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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.
| 951. |
What is the net charge on ferrous ion ?A. ` +2`B. `+3`C. `+4`D. ` +5` |
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Answer» Correct Answer - A Ferrous ion is described as `Fe^(2+)` . Thus, net change is `2+`. |
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| 952. |
The oxidation number of carbon in .` CH_2Cl_2` is .A. `0`B. `2`C. `3`D. `5` |
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Answer» Correct Answer - A Let x be the oxidation number of C in `CH_2CI_2`. The algeraic sum fo the oxidation number of all the atoms in a compoune must be zero. Thus, ` x+ 2 (+1) + 2 (-1) =0` or ` x=0` |
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| 953. |
Which substance is serving as a reducing agent in the following reaction? `14H^(+)+Cr_(2)O_(7)^(2-)+3Ni rarr 2Cr^(3+)+7H_(2)O+3Ni^(2+)`A. `H_(2)O`B. `Ni`C. `H^(+)`D. `Cr_(2)O_(7)^(2-)` |
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Answer» Correct Answer - B The oxidation number of `Ni` changes from `0` to `+2` |
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| 954. |
The oxidation number of sulphur in `H_(2)SO_(4),H_(2)S_(2)O_(4) and H_(2)S_(2)O_(6)` are respectivelyA. `+3,+4,+5`B. `+5,+4,+3`C. `+6,+3,+5`D. `+3,+5,+4` |
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Answer» Correct Answer - C O.N of S in `H_(2)SO_(4)` : Let O.N. of S = x `:. 2(+1)+x+4(-2)=0` `rArr x = +6` O.N of S in `H_(2)S_(2)O_(4)` Let O.N. of S = x `:. 2 (+1)+2x+4(-2)=0` `rArr x = +3` O.N of S in `H_(2)S_(2)O_(6)` Let O.N of S = x `:. 2(+1)+2x+6(-2)=0` `rArr x = +5`. |
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| 955. |
Small quantities of compounds TX, TY and TZ are put into separate test tubes containing X, Y and Z solutions. TX does not react with any of these. TY reacts with both X and Z. TZ reacts only with X. The decreasing order of ease of oxidation of the anions `X^(-),Y^(-) and Z^(-)` isA. `Y^(-),Z^(-),X^(-)`B. `Z^(-),X^(-),Y^(-)`C. `Y^(-),X^(-),Z^(-)`D. `X^(-),Z^(-),Y^(-)` |
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Answer» Correct Answer - A `TX + Y rarr ` No reaction `TX + Z rarr` No reaction `TY + X rarr TX +Y` `TY + Z rarr TZ + Y` `TZ + X rarr TX + Z` Ease of oxidation `Y^(-) gt Z^(-) gt X^(-)` |
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| 956. |
Equivalent mass of oxidizing agent in the reaction, `SO_(2)+2H_(2)S rarr 3S+2H_(2)O` isA. `32`B. `64`C. `16`D. `8` |
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Answer» Correct Answer - C `4e+S^(4+) rarr S^(0)` `SO_(2)` is oxidant, `E_(SO_(2))=(64)/(4)=16` |
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| 957. |
Equivalent mass of oxidising agent in the reaction `SO_(2)+2H_(2)Srarr3S+2H_(2)O` isA. 321B. 64C. 16D. 8 |
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Answer» Correct Answer - C `overset(+4)SO_(2)^(-2)+overset(+1)2H_(2)overset(-2)Srarr3overset(0)S+overset(-2)2H_(2)O` In this reaction , `SO_(2)` acts as oxidising agent `overset(+4)SO_(2)rarroverset(0)S` `SO_(2)+4e^(-)rarrS` therefore Eq mass of `SO_(2)`=`("Molecular mass")/(4)` `=(32+32)/(4) =16` |
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| 958. |
The oxidation number of `Pr` in `Pr_(6)O_(11)` isA. `(22)/(6)`B. `(20)/(6)`C. `3`D. `4` |
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Answer» Correct Answer - A `Pr_(6)O_(11)`: `6x-22=0` `:. x= (22)/(6)` |
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| 959. |
When `SO_(2)` is passed through an acidified `K_(2)Kr_(2)O_(7)` solution, the oxidation state of sulphur changes fromA. `+4` to `0`B. `+4` to `+2`C. `+4` to `+6`D. `+6` to `+4` |
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Answer» Correct Answer - C `6e+(Cr^(6+))_(2)rarr2Cr^(3+)` `S^(4+)rarrS^(6+)rarr2e` |
