1994 PAPER I QUESTIONS 7. Show using a labeled diagram, the SECTION A effect of dissolved 25g of CaC12 in 1(a). What are isotopes? l00g of water on the boiling point of (b). Five atoms have the following water. composition 8. The following reactions occur Atom Protons Electrons readily with a negative sign of ΔG. Neutrons suggest the reactive value (or sign) A 8 8 8 of the other thermodynamic B 8 7 8 functions in each ease C 8 8 7 a) A perfect gas expanding into D 7 8 8 vacuum E 8 8 9 b) The burning of petrol (octane) in a Which of these are isotopes? car 2. 1.32g of magnesium metals was c) The mixing of two gases at dissolved in dilute hydrochloric acid constant total volume (no chemical and the solution heated in a stream of reaction). hydrogen chloride. Find the simplest 9. Distinguish between a π- bond and a formula for the metal chloride sigma-bond using a molecule of formed if 5.26g of the anhydrous ethylene. metal chloride remained. 10. (a). Explain each of the terms in the 3. Consider the data given below expression Atom N 1 O PV 3nmu2 Atomic number 7 (b). Show how the Avogadro’s 8 hypothesis may be deduced from this 1st ionization energy KJ/mol 1450 expression. 1360 Explain the observed trend in the SECTION B ionization of the elements. 11. (a) Explain the term “mechanism of 4. Explain why the alkali metals are a reaction” softer and have lower boiling (b). In the reaction aX + bY à temperature than the alkali earth products, the rate of the reaction metals. expressed as K (X)m (Y)n from the 5. In an experiment where a current of following results; 0.65 A was passed through copper, water and silver calorimeter, the X Y Rate following masses of elements were Moles lit-1 Moles lit- Moles lit 1sec1 respectively liberated: 0.0143g of 4 hydrogen; 0.114g of oxygen: 1.54g 1.4 x 10-2 2.5 x 10-2 7.49 x 10-9 of silver and 0.449g of copper. Show 2.8 x 10-2 4.6 x 10-2 5.92 x 10-8 that the above results agree with 2.8 x 10-2 4.6 x 10-2 5.92 x 10-6 faraday’s second law. 6. Explain the observation that: An i. Deduce the order ‘m’ with respect to X aqueous solution of KCN changes ii. Deduce the order ‘n’ with respect to Y red litmus paper blue whereas iii. What is the overall order of the reaction? aqueous solution KC1 does not. iv. If the rate constant of the above ii. The pH of the solution. reaction 2200C is found to be exactly b.(i) What is buffer solutions? double its value at 2300C. Calculate ii. If 3.28g of sodium ethanoate was the activation energy of ‘1’ reaction dissolved in 1dm3 of the solution in and give a reason for the increase in (a) above what would be the pH of the rate constant. the resulting solution? 12. Describe the electrolysis of CuSO4 iii. Suppose 1cm3 of a molar solution of using hydrochloric acid is now added to a. Platinum electrodes 1dm3 of the solution in b (ii) above, b. Copper electrodes what would be the new pH? State the reaction at the cathode iv. Comment on the pH values obtained and anode respectively giving reason between (ii) and b (iii) above. in case. Also 1 5 . ( a ) S t a t e t h e m a jor contribution of each state what is observed. of the following scientists to the c. Calculate the e.m.f. of a cell elucidation of atomic structure containing half-cells. The K.E. of the half-cell, half-cell that undergoes i. Rutherford ii. Pauli reduction is 0.01. Given that the iii. de Broglie’s iv. Faraday reduction takes place at 250C and v. Moseley latm. (R = 8.314 J/K/mol) b. Using the above stated and defined 13a. Write the electronic configuration of rules or principles show how the the valence shell of the following electrons are arranged in manganese. i. Group IIA ii. Group IV c. In the reaction: iii. Group VII MnO - + H+ + Fe2+ Mn2+ + H20 + 4 b. Using a named member of group II Fe3+ A i. Balance the redox reaction i. Write the formula of an oxide of a ii. What is the oxidation number of group IIA element. manganese in Mn04 ion? ii. Write the bond type of the oxide iii. What species is reduced and which is in (i) above. oxidized iii, Write the chemical equation for iv. Explain using its electron the reaction of the oxide with water. configuration how manganese is able iv. Write an equation for the effect of to show variable oxidation states in adding aqueous hydroxide obtained this reaction. in (iii) above with HCI. v. What role is the H+ playing in the Ci. Give four distinct features common reaction. to d-block elements 16(a). What is the kinetic theory? ii. Which of Sc3+ or Cu2 is (b). How can the kinetic theory be used paramagnetic? to explain: iii. Which of Zn or Cu is a transition i. The Boyle’s law metal? ii. The charle’s law 14.(a) Given a 0.01 molar solution of iii. The dalton’s law of partial pressure ethanoic acid calculate (c). What is the pressure of a mixture of 2g of hydrogen and 3.2g of oxygen i. The hydrogen ion concentration of stored in a the solution (KA for CH3COOH = 15dm-3 vessel at 270C. 1.84 X 10-5) 1994 PAPER 1 SOLUTION filled electronic structure. Hence N SECTION A has more ionization energy than O. 1(a). Isotopes are atoms of the same 4. The alkali metals have only one element with the same atomic valence electron compared to the two number but different atomic mass of the alkali earth metal. The alkali due o difference in the number of earth metals therefore have the neutrons in their bonding electron per metal atom and nucleus e.g3 5CI and 3 7CI are smaller (due to increase in 1 7 1 7 nuclear charge) in size and have b. 16 A, 16B, 15C, 15D, 15E,From the 8 8 8 7 8 higher bonding energies. They also definition of isotopes. A, C, and E have high mass to volume ratio are isotopes hence, higher density which makes 2. Mass of the Mg metal = l.32g them very harder than alkali metals Mass of metal chloride = that have low mass to volume ratio 5.26g and relatively lower charge. Mass of chlorine = 5. Faraday’s second law of electrolysis 3.94g states that if the same quantity of Divide by Mg electricity is passed through solution CI of different electrolytes, the number Atomic mass 24 1.32 of moles of the different substances 3.94 liberated is proportional to their 35.5 electrochemical equivalent. Divide through 0.055 0.111 of mass of H O Cu 0.055 0.111 vide Ag 0.055 through 0.0143 0.114 0.449 0.055 1.542 1 0.0143 0.114 0.449 1.542 2 0.0143 0.0143 0.0143 0.0143 Therefore, the simples formula of the 0.0143 chloride of Mg is MgC1 2 1 : 7.47 : 31.40 3. N = 1s2 2s2 2p3 107.8 Which are similar to the ratios of the O = 1s2 2s2 2p3 equivalent 6. KC1 is a salt of strong acid and strong base and exists in aqueous solution as K+ and C1-ions. The ions The ionization energy increases are not hydrolyzed and hence the generally across the period. solution is neutral. KCN on the other Ionization energy increases from N hand is a salt of a very weak acid and to O due to special stability, which is a strong base. In solution, only CN is as a result of their electronic hydrolyzed as arrangement. The half - filled CN-+H O HCN+OH- 2 electronic structure is more stable and need more energy to be removed than where there is no such half - The solution is alkaline due to the m= Mass of a molecule presence of OH- and hence turns red u= Root Mean Square velocity litmus blue. (b) Consider equal volumes of two gases ^ 7. n A$B at same temperature and D pressure. P ∴ PV for A = PV for B 1 Kinetic theory equation = PV = P2 1 nmu2 3 K.E = 1 mv2 2 T T Temp. (Boiling point) PV = 2 n K.E = nRT 1 2 3 A = Pure solvent with high pressure ∴ n = n => (equal volumes of gases A B and low boiling point. at same temperature and pressure B = Solvent with dissolved boiling contains equal number of molecules) have reduced pressure and elevated which is Avogadro’ s statement. point. 8. ΔG = ΔH + TΔS SECTION B For ΔG to be negative, 11(a). Refer to 2003 Paper 1 question 16 ΔH must be negative and ΔS positive (a) ii (b) Let Rate R = K (x)m (y)n c. Here, ΔS, is zero for ΔG to be i. negative, ΔH will be negative. 