Q. No. 1 Which of the following has been universally accepted as a reference electrode at all temperatures and has been assigned a value of zero volt? Platinum electrode Copper electrode Graphite electrode Standard hydrogen electrode Standard hydrogen electrode Q. No. The reaction 1 H (g) + AgCl (s) = H + (aq) + Cl (aq) + Ag(s) occurs in the galvanic cell : Ag AgCl(s) KCl(sol.)(AgNO 3)(sol.) Ag Pt H (g) HCl(sol. AgNO 3(sol) Ag Pt H (g) HCl(sol.) AgCl(s) Ag Pt H (g) KCl(sol.) AgCl(s) Ag Cell representation is (when ani electrode like platinum is involved in the construction of the cell. Cell representation is : Anodic chamber salt bridge cathodic chamber Q. No. 3 The equation representing the process by which standard reduction potential of zinc can be defined is : Zn + (s) + e Zn Zn (g) Zn + (g) + e Zn + (g) + e Zn Zn + (aq.) + e Zn (s) Reduction is (gain of e )
Q. No. 4 Which of the following statement is wrong about galvanic cell? Cathode id positive charged Anode is negatively charged Reduction takes place at the anode Reduction takes place at the cathode Wrong statement Reduction takes places at the anex Q. No. 5 Which are used as secondary reference electrodes? Calomel electrode Ag/AgCl electrode Hg/Hg Cl KCl electrode All of the above Q. No. 6 The standard electrode potentials (reduction) of Pt/Fe 3+, Fe + and Pt/Sn 4+, Sn + are + 0.77 V and 0.15 V Respectively at 5 0 C. the standard EMF of the reaction Sn 4+ + Fe + Sn + + Fe 3+ is 0.6 V 0.9 V + 0.31 V + 0.85 V As stated Standard reduction potentials of, 3+ + Pt Fe,Fe 4+ + Pt Sn,Sn +0.77 V +0.15V
EMF of cell = tendency of reaction to occure = st. oxidation point of Fe + + st. reduction pt of Sn 4+ =.77 + 0.15 =.6 V Q. No. 8 Strongest reducing agent is : K Mg Al I Reducing agent is one which undergo oxidation easily and the one with lesser reduction potential get oxidised easily and is strongest reducing agent. Q. No. 9 Zn can not displace following ions from their aqueous solution : Ag + Cu + Fe + Na + For a metal to displace other metal. Metal whose reduction potential is less reaction. If we want to remove a metal from is salt chase a metal which is more active/reactive that it means whose red. potential is d than i. Zn has reduction potential more than Na + so can t displace it. Q. No. 10 Which of the following displacement does not occur : Zn + H + Zn + + H Fe + Ag + Fe + + Ag Cu + Fe + Cu + + Fe Zn + Pb + Zn + + Pb For this displacement to occur the red.pot a Cu must be less than Fe. Q. No. 11 The oxidation potential of Zn, Cu, ag, H and Ni are 0.76, 0.34, 0.80, 0, 0.55 volt respectively. Which of the following reaction will provide maximum voltage? Zn + Cu + Cu + Zn +
Zn + Ag + Ag + Zn + H + Cu + H + + Cu H + Ni + H + + Ni Correct Answer Given oxidation potential of Zu, Cu, Ag, H of Ni 0.76, 0.34,.80, 0, 0.55 volt resp. Q. No. 1 The position of some metal s in the electrochemical series in decreasing electropositive character is given as Mg Al Zn Cu Ag. What will happen if a copper spoon is used to stir a solution of aluminums nitrate? The spoon will get coated with aluminum An alloy of copper and aluminum is formed The solution becomes blue There is no reaction There will be no reaction b/c reverse as stated decreasing electropositive character is mg > Al > Zn > Cu > Ag Cu + Al(NO 3 ) 3 No reaction For reaction to occur Cu has to be reactive. Q. No. 13 The standard reduction electrode potential values of the element A, B and C are + 0.68,.50, and 0.50 V respectively. The order of their reducing power is : A B C A C B C B A B C A A +.68 V B.50 C.50 V