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| 960. |
Assertion: Conversation of black lead painting is made to white by the action of `H_(2)O_(2)`. Reason: Sulphur is oxidised to `SO_(4)^(2-)`A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explantion of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A `PbS+4H_(2)O_(2) rarr PbSO_(4)+4H_(2)O` |
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| 961. |
`8 g` of sulphur are burnt to form `SO_(2)`, which is oxidised by `Cl_(2)` water. The solution is treated with `BaCl_(2)` solution. The amount of `BaSO_(4)` precipitated is:A. `1 "mole"`B. `0.5 "mole"`C. `0.24 "mole"`D. `0.25 "mole"` |
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Answer» Correct Answer - D `S rarr SO_(2) rarr SO_(4)^(2-)` `1` mole `S` gives `1` mole `BaSO_(4)`. |
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| 962. |
A 2.5 mol of hydrazine `N_(2)H_(4)` loses 25 mole of electrons is being converted to a new compound X. Assuming that all of the nitrogen appears in the new compound, what is the oxidation state of nitrogen in compound X ?A. `-1`B. `-2`C. `+3`D. `+4` |
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Answer» Correct Answer - C Let the O.N of N in new compound X be be x `overset(+6)(N_(2))H_(4)rarr overset(+x)(2N)+n e^(-)` Let the no. of electron lost per mole of = n `:. 2(-2) = 2x + n (-1)` or `n = 2x + 4 = 2 (x+2)` `:.` No. of electron lost per mole `= 2(x+2)` no. of electrons lost by 2.5 mols `= 2.5 xx 2(2+x)` Now, `2.5 xx 2(2+x) = 25` `5(2+x) = 25` or `2 + x = 5 or x = +3` |
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| 963. |
In the reaction `2CuSO_(4)+4KI rarr 2Cu_(2)I_(2)+I_(2)+2K_(2)SO_(4)` the equivalent weight of `CuSO_(4)` will be:A. 31.75B. 63.5C. 127D. 15.88 |
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Answer» Correct Answer - b |
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| 964. |
A sample weighing `2.198 g` containing a mixture of `AO` and `A_(2)O_(3)` takes `0.015 "mole of" K_(2)Cr_(2)O_(7)` to oxidise the sample completely to form `AO_(4)^(-)` and `Cr^(3+)`. If `0.0187` `"mole of" AO_(4)^(-)` is formed, what is `at.wt`. of `A`? |
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Answer» Let `M.wt. "of" AO` and `A_(2)O_(3)` be `m` and `n` respectively. `therefore m=a+16`…..(1) and `n=2a+48`……(2) where `a` is atomic weight of `A`. Now suppose `X` and `Y g` of `A_(2)O_(3)` are present in mixture. Then `X+Y=2.198`.....(3) Also Meq.of `AO+"Meq.of"A_(2)O_(3)="Meq.of" K_(2)Cr_(2)O_(7)` ` `=0.015xx6xx1000` ....(4)` `{(because, ,A^(2+),rarr,A^(7+)+5e ),(,,A_(2)^(3+),rarr,2A^(7+)+8e),(,6e+,Cr_(2)^(6+),rarr,2Cr^(3+)):}` `therefore` By eqs. (4), `(5X)/(a+16)+(8Y)/(2a+48)=0.09`.....(5) Also mole `AO_(4)^(-)` by `AO+"mole of" AO_(4)^(-)` by `A_(2)O_(3)=0.0187` `(X)/(a+16)+(2Y)/(2a+48)=0.0187`.....(6) `because"Moles raio of "AO:AO_(4)^(-)::1:1,A_(2)O_(3):AO_(4)^(1-)::1:2` Solving eqs. (3), (5) and (6) `a=100` |
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| 965. |
Arrange the following in order of: (a) Increasing oxidation no: `MnCl_(2),MnO_(2),Mn(OH)_(3),KMnO_(4)` (b) Decreasing oxidation no: `HXO_(4),HXO_(3),HXO_(2),HXO` (c ) Increasing oxidation no.: `I_(2),HI,HIO_(4),ICI` |
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Answer» (a) `{:((a),MnCl_(2)lt,Mn(OH)_(3)lt,MnO_(2)lt,KMnO_(4)),("Ox.No".,+2,+3,+4,+7):}` `{:((b),HXO_(4)gt,HXO_(3)gt,HXO_(2)gt,HXO),("Ox.No",+7,+5,+3,+1):}` `{:((c ),HIlt,I_(2)lt,ICIlt,HIO_(4)),("Ox.No",-1,0,+1,+7):}` |
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| 966. |
When the ion `Cr_(2)O_(7)^(2-)` acts as an oxidant in acidic aqueous solution the ion `Cr^(3+)` is formed. How many mole of `Sn^(2+)` would be oxidised to `Sn^(4+)` by one mole `Cr_(2)O_(7)^(2-)` ion:A. `2//3`B. `3//2`C. `2`D. `3` |
