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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.
| 201. |
Which of the following species can functon as an oxidising as well as reducing agent ?A. `Cl^(-)`B. `ClO_(4)^(-)`C. `ClO^(-)`D. `MnO_(4)^(-)` |
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Answer» Correct Answer - C Since Cl in `ClO^(-)` has ON of +1 which is higher than the lowest O.N of -1 and lower than the highest O.N of +5, it can act both as an oxidising agent as well as reducing agent. |
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| 202. |
An element `A` in a compound `ABD` has oxidation number `A^(n-)`. It is oxidised by `Cr_(2)O_(7)^(2-)` in acid medium. In the experiment `1.68xx10^(-3)` moles of `K_(2)Cr_(2)O_(7)` were used for `3.26xx10^(-3)` moles of `ABD`. The new oxidation number of `A` after oxidation is:A. `3`B. `3-n`C. `n-3`D. `+n` |
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Answer» Correct Answer - B Meq. Of `K_(2)Cr_(2)O_(7)`= Meq. Of `ABD` `n-` factor of `K_(2)Cr_(2)O_(7)` in acidic medium `=6` `6xx1.68xx10^(-3)=x xx3.26xx10^(-3)` `x=3` `implies` New oxidation state of `A^(-n)` will be `= -n+3` |
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| 203. |
The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium isA. One-fifthB. fiveC. OneD. Two |
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Answer» Correct Answer - D In alkaline medium `2KMnO_(4)+KI+H_(2)O rarr 2MnO_(2)+2KOH+KIO_(2)` |
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| 204. |
The charge on cobalt in `[Co(CN)_(6)]^(-3)` is -A. `-6`B. `-3`C. `+3`D. `+6` |
| Answer» Correct Answer - C | |
| 205. |
In which of the following compounds of Cr, the oxidation number Cr is not `+6` :-A. `CrO_(3)`B. `CrO_(2)Cl_(2)`C. `Cr_(2)O_(3)`D. `K_(2)Cr_(2)O_(7)` |
| Answer» Correct Answer - C | |
| 206. |
A sulphur containing species that can not be a reducing agent is :-A. `SO_(2)`B. `SO_(3)^(-2)`C. `H_(2)SO_(4)`D. `S_(2)O_(3)^(2-)` |
| Answer» Correct Answer - C | |
| 207. |
Question 1 and 2 should be answered using the unbalanced equation, `ClO_(3)^(-)+Br^(-) to Cl_(2)+Br_(2)` When this equation in balanced, what is the `Br^(-)//ClO_(3)^(-)` ratio?A. `1(1)`B. `2/(1)`C. `(3)/(1)`D. `5/(1)` |
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Answer» Correct Answer - d |
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| 208. |
The oxidation state of sulphur in `Na_(2)S_(4)O_(6)` isA. 1.5B. 2.5C. 3D. 2 |
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Answer» Correct Answer - B `overset(+1)(Na_(2))overset(x)(S_(4))overset(-2)(O_(6))" ":. 2+4x-12=0or4x=10` |
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| 209. |
Oxidation state of cobalt in `[Co(NH_(3))_(4)(H_(2)O)Cl]SO_(4)` is |
| Answer» Correct Answer - D | |
| 210. |
Oxidation number of cobalt in `[Co(NH_(3))_(6)]Cl_(2)Br` is -A. `+6`B. ZeroC. `+3`D. `+2` |
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Answer» Correct Answer - C Let the oxidation number of Co be x Oxidation number of `NH_(2)` is zero Oxidation number of `Cl` is `-1` Oxidation number of Br is `-1` Hence, `x+6(0)-1xx2-1=0` `therefore" "x=+3` So, the oxidation number of corbalt in the given complex compound is `+3`. |
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| 211. |
Determine the oxidation number of the element as indicated (ix) Cr in `[Cr(NH_(3))_(6)]Cl_(3)` |
| Answer» Correct Answer - `+3` | |
| 212. |
The average oxidation state of sulphur in sodium tetrathonate `(Na_(2)S_(4)O_(6))` is : |
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Answer» Correct Answer - c |
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| 213. |