CTM One with Red.pot with get reduced easily Red pot reactivity Reducing poverian to deal with its reactivity, that is the ease at the atom to be oxidized to form cation. So the order of reducing power is neastreactive. Q. No. 14 A metal having negative reduction potential when dipped in the solution of its own ions, has a tendency : To pass into the solution To be deposited from the solution To become electrically positive To remain neutral 0 Q. No. 15 The E 3+ Red. Potential means easy to be oxidized. So the metal will have tendency to pass into the solution leaving surplus is behind. M + M values for Cr, Mn, Fe and Co are 0.41, + 1.57, + 0.77 and + 1.97 V respectively. For which one of these metals the change in oxidation state from + to + 3 is easiest Co Mn Fe Cr Given, E 3+ M /M + (Reduction potential values) Cr.41 V (least red.pot.value) M n + 1.57 V Fe + 0.77 V Co + 1.97 V ease to change to + + 0 + 3 means ease of oxidation easily. (One with R potential value) Q. No. 16 E 0 for the half cell reactions are as, (a) Zn = Zn + + e; E 0 = + 0.76 V
(b) Fe = Fe + + e ; E 0 = + 0.41 V The E 0 for half cell reaction, + + Fe Zn Zn + Fe is 0.35 V + 0.35 V + 1.17 V 0.17 V Correct Answer E 0 for half cell reaction are, E 0 = oxidation potential of Zn + Redox potential of F + e =.76 + (.41) = +.35 V. Q. No. 17 A aqueous solution containing 1 M each of Au 3+, Cu +, Ag+, Li + is being electrolyzed by using inert electrodes. The value of standard potentials are : 0 0 0 0 E + = 0.80 V, E + = 0.34 V and E 3+ = 1.50 V, E Ag Cu Au + = 3.03 V Li Ag Cu Au Li With increasing voltage, the sequence of deposition of metals on the cathode will be : Li, Cu, Ag, Au Cu, Ag, Au Au, Ag, Cu Au, Ag, Cu, Li Given : [deposition of cathode will be maximum for me with reduction potential.] So order will be Au Ag Cu. Q. No. 18 The standard electrode potential for the reaction Ag + (aq) + e Ag (s) Sn + (aq) + e Sn (s) At 5 0 C are 0.80 volt and 0.14 volt, respectively. The emf of the cell. + + SnSn (1M) Ag (1M)Agis
0.66 volt 0.80 volt 1.08 volt 0.94 volt + Ag aq + e Ag s. 80 V.08 + Sn aq +e Sn s 0.14 V +.11 emf for cell Anodic chamber Cathodic chamber + + 1m 1m Sn Sn Ag Ag emf = oxidation potential of Sn + red potential of Ag + =.14 +.80 =.94 V Q. No. 19 The standard free energy change for the following reaction is 10 kj. What is the standard cell potential? H O (aq) H O(l) + O (g) + 0.75 + 1.09 + 0.40 + 0.640 Correct Answer n = F = 96 G= nfecell 1 CV = 1J E 1 10kJ mol = 96500C.mol cell 1 10 =1.088V 193 Q. No. 0 Calculate the standard free energy change for the reaction, Ag + H + H + Ag +, E 0 for Ag + + e Ag is 0.8 V + 154.4 kj + 308.8 kj 154.4 kj 308.8 kj E 0 for Ag + + e Ag = 0.80 V E cell = oxidation potential of Ag + Reduction potential of h =.80 V G= nf Ecell n = f = 96500 C mol 1
