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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.
| 201. |
if `^^_(c)` of `NH_(4)OH` is `11.5Omega^(-1) cm^(2) mol^(-1)`, its degree of dissociation would be (Given. `lambda_(NH_(4)^(+))^(infty)=73.4Omega^(-1) cm^(2) mol^(-1)` and `lambda_(OH^(-))^(infty)=197.6Omega^(-1) cm^(2) mol^(-1))`A. `0.157`B. `0.058`C. `0.0424`D. `0.0848` |
| Answer» Correct Answer - C | |
| 202. |
The relationship `lambda_(m)=lambda_(m)^(0)-BsqrtC` will not hold good for the electrolyte?A. HClB. KClC. `BaCl_(2)`D. HCN |
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Answer» Correct Answer - D Not applicable for weak electrolytel |
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| 203. |
Which of the following solutions has the highest equivalent conductance?A. `0.5` M NaClB. `0.05` M NaClC. `0.005` M NaClD. `0.02` M NaCl |
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Answer» Correct Answer - C Equivalent conductance Increases with Increases in dilution `therefore 0.005M` NaCl solution has highest equivalent conductance |
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| 204. |
Two platinum electrodes were immersed in a solution of `CuSO_(4)` and electric current was passed through the solution. After some time, it was found that colour of `CuSO_(4)` disappeared with evolution of gas at the electrode. The colourless solution contains.A. Platinum sulphateB. Copper sulphateC. Copper hydroxideD. Sulphuric acid |
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Answer» Correct Answer - D `2Cu^(+2)overset(+4e^(-))rarr2Cu` (cathode) `2H_(2)O overset(-4e^(-))rarrO_(2)uparrow+4H^(+)("anode")2H^(+)+SO_(4)^(-2) rarrH_(2)SO_(4)` |
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| 205. |
Which of the following on electrolysis would, not evolve oxygen at the anode ?A. Dilute `H_(2)SO_(4)` with Pt electrodesB. Aqueous silver nitrate using Pt electrodesC. Aqueous `Na_(2)SO_(4)` with Pt electrodesD. 50 % `H_(2)SO_(4)` with Pt electrodes |
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Answer» Correct Answer - D On electrolysis of 50 % `H_(2)SO_(4)` `H_(2)SO_(4)rarrH^(+)+HSO_(4)^(-)2H^(+)+2e^(-)rarrH_(2)uparrow("cathode")` `2HSO_(4)^(-)rarrH_(2)S_(2)O_(8)+2e^(-)("anode")` |
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| 206. |
On passing a current through a molten aluminium chloride for some time, produced `11.2` lit of `Cl_(2)` at NTP at anode, the quantity of aluminium deposited at cathode isA. 27 gramsB. 18 gramC. 9 gramD. 36 gram |
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Answer» Correct Answer - C `11.2" lit"Cl_(2)" at STP"=35.5gm=1GEW` :.1 GEW of Al will be deposited which is 9 gm |
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| 207. |
Number of electrons required to deposit one mole of `Mg^(2+)` ions isA. `6.023xx10^(23)`B. `12.046xx10^(23)`C. `18.069xx10^(23)`D. `3.012xx10^(23)` |
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Answer» Correct Answer - B No. of moles of electrons (or) No. of faraday required to deposit 1 Mole of an element is equal to its charge. |
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| 208. |
A current of 2A passing for 5 hours deposits `22.2` g of tin (at.wt=119), the oxidation state of tin isA. zeroB. threeC. twoD. four |
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Answer» Correct Answer - C `m=(Mct)/(Zf)` find Z |
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| 209. |