9. π - bond is formed between 5.92x10−6 K(2.8x10−4)m (4.6x10−2)n = unhybridized orbitals while σ - bond 5.92x10−8 K(2.8x10−2)m x(4.6x10−2)n is formed from hybrid orbital. In ethylene, the double bonds between ∴ 102 = 10m the two carbon atoms contain one π - ∴ m = 2 bond is formed between ii. unhybridized p – orbital on the C – atoms involved, while the σ + bond in formed between sp2 hybrid 5.92x10−8 K(2.8x10−2)m x(4.6x10−2)n = orbitals, of the two carbon atoms 7.49x10−9 K(7.49x10−9)m x(2.5x10−2)n involved. Also, the for C + H bonds are σ - ∴ 8 = 4(2)n ∴ 2 = 2n bonds formed from sp2hybrid ∴ n = 1 orbitals and s + orbitals of hydrogen iii. Overall order m + n = 2 + 1= 3 atom. π - bond is formed from lateral iv. T = 220C = 493K 1 overlapping and hence weaker than T = 2300C = 503K 2 σ bond, which is formed from head- Ea = ? R = 8.31 on approach of the hybrid orbitals. σ K = X, K = 2X 1 2 bonds are formed from true overlap K Ea ⎡ 1 1 ⎤ of orbitals between S and P, Px and log 2 = ⎢ − ⎥ T 2.303R T T Px orbitals. 1 ⎣ 1 2⎦ 10(a). P = Pressure of the gas molecules Ea ⎡T −T ⎤ = = 2 1 V= Volume of the container ⎢ ⎥ 2.303R TT ⎣ 1 2 ⎦ N= Number of molecules 2X Ea 503−493 0.059 log =log2 E11 = E0 log 0.001 X 2.303x8.31503x493 2 E11 – E0 = - 0.0295 (log 0.001 – Ea ⎡ 10 ⎤ 0.3010 = log0.5) ⎢ ⎥ 2.303x8.31 503x493 ⎣ ⎦ = - 0.0295 x – 2.69 = 0.3010x2.303x8.31x053x493 0.079355 Ea= 13. (a) (i) ns2 (ii) ns2np2 (iii) ns2np5 10 Ea = 14284.72 J. (b). A name member of group 11A is Ca or any of the Kinetic Energy (K.E) increases with temperature. The number of effective i. Oxide: CaO ii. Bond type: electrovalent or collisions and the number of collisions with minimum activation ionic iii. CaO + H O à Ca (OH) energy of reaction also increases. 2 2 12(ai). Platinum electrodes; iv. Ca (OH)2 + 2HC1 à CaC12 CuSO4 Cu2+ + SO 2- + H2O 4 H O H+ + OH- (c)(I)(i) They have high melting and boiling 2 points Cathode: Cu2+ + 2eà Cu discharged because (ii) They form coloured salts (iii) They form complex ions it has lower discharge potentials. Anode: (iv) They have variable oxidation states (v) They can act as catalyst O is given off (liberated) because 2 OH- also has lower discharge II 4S2 3d SC potentials and is discharged (ii) Copper electrodes: 4S0 3d Cathode: Cathode discharges Copper: Se3+ Cu2 + + 2e+à Cu H+ e+ 2 2+ 4S2 3d 2H+ + 2e+ H2 because Cu2+ has lower Cu discharge potential. 4S0 Anode: SO42-à SO + 2e- Cu2 4 OH- àOH + e- Cu à Cu2+ + 2e- Para magnetism is a function of unpaired electron; for an element to Because anode is Cu, and needs the be paramagnetic it must have at least lowest e.m.f therefore no ions are an unpaired electron, Sc3+ has no discharge instead copper dissolves unpaired electron hence is not from the anode or copper dissolves paramagnetic while Cu2+ has one from the anode. unpaired electron hence is (c). Using Nernst equation paramagnetic. 0.059 E1 = E0 - log Ke……….. (l)\ (iii) Cu2+ is a transition metal because it 2 has incompletely filled d – orbitals, For 0.5 while Zn is completely filled hence 0.059 E1 = E0 log 0.5 not a transition metal. 2 14(a) CH COOH CH COO- + H+ 3 3 For 0.001 H+ = K C 15(i). Rutherford’s major contribution to a the atomic structure in the discovery = 1.84zx10−5x0.01 of nucleus and proposition of atomic model. = 1.84zx10−7 (ii) De Broglie’s major contribution was = 4.28 x 10-4mol/= / dm3 his suggestion that the particles- pH = - log (H+) wave-phenomenon is restricted to = - log (4.28 x 10-4) light. But that all particles includes - 3.37 atom must have an associated waves. This is because for a weak acid, it is (iii) Faraday’s major contribution was only little CH COOH that will 3 evidence that electrical charge was dissolve not continuous but existed in the B(i). A buffer solution is a solution which form of diserete particle. He resist large changes in PH if it is discovered that the amount of diluted or if either a strong acid or a element deposited is proportional to strong base was added to the the quantity of electricity needed to solution. deposit one mole of he substance. ii. Concentration of either in