Ecell =.80 h= 154.4kJ Q. No. 1 The standard EMF of Daniel cell is 1.10 volt. The maximum electrical work obtained from the Daniel cell is 1.3 kj 175.4 kj 106.15 kj 53.07 kj Emf = 1.10 V n= nf Ecell n = F = 96500 C mol 1 = 1.300 kj mol 1 maximum work done Q. No. What is the free energy change for the half reaction Li + + e Li? Given E 0 Li+ = 3.0V, F = 96500 C mol 1 and T = 98 K. Li 89.5 kj mol 1 98.5 kj mol 1 3.166 CV 1 mol 1 89500 CV mol 1 h= nf Ecell n = 1 = 89500 J mol 1 f = 90500 C mol 1 = 89.5 kj mol 1 E cell n = 3.0 V Q. No. 3 The emf of Daniel cell is 1.1 volt. If the value of Faraday is 96500 coulombs per mole, the change in free energy in kj is 1.30 1.30 106.15 106.15 Correct Answer h= nf Ecell n = = 89500 J mol 1 f = 96500 C mol 1 E cell = 1.1 V Q. No. 4 Which of the following represents the potential of silver wire dipped in to 0.1 M AgNO 3 solution at 5 0 C? E 0 red (E 0 red+ 0.059) (E 0 ox 0.059) (E 0 red 0.059) + Ag + e Ag Re duction 0.059 E=Ered log10 1
0 red =E.0591 Q. No. 5 The reduction electrode potential E, of 0.1 M solution of M + ions (E 0 RP =.36 V) is :.41 +.41 4.8 None 0.059 1.36 log 1 m + 1.41log m + Q. No. 6 + + H (Pt)H3O (aq) Ag Consider the cell 1atm ph=5.5 xm what is the value of x? E 0 Ag +.Ag = + 0.799 V [T = 5 0 C] 10 M 10 3 M 1.510 3 M 1.510 M 55 10 1.03 = 0.799 0.059 log x x 3.79 5.5 3.79=log =10 10 = x 5.5 10 10 1.703 + = x 0.9 10 10 = x 10 = x Ag. The measured EMF of the cell is 1.03 V. Q. No. 7 The emf of the cell Ti + Ti + (0.0001 M) Cu + (0.01 M)/Cu is 0.83 V The emf of this cell will be increased by : Increase the concentration of Cu ++ ions Decreasing the concentration of Ti + Increasing the concentration of both (1) and () both
anodic chamber cathodic chamber Oxidation reduction to increases the emf the tendency of oxidation in oxidic chamber and tending of red in cathodic chamber should increases. Increasing the conc. of Cu 4 ions of decreasing the conc. of Ti + both will far. Q. No. 8 + + CoCo (C ) Co (C 1) Co for this cell, G is negative if : C C 1 C 1 C C = C 1 Unpredictable Correct Answer 0.0591 C1 E = log C cell n 1 C 1 > C for h n1 Q. No. 9 What will be the emf for the given cell? Pt H (g,p 1 ) H + (aq) H (g, P ) Pt RT P1 In F P RT P In 1 F P RT P In F P None of these Correct Answer RT P1 n = emf= ln F P 1 Q. No. 30 If the pressure of hydrogen gas is increased from 1 atm. To 100 atm., keeping the hydrogen ion concentration constant at 1 M, the voltage of the hydrogen half cell is at 5 0 C will be 0.059 V 0.059 V 0.95 V 0.118 V. Correct Answer emf of hydrogen half cell depends upon its concentration. So answer will be 0.059 V. Q. No. 31 The EMF of the cell Mg Mg + (0.01 M) Sn + (0.1M) Sn at 98 K is (Given E 0 Mg +,Mg =.34 V, E 0 Sn +,Sn = 0.14 V).17 V.3 V.51 V.45 V