Electrolysis can be used to determine atomic masses. A current of `0.550A` deposits `0.55g` of a certain metal in 100 minutes. Calculates the atomic mass of the metal if `n=3 :`A. 100B. 45C. 48.25D. 144.75 |
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Answer» Correct Answer - 3 `(0.55)/(M)xx3=(0.55xx100xx60)/(96500)` `rArr" "M=48.25g//mol` |
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| 210. |
When one faraday of current is passed, which of the following would deposit one gram atomic weight of the metalA. `BaCl_(2)`B. NaClC. `AlCl_(3)`D. `CuCl_(2)` |
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Answer» Correct Answer - B For monovalent ion, 1F will deposit 1 g atwt. |
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| 211. |
Calculate the current ( in `Ma)` required to deposit `0.195 g` of platinum metal in `5.0` hours from a solution of `PtCl_(6)^(2-) : (` Atomic weight `: pt = 195 )`A. 310B. 31C. 21.44D. 5.26 |
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Answer» Correct Answer - 3 `(W)/(E)=(Ixxt)/(96500),(0.195)/(195)xx4=(5xx60xx60xxI)/(96500)` `I=21.44` |
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| 212. |
The best conductor of electricity is a 1M solution ofA. Boric acidB. Acetic acidC. `H_(2)SO_(4)`D. Phosphoric acid |
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Answer» Correct Answer - C Strong acids are strong electrolytes |
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| 213. |
Amount of electricty that can depostit `108 g` of silver from `AgNO_3` aikyruib uaA. FaradayB. 1 AmpereC. 1 CoulombD. None |
| Answer» Correct Answer - A | |
| 214. |
At equimolar concentration of `Fe^(+2)//Fe^(+3)`, what must `[Ag^(+)]` be so that voltage of the galvanic cell made from `Ag^(+)//Ag` and `Fe^(3+)//Fe^(2+)` electrodes equals zero ? The reaction is `Fe^(+2)+Ag hArr Fe^(3+)+Ag`. Determine the equilibrium constant at `25^(@)C` for the reaction. Given `E_(Ag^(+)//Ag)^(@)=0.799` volt and `E_(Fe^(3+)//Fe^(2+))^(@)=0.771` volt. |
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Answer» Correct Answer - 3 `E_(cell)^(@) = E_(Fe^(2+)//Fe^(3+))^(@) +E_(Ag^(+)//Ag)^(@) =` `- 0.771 +7999 = 0.028` volt, Atequilibrium `E_(cell) = 0 0 = E_(cell)^(@) - (0.0591)/(1)log. ([Fe^(3+)])/([Fe^(2+)][Ag^(+)])` `E_(cell)^(@) - 0.5911 log. (1)/([Ag+])` `[Ag^(+)] = 0.34, log K = (nE^(@))/(0.0591) K = 3.0` |
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| 215. |
How much electricity in terms of Faraday is required to produce `20 g` of `Ca` from molten `CaCl_(2)` ?A. 1.5B. 2C. 1D. 4 |
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Answer» Correct Answer - C `"Eq. of Ca",(w)/(E)=(ixxt)/(96500),Ca^(2+)+2erarrCa` `thereforeE_(Ca)=(40)/(2)or (20)/(40//2)=(ixxt)/(96500)` `ixxt=1xx96500=1F` |
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| 216. |
During the working of the following cell `Pb-Hg(1M)|Pb^(2+)(aq)(0.1M)|` `|Pb^(2+)(aq)(0.1M)|Pb-Hg(0.5M)`A. `[Pb^(2+)]` decreases in cathodic half cellB. `[Pb^(2+)]` increases in anodic half cellC. `[Pb^(2+)]` increases in cathodic half cellD. Conc. Of lead amalgam increases in cathodic half cell whereas decreases in anodic half-cell |
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Answer» Correct Answer - A During the working of cell, Pb is oxidized at anode and `Pb^(+2)` is reduced at cathode. |
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| 217. |