The difference in the oxidation numbers of two types of sulphul atoms in `Na_(2)S_(4)O_(6)` is…..A. 5B. 4C. 3D. 2 |
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Answer» Correct Answer - a |
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| 214. |
What is the oxidation number of `Co` in `[Co(NH_(3))_(4)ClNO_(2)]`?A. `+5`B. `+3`C. `+4`D. `+2` |
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Answer» Correct Answer - D `[overset(**)(C )(NH_(3))_(4) CINO_(2)]` `x+4(0)+1(-1)+1(-1)=0` `x+0-1-1=0` `x-2=0`, `x=+2` |
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| 215. |
What happen when a solution of potassium chromate is treated with an excess of dil. Nitic acid?A. Cr reduces in the oxidation state `+3` from `CrO_(4)^(-2)`B. Cr oxidies in the oxidation state `+7` from `CrO_(4)^(2-)`C. `Cr^(+3)` and `Cr_(2)O_(7)^(2-)` will be formedD. `Cr_(2)O_(7)^(2-)` and `H_(2)O` will be formed |
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Answer» Correct Answer - d |
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| 216. |
`1` litre solution of unknown molarity is titrated by taking its `50 mL` solution against `KI` solution is strong acidic medium of excess `HCl`. The equivalence point was detected when `10 mL` of `0.1 M KI` was consumed The molarity of `KIO_(3)` solution is:A. `4xx10^(-4)M`B. `2xx10^(-2)M`C. `4xx10^(-3)M`D. `2xx10^(-3)M` |
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Answer» Correct Answer - C Meq.of `KIO_(3)="Meq.of" KI` `{:(2I^(5+)+,10e,rarr,I_(2)^(0),),(,2I^(-),rarr,I_(2)^(0)+,2e):}` `Mxx5xx50=10xx0.1xx1 rArrM=4.0xx10^(-3)` |
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| 217. |
What happen when a solution of potassium chromate is treated with an excess of dil. Nitic acid?A. `Cr_(2)O_(7)^(2-)` and `H_(2)O` are formedB. `Cr_(2)O_(4)^(2-)` is reduced to `0` state of `Cr`C. `CrO_(4)^(2-)` is reduced to `+3` state of `Cr`D. `Cr^(3+)` and `Cr_(2)O_(7)^(2-)` are formed |
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Answer» Correct Answer - A `2CrO_(4)^(2-)+2H^(+)rarrCr_(2)O_(7)^(2-)+H_(2)O` |
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| 218. |
Equivalent weight of `MnO_(4)^(ɵ)` in acidic neutral and basic media are in ratio of:A. `3:5:15`B. `5:3:1`C. `5:1:3`D. `3:5:5` |
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Answer» Correct Answer - D `KMnO_(4)` can act as an O.A in acidic, basic and neutral mediums as follows : In acidic medium, `MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O` In basic medium `MnO_(4)^(-)+2H_(2)O+3Fe^(-)rarrMnO_(2)+4OH^(-)` In neutral medium `MnO_(4)^(-)+2H_(2)O+3Fe^(-)rarrMnO_(2)+4OH^(-)` Let molecular mass of `KMnO_(4)=M` `:.` Ratio of Eq mass of `KMnO_(4)` in acidic, basic and neutral mediums are `(M)/(5):(M)/(3):(M)/(3)or3:5:5`. |
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| 219. |
Oxidation numbers of `Mn` in `K_(2)MnO_(4)` and `MnSO_(4)` are respectivelyA. `+7`,`+2`B. `+5`,`+2`C. `+6`,`+2`D. `+2`,`+6` |
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Answer» Correct Answer - C `{:(K_(2)overset(**)(M)nO_(4),,overset(**)(M)nsO_(4)),(2+x-8=0,,x+6-8=0),(x=+6,,x=+2):}` |
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| 220. |
Determine the oxidation number (O.N) of Fe in `[Fe(CN)_(6)]^(4-)` |
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Answer» `{:([Fe" " (CN)_(6)]^(4-)),(darr" "darr),(x" "-1):}` `because` O.N of CN is -1 `:. x-6 = -4 or x = -4 + 6 = +2` |
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| 221. |
Assertion :- `MnO_(4)^(-)` is always reduced to `Mn^(+2)`. Reason :- Decrease in oxidation number or gaining of electron means oxidation.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are False. |
