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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2101. |
A cell, as shown below, consists of there compartments separated by porous pots. The first contains a cobalt eletrode in `5.0L` of `0.10M Co(NO_(3)),` the second contains `5.0L` of `0.10M KNO_(3),` the third contains an `Ag` electrode in `5.0L` of `0.10M AgNO_(3)`. Assuming that current with in the cell iis carried equally by the negative and positive ions by passign `0.1F` of electricity. Given `: Co^(2+)+2e^(-) rarr Co" "E^(c-)=-0.28V` Ag^(o+)+e^(-) rarr Ag" "E^(c-)=0.80V` `2Ag^(o+)+Corarr Co^(2+)+2Ag" "E^(c-)._(cell)=1.08V` Finally, the `I` compartment contains `I. Co^(2+)" "II. NO_(3)^(c-)" "III.K^(o+)" "IV.Ag^(o+)`A. `I`B. `I,II`C. `I,II,III`D. `II,III,IV` |
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Answer» Correct Answer - b There is movement of `NO_(3)^(c-)` ions from `III` to `II` and `II` to `I` compartment to maintain electrical neutrality in `I` compartment since there is increase in positive charge due to the oxidation of `Co` to `Co^(2+)` in `I` compartment. Hence, ions present in `I` compartment are `:Co^(2+)` and `NO_(3)^(c-)`. |
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| 2102. |
The `E^(@)` at `25^(@)`C for the following reaction is 0.55 V. Calculate the `DeltaG^(@)` in kJ`//`mol : `4BiO^(+)(aq)+3N_(2)H_(5)^(+)to4Bi(s)+3N_(2)(g)+4H_(2)O(l)+7H^(+)`A. `-637`B. `-424`C. `-106`D. `-318.5` |
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Answer» Correct Answer - A |
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| 2103. |
Calculate the stability constant of the complex `[Zn(NH_(3))_(4)]^(2+)` formed in the reaction `Zn^(2+)+4NH_(4)hArr[Zn(NH_(3))_(4)]^(2+)`. Given that `E_(Zn^(2+)//Zn)^(@)=-0.76V and E_((Zn(NH_(3))_(4)]^(2+))^(@)//Zn,4NH_(3))^(@)=-1.03V` |
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Answer» We are given `Zn^(2+)(aq)+2e^(-)hArrZn(s)" "E^(@)=-0.76V` `[Zn(NH_(3))_(4)]^(2+)+2e^(-)hArrZn(s)+4NH_(3),E^(@)=-1.03V` To get the overal reaction, we should write `Zn^(2+)(aq)+2e^(-)hArrZn(s)," "E^(@)=-0.76V` `Zn(s)+4NH_(3)hArr[Zn(NH_(3))_(4)]^(2+)+2e^(-)," "E^(@)=1.03V` Adding, we get `E_(cell)^(@)=1.03-0.76=0.27V` `E_(cell)^(@)=(0.0591)/(n)logK` or `logKk=(0.27xx2)/(0.0591)=9.1371` or `K="Antilog "9.1371=1.371xx10^(9).` |
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| 2104. |
On the basis of the following data, explain why Co(III) is not stable in aqueous solution? `Co^(3+)+e^(-)toCo^(2+),E^(@)=+1.82V` `2H_(2)OtoO_(2)+4H^(+)+4r^(-),E^(@)=1.23V`. |
| Answer» Adding the two half reaction, EMF comes out to +ve. This means that Co(III) in aqueous solution has the tendency to change to Co(II). Hence, Co (III) is not stable in aqueous solution. | |
| 2105. |
The standerd potential at `25^(@)` for the following Half rection is given : `Zn^(2+)+2e^(-)toZn,E^(@)=-0.762 V` `Mg^(2+)+2e^(-)toMg,E^(@)=-2.37 V` When Zinc dust is added to the solution of `MgCl_(2)`.A. `ZnCl_(2)` is formedB. Mg is percipitaedC. Zn dissolved in the solutionD. No reaction take place |
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Answer» Correct Answer - D |
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| 2106. |
Asseration : A small amount of acid or alkali is added before electrolysis of water. Reason : Pure water is weak electrolyte. |
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Answer» Correct Answer - A Small amount of acid increase the conductivity of water by increasing ionisation. |
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| 2107. |
Mercury cell (Ruben-Mallory cell) gives a constant voltage of____. |