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Answer» Correct Answer - D `[(Cr^(6+))_(2) + 6e rarr 2Cr^(2+)]xx1` `[Sn^(2+) rarr Sn^(4+) + 2e] xx 3` |
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| 967. |
In the chemical reaction, `K_(2)Cr_(2)O_(7)+xH_(2)SO_(4)+ySO_(2)rarrK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+zH_(2)O` `x,y, and z` areA. `1,3,1`B. `4,1,4`C. `3,2,3`D. `2,1,2` |
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Answer» Correct Answer - A The balanced equation is `{:(14H^(o+)+6e^(-)+Cr_(2)O_(7)^(2)rarr2Cr^(3)+7H_(2)O),(6H_(2)O+3SO_(2)rarr3SO_(4)^(2-)+6e^(-)+12H^(o+)),(ulbar(Cr_(2)O_(7)(2-)+2H^(o+)+3SO_(2)rarr2Cr^(3+)+3SO_(4)^(2-)+H_(2)O)):}` or lt brgt `K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+3SO_(2)rarrCr_(2)(SO_(4))_(3)+K_(2)SO_(4)+H_(2)O` `:. x=1, y=3, z=1` |
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| 968. |
`4.9 g` of `K_(2)Cr_(2)O_(7)` is taken to prepare `0.1 L` of the solutio. `10 mL` of this solution is further taken to oxidise `Sn^(2+)` ion into `Sn^(4+) ion` so produced is used in second reaction to prepare `Fe^(3+)` ion then the millimoles of `Fe^(3+)` ion formed will be (assume all other components are in sufficient amount)[Molar mass of `K_(2)Cr_(2)O_(7)=294 g`].A. `5`B. `20`C. `10`D. none of these |
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Answer» Correct Answer - C `K_(2)Cr_(2)O_(7)+Sn^(2+) rarr Sn^(4+)+Cr^(3+)` |
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| 969. |
Predict the products of electrolysis in eaCHM of the following `:` `a.` An aqueous solution of `AgNO_(3)` with silver electrodes. `b.` An aqeous solution of `AgNO_(3)` with platinum electrodes, `c.` A dilute solution of `H_(2)SSO_(4)` with platinum electrodes. `d.` An aqueous solution of `CuCl_(2)` with platinum electrodes. |
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Answer» (i) `AgNO_(3)` ionizes in aqueous solutions to form `Ag^(+)` and `NO_(3)^(-)` ions . On On electrolysis, either `Ag_(+)` ions or `H_(2)O` molecules can be reduced at the cathode. But the reduction potential of `Ag_(+)` ions is higher than that of `H_(2)O`. `Ag_((aq))^(+) + e^(-) to Ag_((s)) , E^(@) = +0.80 V ` `2H_(2)O_((l)) + 2 e^(-) to H_(2 (g)) + 2 OH_((aq))^(-) , E^(@) = - 0.83 V` Hence `Ag^(+)` ions are reduced at the cathode . Similarly , Ag metal or `H_(2)O` molecules can be oxidized at the anode . But the oxidation potential of Ag is higher than that of `H_(2)O ` molecules . `Ag_((s)) to Ag_((aq))^(+) + e^(-) , E^(@) = - 0.80 V` `2H_(2)O_((l)) to O_(2 (g)) + 4 H_((aq))^(+) + 4 e^(-) , E^(@) = -1.23 V` Therefore , Ag metal gets oxidized at the anode . (ii) Pt cannot be oxidized easily . Hence , at the anode , oxidation of water occurs to liberate `O_(2)` . At the cathode , `Ag^(+)` ions are reduced and get deposited . (iii) `H_(2)SO_(4)` ionizes in aqueous solutions to give `H^(+)` and `SO_(4)^(-)` ions . `H_(2)SO_(4 (aq)) to 2H_((aq))^(+) + SO_(4 (aq))^(2-)` On electrolysis , either of `H^(+)` ions or `H_(2)O` molecules can get reduced at the cathode . But the reduction potential of `H^(+)` ions is higher than that of `H_(2)O` molecules . `2H_((ag))^(+) + 2e^(-) to H_(2 (g)) , E^(@) = 0.0V` `2H_(2)O_((aq)) + 2e^(-) to H_(2 (g)) + 2OH_((aq))^(-) , E^(@) = -0.83 V` Hence , at the cathode , `H^(+)` ions are reduced to liberate `H_(2)` gas . On the other hand , at the anode , either of `SO_(4)^(2-)` ions or `H_(2)O` molecules can get oxidized . But the oxidation of `SO_(4)^(2-)` involves breaking of more bonds than that of `H_(2)O` molecules . Hence , `SO_(4)^(2-)` ions have a lower oxidation potential than `H_(2)O` . Thus `H_(2)O` is oxidized at the anode to liberate `O_(2)` molecules . (iv) In aqueous solutions , `CuCl_(2)` ionizes to give `Cu^(2+)` and `Cl^(-)` ions as : `CuCl_(2 (aq)) to Cu_((aq))^(2+) + 2Cl_((aq))^(-)` On