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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.
| 3151. |
Chemical reactions involve interation of atoms and molecules. A large number of atoms `//` molecules `(` approximately `6.023xx10^(23))` are present in a few grams of any chemical compound varying with their atomic `//` molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry, and radiochemistry. The following example illustrates a typical case, involving chemical `//` electrochemical reaction, which requires a clear understanding of the mole concept. A `4.0M` aqueous solution of `NaCl` is prepared and `500mL` of this solution is electrolyzed. This leads to the evolution of chlorine gas at one of the electrodes `(` atomic mass of `Na ` is 23 and `Hg` is `200)(1F=96500C)`. The total number of moles of chlorine gas evolved isA. 0.5B. 1C. 2D. 3 |
| Answer» Correct Answer - b | |
| 3152. |
If `K_c` for the reaction `Cu^(2+)(aq)+Sn^(2+)(aq)toSn^(4+)(aq)+Cu(s)` at `25^(@)` C is represented as `2.6xx10^y` then find the value of y. (Given:`E_(Cu^(2+)"|"Cu)^(@)=0.34V,E_(Sn^(4+)"|"Sn^(2+)^(@)=0.15V)` |
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Answer» The cell reaction : `Sn^(2+)|Sn^(4+)||Cu^(2+)|Cu` At the equilibrium point, `E_(cell)=0` `E_(cell)=E Cu^(2+)|Cu-E_(Sn)^(4+)|Sn^(2+)=0` `"or"[E^(@)Cu^(2+)|Cu+(0.0591)/(2)" log "[Cu^(2+)]]-[E^(@)Sn^(4+)|Sn^(2+)+(0.0591)/(2)"log"(Sn^(4+))/([Sn^(2+)])]=0` `"or"[E_(Cu^(2+))^(@)|Cu-E_(Sn^(4+)|Sn^(2+))^(@)]+[(0.0591)/(2)"log"([Cu^(2+)][Sn^(2+)])/([Sn^(4+)])]=0` `"or"(+0.34-0.15)-(0.0591)/(2)"log"([Sn^(4+)])/([Cu^(2+)][Sn^(4+)])=0` `or 0.19-(0.0591)/(2)" log "K_(c )=0` `log K_(c )=(0.19xx2)/(0.0591)=6.43` `K_(c )="Anti "log (6.43)=2.69xx10^(6)(x=6)` |
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| 3153. |
The nature of curve of `E_(cell)^(@)` vs. log `K_c` is :A. straight lineB. parabolaC. hyperbolaD. elliptical curve |
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Answer» Correct Answer - A |
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| 3154. |
e.m.f. diagram for some ions is given as : `FeO_(4)^(2-)overset(E^(@)=+2.20V)rarrFe^(3+)overset(E^(@)=+0.77V)rarrFe^(2+)overset(E^(@)=-0.445V)(rarr)Fe^(0)` Datermine the value of `E_(FeO_(4)^(2-)//Fe^(2+))^(@)`. |
| Answer» Correct Answer - `1.84 V ;` | |
| 3155. |
When 96500 coulombs of electricity are passed through nickel sulphate solution, the amount of nickel deposited will beA. 1 molB. 0.5 molC. 0.1 molD. 2 mol |
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Answer» Correct Answer - B 1 faraday deposits 1 equivalent of nickel 1 equivalent of `Ni = 1//2` mole of nickel |
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| 3156. |
The required charge for one equivalent weight of silver deposited on cathode isA. `9.65xx10^(7)C`B. `9.65xx10^(4)C`C. `9.65xx10^(3)C`D. `9.65xx10^(5)C` |
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Answer» Correct Answer - B for deposition of one equivalent silver required charged is 96500C. |
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| 3157. |
If ` 0.5 ` amp current is passed through acidified silver nitrate solution for `10` minutes. The mass of silver deposited on cathode, is (eq. wt. of silver nitrate = `108`).A. ` 0. 235 g`B. ` 0.336 g`C. ` 0. 536 g`D. ` 0. 636 g` |
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Answer» Correct Answer - C Faraday constant depends upon the current passed. |
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| 3158. |