| Answer» Correct Answer - 1.35 V | |
| 2108. |
The energy of one joule per second given out by a source is called_________. |
| Answer» Correct Answer - One Watt. | |
| 2109. |
Which of the following is non- electrolyteA. NaClB. `CaCl_2`C. `C_12 H_22 O_11`D. `CH_3COOH` |
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Answer» Correct Answer - C `C_(12)H_(22)O_(11)` is a non-electrolyte. |
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| 2110. |
On electrolysis of an aqueous solution NaCl, why `H_(2)` and not Na is liberated at the cathode? |
| Answer» This is because `H^(+)` ions produced from ionization of water have lower discharge potential that `Na^(+)` ions produced from ionization of NaCl or reduction potential of water is greater than that of sodium ions. | |
| 2111. |
During electrolysis of NaCl solution, part of the reaction is `Na^(+)e^(-) Na`. This is termed asA. OxidationB. ReductionC. DepositionD. Cathode reaction |
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Answer» Correct Answer - B `underset(+1)(Na^(+))+2e^(-)tounderset(0)(Na)`, means oxidation number is decreased so the reaction is reduction. |
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| 2112. |
The addition of a polar solvent to a solid electrolyteA. PolarizationB. AssociationC. IonizationD. Non- liberation of heat |
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Answer» Correct Answer - C When polar solvent added in to solid electrolyte than it ionised. |
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| 2113. |
In a galvanic cell, the electrons flow from : |
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Answer» Correct Answer - C In a galvanic cell, the electrons flow from anode to cathode through the external circuit. At anode (-ve pole) oxidation and at cathode (+ve pole) reduction takes place. |
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| 2114. |
The cathodic reaction of a dry cell is represented as `2MnO_(2)(s)+Zn^(2+)+2e^(-) rarr ZnMn_(2)O_(4) (s)` If there are 8 g `MnO_(2)` in the cathodic compartment then the time for which the dry cell will continue to give current of 2 milliampere, isA. 25.675 dayB. 51.35 dayC. 12.8 dayD. 6.423 day |
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Answer» Correct Answer - B According to Faraday,s law of electrolysis `m=((it)/F)M/Z` or, `t=(mZF)/(IM)` the factor of equivalence can be found dividing the number of electrons by the number of reacted particles. In case of `MnO_(2)`, the factor is `Z=2/2=1` the molar mass of `MnO_(2)` is M `(MnO_(2))=55+16xx2` `=87 g//mol` Now, `t=(8xx1xx96500)/(0.002xx87) =4.436xx10^(6) s=1232=51.3` days |
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| 2115. |
The standard reduction potential of the reaction `H_(2)O+e^(-) rarr 1/2 H_(2) +OH`, at 298 K, isA. `E^(@)=(RT)/(F) ln K_(w)`B. `E^(@)=(RT)/(F) "ln "[p_(H_(2))]^(1//2) [OH^(-)]`C. `E^(@)=(RT)/(F) "ln "([p_(H_(2))]^(1//2))/([H^(+)])`D. `E^(@)=- (RT)/(F) ln K_(W)` |
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Answer» Correct Answer - A The standard reduction potential of the cell reaction is `E^(0)=(RT)/(F) ln K_(w)` |
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| 2116. |
The standard potentials at `298 K` for the following halfreactions are as given ` Zn^(2+) + 2e underset(larr)(rarr) Zn E^o =- 0/ 762 V`. `2 H^+ + 2 Erarr H_2 (g) E^o = 0.000 V` `Cr^(3+) + 3e underset(larr)(rarr) Cr E^o =- 0.740 V` `Fe^(3+) 2e rarr Fe^(2+) E^o =0.772 V` Which of the following is the strongest reducing agent ?A. `Zn`B. CrC. H2D. Fe^2+ |