Correct Answer 0 0.0591.01 E=Ecell log n.1 =. +.09 =.9 n = Q. No. 3 The potential of the cell containing two hydrogen electrodes as represented below Pt, H (g) H + (10 6 M) H + (10 4 M) H (g), Pt at 98 K is 0.118 V 0.0591 V 0.118V 0.0591 V 16 10 E=0+.0591 log 10 4 =.0591 =.118 V Q. No. 33 The emf of the cell H (1 atm)pt H + (a = x) H + (a = 1) H (1 atm) Pt at 5 0 C is 0.59 V. the ph of the solution is 1 4 7 10.59 ph= =10.059 0 E E ph=.059 Q. No. 34 The hydrogen electrode is dipped in a solution of ph = 3 at 5 0 C. the potential of the cell would be (the value of.303 RT/F is 0.059 V) 0.177 V 0.087 V 0.177 V 0.059 V 0.0591 1 E=E log =.05913=.177V 1 3 10 Q. No. 35 When two half cells of electrode potential of E 1 and E are combined to form a cell of electrode potential E 3, then (when n 1, n and n 3 are no. of electrons exchanged in first, second and combined half cells) : E 3 = E E 1 En 1 1+En E 3= n 3 E n E n E 3= n 1 1 E 3 = E 1 + E Correct Answer 3
n 3 E 3 = E 1 n 1 + E n E +E E = 1 n 3 1 n n 3 0 0 0 G 1 = G 1 + G 0 0 0 3 3 1 1 n E = n FE n FE Q. No. 36 If E 0 is 1.69 V and 0 + E is 1.40 V, then 3+ Au / Au Au / Au / 0.19 V.945 V 1.55 V None of these G = G + G 0 0 0 3 1 3 n3 3 0 n1 1 0 n 0 0 0 0 3E 3 = 1E 1 + E FE = FE FE n n n E 1 = 1.69 n = 1 E = 1.40 n = 3 0 0 0 1 1 n E + n E E 3 = n 0 1.69+1.403 E 3 = =.945 3 E 0 Au + Au 3+ Q. No. 37 Which reaction occur at cathode during electrolysis is fused lead bromide? Pb Pb +e Br+e Br Br Br +e + Pb +e Pb + At cathode, Pb +e Pb PbB Pb + +Br cathode cation (Red.) anode anion (oxi.) Q. No. 38 By the electrolysis of aqueous solution of CuSO 4, the products obtained at both the electrodes are O at anode and H at cathode H at anode and Cu at cathode O at anode and Cu at cathode H S O 8 at anode and O at cathode + CuSO Cu + SO 4 4 H O H +OH +
At cathode, + Cu e Cu At anode, 0H H O+O+e O+O O Q. No. 39 During the electrolysis of fused NaCl, the reaction that occurs at the anode is : Chloride ions are oxidized Chloride ions are reduced Sodium ions are oxidized Sodium ions are reduced NaCl Na e At cathode cation (Red.) At anode anion (Oxi.) Q. No. 40 In electroplating the article to be electroplated is made : Cathode Anode Either cathode or anode Simply suspended in the electrolytic bath For electroplating other metal has to get oxidised. Q. No. 41 On electrolyzing a solution of dilute H SO 4 between platinum electrodes, the gas evolved at the anode is SO SO 3 O H H SO H + SO 4 4 + HOH +OH Q. No. 4 During the electrolysis of fused NaCl, which reaction occurs at anode? Chloride ions are oxidised Chloride ions are reduced Sodium ions are oxidised Soldium ions are reduced
NaCl Na + e At cathode Cation (Red.) At anode anion (Oxi.) Q. No. 43 A spoon to be electroplated with gold should be : Cathode Anode Electrolyte None of these For electroplating other metal has to get oxidised. Spoon cathod Q. No. 44 Three faradays of electricity was passed through an aqueous solution of iron (II) bromide. The mass of iron metal (at. Mass 56) deposited at the cathode is 56 g 84 g 11 g 168 g Correct Answer By unity method, 3 56=84g FeBr + Fe + +Br Fe +e 1male F 56 56 3F 3 = 84 g Q. No. 45 The electric charge for electrode deposition of one gram equivalent of a substance is : One amp/sec 96,500C/sec One amp/hour 96,500 C 1 Faraday (96500 coulomb) current deposits 1 g eg of substance Q. No. 46 Number of electron involved in the electrode position of 63.5 g of Cu from a solution of CuSO 4 is : 6.0 10 3.011 10 1.044 10 6.0 10
