (4) Methyl cyanide (v) In the reaction PCl 3 (g) + Cl (g) PCl 5 (g), the equilibrium will shift in the opposite direction, if: (1) Chlorine is added. () PCl 3 is added (3) Pressure is increased (4) Pressure is reduced. (4) Pressure is reduced. (c) Answer the following questions: [5] (i) Among equimolal aqueous solutions of MgCl, NaCl, FeCl 3 and C 1 O 11, which will show minimum osmotic pressure? Why? C 1 O 11. Osmotic pressure is a colligative property which depends on the number of particles. C 1 O 11 is a nonelectrolyte and so it does not ionize in water. MgCl, NaCl and FeCl 3 are electrolytes and so ionize in water to give more particles. ence osmotic pressure is higher than C 1 O 11. (ii) If K c for the reaction N + 3 N 3 is 1.5 x 10-5 (mol/litre) - write the value of K c for the reaction ½ N + 3 / N 3 [N 3] -5 K c = = 1.5 x 10 [N ] [ ] [N 3] K c = = 1.5 x 10 = x 10 ( mol/litre ) 1/ [N 3/ ] [ ] (iii) The p of acetic acid decreases with dilution.state the law governing this statement. Ostwald s dilution law. The degree of dissociation of a weak electrolyte is directly proportional to the square root of its dilution or inversely proportional to the square root of its concentration. α = KV K α = C K = the dissociation constant or ionization constant (iv) Xenon gives a series of fluorides, but elium and Neon do not. Why? Xenon has large size and less ionization energy. elium and Neon do not contain d-orbitals in their respective valence shells and hence their electrons cannot be promoted to higher energy levels like that in Xenon to form compounds with fluorine. (v) Calculate the number of coulombs required to deposit 0.5 g of aluminium ( atomic mass 7 ) from a solution containing Al 3+ ions x 0.5 = 1715 coulombs coulombs is the quantity of electricity required to deposit 1 gram equivalent of any substance. (v) Calculate the number of coulombs required to deposit 0.5 g of aluminium (Atomic mass 7) from a solution containing Al 3+ ions.

3 96500 x 0.5 = 1715 coulombs coulombs is the quantity of electricity required to deposit 1 gram equivalent of any substance and 9 is the equivalent weight of aluminium. (d) Match the following: (i) CCl 3 + NaO (ii) proteins (iii) carbohydrates (iv) Lewis base (v) KF (a) fluorine (b) starch (c) ammonia (d) peptide linkage (e) isocyanide test (i) CCl 3 + NaO (ii) proteins (iii) carbohydrate (iv) Lewis base (v) KF (e) isocyanide test (d) peptide linkage (b) starch (c) ammonia (a) fluorine PART II (50 Marks) Answer six questions choosing two from Section A, two from Section B and two from Section C. SECTIO) A Answer any two questions. Question a) (i) A certain aqueous solution boils at o C. What is the freezing point? K b for water is 0.5 K/mol and K f is 1.67 K/mol. Tb Kb = Tf Kf = Tf x T f = = T f = = K (ii) a solution containing 1 g of NaCl in 100 g of water freezes at o C.Calculate the degree of dissociation of NaCl.(Na= 3, Cl = 35.5 Kf for water = 1.87 K/mol) Kf w T f = m W 1000 x 1.87 x 1 m = = x i = = (56.5 is the molecular mass of NaCl ie = 58.5) i α = x - 1 ( x = for NaCl ie. NaCl Na + Cl ) α = = The degree of dissociation of NaCl = 88.9% b) (i) Explain graphically how the rate of a reaction changes with every 10 o C rise in temperature. 3