mol / dm3 (iv) Pauli’s major contribution was that 3.28 he made it possible to apply the / 82 = 0.04 mol dm3 quantum number to express of atom (C N COONa) = 0.04 mol / dm3 3 with his exclusion configuration of (CH COOH) = 0.01 mol / dm3 3 atom with his exclusion principle K (CH COOH) (H+) = a 3 known as Paulis Exclusion Principle. (CH COON ) (v) Moseley’s major contribution was 3 a 0.01 the determination of nuclear charges (H+) = 1.84 x 105 x and identification of atomic number 0.94 as the number of protons in the = 4.6 x 10-6mol / dn4 nucleus of atom. PH = - log (H4) = -log (4.6 x 10-4) B(i) Electron fill the lower energy level = 5.34 before moving to higher one, the (iii) The addition of 1cm3 of molar HC1 order of increasing energy level is to 1dm3 of H O would give a solute 2 1s < 2s < 2p < 3 s < 3p < 4p < 3d in which (103) = 0.001 mol dm3 by (Aulbau’s principle) combination between the added H+ (ii) In a degenerate orbital, electron is ions and the reserve of ethanoat ions filled singly before pairing occurs. in the buffer solution (CH COO-) 3 (Hund’s rule) will be reduced form 0.04 to 0.039 (iii) The maximum number of electron in mol dm-4 (CH COOH) will increase 3 each orbital should not be more than from 0.01 to 2.011 mol dm-3 the 2. change in pH likely 0.06 unit (Pauli Exclusion Principle) whereas when the acid is added to (iv) No two electrons in each orbital will 1dm-3 of water the pH chances from have the same quantum number. 7 to 3 (Pauli Exclusion Principle) (iv) The value obtained was that the pH Using the above stated rules and of the solution is only slighty principles the electronic affected. configuration of Mn is written as temperature. Temperature depends shown below on average K.E of molecules i.e. T α Mn = 1s2 2s22p6 3s2 3p64s23d5 ½ MU-2 C(i) Mn O - + H+à Mn2+ + H O + Fe3+ Where M is mass of 1 molecule and 4 2 Mn O - + 8H+ + 5e à Mn2+ + H O U is molecular velocity. The speed 4 2 5Fe2+ à 5Fe3+ + 5e- or molecules (U) will increase with MnO + 5Fe2+ + 8H+ à Mn2+ + temperature and vice versa. 4 5Fe2+ + H O Since an increase in U results in 2 (ii) +7 more molecules striking the walls of (iii) MnO - is reduced, Fe2+ is oxidized the container per second, the p of a 4 Mn 4s2 3d5 (valence shell electronic gas will increase in temperature at configuration.) constant volume. Mn+7 4s0 3d0 to Mn2+ 4s0 3d5 (iii) Dalton’s law of partial pressure: In H+ proves the acidic medium for the an ideal mixture in a closed vessel, reaction. the p of mixture is caused by impacts 16Ia) Kinetic theory postulates that toe of the molecules acting molecules of gas in a state of independently against the walls of movement. This assumes the the vessel. The individual pressure of following: each gas will be the same as if it à An ideal gas consists of large no of were alone. particles moving at random (c) no of moles of H gas = 2/2 = 1. 2 à The articles collide with each other 0mole and with the walls of the container no of moles of O gas = 30/32 = 2 à The collision particles is perfectly 0.1mole elastic. : total no of moles of the gases = à No energy is lost when they collide 1.10mole à They mean radius of gas molecules PV = nRT (1 m3 = 1000dm3) is negligible compared to their T = 270C = 300K distance apart 1.1x8.31x300 P = à Increase in temperature causes the 15x10−3 motion of particles will collide more 1.83105Nm−2 frequently and the pressure of the P = container will be increased. 1.5x10−3 B(i) Boyle’s law the pressure is 1995 PAPER QUESTIONS proportional to the no of molecules SECTION A in a given volume or concentration 1.(a) What is a mole of a substance? of the gas. Since P is force per unit (b) How many particles are contained in: area P depends on mass or no of (i) 64g of Sulphur (IV) oxide strikes a unit area has per second. If (ii) 11 .5g of Sodium ions the volume of a given mass of gas is (c) What is the mass of half of Sulphur halved, at constant temperature no of (IV) oxide? molecules hitting unit area of wall in 2. A gas X diffuses four times as 1 sec is doubled. rapidly as Sulphur (iv) oxide under (ii) Charles’ Law: At a given p. the the same conditions. If the density of volume of a given mass of gas is Sulphur (iv) oxide at the given directly proportional to the absolute temperature and pressure is 2.88 x 8. Explain why an aqueous solution of 103gm what is the density of X? NH C1 is acidic whereas an aqueous 4 3.