CuSO Cu + SO + + 4 4 Cu + e 1molecu 1mole mole 63gF To deposit 63g F mole e 1g 63.5 63 63.5 molee= 6.03 10 3 1F1molee Q. No. 47 When one coulomb of electricity is passed through an electrolytic solution the mass deposited on the electrode is equal to : Equivalent weight Molecular weight Electrochemical equivalent One gram If (96500 C) current deposits 1 g eg. of substance. Q. No. 48 W g of copper deposited in a copper voltameter when an electric current of ampere is passed for hours. If one ampere of electric current is passed for 4 hours in the same voltmeter, copper deposited will be : W W W 4 W W = eit W 1 = 4 e W1 e= 4 W =4 W 1 = W W 4 1 Q. No. 49 When the same electric current is passed through the solution of different electrolytes in series the amounts of elements deposited on the electrode are in the ratio of their : Atomic number Atomic masses Specific gravities Equivalent masses Equivalent masses
Q. No.50 13.5 g of Al get deposited when electricity is passed through the solution of AlCl 3. The number of faradays used are : 0.50 1.00 1.50.00 Al Cl Al + 3Cl 3+ 3 Al + 3e 1mole 3mole Al 3F 13.5 g of Al 13.5 =15moleAl 7 No. of F rep. for deposit of 1 mole Al 3F.5 3x.5R = 1.5 F Q. No.51 The ratio of weights of hydrogen and magnesium deposited by the same amount of electricity from aqueous H SO 4 and fused MgSO 4 are : 1 : 8 1 : 1 1 : 16 None of these Correct Answer F g,h H SO H +SO F 4 g,mg Cathode, + H +e H MgSO 4 + 4 Mg +SO4 Mg +e Mg :4 1 : 1 4 4 + HOH +OH Q. No.5 A current of ampere was passed through solutions of CuSO 4 and AgNO 3 in series. 0.635 g of copper was deposited. Then the weight of silver deposited will be : 0.59 g 3.4 g 1.08 g.16 g 1 ampere gives mole 100 + ampere gives mole Ag 100 108 =.16g 100 Faraday s law,
W1 E1 1 = = W E Q. No.53 An ion is reduced to the element when it absorbs 610 electrons. The number of equivalents of the ion is : 0.10 0.01 0.001 0.0001 Unitary method, NA ion reduce 1 equivalent of ion NA 10 3.001 of ion Q. No.54 Electrolysis can be used to determine atomic masses. A current of 0.550 A deposits 0.55 of a certain metal in 100 minutes. Calculate the atomic mass of the metal if n = 3 100 45.0 48.5 144.75 atomic mass Z= esexchanged 96500 also w = zit atomicmass =.0001 3 96500 W Z= it = 48.5.55 Z=.550 6000 Z =.0001 Q. No.55 How many minutes it take to plate out 0.50 g of Cr from a Cr (SO 4 ) 3 solution using a current of 1.50 A?(Atomic weight : Cr = 5.0) 54 30 15 103 Correct Answer 3+ d Cr SO Cr +3SO 4 3 4 3+ Cr +6e Cr 6 96500C 5g of Cr 104g 6 96500C 696500 0.50g=.50=783.6C 104 charg e 783.6C time = = =1855 1 1 1.50es 1.5Csec = 30