If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes the mass of silver deposited on cathode, is (eq. wt. of silver nitrate=108) :A. 2.3523 gB. 3.3575 gC. 5.3578 gD. 6.3575 g |
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Answer» Correct Answer - B (b) Current (I) =0.5 amp. Time (t)=100 mm `=100xx60 sec.=6000 s`. Eq. mass of Ag (E ) =108 `W=(EIt)/(96500)=(108xx0.5xx6000)/(96500)=3.3575 g`. |
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| 3159. |
The requrierd charege for one equivalent weifht of silver deposited on cathode is.A. ` 9. 65 xx 10^7 C`B. ` 9. 65 xx 10^4 C`C. ` 9. 65 xx 10^3 C`D. ` 9.65 xx 10^5 C` |
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Answer» Correct Answer - B For deposition of one equivalent silver required charged is `96500 C`. |
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| 3160. |
How many amperes would be needed to produce 60.0 g of magnesium during the electrolysis of molten `MgCl_(2)` in 2 hours ? |
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Answer» Correct Answer - 67 A `Mg^(2+)(aq)+underset(2F)underset(1mol)2e^(-) to underset(24" g.")underset(1" mol")(Mg(s))` 24.0 g of Mg are produced by passing charge `=2xx96500" C"` 60.0 g of Mg are produced by passing charge `=((60g))/((24.0g))xx(2xx96500C)=482500" C"` `l=(Q)/(t)=((482500" C"))/((2xx60xx60s))=67.0" amp"` |
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| 3161. |
A metal wire carries a current of `4` ampere. How many electrons pass through a point in the wire in one second? |
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Answer» Total charge passed in `1` sec. `= 4 xx 1` `= 4` coulomb `(because Q = i xx t)` `:. 1` Faraday or `96500 C` current carried by `= 6.023 xx 10^(23)` electrons `:. 4` coulomb current carried by `= (6.023 xx 10^(23) xx 14)/(96500)` `= 2.5 xx 10^(9)` electrons |
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| 3162. |
How much electricity in terms of Faraday is required to produce `40.0 g` of `Al` from molter `Al_(2)O_(3)`? |
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Answer» Eq. of `Al = (i.t)/(96500)` `(6e+Al_(2)^(3+) rarr 2Al)` `:.` Eq. `wt.` of `Al = (27)/(6//2)` or `(40)/(27//3) = (i.t)/(96500)` or i.t `= (120)/(27) xx 96500 = 4.44 F` |
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| 3163. |
How many electrons per second pass through a cross-section of copper wire carrying `10^(-16)` A current ? |
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Answer» Correct Answer - `624 es^(-1)` Charge carried by one electron `=1.602xx10^(-19)C` Charge carried by cross-section of copper wire`=10^(6)A=10^(-16)Cs^(-1)` Number of electrons passing per second through the copper wire`=((10^(-16)Cs^(-1)))/((1.602xx10^(-19)C))` `=0.624xx10^(3)=624es^(-1)` |
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| 3164. |
A current of 12 ampere is passed through an electrolytic cell containing aq. `NiSO_(4)` solution . Both Ni and `H_(2)` gas are formed at the cathode . The current efficiency is 60% . What is the mass of nickel deposited on the cathode per hour ? (At.mass of Ni = 98.7)A. 5.91 gB. 3.941 gC. 7.883 gD. 2.645 g |
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Answer» Correct Answer - C `Q = I xx t (s) 12 xx 60 xx 60 = 43200` C . `Ni^(2+) + 2e^(-) to Ni` . Theortically , `2 xx 96500` C will deposit Ni = 58.7 g . Hence , 43200 C will deposit Ni = 13.137 g . As current efficiency is 60 % , mass actually deposited = `(60)/(100) xx 13.137` = 7.88 g . |
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| 3165. |