The specific conductivity of a saturated solution of AgCl is `3.40xx10^(-6) ohm^(-1) cm^(-1)` at `25^(@)C`. If `lambda_(Ag^(+)=62.3 ohm^(-1) cm^(2) "mol"^(-1)` and `lambda_(Cl^(-))=67.7 ohm^(-1) cm^(2) "mol"^(-1)`, the solubility of AgC at `25^(@)C` is:A. `2.6xx10^(5)M`B. `3.73xx10^(-3)g//L`C. `3.7xx10^(-5)M`D. `2.6xx10^(-)g//L` |
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Answer» Correct Answer - A::B `A^(oo)=62.3+67.7=130Scm^(2)mol^(-1)` `rArrDelta^(oo)=(K1000)/(S)rArrS=(3.4xx10^(-6)xx1000)/(130)` `=2.6xx10^(-5)"molL"^(-1)` In `gL^(-1)` unit , solubility =`2.6xx10^(-5)xx143.5` `=3.7xx10^(-3)gL^(-1)` |
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| 218. |
At what `[Br^(-)]/sqrt(CO_(3)^(2-)]` does the following cell have its reaction at equilibrium? `Ag(s) | Ag_(2)CO_(3)(s)| Na_(2)CO_(3)(aq)|| KBr(aq) | AgBr(s)| Ag(s)` `K_z9sp) = 8 xx 10^(-12)` for Ag_(2)CO^(3)` and `K_(sp) = 4 xx 10^(-13)` for AgBrA. `sqrt(1)xxx10^(-7)`B. `sqrt(2)xx10^(-7)`C. `sqrt(3)xx10^(-7)`D. `sqrt(4)xx10^(-7)` |
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Answer» Correct Answer - B `Ag_(("cathod"))^(+)overset(1e^(-))rarr Ag_(("anode"))^(+)` `E_("cell")=E_("cell")+(0.059)/(1)"log"(K_(sp)(AgBr)//[Br^(-)])/(sqrt(K_(sp)(Ag_(2)CO_(3))//[CO_(3)^(--)])` `rArr ([Br^(-)])/(sqrt([CO_(3)^(--)]))=sqrt(2)xx10^(-7)` |
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| 219. |
A standard solution of `KNO_(3)` is used to make salt bridge, becauseA. Velocity of `K^(+)` is greater than that of `NO_(3)^(-)`B. Velocity of `NO_(3)^(-)` is greater than that of `K^(+)`C. Velocities of both `K^(+)` and `NO_(3)^(-)` are nearly the sameD. `KNO_(3)` is highly soluble in water |
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Answer» Correct Answer - C The salt bridge possesse, the electrolyte having nearly same ionic mobilities of its cation and anion. |
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| 220. |
For the calomel half-cell, `Hg, Hg_(2)Cl_(2)|Cl^(-)(aq)` values of electrode potentials are plotted at different log `[Cl^(-)]`. Variation is represented byA. B. C. D. |
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Answer» Correct Answer - D Half-cell reaction is : `2Hg(l)+2Cl^(-)(aq)rarr Hg_(2)Cl_(2)(s)+2e^(-)` `therefore K (1)/([Cl^(-)]^(2))` `E=E^(@)-(0.0591)/(2)"log"(1)/([Cl^(-)]^(2))` `E=E^(@)+0.0591 log [Al^(-)]` Thus, E increase as `[Cl^(-)]` increases. |
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| 221. |
Zinc granules are added in excess to 500 mL of 1M `Ni(NO_(3))_(2)` solution of `25^(@)C` untill the equilibrium is reached. If `E_(Zn^(2+)//Zn)^(@)` and `E_(Ni^(2+)//Ni)^(@)` are `-0.75V` and `-0.24V` respectively, find out the `[Ni^(2+)]` at equilibrium. |
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Answer» `{:("The redox change is "Zn+ni^(2+)hArrZn^(2+)+Ni),("mM before equilibrium 500 0"),("mM at equilibrium a (500-a)"):}` `E_(cell)^(0)=E_(OP_(ZN//Zn^(2+)))+E_(RP_(Ni^(2+)//Ni))` `E_(cell)^(0)=E_(Nn//Zn^(2+))^(0)+E_(RP_(Ni^(2+)//Ni))^(0)+(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)])` At equilibrium `E_(cell)=0` `thereforeE_(OP_(Zn//Zn^(2+)))^(0)+E_(RP_(Ni^(2+)//Ni))^(0)=-(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)])` `or 0.75+(-0.24)=-(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)])` `([Ni^(2+)])/([Zn^(2+)])="anti log"((-0.51xx2)/(0.059))=5.15xx10^(-18)` `therefore(a)/(500-a)=5.15xx10^(-18)` `thereforea=500xx5.15xx10^(-18)` `therefore[Ni^(2+)]=(mV)/(V)=(500xx5.15xx10^(-18))/(500)` `=5.15xx10^(-18)M` |
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| 222. |