electrolysis , either of `Cu^(2+)` ions or `H_(2)O` molecules can get reduced at the cathode . But the reduction potential of `H^(+)` ions is higher than that of `H_(2)O` molecules . `2H_((aq))^(+) + 2e^(-) to H_(2(g)) , E^(@) = 0.0V` `2H_(2)O_((aq))k + 2e^(-) to H_(2 (g)) + 2OH_((aq))^(-) , E^(@) =- 0.83 V` . Hence , at the cathode , `H^(+)` ions are reduced to liberate `H_(2)` gas . On other hand , at the anode , either of `SO_(4)^(2-)` ions or `H_(2)O` molecules can get oxidized . But the oxidation of `SO_(4)^(2-)` involves breaking of more bonds than that of `H_(2)O` molecules . Hence , `SO_(4)^(2-)` ions have a lower oxidation potential than `H_(2)O` . Thus , `H_(2)O` is oxidized at the anode to liberate `O_(2)` molecules . (iv ) In aqueous solutions , `CuCl_(2)` ionizes to give `Cu^(2+)` and `Cl^(-)` ions as : `CuCl_(2(aq)) to Cu_((aq))^(2+) + 2Cl_((aq))^(-)` On electrolysis , either of `Cu^(2+)` ions or `H_(2)O` molecules can get reduced at the cathode . But the reduction potential of `Cu^(2+)` is more than that of `H_(2)O` molecules . `Cu_((aq))^(2+) + 2e^(-) to Cu_((aq)) , E^(@) = - 0.34 V` `H_(2)O_((l)) + 2e^(-) to H_(2(g)) + 2OH^(-) , E^(@) = - 0.83 V` Hence `Cu^(2+)` ions are reduced at the cathode and get deposited . Similarly , at the anode , either of `Cl^(-)` or `H_(2)O` is oxidized . The oxidation potential of `H_(2)O` is higher than that of `Cl^(-)` . `2Cl_((aq))^(-) to Cl_(2 (aq)) + 2e^(-) , E^(@) = - 1.36V` `2H_(2)O_((l)) to O_(2 (g)) + 4H_((aq))^(-) + 4e^(-) , E^(@) = -1.23 V ` But the oxidation of `H_(2)O` molecules occurs at a lower electrode potential than that of `Cl^(-)` ions because of over- voltage (extra voltage required to liberate gas ). As a result , `Cl^(-)` ions are oxidized at the anode to liberate `Cl_(2)` gas . |
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| 970. |
0.0026 M `I_(2)` solution having unknown volume is reacted with excess of ferrous thiocynate solution to form `Fe_(2)O_(3),SO_(4)^(2-),CN^(-)` along with `I^(-)` . If all the sulphate ions formed are precipitated using `BaCl_(2)` such that 16776 gm of `BaSO_(4)` is obtained, calculate volume of `I_(2)` consumed in litre. (At. mass : Ba=137) |
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Answer» Correct Answer - 9 |
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| 971. |
Assertion (A) In the reaction between potassium permanganate and potassium iodide, permanganate ions acts as oxidising agent. Reason ( R) Oxidation state of manganese changes from +2 and +7 during the reaction.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not the correct explanation of AC. A is true but R is falseD. Both A and R are false |
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Answer» Correct Answer - c Assertion is true but reason is false `10KI+2Koverset(+7)(M)nO_(4)+8H_(2)SO(4)to2overset(+2)(M)nSO_(4)+6K_(2)SO_(4)+8H_(2)O+5I_(2)` Oxidation state of Mn decreases from +7 to +2 |
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| 972. |
Arrange the following metals in the order in whiCHM they displace eaCHM other from the solution of their salts. `Al, Cu, Fe, Mg, ` and `Zn`.A. Cu, Fe, Zn, Al, MgB. Fe, Zn, Cu, Al, MgC. Mg, Cu, Fe, Zn, AlD. Mg, Al, Zn, Fe, Cu |
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Answer» Correct Answer - D Since a metal with lower electrode potential is a stronger reducing agent, Mg can displace all the given metals, Al can displace all metals except Mg. Zn can displace all metals except Mg and Al. Fe can displace only Cu. The order in which they can displace each other from their salt solutions is Mg, Al, Zn, Fe, Cu. |
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| 973. |