`0.2864 g` of `Cu` was deposited on passage of a current of `0.5` ampere for `30` minutes through a copper sulphate solution. The electrochemical equivalent of copper isA. `0.32 g`B. `0.00032 g`C. `0.032 g`D. `0.0032 g` |
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Answer» Correct Answer - B In electrolysis, the mass of a substacne deposited or libereated when `1` ampere of current is passed through an electrolyte for `1` second (`1` coulomb of electricity) is known as its electrochemical equivalent `(Z)`. Therefore `Z = ("Mass of Cu deposited")/("No. of Coulombs")` `= (0.2864 g)/((0.5 A)(30 xx 60 s))` `= 0.00032 g//C` |
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| 3166. |
Calculate the mass of Ag deposited at cathode when a current of 2 ampere was passed through a solution for 15 minutes. |
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Answer» Correct Answer - 2.014 g Reaction at cathode is : `underset(1F(1xx96500" C"))(Ag^(+)(aq))+e^(-) to underset(1" mol"(108 g))(Ag(s))` 96500C of charge is required to deposit Ag=108 g. 1800 C of charge required to deposit `Ag=(108 g)xx((1800 C))/((96500C))=2.014 g`. |
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| 3167. |
How many electrons per sec pass through a cross-section of a copper wire carrying `10^(-15)` ampere? |
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Answer» Current strength `(i) = ("Total charge")/("Time (in sec.)")=(C)/(s)` Also Total charge = Number of electron `xx` Change on one electron `:. i = ("Number of electron" xx "Charge on one electron")/("Time (in sec.)")` `:.` No. of electron passed in `1` sec. `= (i)/("Charge on 1 electron") = (10^(-15))/(1.6 xx 10^(-19)) = 6250` |
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| 3168. |
How many copper will be deposited at cathode of an electrolytic cell containing `Cu^(2+)` ions by passing 2 ampere of current for 60 minutes. |
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Answer» Correct Answer - 2.37 g `Cu^(2+)(aq)+underset(2F)underset(2" mol")(2e^(-)) to underset(63.5 g)underset(1" mol")(Cu(s))` Quantity of charge (Q) passed `=(2"amp".)xx(60xx60s)=7200"C"` `2xx96500C` of charge deposit `Cu=((63.5 g))/((2xx96500C))xx(7200C)=2.37 g`. |
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| 3169. |
On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of `CuCl_(2)` , all copper of the solution of `CuCl_(2)` solution was (Molar mass of Cu = 63.5 , Faraday constant = `96500 C mol^(-1)`).A. 0.07 MB. 0.2 MC. 0.005 MD. 0.02 M |
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Answer» Correct Answer - C `Q = 1 xx 965 = 965` C `Cu^(2+) + 2e^(-) to Cu`, `2 xx 96500` C decompose `CuCl_(2) = 1 ` mole `therefore` 965 C decompose `CuCl_(2) = (1 xx 965)/(2 xx 96500) =(1)/(200)` mole This was present in one litre of `CuCl_(2)` solution . Hence , strength = 0.005 M. |
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| 3170. |
Sodium cannot be obtained by the electrolysis of aqueous solution of `NaCl` using `Pt` electrodes. |
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Answer» Correct Answer - T Correction `:` in question sodium cannot be obtained . In aqueous solution, it is `H_(2)O` which is reduced. `2H_(2)O+2e^(-) rarr 2H_(2)+2overset(c-)(O)H` |
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| 3171. |
A current liberates 0.504 g of hydrogen in 2 hours, the amount of copper lib erated from a solution `CuSO_(4)` by the same current flowing for the same time would beA. 31.8gB. 63.6gC. 15.9gD. 6.36g |
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Answer» Correct Answer - A g eq. of hydrogen =g eq. of Cu `(0.504)/(1)=(W)/(63.5//2) therefore W=15.9g` |