| Answer» Correct Answer - D | |
| 222. |
`MnO_(4)^(-)` ions are reduced in acidic conditions to `Mn^(2+)` ions whereas they are reduced in neutral condition to `MnO_(2)`. The oxidation of 25 mL of a solution `x` containing `Fe^(2+)` ions required in acidic condition 20 mL of a solution y containing `MnO_(4)` ions. What value of solution y would be required to oxidize 25 mL of solution x containing `Fe^(2+)` ions in neutral condition ?A. 11.4 mLB. 12.0 mLC. 33.3 mLD. 35.0 mL |
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Answer» Correct Answer - C `overset(+7)(MnO_(4)^(-))+5e^(-)overset(H+)(rarr)overset(+2)(Mn^(2+))` `overset(+7)(MnO_(4)^(-))+3e^(-)overset(("neutral"))(rarr)overset(+4)(MnO_(2))` `:.` 20 mL of `MnO_(4)^(-)(H^(+))` `-=(20xx5)/(3)"mL of " MnO_(4)^(-)` (neural) `:. y=33.33 mL` |
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| 223. |
Oxidation numbers of `Mn` in `K_(2)MnO_(4)` and `MnSO_(4)` are respectivelyA. `+5`B. `+7`C. `+4`D. `+2` |
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Answer» Correct Answer - D As the `SO_(4)^(2-)` ion has O.N = charge on the ion `:.` O.N of Mn = +2. |
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| 224. |
The oxidation number of Fe in `[Fe(CN)_(6)]^(3-)` ion isA. `+2`B. `+3`C. `-2`D. `-3` |
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Answer» Correct Answer - B Let O.N. of Fe in `[Fe(CN)_(6)]^(3-)` is `x` (O.N. of `CN^(-1)=-1` ) `:. X +6(-1) = -3` or `x =+3` |
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| 225. |
The oxidation number of Mn is `+7` inA. manganese dioxideB. manganese chlorideC. manganese sulphateD. potassium permanganate |
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Answer» Correct Answer - D In `KMnO_(4)`, O.N. of Mn is +7 Let O.N. of Mn = x `1 + x+4 (-2)=0` or `x = 7`. |
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| 226. |
Match the following |
| Answer» Correct Answer - A(r),B(p),C(s),D(q) | |
| 227. |
Match the following |
| Answer» A(q,s),B(ps),C(p,r),D(p,r,t) | |
| 228. |
`aMnO_(4)^(-)+bI^(-)+cH^(+)rarrdMn^(2+)+eI_(2)+fH_(2)O` In above balance reaction, value of `((c)/(d))` will be :A. 1.3B. 1.2C. 8D. 5 |
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Answer» Correct Answer - c |
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| 229. |
Assertion : Insert electrolytes like KCl, `KNO_(3)` are used in salt bridge. Reason : Salt bridge provides an electric contact between the two solutions without allowing them to mix with each other.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 230. |
Reduction is a process wich involves (i) decrease in the oxidation number of one fo the atoms (ii) loss of oxygen or an electronegative element (iii) addition of hydrogen or an electropositive element . (iv) gain of electrons .A. (i), (ii), (iiii), (iv)B. (i), (ii) (iii)C. (ii), (iii), (iv)D. (i), (iv) |
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Answer» Correct Answer - A Reduction is a decrease in oxidation number and correspons to a gain, or apparent gain, of electrons (electronation). In biological system, reduction often corresponds to the addition of hydrogen to molecules or polyatomic ions and oxidation often corresponds to the removal of hydrogen. |
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| 231. |
