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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.
| 301. |
The oxidation number of sulphur in `Na_2S_4O_6` is .A. ` 1.5`B. ` 2.5`C. ` 3.0`D. ` 2.0` |
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Answer» Correct Answer - B The algebraic sum of the the oxidation numbers of all the atoms in a compound must be zero . Let the the oxidation number of S be x . Then Note that this is the average oxidation number of S in ` Na_2S_4O_6`. In fact, two S atoms have an oxidation state of zero while other two S atoms have an oxidation state of `+5` each as shown by its structure : `Na^(+)overset(-)(O)-underset(O)underset(||)overset(O)overset(||)(S)overset(+5)(-)overset(0)(S)-overset(0)(S)overset(+5)(-)underset(O)underset(||)overset(O)overset(||)(S)-overset(-)(O)Na^(+)` |
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| 302. |
Classify the reaction `(i) 2NO_(2) + 2OH^(-) to NO_(2)^(-)+H_(2)O` (ii) `2Pb(NO_(3))_(2) to 2PbO+ 2NO_(2) + (1)/(2) O_(2)` |
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Answer» (i) Disproportionation redox reaction (ii) Decompositon redox reaction. |
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| 303. |
Which of the following satements is`//`are correct?A. In the reaction `H_(2)O_(2)+I_(2)rarrI^(ө)+?` the missing product is `O_(2)`.B. In the above reaction `(a)`, the missing product is `H_(2)O`C. In the reaction `H_(2)O_(2)+Sn^(2+)rarrSn^(4+)+?`, the missing product is `O_(2)`D. In the above reaction `(c )`, the missing product is `H_(2)O` |
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Answer» Correct Answer - A::D Since `I_(2)` is reduced to `I^(ө)`, so `H_(2)O_(2)` must be oxidised to `O_(2)` (Oxidation state of `O` is zero). `H_(2)O_(2)+I_(2)rarr2I^(ө)+O_(2)+2H^(o+)` d. `Sn^(2+)` is oxidised to `Sn^(4+)`, so `H_(2)O_(2)` must be reduced to `H_(2)O`. (Oxidation state of `O` is `-2`). |
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| 304. |
Maximum oxidation state is present inA. `CrO_(2)Cl_(2)` and `MnO_(4)^(ө)`B. `MnO_(2)`C. `[Fe(CN)_(6)]^(3-)` and `[Co(CN)_(6)]^(3-)`D. `MnO` |
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Answer» Correct Answer - A `overset(+6)(Cr)overset(-2xx2)(O_(2))overset(-1xx2)(Cl_(2)) and overset(+7)MnO_(4)^(ө) (+6 and +7)` b. `overset(+4)Mnoverset(-2xx2)(O_(2))` `[overset(+3)Feoverset(-1xx6)((CN))_(6)]^(3-) and [overset(+3)(Co)overset(-1xx6)((CN)_(6))]^(3-)` d. `overset(+2)Mnoverset(-2)O` |
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| 305. |
Which of the following statements is`//`are correct?A. In the reaction `MnO_(4)^(2-)+H^(o+)rarrMn^(2+)+?` the missing product is `MnO_(4)^(ө)`.B. In the above reaction `(a)`, the missing product is `MnO_(2)`.C. In the reaction `NO_(2)+H_(2)OrarrNO+` ? the missing product is `NO_(3)^(ө)`.D. In the above reaction `(c )`, the missing product is i`NO_(2)^(ө)`. |
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Answer» Correct Answer - A::C The `MnO_(4)^(2-)` is reduced to `Mn^(2+)`, so it must also be oxidised to `Mn^(7+) (MnO_(4)^(ө))` since `H^(o+)` is already in its maximum oxidation state. `5MnO_(4)^(2-)+H_(2)OrarrMn^(2+)+4MnO_(4)^(ө)+4H_(2)O` c. `NO_(2)` disproportionation to `NO` and `NO_(3)^(ө)`, (Oxidation state of `N is +5`) `3NO_(2)+H_(2)OrarrNO+2NO_(3)^(ө)+2H^(o+)` |
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| 306. |
Which of the following will not be oxidised by `O_(3)` ?A. `KI`B. `FeSO_(4)`C. `KMnO_(4)`D. `K_(2)MnO_(4)` |
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Answer» Correct Answer - C Since in `KMnO_(4)`, the highest oxidation state of `Mn` is `+7`, so it will not be oxidised. |