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| 3172. |
How much electricity is required in coulomb for the oxidation of : (a) `1` mol of `H_(2)O` to `O_(2)`, (b) `1` mole of `FeO` to `Fe_(2)O_(3)` ? |
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Answer» (i)` " " underset(1"mol")(H_(2)O)to 2H^(+)+1//2O_(2)+underset(2F)2e^(-)` Electricity (charge) required =2F`=2xx96500" C"=1.93xx10^(5)" C"` ltbr. (ii) `" " 2FeO+1//2O_(2) to Ni+2HNO_(3)`. `underset(2" mol")(Fe^(2+)) to Fe_(2)^(3+)+underset(2F)2e^(-)` For the oxidation of two moles of FeO, charge required =2F For the oxidation of one mole of FeO, charge required =1 F =96500 C. |
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| 3173. |
In an electrolytic cell :A. anode is positively chargedB. cathode is negatively chargedC. oxidation takes place at anodeD. reduction takes at cathode |
| Answer» Correct Answer - A::B::C::D | |
| 3174. |
The number of electrons passing per second through a cross section of Cu wire carrying 10 ampere isA. `6 xx 10^(19)`B. `8 xx 10^(19)`C. ` 1 xx 10^(19)`D. `1.6 xx 10^(19)` |
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Answer» Correct Answer - C Charge = `10 xx 1 = 10 C ` 1 F = 96500 C = `6 xx 10^(23)` electron `therefore` 10 C = n electrons `therefore n = (6 xx 10^(23) xx 10)/(96500) = 6.2 xx 10^(19)` |
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| 3175. |
An unknown metal M displaces nickel from nickel (II) sulphate solution but does not displace magnese from magnese sulphate solution which order represents the correct order of reducing power?A. `MngtNigtM`B. `NigtMngtM`C. `MngtMgtNi`D. `MgtNigtMn`. |
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Answer» Correct Answer - C As M can displace `Ni^(2+)`. As M cannot displace `Mn^(2+)` from `MnSO_(4)` O.P. of M is less than Mn. `therefore` Order of O.P. is `NiltMltMn` More the O.P. ester to oxidise stronger is the reducing power. `therefore` Correct order of reducing power. `MngtMgtNi`. |
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| 3176. |
On the electrolysis of very dilute aqueous solution of `NaOH` using Pt electrodesA. `H_(2)` i s evolved at cathodeB. Na is evolved at cathodeC. `O_(2)` is evolved of anodD. `H_(2)` is evolved at anode. |
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Answer» Correct Answer - A,C On the electrolysis of very dilute aqueous solution of NaOH using inert electrodes `H_(2)` is evolved at cathode while `o_(2)` is evolved at anode. |
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| 3177. |
A certain current liberated 0.504 g of hydrogen in 2 hours. How many gram of copper can be liberated by the same current flowing for the same time in `CuSO_(4)` solution ?A. 31.8 gB. 63.6 gC. 15.9 gD. 6.36 g |
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Answer» Correct Answer - C `("Wt. of Cu")/("Wt. of" H_(2)) = ("Eq. mass of Cu")/("Eq. of mass of H") ` , i.e., `("Wt. of Cu")/(0.50) = (63.6//2)/(1)` or Wt. of Cu = 15.9 g . |
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| 3178. |
How many electrons are there in one coulomb of electricity?A. `6.02 xx 10^(21)`B. `6.24 xx 10^(18)`C. `6.24 xx 10^(15)`D. `6.02 xx 10^(16)` |
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Answer» Correct Answer - B 96500 C = `6 xx 10^(23)` electron `therefore 6 xx 10^(20) = 1 xx 10^(-3)` eq. `therefore 1 C = 6.24 xx 10^(18)` |
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| 3179. |
Cations absorb `6.023xx10^(22)` electrons for their reduction. How many equivalents of the ion are reduced ?A. 0.1B. 0.01C. 0.001D. 0.0001 |