Given that `E_(Fe^(2+)//Fe)^(.)=-0.44V,E_(Fe^(3+)//Fe^(2+))^(@)=0.77V` if `Fe^(2+),Fe^(3+)` and `Fe` solid are kept together thenA. `Fe^(3+)` IncreaseB. `Fe^(3+)` decreaseC. `Fe^(2+)//Fe^(3+)` remains unchangedD. `Fe^(2+)` decreases |
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Answer» Correct Answer - B `{:("Anode",,Ferarr Fe^(@+)+2e^(-),,),("Cathode",,[Fe^(3+)+e^(-)rarrFe^(2+)]xx2,,),("Cell",,Fe+2Fe^(3+)rarr3Fe^(2+),,),(E^(2)=0.77-(-0.44)=1.21V,,,,):}` `DeltaG^(@)=( - )ve" "` Reaction is spontaneous. |
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| 223. |
The charge required for the reduction of 1 mole of `Cr_(2)O_(7)^(2-)` ions to `Cr^(3+)` isA. 3FB. 3 coulombC. 6FD. `2xx6.023xx10^(23)e^(-)` |
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Answer» Correct Answer - C `Cr_(2)O_(7)^(-2)rarr2Cr^(3+)` Change in O.S.=6` `therefore`Charge required = 6F |
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| 224. |
Calculate the equilibrium constant for the reaction : `Fe^(2+)+Ce^(4+)hArr Fe^(3+)+Ce^(3+)` Given, `E_(Ca^(4+)//Ce^(3+))^(@)=1.44V` and `E_(Fe^(3+)//Fe^(2+))^(@)=0.68V` |
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Answer» We know, `E_("cell")^(@)=(0.059)/(1)log_(10)K_(C )` `E_("cell")^(@)=E_(OP_(Fr^(2+)//Fe^(3+)))^(@)+E_(RP_(Ce^(4+)//Ce^(3+)))^(@)` `=-0.68+1.44=0.76V` `therefore log_(10)K_(C )=(0.76)/(0.059)=12.8814` `therefore K_(C )=7.6xx10^(12)` |
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| 225. |
A solution containing 4.5mM of `Cr_(2)O_(7)^(2-)` and 15mM of `Cr^(3+)` shows a pH of 2.0. Calculate the potential of half reaction. (Standardd potential of the reaction `Cr_(2)O_(7)^(2-)rarrCr^(2+)` is 1.33 V) |
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Answer» We know, `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr 2Cr^(3+)+7H_(2)O` `E=E_(Cr_(2)^(6+)//Cr^(3+))^(@)+(0.059)/(6)"log"([Cr_(2)O_(7)^(2-)][H^(+)]^(14))/([Cr^(3+)]^(2))` `E=1.33+(0.059)/(6)"log"([(4.5)/(1000)]xx[10^(-2)]^(14))/([(15)/(1000)]^(2))` `=1.33+(0.059)/(6)log 20xx10^(-28)` `=1.33+(0.059)/(6)[log 20-28log 10]` `=1.33+(0.059)/(6)[1.3010-28]` `=1.33-0.26=1.07V` |
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| 226. |
Compute standard e.m.f. of cell `Zn_((s))+CO^(2+)rarr CO_((s))+Zn^(2+)` Given `E_(Zn//Zn^(2+))^(@)=+0.76V` and `E_(CO //CO^(2+))^(@)=+0.28V` |
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Answer» We know, `E_("Zn//Zn2^(+))^(@)=+0.76, E_(CO //CO^(+2))^(@)=+0.28V` `E_("cell")^(@)=E_(Zn//Zn^(2+))^(@)+E_(CO^(2+)//CO)^(@)=0.48V` |
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| 227. |
How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 exidation state for a period of 8.0 hour at a current of 8.46 ampere ? What is the area of the tray if the thickness of silver plating is `0.00254cm` ? Density of silver is `10.5g//cm^(3)`. |
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Answer» We, Know, `w_(Ag)=("E.i.t")/(96500)=(107.8xx8.46xx8xx60xx60)/(96500)=272.18g` Volume of `Ag=(272.18)/(10.5)=25.92mL` `E_("cell")^(@)=E_(OP_(Zn//Zn^(2+)))^(@)+E_(RP_(CO^(2+)//CO))^(@)` `=0.76-0.28` `E_("cell")^(@)=0.48V` |
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| 228. |
The electolysis of acetate solution produces ethane according to reaction: `2CH_3COO^-)toC_6H6(g)+2CO_2(g)+2e^-` The current efficiency of the process is 80% . What volume of gases would be produced at `27^(@)` C and 740 torr, if the current of 0.5 amp is used though the solution for 96.45 min?A. `6.0L`B. `0.60 L`C. `1.365 L`D. `0.91 L` |