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| 307. |
Which of the following statements is`//`are correct? In the reaction `xCu_(3)P+yCr_(2)O_(7)^(2-)rarrCu^(2+)+H_(3)PO_(4)+Cr^(3+)`A. `Cu` in `Cu_(3)P` is oxidised to `Cu^(2+)` whereas `P` in `Cu_(3)P` is also oxidised to `PO_(4)^(3-)`.B. `Cu` in `Cu_(3)P` is oxidised to `Cu^(2+)` whereas `P` in `Cu_(3)P` is reduced to `H_(3)PO_(4)`.C. In the conversion of `Cu_(3)P` to `Cu^(2+)` and `H_(3)PO_(4),11 "electrons" are involved.D. The value of `x` is `6` |
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Answer» Correct Answer - A::C::D `4H_(2)O+overset(+1xx3-3)(Cu_(3)p)rarr3cu^(2+)+H_(3)PO_(4)+11e^(-)+5H^(o+)]xx6` `6e^-)+14H^(o+)+Cr_(2)O_(7)^(2-)rarr2Cr^(3+)+7H_(2)O]xx11` `ulbar(6Cu_(3)P+124H^(o+)+11Cr_(2)O_(7)^(2-)rarr18Cu^(2+)+6H_(3)PO_(4)+22Cr^(3+)+53H_(2)O)` |
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| 308. |
In the following equation: `CIO_(3)^(-)+6H^(+)+.XrarrCl^(-)+3H_(2)O`, then `X` isA. `O`B. `6e^(-)`C. `O_(2)`D. `6e^(-)` |
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Answer» Correct Answer - B `ClO_(3)^(-)+6H^(+)+6e^(-) rarr cl^(-)+3H_(2)O` |
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| 309. |
Oxidation states of the metal in the minerals haematite and magnetite, respectively, areA. `II,III` in haematite and III in magnetiteB. `II,III` in heamatite and II in magnititeC. `II` in haematite and `II,III` in magnetiteD. `III` is haematite and `II, III` in magnetite |
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Answer» Correct Answer - D Haematite: `Fe_(2)O_(3)implies"Oxidation state of"Fe=III` Magnetite: `Fe_(3)O_(4)-=FeO.Fe_(2)O_(3)` Oxidation state of `Fe=II,III` |
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| 310. |
Oxidation states of the metal in the minerals haematite and magnetite, respectively, areA. II,III in haematite and III in magnetiteB. II,III in haemitite and II in magnetiteC. II in a haematite and II, III in magnetiteD. III in haematite and II, III in magnetite |
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Answer» Correct Answer - D Haematite: `Fe_(2)O_(3):2a+3xx(-2)=0` `a=3` |
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| 311. |
Maximum oxidation state is present inA. `CrO_(2)Cl_(2)` and `MnO_(4)^(-)`B. `MnO_(2)`C. `[Fe(CN)_(6)]^(3-)` and `[Co(CN)_(6)]^(3)`D. `MnO` |
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Answer» Correct Answer - A `Cr` has `+6` and `Mn` has `+7` oxidation state. |
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| 312. |
In balancing the half reaction `CN^(ө)rarrCNO^(ө)`(skeltan) The number of electrons that must be added is |
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Answer» Correct Answer - D `CN^(-) to CNO^(-)` Oxidation number of nitrogen in `CN^(-)` `4+x = -1` and in `CNO^(-) 4 + x-2 =-1` `x=-5 , x=-3` `CN^(-)` is thus oxidised by losing two electrons. |
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| 313. |
In balancing the half reaction `CN^(ө)rarrCNO^(ө)`(skeltan) The number of electrons that must be added isA. `0`B. `1` on the rightC. `1` on the leftD. `2` on the right |
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Answer» Correct Answer - D ` {:(,CN^(-),to,CNO^(-)),("O.N. of nitrongen", 4+x=-1,,4+x-2=-1),(,x=-5,,x=-3):}` `CN^(-)` is thus oxidised. Thus, two electrons are lost. |
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| 314. |
In the reaction `CH_(3)OHrarrHCOOH`, the number of electrons that must be added to the right is :-A. 4B. 3C. 2D. 1 |
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Answer» Correct Answer - A `CH_(3)OHrarrHCOOH` `C^(2-)rarrC^(+2)+4e^(Theta)` |