Oxidation reaction involves loss of electrons, and reduction reaction involves gain of electrons. The reaction in which a species disproportinates into two oxidation states ( lower and higher) is called disproportionation reaction. Which of the following statements is correct?A. An element in the lowest oxidation state acts only as a reducing agent.B. An element in the highest oxidation state acts only as a reducing agent.C. The oxidation number of `V` is `Rb_(4)K(HV_(10)O_(28)) is +4`.D. The oxidation number and valency of `Hg` in calomel is `+1` |
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Answer» Correct Answer - A a. An element in the lowest oxidation state can only attain higher oxidation state. So it is a reductant and undergoes oxidation. b. An element in the highest oxidation state only attain lower oxidation state. So it si an oxidant and undergoes reduction. c. The oxidation state of `Rb` and `K` is `+1` ( first group element) `overset(+1xx4+1)(Rb_(4)K(HV_(10)O_(28))^(-5)` `HV_(10)O_(28)^(-5)=1+10x-2xx28= -5` d. Calomel is `overset(+1xx2-1xx2)(Hg_(2)+Cl_(2))` The oxidation number of `Hg` is `+1` and valency is `2`. |
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| 232. |
Assertion (A): In the process of drying dishes with a towel, the wetting agent is the dish and the drying agent is the towel. Reason (R ): The wetting agent gets wet during the process.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)` and `(R )` are incorrect. |
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Answer» Correct Answer - C `(A)` is correct and `(R)` is incorrect. Correct (R): The drying agent gets wet during the process, since water must go into it. In redox reaction, the oxidant is reduced, the reducing agent is oxidised. The electrons are transferred in a manner similar to the water in drying analogy. |
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| 233. |
The pair of compounds that can exist together is:A. `FeCl_(3)`, `SnCl_(2)`B. `HgCl_(2)`, `SnCl_(2)`C. `FeCl_(2)`, `SnCl_(2)`D. `FeCl_(3)`, `KI` |
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Answer» Correct Answer - C The compounds with lower oxidation number and which cannot reduced by one another can exist together. Thus `FeCl_(2)` and `SnCl_(2)` can exist together as `Fe^(2+)` cannot be reduced by `Sn^(2+)`. |
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| 234. |
Oxidation numbers of P in ` PO_4^(3-)`, of S in ` SO_4^(2-)`, and that of `Cr` in ` Cr_2O_7^(2-)` are respectively ,A. ` +3, +6` and `+6`B. ` +5, +6` and `+6`C. ` +3. +6`, and `+5`D. ` +5. +3`, and `+6` |
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Answer» Correct Answer - B Let x be the oxidation number of P in ` PO_4^(3-)` ` x+4 (-2) =-3` ` x-8 =-3` or ` x= + 5` Let x be the oxidation number of S in ` SO_4^(2-)` : ` x +4 (-2) =-2` ` x-8 =-2` ` x = +6` Let x be the oxidation number of ` Ci` in ` Cr_2O_7^(2-) `: ` 2x+7 (-2) =-2` ` 2x -14 =-2` ` 2x = 12` ` x= +6 ` In all the three cases, we have used the rule that the sum of all oxidation numbers in a polyatomic ion must equal the charge on the ion. |
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| 235. |
Which of the following staement (s) `is//are` not true about the following decomposition reaction `2KCIO_(3)rarr2KCI+3O_(2)`A. Potassium is undergoing oxidationB. Cholorine is undergoing oxidationC. Oxygen is reducedD. None of the species are undergoing oxidtion or reduciton |
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Answer» `2overset(+1)Koverset(+5)Cioverset(-2)O_(3)rarr2overset(+1)Koverset(-1)CI+3overset(0)O_(2)` (a) O.N of K does not change (b) Since O.N of CI decreases from +5 to -1 therefore chlorine undergoes reduction and not oxidation (c )Since O.N of O increases from -2 to O therefore oxygen is undergoing oxidation and not reduction (d) This statement is also not correct Thus all the option (a,b,c,d) are not correct |