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Answer» Correct Answer - D Equivalents of `CO_(2)` produced `=((Ixx eta)xxt)/(96500)` `=(0.5xx0.8xx96.5xx60)/(96500)=0.024` moles of `CO_(2)(n=1)` produced `= 0.024` moles of `C_(2)H_(6)(n=2)` produced `=(0.024)/(2)=0.012` Total moles of gases produced `rArr 0.036` `V_("gases")=(nRT)/(P)=(0.036xx0.0821xx300)/(((740)/(760)))` |
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| 229. |
When an electric current is passed through acidified water, `112ml` of `H_(2)` gas at `NTP` is collected at the cathode is `965` seconds. The current passed in amperes isA. 1B. 0.5C. 0.1D. 2 |
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Answer» Correct Answer - 2 `H_(2)O+e^(-)rarr(1)/(2)H_(2)+OH^(-)` mole of `H_(2)=(112)/(22400)=5xx10^(-3)` mole of `H_(2)=(E)/(F)It` `I=0.5` |
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| 230. |
Assertion (A): During electrolysis 48250 coulombs of electricity will deposit `0.5` gramequivalent of silver metal from `Ag^(+)` ions Reason (R): One Faraday of electricity will be required to deposity `0.5`gram - equivalent of any substanceA. A and R are correct R is the correct explanation of AB. A and R are correct R is not the correct explanation of AC. A is correct, but R is worngD. A is wrong, but R is correct |
| Answer» Correct Answer - C | |
| 231. |
The reduction potential of hydrogen electrode containing a solution of pH=4 isA. `0.236 V`B. `4.059 V`C. `-0.236V`D. `3.941V` |
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Answer» Correct Answer - C R.P of hydrogen electrode `=-0.059 xx p^(H)` |
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| 232. |
Enf of a cell in terms of reduction potential of its left and right electrodes is :A. `E=E_("left")-E_("right")`B. `E=E_("left")+E_("right")`C. `E=E_("right")-E_("left")`D. `E=-[E_("right")-E_("left")]` |
| Answer» Correct Answer - C | |
| 233. |
The reduction potential of hydrogen half cell will be negative if :A. `p(H_(2))"= 1 atm and "[H^(+)]=0.1M`B. `p(H_(2))"= 2 atm and "[H^(+)]=0.1M`C. `p(H_(2))"= 2 atm and "[H^(+)]=2.0M`D. `p(H_(2))"= 1 atm and "[H^(+)]=2.0M` |
| Answer» Correct Answer - B | |
| 234. |
If ` Theta` denotes standard reduction potentical, which is true:A. `E_(cell)^(0)=Phi_(R)-Phi_(L)`B. `E_(cell)^(0)=Phi_(L)+Phi_(R)`C. `E_(cell)^(0)=Phi_(L)-Phi_(R)`D. `E_(cell)^(0)=-(Phi_(L)+Phi_(R))` |
| Answer» Correct Answer - A | |
| 235. |
Conductivity (unit siemens) is directly propotional to area of the vessel and the concentration of the solution it and is inversely proportional to the length of the vessel then the unit of constant of proportionality isA. `"S m mol"^(-1)`B. `"S m"^(2)mol^(-1)`C. `S^(-2)"m mol"`D. `S^(-2)m^(2)mol^(-2)` |
| Answer» Correct Answer - B | |
| 236. |
`{:(Column-I,Column-II),((A)"Conductance",(p)Cm^(-1)),((B)"Specific condutance",(q)Ohm^(-1)cm^(2)mol^(-1)),((C)"Cell constant",(r)Ohm^(1-)),((D)"Equivalent conductance",(s)Ohm^(-1)cm^(-1)),((E)"Molar conductance",(u)Ohm^(-1)cm^(2) "equivalent"^(-1)):}`A. (a-q)(b-r)(c-s)(d-p)B. (a-r)(b-q)(c-p)(d-s)C. (a-r)(b-q)(c-s)(d-p)D. (a-p)(b-r)(c-s)(d-q) |
| Answer» Correct Answer - A | |
| 237. |
Specific conductance of 0.1 M `CH_(3)COOH` at `25^(@)` C is `3.9xx10^(-4) " ohm"^(-1) "cm"^(-1)` If `lambda^(infty)(H_(3)O^(+))` and `lambda^(infty)(CH_(3)COO^(-))` at `25^(@)`C are `349.0 and41.0 " ohm"^(-)"cm"^(2) "mol"^(-1)` respectively degree of ionisation of `CH_(3)COOH` at the given concentration isA. `1.0 %`B. `4.0 %`C. `5.0%`D. `2.0%` |