The standard reduction potentials at 298K, for the following half cells are given: `Zn^(2+)(aq)+2e^(-)hArrZn(s):E^(@)=-0.762V` `Cr^(3+)(aq)+3e^(-)hArrCr(s):E^(@)=-0.740V` `2H^(+)(aq)+2e^(-)hArrH_(2)(g), E^(@)=0.000V` `Fe^(3+)(aq)+e^(-)hArrFe^(2+)(aq),E^(@)=0.770V` Which is the stronget reducing agent?A. Zn (s)B. Cr (s)C. `H_(2)(g)`D. `Fe^(+2)(aq)` |
| Answer» Correct Answer - A | |
| 974. |
In the reacion , `3Br_(2)+6CO_(3)^(2-)+3H_(2)Orarr5Br^(-)+BrO_(3)^(-)+6HCO_(3)^(-)`A. Bromine is reduced and carbonate ion is oxidised.B. Bromine undergoes disproportionation.C. Bromine is reduced and water is oxidised.D. Only water is oxidised to carbonic acid. |
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Answer» Correct Answer - B `3Br_(2)+6CO_(3)^(2-)+3H_(2)Orarr 5 Br^(-)+BrO_(3)^(-)+6HCO_(3)^(-)` Here `Br_(2)` is both reduced and oxidised. `Br_(2)(0)rarr Br^(-)(-1)` (reduction) `Br_(2)(0) rarr BrO_(3)^(-) (+5)` ( oxidation) |
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| 975. |
`5 mL` of `N-HCl`, `20 mL` of `N//2 H_(2)SO_(4)` and `30 mL` of `N//3` `HNO_(3)` are mixed together and the volume is made to `1 L`. The normality of the resulting solution isA. 1gmB. 0.5gmC. 0.1gmD. 21.8gm |
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Answer» Correct Answer - d |
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| 976. |
10mL of 0.5NHCl, 30mL of `0.1NHNO_(3)` and 75mL of 0.1M `H_(2)SO_(4)` are mixed together. The normality of the resulting solution will be :A. 0.2NB. 0.1NC. 0.4ND. 0.5N |
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Answer» Correct Answer - a |
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| 977. |
Assertion (A): `SO_(2)` and `Cl_(2)` are both bleaching agents. Reason (R ): Both are reducing agents.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 Correct `(R )` : `Cl_(2)` is an oxidising agent while `SO_(2)` is a reducing agent. |
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| 978. |
0.2828 g of iron wire was dissolved in excess dilute `H_(2)SO_(4)` and the solution was made upto 100mL. 20mL of this solution required 30mL of `N/(30)K_(2)Cr_(2)O_(7)` solution for exact oxidation. Calculate percent purity of Fe in wire. |
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Answer» Correct Answer - 99 |
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| 979. |
`5.7 g` of bleaching powder was suspended in `500 mL` of water. `25 mL` of this suspension on treatment with `KI` and `HCl` liberated iodine which reacted with `24.35 mL "of" N//10Na_(2)S_(2)O_(3)`. Calculate `%` of available `Cl_(2)` in bleaching powder. |
| Answer» `%` of chlorine `=30.33%` | |
| 980. |
What is molarity and normality of a `MnO_(4)^(-)` solution if `32.00 mL` of the solution is required to titrate `40.00mL` of `0.400N Fe^(2+)`? ` MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarrMn^(2+)+5Fe^(3+)+4H_(2)O` |
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Answer» Meq of `MnO_(4)^(-)=Meq.of fe^(2+)` `(Mn^(7+)+5erarrMn^(2+))` `:. Nxx32=40xx0.400` `(Fe^(2+)rarrFe^(3+)+e)` ` N=0.5` or `M_(KMnO_(4))=N_(KMnO_(4))//"valence factor" ` Also, `M_(KMnO_(4))=(0.5)/(5)=0.1M` |
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| 981. |
1mL of unknown solution of `H_(2)SO_(4)` is diluted upto 100mL and then its 25mL is titrated with 10mL of 0.2 M NaOH. The excess acid required 10mL of 0.1 M `Ba(OH)_(2)` solution. Find molarity of original `H_(2)SO_(4)` solution. |
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Answer» Correct Answer - 8 |
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| 982. |
2 g of brass containing Cu and Zn only reacts with 3M `HNO_(3)` solution. Following are the reactions taking place `Cu(s) + HNO_(3) (aq) to Cu^(2+) (aq) +NO_(2)(g) + H_(2)O(I)` `Zn(s) + H^(+)(aq) + NO_(3)^(-)(aq) to NH_(4)^(+) + Zn^(2+)(aq) + H_(2)O(l)` The liberated `NO_(2)(g)` was found to be 1.04 L at `25^(@)`C and 1 atm `[Cu=63.5 , Zn=65.4]` The percentage by mass of Cu in brass wasA. `67%`B. `70%`C. `80%`D. `90%` |