Q. No.56 An electrolysis of a oxytungsten complex ion using 1.10 A for 40 min produces 0.838 g of tangsten. What is the charge of tungsten in the material? (Atomic weight: W = 184 ) 6 4 1 Q = it = 1.10 C sec 1 40 60 sec Q = 640 C.830 g = 640 C 184 g = 640 184 C.838 = 579665.8 C 579665.8 = = 6F 96500 Q. No.57 When molten lithium chloride (LiCl) is electrolyzed, lithium metal is formed at the cathode. If current efficiency is 75 % then how many grams of lithium are liberated when 1930 C of charge pass through the cell : (Atomic weight : Li = 7) 0 : 105 0 : 10 108 : 9 3 : 8 current passed actualy current efficiency = 100 current passed experimently E c t wt metal = 96500 Applied charge = 1930 C Efficiency = 75% Word charged =1930 75% = 1447.5 C 96500 C liberate 1 mole 1 1C = moles 96500 1447.5 1447.5C = moles 96500 weight =1447.5 7g Q. No.58 The weight ratio of Al and Ag deposited using the same quantity of current is : 9 : 108 : 1 108 : 9 3 : 8 W E Z = = W E Z
Q. No.59 The weight of silver (eq. wt = 108) displaced by that quantity of current which displaced 5600 ml of hydrogen at STP is : 54 g 108 g 5.4 g None of these By definition : displaced 11.ml 1gH 1F is displaced 11.ml 1F 1 5600ml 5600ml 11. = 500 F 108g 1F 500 108g 500F = 54 g Q. No.60 A current of 9.65 ampere is passed through the aqueous solution NaCl using suitable electrodes for 1000 s. The amount of NaOH formed during electrolysis is.0 g 4.0 g 6.0 g 8.0 g Correct Answer Due to electrolysis + OH formed=naoh formed= Zi 40 = 9.65 1000 96500 = 4.0 g Q. No.61 How many electrons are delivered at the cathode during electrolysis by a current 1A in 60 seconds? 3.7410 6.010 7.4810 6.010 Quantity of electricity = C t 1 = 1Csec 60sec Q = 60 C 96500C 1molee
1 60C = 60 96500 =.00061 mole el 3 =.00061 6.03 10 el 3 =.0038710 = 3.8710 0 el Q. No.6 The moles of electrons required to deposit 1 gm equivalent aluminium (at. wt. = 7) from a solution of aluminium chloride will be 3 1 4 3+ A l + 3e Al g 1mole 3mole Small of e required to deposit 1 gm equivalent Al. Q. No.63 Time required to deposit one millimole of aluminium metal by the passage of 9.65 amperes through aqueous solution of aluminium ion is : 30 s 10 s 30,000 s 10,000 s Charge required For mole Al B F 3 For 10 3 mole 3 10 F 3 = 3 10 96500 C 3 3 10 96500C t = = i 1 9.65Csec = 30, 000 10 3 sec = 30 sec Q. No.64 How many coulomb of electricity are consumed when 100 ma current is passed through a solution of AgNO 3 for 30 minute during an electrolysis experiment. 108 18000 180 3000 + AgNO A g +NO 3 3 + A g +1e Ag 3 1 Q = it =1930 10 CS 30 60sec
= 180 C Q. No.65 A current of 9.65 amp. Flowing for 10 minute deposit 3.0 g of a metal. The equivalent wt. of the metal is : 10 30 50 96.5 m = Q C 95600 1 3 = it = 9.65CSec 10 60 sec 95600 96500 3 = = 50.06 Q. No. 67 When a lead storage battery is discharged PbSO 4 is formed Pb is formed SO is consumed H SO 4 is formed When the cell is discharging (Providing current) there reaction one Q. No.68 A fuel cell is : The voltaic cells in which continuous supply of fuels are send at anode to give oxidation The voltaic cell in which fuels such as : CH 4, H, CO are used up at anode It involves the reactions of H O fuel cell such as Anode : H + 4OH 4HO( l)+ 4e Cathode : O +H O( )+4e 4OH l All of the above A fuel cell is a galvanic cell in which one of the reactants is a traditional fuel such as CH 4 or H or CO. the fuel (H g) and oxidising agent (O g) do not react directly but instead of flow into separate cell compartments where H is oxidised at the anode + O is reduced at the cathode.