(a) Define a Brownsted-Lowry acid and solution of Na CO is alkaline. 2 3 base theory. 9. Silicon and Phosphorus belong to the (b) State the conjugate acid pair of the period in the periodic table. Both following bases. form hydrides, but silane (SiH4) (i) SO42- (ii) H O (iii)NH (iv)OH- reacts readily with water whereas 2 3 4. In a hypothetical chemical reaction phosphine (PH3) does not suggest a A + B + C à X + Y H = - ve reason. (g) (g) (g) (g) (g) Discuss the effect of temperature on 10. Calculate the relative molecular mass the thermodynamic feasibility of the of Chloroform from the following reaction. data obtained by Victor Meyer’s 5.(a) Differentiate between a Salt and a method. 0.22g of chloroform Base displaced 45 cm3 of air at 200C and (b) Identify with reason(s) which of 100.7 kNm-2 pressure (the saturated these is vapour pressure of water at 200C is a salt or a base. 2.3 19 kNm-2). Sodium chloride Sodium hydroxide SECTION B Aluminium Chloride anhydrous 11.a(i) Name the radiations emitted by Trioxocarbonate (iv) ion. naturally occurring radioactive 6. Classify the following statements as elements. TRUE OR FALSE (ii) Describe with the aid of a sketch, (i) A Daniel cell is a combination of how the respective radiations can be copper and Zinc electrodes immersed separated. in a standard solution of their ions. (iii) What changes in atomic number and (ii) Any metal above hydrogen in the mass number accompany a nucleus electrochemical series will displace that emits each of the naturally hydrogen from an aqueous solution occurring radioactive radiations? of an acid. (b) (i) What are isotopes? (iii) The attraction of metals for oxygen (ii) The isotopic mass of two isotopes as decreases down the electrochemical measured by a mass spectrometer are series. 34.969 and 36.968 a.m.u (iv) Iron will not rust if oxygen is absent respectively. Estimate the relative (v) The tendency for the halogens in atomic mass of the element if the their normal states to form negative percentage abundance of the less ions in solutions increase as you isotope is 75.53%. moves down the group. (c) The isotope 42K undergoes Beta 19 7. Balance the following redox decay to produce 42Ca. If after 62.0 equations by first writing an 29 appropriate balanced ion-electron hours, 96.88% of 42K was found to 19 half equation. have undergone transformation, (a) C1- + MnO4- + H+à Mn2+ + H2O determine the half-life of the isotope. + Cl2 12a) ‘Illustrate the diagonal relationship (b) Cu + NO3- à Cu2+ + NO + H2O between Lithium and Magnesium. (b) List four general features of the (iii) CH COOC H + H O 3 2 5(1) 2 (l) group.1A elements of the periodic CH COOH + C H OH 3 (l) 2 5 (g) table. (iv) H + S ‘ H S 2(g) (s) 2 (g) (c) Explain the following observations (v) CaCO CaO + CO 3(g) (s) (g) (i) The boiling points of group 1A (c) A mixture of 2 moles of ethanol and elements decreases down the group one mole of ethanoic acid are whereas those of VII B elements allowed to attain reaction, how much increases. ethyl ethanoate is present There is a general increase in size equilibrium. down the group 1A element 15(a) State the first and second laws of The various radii of an element in thermodynamics. group 1A follows the trend rx +< rx (b) In the thermodynamic equation ΔG = +< rx- ΔH - TΔS where ΔG is change in How can the electronegativity free energy, ΔH is change in entropy, difference between the constituent of and t is temperature of the system, a the following compounds or desirability or the system is the molecules be used to predict their attainment of a negative free energy. bond types? MgO, Br2 and Propose four (4) ways in which Differentiate between the concentration of chemical reactions can apply the the strength of that acid the following acids variations in values of the enthalpy with the dissociation constant, in order of and entropy to achieve this negative H3BO3 K1 = 10-9 free energy. HC1O4 K1 = 108 (c) From the following data, determine H3PO4K1 = 10-2 the thermodynamics feasibility of the Tetraoxsulpha(iv)acid H3SO4 K1 = 10 reaction Trioxonitrate (v) acid. HNO3 K1 = 101 C2H4(g) + H2(g) à C2H6(g) (c) Explain the observation the hydride If the standard heat of combustion in of iodine than the hydride of aqueous KJmol-1 are solution. C H => - 1410.0, H => - 286.0, 2 4(g) 2 (d) The dissociation constant is 1.8 x 10- C 2- 5 mol dm-3. For a 0.001 molar H => -1560.0 and if the standard 6(g) solution of the molecule, ca1culate entropy in JK mol are C H => -1 -1 2 4(g) (i) The degree of dissociation 219.0, H (g) => 130.6 and C H > 2 2 6(g) (ii) The hydrogen ion concentration 229.5. (iii) 16.(a)(i) What is colligative property? 14(a) Briefly explain the following terms (ii) Illustrate with two examples Reversible reaction (b) Show by using a labeled phase Equilibrium constant diagram, the effect of dissolving 2g Specific rate constant and 6g respectively of a compound Y (b) Write the concentration equilibrium in l30g of water in the freezing point constant Kc expression for the of water. following reactions. 1995 PAPER 1 SOLUTION (i) CO(g) + Cl2(g) COCI2(g) SECTION A (ii) 3Fe(S) + 4H2O(g) Fe3O4(s) + 4H2(g) la. A mole of a substance is the amount of that substance which contains the same number of particles as there are point, which is 0.250C lower than in 12g of 12C that of the pure solvent. Using the 6 above data, determine he relative bi. 46g of SO 2 molecular mass of solute Y. (relative molar mass of SO = 32 + (2 x 16) = 2 molar mass of naphthalene = 128g, 64g mass of Y in solution is 2g). No. of moles of SO = 2 Since from the equation we have mass 64 = = 1 mol been told that ΔH is negative, and the molarmass 64 more negative the free energy is, the No. of particles = 6.020 x 1023 more spontaneous is the reaction. particles. When there is increase in (ii) 11.5g of Sodium temperature, it will be observed that 11.5 reaction will be very fast because it No. of mole of Na = 0.5 mol 23 will make free energy more negative. No. of particles = 0.5 x 6.02 x 1023 When there is decrease in 1 mole of SO = 64g temperature, there will be decrease in 2 Half mole of SO =0.5 x 64g =32g. the rate of formation of the product 2 2. V.D of SO = 2.88 x 103 g/dm3, V.D because of less negativity of free 2 of energy. X= ?, Rate of SO =R, Rate of X=4R 5(a) A Salt is a product of neutralization 2 R V.Dso 4R 2.88x103 reaction while a Base is a substance, x = 2 = which reacts with an Acid during Rso V.D R V.D 2 x x neutralization reaction to form salt. 4 53.66 ÷√V.Dx b (i) NaC1 à It is a salt because it is the √V.Dx = 53.66 ÷ 4 = 13.42, V.Dx = product of the reaction between HC1 13.422 V.Dx = 180.09g/m3 and NaOH i.e. 3.(a) Brownsted - Lowry acid are proton (ii) NaOH à It is a base because it donors. Browñsted - lowry base are contains OH group. proton acceptor. (iii) AICI à It is a salt because it is the 3 (b) product of reaction between an acid Bases Acids and a base i.e. SO2− HSO4 AICI3 + 3H2O à AI (OH)3 + 3HCI 4 (iv) CO 2- à It is a Brownsted-lowry NH4 NH+4 base3 because it accepts H+ when OH- H O 2 added to it. 6. (i) True 4. Using ΔG = ΔH - TΔS (ii) True where ΔG = Gibbs free energy, Δ11 (iii) True = Enthalpy change, T = (iv) True Temperature, S = Entropy change. (v) False (c) A solution of 6.0g of naphthalene in 7(a) CI + Mno4 + H+ à H O + CI 2 2 125g of a certain solvent had a CI à CI --------------------- 2 freezing point, which is 1.500c lower CI- à CI + 2e ----------------- 2 than that of the pure solvent. Another Mno à Mn ------------------- 4 2 solution of compound Y in 130g of 2MnO - + 5e- + 8H à Mn2 4H O 4 + + 2 the same solvent has a freezing + 5e- --------------