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| 236. |
`NH_(3)+Ocl^(ɵ)toN_(2)H_(2)+Cl^(ɵ)` On balancing the above equation in basic solution, using integral coefficient, which of the following whole number of will be the coefficient of `N_(2)H_(4)`?A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A `2NH_(3)+OCl^(-)rarrN_(2)H_(4)+Cl^(-)+H_(2)O` |
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| 237. |
A certain weioght of pure `CaCO_(3)` is made to react completely with 200mL of a HCl solution to given 227mL of `CO_(2)` gas at STP. The notmality of the HCl` solution is :A. `0.05N`B. `0.1N`C. `1.0N`D. `0.2N` |
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Answer» Correct Answer - b |
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| 238. |
Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance. Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. From the reaction `M^(x+)+MnO_(4)^(-)rarrMO_(3)^(-)+Mn^(2+)+1//2O_(2)` If one mole of `MnO_(4)^(-)` oxidizes 1.67 moles of `M^(x+)` to `MO_(3)^(-)`, then the value of x in the reaction isA. 5B. 3C. 2D. 0 |
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Answer» Correct Answer - C `overset(+7)(MnO_(4)^(-))+5e^(-)rarroverset(+2)(Mn)` Since 1 mole of `MnO_(4)^(-)` accepts 5 moles of electrons, therefore 5 moles of electron are lost by 1.67 moles of `M^(x+)` 1 mole of `M^(x+)` will lose electrons `= 5//1.67 =3` mol (approx) Since `M^(x+)` changes to `MO_(3)^(-)` (where O.N of M = 5) by accepting 3 electrons `:.` oxidising state of `M^(x+)` i.e., `x = +5 - 3 = +2` `overset(+2)(M^(2+))rarroverset(+5)(M)O_(3)^(-)+3e^(-)`. |
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| 239. |
Compound ` CrO_5` has structure as shown ltbtgt The oxidation number fo Cr in the above compound is .A. `10`B. `5`C. `4`D. `6` |
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Answer» Correct Answer - D In the given compound, there are four oxygen atoms in the peroxidge while the fifth oxygen is in the oxide state. If we assume that the oxidation state fo Cr in ` CrO_5` is `x` then ` x+ (-2) + 4 (-4) =0` ltbgt ` x-6 =0` ` x-=+6` Alternatively, we can work out the oxidation number by breaking the bonds heterolytically. Since Cr is forming six bonds whith O atoms, it loses `6` electrons when bonding electrons are given to oxgne. Thus, its oxidation number is `+6`. |
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| 240. |
Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance. Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. For the redox reaction `xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+)rarrx,y and z` areA. `"2 5 16"`B. `"16 5 2"`C. `"5 16 2"`D. `"2 16 5"` |
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Answer» Correct Answer - A The balanced redox reaction is `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)rarr2Mn^(2+)+10CO_(2)+8H_(2)O` Thus the coefficients of `MnO_(4)^(-), C_(2)O_(4)^(2-) and H^(+)` respectively are 2,5 and 16. |
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| 241. |
Consider the salt `K_(x)H_(y)(C_(2)O_(4))_(z).2H_(2)O` . The relationship between x, y and z is :A. `x+y-z=0`B. `x+y=2z`C. `x+y+z=0`D. None of these |
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Answer» Correct Answer - b |
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| 242. |
Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance. Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. A mole of `N_(2)H_(4)` loses 10 moles of electrons to forn a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation rate of nitrogen in Y ? (There is no change in the oxidation number of hydrogen)A. `-1`B. `-3`C. `+3`D. `+5` |