4 Energy distribution curve of a gaseous reacting system at a temperature t and and t With rise in temperature the kinetic energy of the molecules increases. The energy distribution curve flattens out and shifts towards higher enrgy regions. At (t + 10) collisions of a much larger fraction of molecules become effective to form products. This is why the rate of reaction becomes nearly two times for every 10 o C rise in temperature. (ii) ow is the activation energy related to its rate constant? Arrhenius equation k = Ae / or -Ea RT ln k = ln A - / or Ea RT log k = log A - / Ea.303RT A is a constant called frequency factor k = rate constant Ea = Activation energy of a reaction R = gas constant T = temperature of the reaction on absolute scale. (iii) The half life period for the decomposition of of a substance is.5 hours.if the initial weight of the substance is 160 g how much of the substance will be left after 10 hours? Quantity left after n half life period = Number of half life for 10 hours = 4 So the quantity left after 4 half life = 1 n x initial quantity 1 4 x 160 = 10 g Question 3 (a) (i) Define Frenkel defects of an ionic crystal Frenkel defect is due to shifting of an ion from its normal position to an interstitial in the crystal lattice thus causing vacancy in the original position. The overall density is not affected by this change. These defects are found in silver halides. (ii) Iron has an edge length 88pm. Its density is 7.86 gm cm -1 Find the type of cubic lattice to which the crystal belongs (atomic mass of iron is 56) z x = 6.03 x 10 x (88 x 10 ) -10 ( 1 pm = 10 cm ) x 6.03 x 10 x (88 x 10 ) 3 z = = Number of particles in a body centred cubic unit cell = No of corners x 1 / 8 + centre particle = 8 x 1 / =. So the crystal belongs to body centred unit cell. 4

5 (b) Explain giving reasons why: (i) Mg(O) is sparingly soluble in water but highly soluble in ammonium chloride solution. Mg(O) is a weak base. N 4 Cl is acidic due to hydrolysis. So neutralization takes place and dissolves. (ii) When hydrogen sulphide is passed through acidified zinc sulphate solution, white ppt of zinc sulphide is not formed. The solubility product of zinc sulphide is high and is not exceeded by ionic product in acid solution and so remains in solution. In acid solution dissociation of S is suppressed due to common ion effect of +. It gives less S - ions. (c) The equilibrium constant for the reaction (g) + I (g) I(g) is 49.5 at 440 o C. If 0. mole of and 0. mole of I are allowed to react in a 10 litre flaks at this temperature, calculate the concentration of each at equilibrium. 4x 49.5= (0. - x) (x) 49.5 = (0. - x) x 49.5 = (0. -x) 0.x 7.03 = (0. -x) x = 7.03 x x x = x x = = [ ] = = [I ] = [I] = x = x = Question 4 (a) (i) What is specific conductance of a solution and what is its unit? ow is it related to the equivalent conductance of the solution? The reciprocal of specific resistance is called specific conductance of a solution. 1 specific conductance = ρ It is defined as the conductance of a solution taken in a cell when electrodes are at unit distance from each other and each has area equal to 1 cm Or It is defined as the conductance of one centimeter cube of the solution of the electrolyte. Unit: ohm -1 m -1 (S I Unit) ohm -1 cm -1 (C G S Unit) Equivalent conductnce Λeq = k x 1000 C C is the concentration of the solution in gram equivalent per litre k = specific conductance (ii).5 ampere current is passed through copper sulphate solution for 30 minutes. Calculate the number of copper atoms deposited at the cathode (Cu = 63.5) Eiz x.5 x 30 x 60 3 w = = = x 10 = x 10 atoms x x (iii) Four metals W,X,Y and Z have the following values of E o red W = V X = -.93 V Y = V Z = V Arrange them in the increasing order of reducing power. 5