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| 315. |
Molecular weight of `KMnO_(4)` in acidic medium and neutral medium will be respecitvely -A. 7 `xx` equivalent weight and 2 `xx` equivalent weightB. 5 `xx` equivalent weight and 4 `xx` equivalent weightC. 4 `xx` equivalent weight and 25`xx` equivalent weightD. 2 `xx` equivalent weight and 4 `xx` equivalent weight |
| Answer» Correct Answer - B | |
| 316. |
Assertion :- O.N. of carbon in `H-C-=N` is `+4`. Reason :- Carbon always shows an O.N of +4.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 | |
| 317. |
The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions isA. `2//3`B. `1//6`C. `1//3`D. `1` |
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Answer» Correct Answer - C `undersetulbar(Cr(2)O_(7)^(2-)+14H^(+)2Sn^(2+) rarr 3Sn^(4+)+2Cr^(3+)+7H_(2)O)(Cr_(2)O_(7)^(2-)+14H^(+)+underset((Sn^(2+) rarr Sn^(4+)+2e^(-))xx3)(6e^(-)to2Cr^(3+)+7H_(2)O)` It is clear form this equation that `3` moles of `Sn^(2+)` reduce one mole of `Cr_(2)O_(7)^(2-)`, hence `1` mol. Of `Sn^(2+)` will reduce `(1)/(3)`moles of `Cr_(2)O_(7)^(2-)`. |
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| 318. |
Which of the following are correct about the reaction, `FeS_(2)+O_(2)rarrFe_(2)O_(3)+SO_(2)`A. Eq.wt.of `FeS_(2)` is `M//11`B. Eq.wt.of `SO_(2)` is `M//5`C. `S` has `-2` oxidation state in `FeS_(2)`D. `1` mole of `FeS_(2)` requires `7//4` mole of `O_(2)` |
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Answer» Correct Answer - A::B `2Fe^(2+)rarr(Fe^(3+))+2e`, Eq.wt of `FeS_(2)=(M)/(22//2)=(M)/(11)` `2(S^(-))_(2)rarr4(S^(4+))+20e` Eq.wt. of `SO_(2)=(M)/(20//4)=(M)/(5)` `O_(2)^(0)+4erarr2(O^(2-))` `4FeS_(2)+11O_(2)rarr2Fe_(2)O_(3)+16SO_2` `S` has `-1` ox.state. |
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| 319. |
Assertion :- Equivalent weight of `KMnO_(4)` in acidic medium is `M//5` (M=molecular weight) while in basic medium, it is equal of M/3. Reason :- In acidic medium, 1 mol of `MnO_(4)^(-)` gains 5 mole electrons while in basic medium it gains 3 mole electrons.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 - A | |
| 320. |
Which of the following statements is corrected about equivalent weight of `KMnO_(4)`?A. Equivalent weight is `(1)/(3)` of molecular mass in neutral and weak basic mediumB. Equivalent weight is `(1)/(5)` of molecular mass in basic mediumC. Equivalent weight is equal to molecular mass in acidic mediumD. Equivalent weight is `(1)/(3)` of molecular mass in acidic medium |
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Answer» Correct Answer - a |
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| 321. |
The no.of electrons involved in the change, `Fe_(3)O_(4)rarrFe_(2)O_(3)`:A. `2`B. `8`C. `6`D. `4` |
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Answer» Correct Answer - A `2e +(Fe^(8//3+))_(3)rarr(Fe^(3+))_(2)` |
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| 322. |
Which of the following reactions does not involve either oxidation or reduction ?A. `VO^(2+)rarrV_(2)O_(3)`B. `NararrNa^(o+)`C. `CrO_(4)^(2-)rarrCr_(2)O_(7)^(2-)`D. `Zn^(2+)rarrZn` |
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Answer» Correct Answer - C `overset(+6)(Cr)O_(4)^(2-)rarroverset(+6)(Cr_(2))O_(7)^(2-)` Since oxidation state of `Cr` in both reactant and product is same. |
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| 323. |
In which of the following processess is nitrogen oxidised ?A. `NH_(4)^(o+)rarrN_(2)`B. `NO_(3)^(ө)rarrNO`C. `NO_(2)rarrNO_(2)^(ө)`D. `NO_(3)^(ө)rarrNH_(4)^(ө)` |
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Answer» Correct Answer - A `NH_(4)^(o+)rarrN_(2)` Oxidation state of `N = -3` Oxidation state of `N=0` In the other reaction, the oxidation state of `N` decreases. |