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Answer» Correct Answer - D higher `E_("reduction")^@` means good oxidizing agent . |
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| 2117. |
The standard potentials at `298 K` for the following halfreactions are as given ` Zn^(2+) + 2e underset(larr)(rarr) Zn E^o =- 0/ 762 V`. `2 H^+ + 2 Erarr H_2 (g) E^o = 0.000 V` `Cr^(3+) + 3e underset(larr)(rarr) Cr E^o =- 0.740 V` `Fe^(3+) 2e rarr Fe^(2+) E^o =0.772 V` Which of the following is the strongest reducing agent ?A. ` Zn(s)`B. `Cr(s)`C. ` H_2(g)`D. ` Fe^(2+)(aq)` |
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Answer» Correct Answer - A More negative is the reduction poptential, higher will be the reducing property , i.e. power to give up elctrons . |
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| 2118. |
In a gavlanic cell.A. Chemical reaction produces electrical energyB. electrical energy produces chemical reactionC. reduction occurs at anodeD. oxidation osccurs at cathode |
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Answer» Correct Answer - A In galvanic cell//electochemical cell electrical energy is produced due to same chemical reaction. |
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| 2119. |
Which one of the following will increase the voltage of the cell ? `(T = 298 K)` ` Sn+ 2Ag^+ rarr Sn^(2+) +2 Ag`.A. increase in the size of silver rodB. increase in the concentrationnnnn of `Sn^(+2)` ionsC. increase in the concentrationnnnn of `Ag^(+2)` ionsD. none of the above |
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Answer» Correct Answer - C `E_(cell)=E_(cell)^@ -(0.059)/2 log .([Sn^(2+)])/([Ag^+]^2)` ` Ag^+` inceases, ` E_(cell)` increases. |
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| 2120. |
The standard potentials at `25^@C` for the following half reactions are given against them ` Zn^(2+) +2e^(-) rarr Zn, E^0 =-0 . 762 V ` `Mg^(2+) +2e^(-) rarr Mg, =- 2.37 V`. When zinc dust is added to a solution of magnesium chloride .`A. No reaction will take plaeB. Zinc chloride is formedC. Zinc dissolve in solutionD. Magnesium is preciptiated |
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Answer» Correct Answer - A ` Zn+ MgCl_2rarr ZnCl_2+Mg` `:.E_("cell")^@ = E_(Zn//Zn^(+2))^@ + E_(Mg^(+2)//Mg)^@ = + 0.762 - 2.37` `-= 1.608 V`. Here, ` E_("cell")^@` is negative so no reaction will take place. |
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| 2121. |
For some reactions the standard reduction potentials are given below : `Zn^(2+) + 2 e to Zn , " " E^(@) = -0.762 V` `Fe^(2+) + 2e to Fe , " " E^(@) = -0.440 V` `Cu^(2+) + 2e to Cu , " " E^(@) = + 0.345 V` `Ag^(+) + e to Ag " " E^(@) = + 0.800V` Which one of the following is most easily oxidised ?A. ZnB. CuC. FeD. Ag |
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Answer» Correct Answer - A Less the reduction potential or greater the oxidation potential , more easily the substance is reduced `E_(Zn)^(@) lt E_(Fe)^(@) lt E_(Cu)^(@) lt E_(Ag)^(@)` . |
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| 2122. |
What is the correct nernst equation for reaction taking place in the following cell`Mg_((s))|Mg_((aq))^(2+)||Cl_((aq))^(-)|Cl_(2(g))(1atm)//Pt`A. `E_(cell)=E_(cell)^(o)-(0.0592)/(n)xxlog(([Cl^(-)]^(2))/([Mg^(2+)]))`B. `E_(cell)=E_(cell)^(o)-(0.0591)/(n)xxlog (([M^(2+)])/([Cl^(-)]))`C. `E_(cell)=E_(cell)^(@)-(0.592)/(n)xxlog[Mg^(2+)][Cl^(-)]^(2)`D. `E_(cell)=E_(cell)^(@)-(0.0591)/(n)xxlog(([Mg^(2+)])/([Cl^(-)]^(2)))` |