| Answer» Correct Answer - A | |
| 3180. |
2.5 faradays of electricity is passed through solution of `CuSO_(4)`. The number of gram equivalents of c opper depsoited on the cathode would beA. 1B. 2C. 2.5D. 1.25 |
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Answer» Correct Answer - C 1F deposits 1g eq. `therefore` 2.5F deposit 2.5g eq. |
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| 3181. |
An ion is reduced to the element when it absords `6xx10^(20)` electrons. The number of equivalents of the ion is:A. 0.1B. 0.01C. 0.001D. 0.0001. |
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Answer» Correct Answer - C `6.02xx10^(23)` electrons reduce 1g eq. of the ions `6xx10^(20)` electrons reduce `(6xx10^(20))/(6.20xx10^ (23))=10^(-3)geq.=0.001g eq` |
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| 3182. |
An ion is reduced to the element when it absords `6xx10^(20)` electrons. The number of equivalents of the ion is:A. `0.10`B. `0.01`C. `0.001`D. `0.0001` |
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Answer» Correct Answer - C `6 xx 10^(23)` electron = 1 eq. (A/C to `"def"^(n)` of Faraday) `6 xx 10^(20)` electrons = 0.001 eq. |
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| 3183. |
`{:(Content,Resistance ,"Volume of gas produced"),(,(at same T & P),),(Cell-1 CuSO_(4)(aq),10Omega,"at anode is" V_(1)),(Cell-2NaCI(aq),2Omega,"at anode is"V_(2)),(Cell-3 ZnSO_(4)(aq),6Omega,"at cathode is"V_(3)):}` |
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Answer» Correct Answer - `2:3:1` `{:(,Cell-1,Cell-2,Cell-3),("Product",O_(2),CI_(2),H_(2)),("n-Factor",4,2,2),("Equivalent from the",1,3//4,1//4),("gives resistance",,,),(Mol,1//4,3//8,1//8),("Volume ratio",2,3,1):}` |
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| 3184. |
`A 500mL` sample of a `0.1M Cr^(3+)` is electrolyzed with a current of `96.5A`. If the remaining `[Cr^(3+)]` is `0.04M` then the duration of process is:- |
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Answer» Correct Answer - `90sec` Initial moles of `Cr^(3+) = 0.5 xx 0.1 rArr 0.05` Final moles of `Cr^(3+) = 0.04 xx 0.5 rArr 0.02` Moles of `Cr^(3+)` reduced `Ϸ 0.05 -0.02 rArr 0.03` `:.` Number of equivalents of `Cr^(3+)` reduced `=(1 xxt)/(96500)` `=(t xx 96.5)/(96500)` `3 xx 0.03 =(t xx 96.5)/(96500), t = 90` sec. |
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| 3185. |
A 250.0 mL sample of a 0.20M `Cr^(3+)` is electrolysed with a current of 96.5 A. If the remaining `[Cr^(3+)]` is 0.1 M, the duration of process is:A. 25secB. 225secC. 150secD. 75sec |
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Answer» Correct Answer - D |
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| 3186. |
For `0.0128 N` solution fo acetic at `25^(@)C` equivalent conductance of the solution is 1.4 mho `cm^(3) eq^(-1)` and `lambda^(oo) = 391` mho `cm^(2)eq^(-1)`. Calculate dissociation constant `(K_(a))` of acetic acid. |
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Answer» Correct Answer - `1.6 xx 10^(-7)` `alpha=(lamda_(N))/(lamda^(infty))=(1.4)/(391)=3.58 xx10^(-3)` `{:(CH_(3)COOH rarr,CH_(3)COO^(-),+,H^(+)),(C,,,),(C-C alpha,,bar(C alpha),bar(C alpha)):}` `K_(a)=C alpha^(2)` `K_(a)=1.28 xx 10^(-2) xx(3.58 xx10^(-3))^(2)=1.6 xx10^(-7)`. |
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| 3187. |
If the aqueous solutions of the following salts are electrolysed for 1 hour with 10 ampere current, which solution will deposit the maximum mass of the metal at cathode ? The atomic weights are : `Fe=56, Zn=65, Ag=108, Hf=178` and `W=184`.A. `ZnSO_(4)`B. `FeCl_(3)`C. `HfCl_(4)`D. `AgNO_(3)` |