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| 324. |
STATEMENT-1: The equivalent mass of of `KMnO_(4)` in acidic medium is `(M)/(5)` where M=molecular mass of `KMnO_(4)` STATEMENT-2: Equivalent mass is equal to product of molecular mass and change in oxidation number. |
| Answer» Correct Answer - 3 | |
| 325. |
STATEMENT-1 `KMnO_(4) overset("acidic medium")to Mn^(2+) `, n factor of `KMnO_(4)` is 5 and STATEMENT-2 Equivalent mass of `KMnO_(4)` in acidic medium is `(M)/(5)` (M=molecular mass of `KMnO_(4)`)A. Statement-1 is True, Statement-2 is True , Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False , Statement-2 is True |
| Answer» Correct Answer - 1 | |
| 326. |
Equivalent weight of `S` in `SO_(3)^(2-)` is `(S=32)`A. `6`B. `8`C. `9`D. `4` |
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Answer» Correct Answer - B Eq. wt. of `S=(At omic wt. of S)/(O.NO. of S)=(32)/(4)=8` |
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| 327. |
A sample of `NaHCO_(3) + Na_(2)CO_(3)` required 20 ml of HCl using phenolphthalein as indicator and 35ml more required if methyl orange is used as indicator . Then molar ratio of `NaHCO_(3)` to `Na_(2)CO_(3)` isA. `1/2`B. `2/3`C. `3/4`D. `1/3` |
| Answer» Correct Answer - C | |
| 328. |
Ethanol reacts with dichromate ion in acid solution according to the equation: `C_(2)H_(5)OH(l)+Cr_(2)O_(7)^(2-)(aq)+H^(+)(aq)rarrCO_(2)(g)+Cr^(3+)(aq)+H_(2)O(l)` What is the coefficient for `H^(+)(aq)` when this equation is balanced with the smallest whole number coefficients?A. 10B. 12C. 14D. 16 |
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Answer» Correct Answer - d |
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| 329. |
Write the half reaction for the reaction `2Fe^(+3) + 2I^(-) to 2Fe^(+2) + I_(2)` |
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Answer» `2Fe^(+3) overset(2e^(-))to 2Fe^(+2)` (Reduction half) `2I^(-) to I_(2) + 2e^(-)` (Oxidation half) |
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| 330. |
identify oxidation and reduction process for the reaction `CH_(2) =CH_(2) + H_(2) to CH_(3) -CH_(3)` |
| Answer» `underset("Reduced")(CH_(2)=CH_(2)) underset("Oxidised")(H_(2))` | |
| 331. |
The fast reaction between water and sodium is the example is -A. OxidationB. ReductionC. RedoxD. none |
| Answer» Correct Answer - C | |
| 332. |
Choose the redox reaction from the following -A. `Cu+2H_(2)SO_(4)rarrCuSO_(4)+SO_(2)+2H_(2)O`B. `BaCl_(2)+H_(2)SO_(4)rarrBaSO_(4)+2HCl`C. `Zn+H_(2)SO_(4)rarrZnSO_(4)+H_(2)`D. `KNO_(3)+H_(2)SO_(4)rarr2HNO_(3)+K_(2)SO_(4)` |
| Answer» Correct Answer - A::C | |
| 333. |
Which of the following not a redox reaction -A. `MnO_(4)^(-)rarrMnO_(2)`B. `Cl_(2)+H_(2)OrarrHCl+HClO`C. `2CrO_(4)^(2-)+2H^(+)rarrZnSO_(4)+H_(2)`D. `MnO_(4)^(-)+8H^(+)+5Ag rarrMg^(+2)+4H_(2)O` |
| Answer» Correct Answer - C | |
| 334. |
In the reaction `6Li+N_(2)rarr2Li_(3)N`A. Li undergoes reductionB. Li undergoes oxidationC. N undergoes oxidationD. None of these |
| Answer» Correct Answer - B | |
| 335. |
When 0.36 gm of iron pyrite `(FeS_(2))` was heated strongly in air following reaction takes place `FeS_(2)(s)+O_(2)(g) to Fe_(2)O_(3)(s)+SO_(2)(g)` The `SO_(2) (g)` produced was titrated with acidified ` K_(2)Cr_(2)O_(7)` solution. Calculate the volume in mL of 1 M `K_(2)Cr_(2)O_(7)` solution used in the redox titration. [Fe-56,S-32] |
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Answer» Correct Answer - 2 |
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| 336. |
The oxidation number of Cr in `Cr(CO_(6))` is : |
| Answer» `overset(x)Croverset(CO_(6))` therefore O.N of Cr=0 | |