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Answer» Correct Answer - A `alpha=(kxx1000//conc.)/(lambda_(CH_(3)COO^(-))^(prop)+lambda_(H_(3)O^(+))^(prop))` |
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| 238. |
Resistance of a decimolar solution between two electrodes `0.02` meter apart and `0.0004m^(2)` in area was fround to be `50 ohm`. Specific conductance `(k)` is `:`A. `0.1 S-m^(-1)`B. `1 S-m^(-1)`C. `10 S-m^(-1)`D. `4xx10^(-4)S-m^(-1)` |
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Answer» Correct Answer - 2 `G^(@)=(l)/(A)=(0.02)/(0.0004)rArr50m^(-1)` `G=(1)/(R)=(1)/(50)rArr0.02S` `k=G,G^(@)=50xx0.02=1 Scm^(-1)` |
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| 239. |
In the plot of `Lambda ` and `sqrtC`, the slope isA. `Lambda^(@)`B. `-b`C. `(-2.303)/(R)`D. `oo` |
| Answer» Correct Answer - B | |
| 240. |
The limiting equivalent conductivity of `NaCl, KCl` and `KBr` are `126.5,150.0` and `151.5 S cm^(2) eq^(-1)` , respectively. The limiting equivalent ionic conductance for `Br^(-)` is `78 S cm^(2) eq^(-1)`. The limiting equivalent ionic conductance for `Na^(+)` ions would be `:`A. 128B. 125C. 49D. 50 |
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Answer» Correct Answer - 4 `Lambda_(m)^(oo)=(NaBr)=Lambda_(m)^(oo)(NaCl)+Lambda_(m)^(oo)(KBr)-Lambda_(m)^(oo)(KCl)` `Lambda_(m)^(oo)(Na^(+))+Lambda_(m)^(oo)(Br)=126.5+151.5-150` `Lambda_(m)^(oo)(Na^(+))=50` |
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| 241. |
The value of molar conductivity of `HCl` is greater than that of `NaCl` at a particular temperature becauseA. Ionic mobility of `H^(+)` is greater than that of `Na^(+)`B. the dipolie moment of NaCl is greater than that of HClC. NaCl is more ionic than HClD. HCl is Bronsted acid and NaCl is a salt of a strong acid and strong base |
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Answer» Correct Answer - A The greater the ionic mobility, the greater the molar conductance. The ionic mobility of `H^(+)` is greater than that of `Na^(+)` |
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| 242. |
A hydrogen electrode is dipped in a solution at `25^(@)C`. The potential of cell is `-0.177 V`. Calcualte the `pH` of the solution. |
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Answer» Correct Answer - C `E=E^(@)-(0.059)/(n)log.(1)/([H^(+)])` `-0.177=0+(0.059)/(1)log[H^(+)]` `log[H^(+)]=-3" "thereforepH=-log[H^(+)]=3` |
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| 243. |
When water is electrolysed, hydrogen and oxygen gas are produced. If 1.008 g of `H_(2)` is liberated at the cathode. What mass of `O_(2)` is formed at the anode? |
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Answer» `because(w_(1))/(w_(2))=(E_(1))/(E_(2)),` where `E_(1)` and `E_(2)` are eq.wt. of `H_(2)` and `O_(2)` respectively `therefore(1.008)/(w_(2))=(1.008)/(8)rArrw_(2)=8g` |
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| 244. |
The same current if passed through solution of silver nitrate and cupric salt connected in series. If the weights of silver deposited is `1.08g`. Calculate the weight of copper depositedA. `0.6454` gB. `6.354` gC. `0.3177` gD. `3.177` g |
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Answer» Correct Answer - C `m_(1)/m_(2)=E_(1)/E_(2)` |
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| 245. |