| Answer» Correct Answer - A | |
| 983. |
In which of the following reactions `H_(2)O_(2)` is a reducing agent?A. `2FeCl_(2)+2HCl+H_(2)O_(2) rarr 2FeCl_(3)+2H_(2)O`B. `Cl_(2)+H_(2)O_(2) rarr 2HCl + O_(2)`C. `2HI + H_(2)O_(2) rarr 2H_(2)O + I_(2)`D. `H_(2)SO_(3)+H_(2)O_(2) rarr H_(2)SO_(4)+H_(2)O` |
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Answer» Correct Answer - B In (B) `H_(2)O_(2)` (O.N of O =-1) is getting oxidised to `O_(2)` (O.N of O = 0) and thus acts as a reducing agent. |
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| 984. |
For the redox reaction `MnO_(4)^(ө)+C_(2)O_(4)^(2-)+H^(o+)rarrMn^(2+)+CO_(2)+H_(2)O` the correct coefficients of the reactions for the balanced reaction areA. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,5,16):}`B. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(16,5,2):}`C. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(5,16,2):}`D. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,16,5):}` |
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Answer» Correct Answer - A The balanced chemical equations `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)rarr2Mn^(2+)+10CO_(2)+8H_(2)O` |
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| 985. |
For the redox reation `MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)rarrMn^(2+)CO_(2)+H_(2)O` The correct stoichiometric coefficients of `Mno_(4)^(-),C_(2)O_(4)^(2-)` and `H^(+)` respectively:A. 2,5,16B. 16,5,2C. 5,16,2D. 2,16,5 |
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Answer» Correct Answer - A The balanced equation is `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+10H^(+)rarr2Mn^(2+)+10CO_(2)+8H_(2)O` |
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| 986. |
For the redox reation `MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)rarrMn^(2+)CO_(2)+H_(2)O` The correct stoichiometric coefficients of `Mno_(4)^(-),C_(2)O_(4)^(2-)` and `H^(+)` respectively:A. 2, 5, 16B. 16, 5, 2C. 5, 16, 2D. 2, 16, 5 |
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Answer» Correct Answer - a |
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| 987. |
Potassium dichromate `(K_(2)Cr_(2)O_(7))` is an orange coloured compound, very frequently used in laboratory as an oxidising agent as well as in a redox titration. It is generally prepared from chromit `(FeCr_(2)O_(4))` ore according to the following reactions: (a) Fusion of chromite ore with sodium carbonate in excess of air. `FeCr_(2)O_(4)+Na_(2)CO_(3)+O_(2) to Na_(2)CrO_(4)+Fe_(2)O_(3)+CO_(2)` (b) Acidifying filetered sodium chromate solution with sulphuric acid. `Na_(2)CrO_(4)+H_(2)SO_(4) to Na_(2)Cr_(2)O_(7)+Na_(2)SO_(4)+H_(2)O` (c) Treating sodium dichromate with potassium chloride. `Na_(2)Cr_(2)O_(7)+KCl to K_(2)Cr_(2)O_(7)+NaCl` Answer the following questions using above information. If you are intially provided with 224 gm of pure chromite ore and 169.6gm of sodium carbonate, the minimum volume of air required at 1 atm and 273 K to consume at least one of the reactant completely, if aire contains 20% by volume of oxygen gas is :A. 156.8LB. 196LC. 28LD. 152.4L |
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Answer» Correct Answer - a |
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| 988. |
In an ore, the only oxidizable material is `Sn^(2+)`. This ore is titrated with a dichromate solution containing `2.5g` of `K_(2)Cr_(2)O_(7)` in `0.5 "litre"`. A `0.40 g` sample of the ore required `10.0 cm^(3)` of titrant to reach equivalence point. Calculate the percentage of tin in ore. |
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Answer» Redox changes are: `Sn^(2+)rarrSn^(4+)+2e` `6e+Cr_(2)^(6+)rarr2Cr^(3+)` Since `Sn^(2+)` is oxidised by `K_(2)Cr_(2)O_(7)`. `therefore` Meq.of `Sn^(2+)="Meq.of" K_(2)Cr_(2)O_(7)` used for tin `=NxxV_("in " mL) ( because N=(2.5)/((294.2)/(6)xx0.5))` `= (2.5)/((294.2)/(6)xx0.50)xx10=1.0197` `therefore (w_(Sn^(2+)))/(118//2)xx1000=1.0197` `therefore w_(Sn^(2+))=0.06g` `therefore % Sn=(0.06)/(0.4)xx100=15%` |