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Answer» Correct Answer - C `Mg_((s))|Mg_((aq))^(2+)||Cl_((aq))^(-)|Cl_(2(g))(1atm)//Pt` Oxidation half reaction, `MgtoMg^(2+)+2e^(-)` Reduction half reaction, `underline(Cl_(2)+2e^(-)to2Cl^(-)" ")` Net cell reaction, `Mg+Cl_(2)toMg^(2+)+2Cl^(-)` Nernst equation, `E_(cell)=E_(cell)^(o)-(0.0591)/(n)"log"[Cl^(-)]^(2)[Mg^(2+)]` Here, n=2 |
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| 2123. |
`Ag_((s))|Ag_((aq))^(+)(0.01M)||Ag_((aq))^(+)(0.1M)|Ag_((s))E_(Ag_((s))//Ag_((aq)))^(@)=0.80`voltA. Cell cannot function as anode and cathode are of same materialB. `E_(cell)=0.0592V`C. `E_(cell)=0.80V`D. `E_(cell)=0.0296V` |
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Answer» Correct Answer - B For an electrolytic concentration cell, `E_(cell)=(0.0591)/(n)"log"(C_(1)(R.H.S.))/(C_(2)(L.H.S.))` `=(0.0591)/(1)"log"(0.1)/(0.01)=0.0591log10=0.0591V`. |
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| 2124. |
The emf of a galvanic cell, with electrode potential of silver=+0.80 and that of copper=+0.34V, isA. `-1.1V`B. `+1.1V`C. `+0.46V`D. `+0.76V` |
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Answer» Correct Answer - C `E^(o)=E_(Ag^(2+)//Ag)^(o)+E_(Ca//Cu^(2+))^(o)=-0.34+0.80=+0.46V` |
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| 2125. |
The standard emf of a galvanic cell involving cell reaction with n=2 found to be 0.295 V at `25^(@)C`. The equilibrium constant of the reaction would beA. `2xx10^(11)`B. `4xx10^(12)`C. `1xx10^(2)`D. `1xx10^(10)` |
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Answer» Correct Answer - D `E_(cell)^(o)=(2.303RT)/(nF)log" "K_(eq)` `0.295=(2.303xx8.314xx298)/(2xx96500)log" "K_(eq)` `log" "K_(eq)=(2xx96500)/(2.303xx3.314xx298)=10` `K_(eq)="antilog "10=1xx10^(10)` |
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| 2126. |
Ammonia is always added in this reaction. Which of the following must be incorrectA. `NH_(3)` combines with `Ag^(+)` to form a complexB. `Ag(NH_(3))_(2)^(+)` is a stronger oxidising reagent than `Ag^(+)`C. In absence of `NH_(3)` silver salt of gluconic acid is formedD. `NH_(3)` has affected the standard reduction potential of glucose/gluconic acid electrode |
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Answer» Correct Answer - D Ammonia has no effect on the standard reduction potential. |
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| 2127. |
A standard hydrogen electrode has zero electrode potential becauseA. Hydrogen is easier to oxidiseB. This electrode potential is assumed to be zeroC. Hydrogen atom has only one electronD. Hydrogen is the lightest element. |
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Answer» Correct Answer - B Standard hydrogen electrode is a reference electrode. Potential of any half cell is calculated with respect to standard hydrogen electrode (SHE). Hence, electrode potential of SHE has been arbitrarily taken to be zero. |
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| 2128. |
The standard reduction potential values of three metallic cations, X,Y and Z are 0.52, -3.03 and -1.18V respectively. The order of reducing power of the corresponding metals isA. `YgtZgtX`B. `XgtYgtZ`C. `ZgtYgtX`D. `ZgtXgtY` |
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Answer» Correct Answer - A More negative or lower is the reduction potential, more is the reducing property. Thus the reducing power of the corresponding metal will follow the reverse order, i.e., `Y gt Z gt X`. |
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| 2129. |