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Answer» Correct Answer - C Total O.N of 2 nitrigen atoms in `N_(2)H_(4)` is -2 since it loses 10 moles of electrons, thereforem the total O.N of two nitrogen atoms in Y increases by 10 i.e., the total O.N of the two introgen atoms in`Y = -4 + 10 = +6` `:.` O.N of each nitrogen in `Y = 6//2 = +3`. |
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| 243. |
Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance. Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation `xCu+uHNO_(3)rarrCu(NO_(3))_(2)+NO+NO_(2)+H_(2)O`. The coefficients of x and y areA. 2 and 3B. 2 and 6C. 1 and 3D. 3 and 8 |
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Answer» Correct Answer - B Balanced equations for producing NO and `NO_(2)` respectively are `3Cu+8HNO_(3)rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O`....(i) `Cu+4HNO_(3)rarrCu(NO_(3))_(2)+2NO_(2)+2H_(2)O` ....(ii) Adding eqn (i) and eqn (ii), `4Cu+12HNO_(3)rarr4Cu(NO_(3))_(2)+2NO_(2)+2NO+6H_(2)O` or `2Cu+6HNO_(3)rarr2Cu(NO_(3))_(2)+NO_(2)+NO+3H_(2)O` Thus, coefficients x and y of Cu and `HNO_(3)` are respectively 2 and 6. |
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| 244. |
How many litres of `Cl_(2)` at STP will be liberated by the oxidation of `NaCl` with `10 g KMnO_(4)` in acidic medium: (Atomic weight: `Mn=55 and K=39)`A. 3.59B. 7.08C. 1.77D. None of these |
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Answer» Correct Answer - a |
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| 245. |
Which of the following reactions involve oxidation and reduction?A. `NaBr+HClrarrNaCl+HBr`B. `HBr+AgNO_(3)rarrAgBr+HNO_(3)`C. `H_(2)+Br_(2)rarr2HBr`D. `Na_(2)O+H_(2)SO_(4)rarrNa_(2)SO_(4)+H_(2)O` |
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Answer» Correct Answer - c |
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| 246. |
The number of mole of oxalate ions oxidised by one mole of `MnO_(4)^(-)` ion is: |
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Answer» The redox reaction is : `2MnO_(4)^(-)+16H^(+)+5C_(2)O_(4)^(2-)rarr2Mn^(2+)+10CO_(2)+8H_(2)O` 2 moles of `MnO_(4)^(-)` oxidise `C_(2)O_(4)^(2-)` ions =5mole 1 mole of `MnO_(4)` oxidises `C_(2)O_(4)^(-)` ions =`5//2` mol Since `x/y`=`5/2`, Therefore" x=5,y=2. or x+y=7 |
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| 247. |
If 25mL of a `H_(2)SO_(4)` solution reacts completely with `1.06g` of pure `Na_(2)CO_(3)` , what is the normality of this acid sotution :A. 1NB. 0.5NC. 1.8ND. 0.8N |
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Answer» Correct Answer - d |
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| 248. |
The number of mole of oxalate ions oxidised by one mole of `MnO_(4)^(-)` ion is:A. `1//5`B. `2//5`C. `5//2`D. `5` |
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Answer» Correct Answer - C `[Mn^(7+)+5e rarr Mn^(2+)] xx 2` `[(C^(3+))_(2) rarr 2C^(4+) + 2e] xx 5` |
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| 249. |
The number of moles of oxalate ions oxidised by one mole of `MnO_(4)^(-)` ion in acidic medium is :A. `(5)/(2)`B. `(2)/(5)`C. `(3)/(5)`D. `(5)/(3)` |
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Answer» Correct Answer - a |
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| 250. |
The mass of oxalic acid crystals `(H_(2)C_(2)O_(4).2H_(2)O)` required to prepare 50 mL of a 0.2 N solution is:A. 4.5gB. 6.3gC. 0.63gD. 0.45g |
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Answer» Correct Answer - c |
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