An electric current is passed through a copper voltmeter and a water voltameter connected in series. If the copper voltameter now weights 16 mg less, hydrogen liberated at the cathode of the water voltmeter measures at STP aboutA. `4.0`mlB. `5.6`mlC. `6.4`mlD. `8.4`ml |
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Answer» Correct Answer - B `(W_(1))/(W_(2))=(E_(1))/(E_(2))` |
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| 246. |
If a spoon of copper metal is placed in a solution of ferrous sulphate .A. Copper will precipitate outB. Iron will precipitateC. Cu and Fe will precipitateD. No reaction takes place |
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Answer» Correct Answer - D Copper electrode SRP is higher than the iron electrode it can not displaces the ion from `FeSO_(4)` |
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| 247. |
In the given figure the electrolytic cell contains `1L` of an aqueous `1M` Copper `(II)` sulphate solution. If `0.4 ` mole of electrons passed through of cell, the concentration of copper ion after passage of the charge will be A. `0.4M`B. `0.8M`C. `1.0M`D. `1.2M` |
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Answer» Correct Answer - 3 Number of moles of `Cu^(2+)` discharged from anode `=` number of moles of `Cu^(2+)` deposited at cathode. |
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| 248. |
`E_(RP)^(@)` and their respective half reactions for some change are given below: i) `Mn^(7+)(aq)+8H^(+)+5erarrMn^(2+)(aq)+4H_(2)O(l),` `E^(@)=+1.52V` ii) `Mn^(7+)(aq)+4H^(+)+3erarrMnO_(2)(s)+2H_(2)O(l),` `E^(@)=+2.26V` iii) `Mn^(7+)(aq)+2erarrMn^(5+)(aq),` `E^(@)=+0.56V("acid med.")` `Mn^(3+)+erarrMn^(2+)`, `E^(@)=+1.51V("acil med.")` v) `Mn^(4+)+erarrMn^(3+),` `E^(@)=+0.95V("acid med.")` The `E^(@)` for, `Mn^(4+)+2erarrMn^(2+),` is equal to:A. `1.23V`B. `2.46V`C. `0.56V`D. `-0.56V` |
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Answer» Correct Answer - A `{:(Mn^(+3)+e^(-)rarrMn^(2+),E^(@)=1.51V),(Mn^(+4)+2e^(-)rarrMn^(3+),E^(@)=0.954V),(Mn^(4+)+2e^(-)rarrMn^(2+),E^(@)=(1.51+0.95)/(2)):}` |
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| 249. |
`E_(RP)^(@)` and their respective half reactions for some change are given below: i) `Mn^(7+)(aq)+8H^(+)+5erarrMn^(2+)(aq)+4H_(2)O(l),` `E^(@)=+1.52V` ii) `Mn^(7+)(aq)+4H^(+)+3erarrMnO_(2)(s)+2H_(2)O(l),` `E^(@)=+2.26V` iii) `Mn^(7+)(aq)+2erarrMn^(5+)(aq),` `E^(@)=+0.56V("acid med.")` `Mn^(3+)+erarrMn^(2+)`, `E^(@)=+1.51V("acil med.")` v) `Mn^(4+)+erarrMn^(3+),` `E^(@)=+0.95V("acid med.")` Which is the least stable oxidation state of Mn?A. `Mn^(2+)`B. `Mn^(4+)`C. `Mn^(7+)`D. `Mn^(6+)` |
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Answer» Correct Answer - C The three case `Mn^(+7)` dissociates to give `Mn^(2+),Mn^(4+)` and `Mn^(5+)` respectively, Thus it is least stable. |
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| 250. |
`E_(RP)^(@)` and their respective half reactions for some change are given below: i) `Mn^(7+)(aq)+8H^(+)+5erarrMn^(2+)(aq)+4H_(2)O(l),` `E^(@)=+1.52V` ii) `Mn^(7+)(aq)+4H^(+)+3erarrMnO_(2)(s)+2H_(2)O(l),` `E^(@)=+2.26V` iii) `Mn^(7+)(aq)+2erarrMn^(5+)(aq),` `E^(@)=+0.56V("acid med.")` `Mn^(3+)+erarrMn^(2+)`, `E^(@)=+1.51V("acil med.")` v) `Mn^(4+)+erarrMn^(3+),` `E^(@)=+0.95V("acid med.")` The most stable oxidation state of Mn is:A. `Mn^(4+)`B. `Mn^(2+)`C. `Mn^(7+)`D. `Mn^(6+)` |
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Answer» Correct Answer - B Most `-ve` value of `DeltaG^(@)` correspond to most stable oxidation state |
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