From the rate expression for the following reactions, determine their order of reaction and the admensions of the rate constants. `i) 3NO (g)to N_(2)O (g)" "Rate=k[NO]^(2)` ii) `H_(2)O_(2)(aq) +3I^(-)(aq)+ 2H^(+) to 2H_(2)O(l)+I_(3)" "Rate =k [H_(2)O_(2)][I^(-)]` iii) `C_(2) H_(5)Cl(g) to C_(2) H_(4)(g) +CO(g)" "Rate =k[CH_(3)CHO]^(3//2)` iv) `C_(2) H_()5) Cl (g) to C_(2)H_(2) to C_(2)H_(4) (g) + HCl(g)" "Rate =k [C_(2)H_(5)Cl]` |
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Answer» Rate `=k [NO]^(2)` order of reaction =2 units (dimensions) of rate constant `[k] =(Rate)/([NO]^(2))=(mol L^(-1)s^(-1))/((mol L ^(-1))^(2))= mol ^(-1) L s^(-1)` ii) Rate `=k[H_(2)O _(2)] [I^(-)]` order of reaction `=1+1=2` Dimensions of `k =(Rate)/([H_(2)O_(2)][I^(-)])=(mol L^(-1)s^(-1))/((mol L ^(-1))[mol L ^(-1]))= mol ^(-1) L s^(-1)` iii) Rate ` =k [CH_(3) CHO]^(3//2)` order of reaction `=3/2` Dimensions of `k=(Rate)/([CH_(3)CHO]^(3//2))=(mol L ^(-1)s^(-1))/((mol L ^(-1))^(3//2))= mol ^(-1//2)L ^(1//2)s^(-1)` iv) Rate `=k [C_(2) H_(5)Cl]` order of reaction =1 Dimensions of `k=(Rate)/(C_(2) H_(5)Cl)=(mol L ^(-1)s^(-1))/(mol L ^(-1))=s^(-1)` |
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| 2130. |
The deposition of different ions at electrodes stops when the supply of electricity is stopped |
| Answer» Correct Answer - True | |
| 2131. |
`Zn^(+2) and Ag^(+)` are present in a solution to be electrolysed. The metal that is deposited first at cathode is ______ |
| Answer» Correct Answer - zinc | |
| 2132. |
The molar conductivity is maximum for the solution of concentration __________.A. 0.004 MB. 0.002 MC. 0.005 MD. 0.001 M |
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Answer» Correct Answer - D Molar conductivity is inversely proportional to molarity |
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| 2133. |
Units of the properties measured are given below. Which of the properties has not been matched correctly?A. molar conductance `= Sm^2 mol^(-1)`B. Cell constant =`m^(-1)`C. Specific conductance = S `m^2`D. Equivalent conductance =S `m^2 (g eq)^(-1)` |
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Answer» Correct Answer - C Specific conductance =`S m^(-1)`C is the correct answer. |
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| 2134. |
According of Kohlrausch law, the limiting value of molar conductivity of an electrolyte `A_(2)B` isA. `lambda^(oo)._(A^(o+))+lambda^(oo)._((B^(-))`B. `lambda^(oo)._(A^(o+))-lambda^(oo)._(B^(-))`C. `2lambda^(oo)._(A^(o+))+(1)/(2)lambda^(oo)._((B^(-))`D. `2lambda^(oo)._(A^(o+))+lambda^(oo)._(B^(-))` |
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Answer» Correct Answer - d Factual statement. |
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| 2135. |
According of Kohlrausch law, the limiting value of molar conductivity of an electrolyte `A_(2)B` isA. `lambda_((A^(+)))^(oo)+lambda_((B^(+))^(oo)`B. `lambda_((A^(+)))^(oo)-lambda_((B^(-)))^(oo)`C. `2lambda_((A^(+)))^(oo)+(1)/(2)lambda_((B^(-)))^(oo)`D. `2lambda_((A^(+)))^(oo)+lambda_((B^(2-)))^(oo)` |
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Answer» Correct Answer - D the electrolyte `A_(2) B` ionises as `A_(2)Brarr2A^(+)+B^(-)` `therefore ^^_(m)^(oo)=2lambda_((A^(+)))^(oo)_+2lambda_((B^(-)))&_(oo)` |
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| 2136. |
According to Kohlrausch law, the limiting molar conductivity of an electrolyte, `A_(m)B_(n)`, can be expressed asA. `Lambda_(m)^(0)=m_(+)lambda_(A^(n+))^(0)-n_lambda_(B^(m-))^(0)`B. `Lambda_(m)^(0)=m_(+)lambda_(A^(m+))^(0)+n_lambda_(B^(n-))^(0)`C. `Lambda_(m)^(0)=n_(+)lambda_(A^(m+))^(0)-m_lambda_(B^(n-))^(0)`D. `Lambda_(m)^(0)=m_(+)lambda_(A^(n+))^(0)-n_lambda_(B^(m-))^(0)` |