6 Ans; Reduction potential more negative is the most reducing. So the increasing order of reducing power is Z,Y,W and X. b) (i) On adding sodium acetate to aqueous solution of acetic acid what happens to the p of the solution? Give reason for your answer. - + C3COO C3COO + Na Due to common ion effect of C 3 COO - ydrogen ion concentration [ + ] decreases. So the p value increases. (ii) Calculate the p of an aqueous solution of ammonium formate assuming complete dissociation. (pka for formic acid = 3.8 and pkb for ammonia = 4.8) Ammonium formate is a salt of weak acid and weak base. (pkw + pka - pkb) p = p = = 6.5 c) Explain autocatalysis with one example. A substance formed in the course of a reaction, sometimes acts as a catalyst. This phenomenon is known as autocatalysis. For example, when acidified potassium permanganate is added to warm oxalic acid the decolourisation of the permanganate does not take place readily in the beginning. But after the first portion of the permanganate is decolourised, the reaction becomes quite fast. The manganese sulphate formed in the reaction acts as an autocatalyst. SECTIO) B Answer any two questions Question 5 (a) (i) State the geometry and magnetic property of tetracarbonyl nickel according to the valence bond theory. Nickel atom: 3d 8,4s [Ni(CO) 4 ] The geometry of tetrahedral. sp 3 hybridization the molecule is The number of unpaired electrons is 0 It is diamagnetic. (ii) What type of structural isomers are [Pt(O) (N 3 ) 4 ]SO 4 and [Pt SO 4 (N 3 ) 4 ] (O)? ow will you identify the isomers with a chemical test? Ionization isomers. [Pt(O) (N 3 ) 4 ]SO 4 gives test for SO4 - ion. With BaCl solution a thick white ppt is formed. [Pt SO 4 (N 3 ) 4 ] (O) will not give a thick white ppt with BaCl. (b) Name the co-ordination compound used for the following: (i) Treatment of cancer. 6

7 (ii) Treatment of lead poisoning (1) Cis [PtCl (N 3 ) ] known as cisplatin is useful in the treatment of cancer. (ii) Ethylenediaminetetraacetic acid [EDTA] Question 6 (a) Explain giving reasons why: (i) The halogens are coloured and the colour deepens from fluorine to iodine. alogen Fluorine Chlorine Bromine Iodine Colour Pale yellow Greenish yellow Reddish brown Violet The colour of halogens is due to the fact that their molecules absorb radiations from visible light and the outer electrons are easily excited to higher energy levels. The amount of energy required for excitation depends upon the size of the atom. Fluorine atom is the smallest and the force of attraction between the nucleus and the outer electrons is very large. As a result it requires large excitation energy and absorbs violet light (high energy) and therefore appears pale yellow. The size of halogen atom increases from fluorine to iodine. Iodine needs less excitation energy and absorbs yellow light of low energy. Thus it appears dark violet. So the colour deepens from fluorine to iodine. (ii) In a given transition series, the atomic radius does not change very much with increasing atomic number. The screening effect caused by 3d electrons decreases the magnitude of nuclear charge and consequently the atomic radii do not change very much. (b) Draw the resonating structures of ozone molecule. Question 7 (a) (i) Give equations to show the use of aqua regia in dissolving platinum. Pt + NO 3 + 8Cl [PtCl 6 ] + NOCl + 4 O (ii) Draw the structure of Xenon hexafluoride molecule and state the hybridization of the central atom and the structure of the molecule. sp 3 d 3 hybridization Distorted octahedral (b) Write balanced equations for the following reactions: (i) Ozone and alkaline potassium iodide. (ii) Sodium sulphite and acidified potassium permanganate (i) Ozone is oxidized to KIO 3 and KIO 4 KI + 3O 3 KIO 3 + 3O KI + 4O 3 KIO 4 + 4O (ii) Acidified KMnO 4 oxidizes sodium sulphite to sodium sulphate KMnO SO 4 [ Na SO 3 + [O] Na SO 4 ] x 5 K SO4 + MnSO O + 5[O] KMnO SO Na SO 3 K SO 4 + MnSO O + 5 Na SO 4 7

8 SECTION C Answer any two questions. Question 8 (a) Write equations for the following reactions and name the reactions: (i) Benzene diazonium chloride is treated with copper and hydrochloric acid. N + Cl - Cu Cl Cl + N This reaction is called Gatterman reaction (ii) Formaldehyde is treated with 50% caustic soda solution. CO + NaO C 3 O + COONa Methanal Methanol sodium formate This reaction is called Cannizzaro reactions. (b) (i) Write the structures of the isomers of 3 phenyl prop--enoic acid. C = C COO C = C COO Cis Trans (ii) What type of isomerism is exhibited by the following pairs of compounds: (1) C(C)3 CO and (C5) CO Position isomerism C 3 -C -C -C -C -O and C 5 C 5 CO Optical isomerism Mirror 8