Q. No.69 Reaction that takes place at graphite anode in dry cell is + Zn + e Zn(s) Zn(s) Zn + e + Mn + e Mn(s) Mn(s) Mn + e +1.5V Correct Answer Dry cell has Anode Zinc metal can Cathode inert graphite rod surrounded by paste at MnO and carbon black. Electrolyte Moist past of NH 4 Cl and ZnCl in starch. Reaction : Q. No. 70 As lead storage battery is charged Lead dioxide dissolves Sulphuric acid is regenerated Lead electrode becomes coated with lead sulphate The concentration of Sulphuric acid decreases Correct Answer When the cell is discharging (Providing current) there reaction one Q. No.71 The specific conductance of a N/10 KCI at 5 0 C is 0.011 ohm 1 cm 1. The resistance of cell containing solution at the same temperature was found to be 55 ohms. The cell constant will be 6.16 cm 1 0.616 cm 1 0.0616 cm 1 616 cm 1 Correct Answer Cell const. = Resistance specific conductance.011 55 =.616
Q. No.7 The specific conductance of a salt of 0.01 M concentration is 1.06110 molar conductance of the same solution will be : 1.06110 1.061 10.61 106.1 1000 specificconductance molarconductance = molarity 4 1000 1.061 10 = =10.61.01 Q. No. 73 Which of the following solutions of NACI will have the highest specific conductance? 0.001 N 0.1 N 0.01 N 1.0 N Specific conductance decreases with dilution is no. of ions decreases w.r.t. volume. Q. No.74 The molar conductance at infinite dilution of AgNO 3, AgCI and NaC are 116.5, 11.6 and 110.3 respectively. The molar conductances of NaNO 3 is : 111.4 105. 130.6 150. Correct Answer = AgNO + NaCl m m 3 m = m AgCl = 116.5 + 110.3 11.6 = 105. Q. No.75 If x specific resistance (in S 1 cm) of the electrolyte solution and y is the molarity of the solution, then m ( in S cm mol 1 ) is given by : 1000x y x 1000 y 1000 xy xy 1000 molar conductance = 1000 specificconductance molarity
1 100 7 1000 = = molarity xy Q. No.76 Resistance of 0.1 M KCI solution in a conductance cell is 300 ohm and conductivity is 0.013 Scm 1. The value of cell constant is : 3.9 cm 1 39 m 1 3.9 m 1 None of these 1 l k = cell constant R 9.013 300 = 3.9cm 1 Q. No.77 The specific conductance of a saturated solution of silver bromide is k S cm 1. The limiting ionic conductivity of Ag+ and Br 1 ions are x and y, respectively. The solubility of silver bromide in gl 1 is : (molar mass of AgBr = 188 ) k1000 x y k x + y 188 k 1000 188 x+ y x + y 1000 k 188 1000 specific conductance solublity = mol C C = x + y 1000k solibility = 188 g / L x + y Q. No.78 The conductivity of 0.1 N NaOH solution is 0.0 S cm 1 When equal volume of 0.1 N HCI solution is added, the conductivity of resultant solution is decreases to 0.0055 Scm 1. The equivalent conductivity in S cm equivalent 1 of NaCi solution is 0.0055 0.11 110 None 1000 specificconductance normality 1000.0055 = = 55.1
Q. No. 79 The specific conductivity of a saturated solution of AgCI is 3.4010 ohm 1 cm 1 at 5 0 C. If Ag + cm mol 1 & CI =67.7 ohm 1 cm mol 1, the solubility of AgCI at 5 0 C is :.610 M 4.510 M 3.610 M 3.610 M 0 = Ag + Cl = 130.0 ohm 1 cm.mol 1 1000 specific conductivity solubility = 1000 3.40 10 ohm = 1 130.0mol 0 6 1 10003.40 6 3 = 10 cm mol 130.0 =.610 5 mol/l Q. No.80 Molar conductance of 0.1 M acetic acid 7 ohm 1 cm mol 1 If the molar cond. of acetic acid at infinite dilution is 380.8 ohm 1 cm mol 1, the value of dissociation constant will be 5 3 610 moldm 3 1 1.6610 moldm 3 1.6610 moldm 5 3 3.4410 moldm C dissociation cast =.1 7 = 380.8 380.8 7 m 5 3 =3.4410 moldm C m m fm Q. No.81 At infinite dilution, the eq. conductances of CH 3 COONa, HCI and CH 3 COOH are 91,46 and 391 mho cm respectively at 5 0 C, The eq. conductance of NaCI at infinite dilution will be : 16 09 391 908 Conductance of NaCl at infinite dilution = 46
=46 + 91 517 391 16 Q. No.8 The equivalent conductivity of 0.1 N CH 3 COOH at 5 0 C is 80 and at infinite dilution 400. The degree of dissociation of CH 3 COOH is 1 0. 0.1 0.5 Correct Answer C = = 80 n =. 400 m