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Answer» Correct Answer - D On the basis of conductivity measurements on a series of strong electrolytes, Kohlrausch discovered that this moalr conductivity at infinite dilutionis the sum of the contributions from each ion. It is thus knowm as the law of independent mifration (or mobility) of ions. Here `m_(+)` and `n_(-)` are numbers of cations and anions per formula unit, respectively and `lambda_(A^(n+))^(0)` and `lambda_(B^(m-))^(0)` are the corresponding molar conductivities of the ions at infinite dilution. |
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| 2137. |
According of Kohlrausch law, the limiting value of molar conductivity of an electrolyte `A_(2)B` isA. `lambda^(@) (A^(+)) + lambda^(@) (B^(-))`B. `lambda^(@) (A^(+)) - lambda^(@) (B^(-))`C. `2 lambda^(@) (A^(+)) + (1)/(2) lambda^(@) (B^(-))`D. `2 lambda^(@) (A^(+)) + lambda^(@) (B^(-))` |
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Answer» Correct Answer - D By definition of Kohlrausch law in terms of molar conductivities . |
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| 2138. |
The correct order of equivalent the electrolyte is dissociated more the correct order of equivalent conductances at infinite dilution in water at room temperature for `H^(+),CH_(3)COO^(-) and HO^(-)` ions isA. `HO^(-) gt H^(+) gt K^(+) gt CH_(3)COO^(-)`B. `H^(+) gt HO^(-) gt K^(+) gt CH_(3)COO^(-)`C. `H^(+) gt K^(+) gt HO^(-) gt CH_(3)OO^(-)`D. `H^(+)gtK^(+)gtCH_(3)COO^(-) gt HO^(-)` |
| Answer» Correct Answer - B | |
| 2139. |
STATEMENT 1 : The molar conductivity of strong electrolyte decreases with increases in concentration and STATEMENT 2 AT high concentration migration of ion is slowA. Statement 1 is ture , Statement 2 is true Statement 2 is correct explanation for Statement 1B. Statement 1 is true Statement 2 is ture Statement 2 is NOT a correct explantion for Statement 1C. Statement 1 is true statement 2 is tureD. Statement 1 is false Statement 2 is true |
| Answer» Correct Answer - A | |
| 2140. |
Which of the following plots represents correctly the variation of molar conductivity with dilution for a strong electrolyte?A. B. C. D. |
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Answer» Correct Answer - A This is the general shape of molar conductivity-dilution curve for a strong electrolyte. The second graph is for a weak electrolyte. The second graph is for a wak electrolyte. Similar graphs are obtained when we plot equivalent conductivity against dilution. For a strong electrolyte, the graph reaches a maximum at low concentrations, and the `Lambda_(m)^(0)` value can be obtained by extrapolation from the experimental curve. For a weak elecrolyte, the graph would also reach a maximum at a low concentration but readings cannot be obtained experimentally, at such low concentrations. The third graph is obtained when we plot molar or equivalent conductivity of a stronfg electrolyte against `sqrtC` while the fourth graph is obtained when we plot molar or equivalent conductivity of a weak electrolyte against `sqrtC`. |
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| 2141. |
Which of the following equations corrects the molar conductivity with concentration for a strong electrolyte?A. `Lambda_(m)^(0) = Lambda_(m) - AsqrtC`B. `Lambda_(m) = Lambda_(m)^(0) - AsqrtC`C. `Lambda_(m) = Lambda_(m)^(0) - AsqrtC`D. `Lambda_(m) = Lambda_(m)^(0) + AsqrtC` |
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Answer» Correct Answer - C This emirical formula was given by Kohlrausch on the basis of his experimental results. A is a constant, its value depends on the nature of the electrolyte rather than on the identity of the specific ions. For example, electrolytes of the type `KCl, NaOH`, etc. have the same value of `A` while `BaCl_(2), CuSO_(4)` etc. gave a different value. |