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| 989. |
Redox is a reaction in which both oxidation and reduction will take place simultaneously . It is obvious that if one substance gives electron there must be another substance to accept these electrons . In some reactions, same substance is reduced as well as oxidised, these reactions are termed as disproportionation reactions. For calculating equivalent mass in redox reaction change in oxidtaion number is realted to n-factor which is reciprocal of molar ratio. This reaction is an example of `Br_(2) + OH^(-) to BrO_(3)^(-) + H_(2)O + Br^(-)`A. Oxidation reaction onlyB. Reduction reaction onlyC. Neutralization reactionD. Disproportionation reaction |
| Answer» Correct Answer - D | |
| 990. |
Redox is a reaction in which both oxidation and reduction will take place simultaneously . It is obvious that if one substance gives electron there must be another substance to accept these electrons . In some reactions, same substance is reduced as well as oxidised, these reactions are termed as disproportionation reactions. For calculating equivalent mass in redox reaction change in oxidtaion number is realted to n-factor which is reciprocal of molar ratio. Which of the following is an example of redox reaction ?A. `2NO_(2) to N_(2)O_(4)`B. `NH_(4)OH to NH_(4)^(+) + OH^(-)`C. `2NO_(2) + H_(2)O to HNO_(3) + HNO_(2)`D. `N_(2)O_(5) +H_(2)O to 2HNO_(3)` |
| Answer» Correct Answer - C | |
| 991. |
Redox is a reaction in which both oxidation and reduction will take place simultaneously . It is obvious that if one substance gives electron there must be another substance to accept these electrons . In some reactions, same substance is reduced as well as oxidised, these reactions are termed as disproportionation reactions. For calculating equivalent mass in redox reaction change in oxidtaion number is realted to n-factor which is reciprocal of molar ratio. The equivalent weight of `Cu_(2)S` in the following reaction is `Cu_(2)S + O_(2) to Cu^(+2) + SO_(3)`A. `(M.wt)/(1)`B. `(M.wt)/(10)`C. `(M.wt)/(8)`D. `(M.wt)/(11)` |
| Answer» Correct Answer - B | |
| 992. |
which of the following is a redox reaction ?A. ` Mg(OH)_2 + 2NH_4Cl rarr MgCl_2 + 2NH_4OH`B. ` Zn+ 2 AgCN rarr 2Ag+ Zn(CN)_2`C. `CaC_2O_4 + 2HCl rarr CaCl_2 + H_2C_2O_4`D. `NaCl +KNO_3 rarr NaNO_3 + KCl` |
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Answer» Correct Answer - B It is a redox reaction as `Zn` is oxidized to ` Zn^(2+)` while ` Ag^(+)` is reduced to Ag. Reactions (1) and (3) are displacement reactions. In the former, a stronger base displaces the weaker base while in the latter a stronger acid displaces the weaker acid. Reaction (4) is a double desplacement reaction. |
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| 993. |
Oxidation numbers fo iodine in ` IO_3^(-), IO_4^(-), Kl`, and `I_3` , respectively, are .A. ` +5, +7, -1, 0`B. ` -1, -5, -1, 0`C. ` -2, -5 , -1, 0`D. ` +3, +5, +7, 0` |
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Answer» Correct Answer - A The algebraic sum of the oxidation number of all the atoms in a compound must be zero and in a polyatomic ion must equal the charge on the ion. Let the oxidation number or iodine be `x`. Then `IO_3^(-) rArr x+ 3(-2) =- 1` `x=+5` ` IO_4^(-) rArr x + 4 (-2) =- 1` ` x =+7` `KI rArr ( +1) + (x) =0` ` x=- 1` ` I_2 rArr 2 x =0` ` x = 0`. |
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| 994. |
125mL of 63% (w/v) `H_(2)C_(2)O_(4).2H_(2)O` solution is made to react with 125mL of a `40%` (w/v) `NaOH` solution. The resulting solution is : (ignoring hydrolysis of ions)A. neutralB. acidicC. strongly acidicD. alkaline |