9 d-form l-form (c) Give one good chemical test to distinguish between the following pairs of compounds: (i) Urea and acetamide On heating gentrly urea form biuret which gives violet colour with copper sulphate solution.acetamide will not give this reaction. (ii) 1-propanol and methyl -propanol. Lucas test: With Lucas reagent (Cl and ZnCl ) cloudiness appears immediately for -methyl--propanol while 1- propanol does not react with Lucas reagent and no cloudiness is produced. (d) Name the monomeric units of Nylon 66. Ans; COO ( C ) 4 COO Adipic acid and N Question 9 (a) Identify the compounds A, B, C, D, E and F. (C ) 6 N examethylene diamine A = C 3 Cl (Friedel Craft s alkylation B = Chromyl chloride (CrO Cl ): Etard s reaction C = NaO D = Benzoic acid ( C 6 5 COO) E = Benzoin O O C 6 5 -C-C-C 6 5 F = K Cr O 7 / SO 4 (b) ow can the following conversions be brought about? (i) Ethanoic acid to ethylamine. eat C 3 COO N 3 C 3 COON 4 C 3- CO-N + O (ii) Aniline to benzoic acid. N C 3 C N N = NCl LiAl 4 C 3 CN CN P O 5 COO NaNO /dil.cl Below 5 o C Cu (CN) KCN alcoholic (c) What is a zwitter ion? Represent the zwitter ion of glycine. O ( + ) Benzoic acid 9

10 Amino acids contain both acidic and an amino group in the same molecule.in aqueous solution the COO group loses a proton while the N group gains a prton to form a dipolar ion having the terminals as COO - and -N 3 +. It is called zwitter ion. 3 N + C COO - Question 10 (a) An organic compound A on treatment with ethanol gives a carboxylic acid B and a compound C. ydrolysis of C under acidic condition gives B and D. Oxidation of D with acidified potassium permanganate also gives B. B on heating with calcium hydroxide gives E with molecular formula C36O. E does not give Tollen s test but reacts with iodine and caustic potash to give a yellow precipitate. (i) Identify A, B, C, D and E. E is Acetone C 3 -CO-C 3 C 3 CO O + C 5 O C 3 COO + C 3 COOC 5 C 3 CO Acetic acid Ethyl acetate Acetic anhydride B C A ydrolysis C 3 COOC 5 C 3 COO + C 5 O Ethyl acetate [ + ] Acetic acid Ethanol C B D [O] C 5 O C 3 COO Ethanol KMnO 4 Acetic acid SO 4 D B OOCC 3 C 3 COO + Ca(O) Ca + O Acetic acid OOCC 3 B OOCC 3 Ca OOCC 3 C 3 -CO-C 3 Acetone E + CaCO 3 (ii) Write balanced equation of E with iodine and caustic potash and name the reaction. C 3 -CO-C 3 Acetone E + 3I + 4KO CI 3 + C 3 COOK + 3KI + 3 O Iodoform Iodoform reaction (b) (i) Name the functional groups that distinguish glucose and fructose. ow will you distinguish between the two compounds? 10

11 Glucose functional group is aldehydes group CO Fructose functional group is keto group CO Glucose forms a brown resinous mass with warm sodium hydroxide solution. Fructose does not form a resin with sodium hydroxide solution. (ii) What are polyesters? Give one example of polyester and the monomers Polyesters are condensation polymers of diols with aromatic dicarboxylic acids and contain ester linkage eg. Terylene or Dacron The monomers are Ethylene glycol C O C O and Terephthalic acid OOC COO 1,4-benzene dicarboxylic acid (c) Give balanced equations for the following reactions: [] (i) Aniline and benzoyl chloride. N NCOC C 6 5 COCl + Cl Aniline benzanilide (ii) Diethyl ether and hydroiodic acid (cold). C 5 - O - C 5 + I cold C 5 O + C 5 I 11