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| 2142. |
Which of the following aqueous solution produces metal after electrolysis?A. `K_2Cr_2O_7`B. `KMnO_4`C. `CH_3COONa`D. `CuCl_2` |
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Answer» Correct Answer - D |
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| 2143. |
How much time is required for complete decomposition of 4 moles of water using 4 ampere?A. `3.86xx10^5`secB. `1.93xx10^5`secC. 96500secD. 48250sec |
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Answer» Correct Answer - B |
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| 2144. |
Passage of a current for 548 seconds through a siver coulometer results in the deposition of 0.746g of silver.What is the current (in A)?A. 1.22B. 1.16C. 1.07D. 1 |
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Answer» Correct Answer - A |
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| 2145. |
How many minutes are required to deliver `3.21xx10^(6)` coulombs using a current of 500 A used in the commercial production of chlorine?A. 8.3B. `5.3xx10^(4)`C. 6420D. 107 |
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Answer» Correct Answer - D |
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| 2146. |
Calculate `^^_(m)^(0)` for `CaCl_(2) and MgSO _(4)` from the data given in Table `3.4.` |
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Answer» We kniow from Kohlrausch law that `^^_(m(CaCl_(2)))^(0)=^^_(Ca^(2+))^(0) +2 lamda_(Cl)^(0) =119.0 S cm^(2) mol ^(-1)+ 2 (76.3) S cm ^(2) mol ^(-1)` `=(119.0+152.6) S cm ^(2) mol ^(-1)` `=271.6 S cm ^(2) mol ^(-1)` `^^_(m(Mg SO_(4)))^(0)=lamda _(Mg ^(2+))^(0)+ lamda _(SO_(4)^(2-))^(0) =106.0 S cm ^(2) mol ^(-1) + 160.0 S cm^(2) mol ^(-1)` `=266 S cm^(2) mol^(-1)` |
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| 2147. |
What would be the equivalent conductivity of a cell in which 0.5 salt solution offers a resistance of 40 ohm whose electrodes are 2 cm apart and 5 `cm^2` in area?A. `10 "ohm"^(-1) "cm"^(2) "eq"^(-1)`B. `20 "ohm"^(-1) "cm"^(2) "eq"^(-1)`C. `30 "ohm"^(-1) "cm"^(2) "eq"^(-1)`D. `25 "ohm"^(-1) "cm"^(2) "eq"^(-1)` |
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Answer» Correct Answer - B `K=1/Rxx1/A=1/40xx2/5` `Lambda_"eq"=Kxx1000/N=1/40xx2/5xx1000/0.5= 20 ohm^(-1) cm^2 eq^(-1)` |
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| 2148. |
The electrical properties and their respective SI units are given below. Identify the wrongly matched pair:-A. Electrical property-Specific conductance, SI-unit-`Sm^(-1)`B. Electrical property-conductance, SI Unit-SC. Electrical property-Equivalent Conductance, S `m^(2)` `"(gm equiv)"^(-1)`D. Electrical property-Cell constant, SI unit-m |
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Answer» Correct Answer - D Cell constant=1/a Unit=m//`m^(2)=m^(-1)` |
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| 2149. |
The conductivity of 0.25 M solution of KCI at 300 K is 0.0275 `cm^(-1)` calculate molar conductivity |
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Answer» `Molar conductivity =(conductivity xx1000)/(molarity)` `A_(m)=(0.0275xx1000)/(0.25)=110 S cm^(2) mol^(-1)` |
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| 2150. |
The units of conductivity of the solution areA. `ohm^(-1)`B. `ohms`C. `ohm^(-1)cm^(-1)`D. `ohm^(-1)eq^(-1)` |
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Answer» Correct Answer - c Factual statement. |
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