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Answer» Correct Answer - a |
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| 995. |
Assertion: If `200 mL` of `0.1 N NaOH` is added to `200 mL` of `0.1 N H_(2)SO_(4)` solution. Then the resulting solution is acidic. Reason: If milliequivalent of acid is greater than milliequivalents of base, then upon mixing the solution is acidic.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explantion of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D Meq of `NaOH=NxxV=0.1xx200=20`, Meq of `H_(2)SO_(4)=NxxV=0.1xx200=20`. `:.` Resulting solution is neutral. |
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| 996. |
In basic medium `CrO_(4)^(2-)` reacts with `S_(2)O_(3)^(2-)` resulting in the formation of `Cr(OH)_(4)^(ɵ)` and `SO_(4)^(2-)` How many " mL of " 0.1 M `Na_(2)CrO_(4)` is required to react with 40 " mL of " 0.25 M `Na_(2)S_(2)O_(3)`?A. 16mLB. 32mLC. 128mLD. 42mL |
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Answer» Correct Answer - c |
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| 997. |
In which of the following pairs in there the greatest difference in the oxidation numbers of the underlined elements?A. `ul(N)O_(2)` and `ul(N_(2))O_(4)`B. `ulP_(2))O_(5)` and `ul(P_(4))O_(10)`C. `ul(N_(2))O` and `ul(N)O`D. `ul(S)O_(2)` and `ul(S)O_(3)` |
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Answer» Correct Answer - D a. `NO_(2) (x=4), N_(2)O_(4)(x=4)` Difference in oxidation state of `N=0` b. `P_(2)O_(5)` and `P_(4)O_(10)` (Difference in oxidation state of `P` is zero). c. `N_(2)O(x=1)` and `NO(x =2)` Difference in oxidation state of `N=1` d. `SO_(2) (x=4)` and `(SO_(3))(x=6)` and Difference in oxidation state of `S=2` |
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| 998. |
Assertion: If `200 mL` of `0.1 N NaOH` is added to `200 mL` of `0.1 N H_(2)SO_(4)` solution. Then the resulting solution is acidic. Reason: If milliequivalent of acid is greater than milliequivalents of base, then upon mixing the solution is acidic.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-6B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-6C. Statement-1 Is True, Statement-2 is False .D. Statement-1 is True False, Statement-2 is True. |
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Answer» Correct Answer - d |
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| 999. |
Statement 1: In the reaction, `MnO_(4)^(-)+5Fe^(2+)+4H_(2)O,MnO_(4)^(-)` acts as oxidising agent. Statement 2: In the above reaction, `Fe^(2+)` is converted to `Fe^(3+)`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-5B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-5C. Statement-1 Is True, Statement-2 is False .D. Statement-1 is True False, Statement-2 is True. |
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Answer» Correct Answer - a |
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| 1000. |
5.5 g of a mixutre of `FeSO_4.7H_2O` and `Fe_2(SO_4)_3.9H_2O` requires 5.4 " mL of " `0.1 N KMnO_4` solution for complete oxidation. Calculate the number of gram moles of hydrated ferric sulphate in the mixture. |
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Answer» Redox changes are: `5e+Mn^(7+)rarrMn^(2+)` `Fe^(2+)rarrFe^(3+)+1e` It is no be noted here that only `FeSO_(4).7H_(2)O` will react with `KMnO_(4)` to bring in redox change. `because` Meq.of `FeSO_(4).7H_(2)O= "Meq.of" KMnO_(4)` `(w)/(E )xx1000=5.4xx0.1` `therefore (w)/((278)/(1))xx1000=0.54` `therefore w=0.150g` `therefore` Weight of `Fe_(2)(SO_(4))_(3).9H_(2)O=5.5-0.150g` `=5.350 g` `therefore` Mole of `Fe_(2)(SO_(4))_(3).9H_(2)O=(5.350)/(562)` `=9.5xx10^(-3)"mole"` `( because "M.wt." of `Fe_(2)(SO_(4))_(3).9H_(2)O=562)` |
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