| Answer» Correct Answer - D | |
| 3188. |
Which of the following units is correctly matched?A. SI units of conductivity is `Sm^(-1)`B. SI units of molar conductivity is `Scm^2mol^(-1)`C. SI unit of conductance is `S^(-1)`D. All of these |
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Answer» Correct Answer - A |
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| 3189. |
Molar conductivity decreases with decrease in concentrationA. For strong electrolytesB. For weak electrolytesC. Both for strong and weak electrolytesD. For non electrolytes |
| Answer» Correct Answer - D | |
| 3190. |
Define limiting molar conductivity. Why does conductivity of an electrolyte decrease with discrease in concentration. |
| Answer» For answer, consult Section 19. | |
| 3191. |
Molar conductivity of 0.15 M solution of KCI at 298 K, if its conductivity is 0.0152 S `cm^(-1)` will beA. `124 Omega^(-1) "cm"^(2) "mol"^(-1)`B. `204 Omega^(-1) "cm"^(2) "mol"^(-1)`C. `101 Omega^(-1) "cm"^(2) "mol"^(-1)`D. `300 Omega^(-1) "cm"^(2) "mol"^(-1)` |
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Answer» Correct Answer - C `Lambda_m=(Kxx1000)/M=(1.52xx10^(-2)xx1000)/0.15` `=101 Omega^(-1) cm^2 mol^(-1)` |
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| 3192. |
Standared reduction electrode potenitals of three metals `A, B` and ` C ` are ` =0.5 V, - 3.0 V`, and ` -1.2 V` respectively. The reducing power of these metals are :A. `A gt B gt C`B. `C gt B gt A`C. `A gt C gt B`D. `B gt C gt A` |
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Answer» Correct Answer - D (d) Higher the nagative value of reductionpotential, more will be the reducing power. Hence `B gt C gt A`. |
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| 3193. |
The ion which has the lowest ionic mobility isA. `Rb^+`B. `K^(+)`C. `Na^(+)`D. `Li^(+)` |
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Answer» Correct Answer - D `Li` ion has the lowest ionic mobility due to the extensive hydration of ion as a result of very high charge density on account of its very small size. The higher charger density causes `Li` ions to be more highly hydrated by ion-diode interactions than the larger ions. Since a hydrated ion has to drag along a shell of water as it moves through the solution, its mobility in naturally less than that of an unhydrated ion. |
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| 3194. |
Assertion `(A):` The mobility of `Na^(o+)` is lower than that of `K^(o+)` ion. Reason `(R): ` The ionic mobility depends upon the effective radius of the ion.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R)` is incorrect.D. If `(A)` is incorrect, `(R)` is correct. |
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Answer» Correct Answer - C `(A)` is correct but `(R)` is incorrect. In aqueous medium, the mobilities of cations depend on`:` `(i)` Charge density `[(` ionic charge `//` ionic size `)` ratio`]` ` (ii)` Extent of hydration Decreasing order of mobility of cations of first groups are `:` `Cs^(o+)gt Rb^(o+), K^(o+)gtNa^(o+)gtLi^(o+)` |
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| 3195. |
Statement : `E_(cell)^(@)` is an intensive property. Explanation : `DeltaG^(@)//n` is also intensive property.A. `S` is correct but `E` is wrongB. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C `DeltaG^(@) = nE^(@)F` `:. E^(@) = (-DeltaG^(@))/(nF)` Since `DeltaG` is negative property then `E^(@)` is also intensive property. |
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