| 337. |
Which of the following is a redox reaction ?A. Reaction of `H_(2)SO_(4)` with NaOHB. In atmosphere, formation of `O_(3)` from `O_(2)` by lighteningC. Formation of oxides of nitrogen from nitrogen and oxygen by lighteningD. Evaporation of `H_(2)O` |
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Answer» Correct Answer - C (a) Neutralisation reaction (b) `3O_(2)rarr 2O_(3)`, allotropic formation (c) `N_(2)+O_(2)rarr 2NO` `N_(2)` is reduced and `O_(2)` is oxidised. (d) In evaporation of `H_(2)O`, state changes only. |
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| 338. |
Fluorine can have an oxidation number of .A. `-1` onlyB. `0` onlyC. `-1,0`D. ` +1` only |
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Answer» Correct Answer - C Being univalent and most electronegative, `F` has an oxidation nmber of `-1` in all of its compounds. In ` F_2` (free uncombined element), the oxidation number of `F` is zero. |
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| 339. |
Which of the following is not a redox reaction?A. `CaCO_(3)rarrCaO+CO_(2)`B. `O_(2)+2H_(2)rarr2H_(2)O`C. `Na+H_(2)OrarrNaOH+1//2H_(2)`D. `MnCI_(3)rarr?MnCI_(2)+1//2CI_(2)` |
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Answer» `(overset(+2)Ca overset(-2)O_(3))^(+4)rarroverset(+2)Caoverset(-2)O+overset(+2)Coooverset(-2)O_(2)` This is not a redo reaction because no element undergoeso changes in oxidation number |
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| 340. |
The sum of oxidation numbers fo all the atoms in the dichromate ion is .A. `-3`B. `-1`C. `-4`D. `-2` |
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Answer» Correct Answer - D Dichromate ion is `Cr_2O_7^(2-)`. In a polyatomic ion, the sum of oxidation numbers fo all the atoms in the ion must be equal to the net charge fo the ion. However, in a neutral molecule , the sum of the oxidation numbers of all the atoms must be zero. |
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| 341. |
Balance the net equtation fro th reaction of potassium dichromate (VI), `K_(2) Cr_(2) O_(7)`, with sodium sulphite, `Na_(2)SO_(3)`, in an acid solution to give chromium (III) ion and and sulphate ion. Strategy : Follow the seven -step proceduce , one step at a time. |
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Answer» Step 1 : The skeletal ionic equation is : `Cr_(2)O_(7)^(2-) (aq) + SO_(3)^(-2) (aq) to Cr^(3+) (aq) + SO_(4)^(2-) (aq)` Step 2: Assign oxidation numbers for Cr and S `overset(+6)(Cr_(2))overset(-2)(O_(7)^(2-)) (aq) + overset(+4)(S)overset(-2)(O_(3)^(2-)) (aq) to overset(+3)(Cr)(aq) + overset(+6)(S)overset(-2)(O_(4)^(2-)) (aq)` This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant. Step 3: Calculate the increase and decrease of oxidation number, and make them equal: from step-2 we can notice that there is change in oxidation state of chromium and sulphur. Oxidation state of chromium changes form +6 to +3. There is decrease of +3 in oxidation state of chromium on right hand side of the equation. Oxidation state of sulphur changes from +4 to +6. There is an increase of +2 in the oxidation state of sulphur on right hand side. To make the increase and decrease of oxidation state equal, place numeral 2 before cromium ion on right hand side and numeral 3 before sulphate ion on right hand side and balance the chromium and sulphur atoms on both the sides of the equation. Thus we get `overset(+6)(Cr_(2))overset(-2)(O_(7)^(2-))(aq) + overset(+4)(3S)overset(-2)(O_(3)^(2-)) (aq) to overset(+3)(2) Cr^(3+) (aq) + 3SO_(4)^(2-)(aq)` Step 4 : As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add 8H+ on the left to make ionic charges equal `Cr_(2)O_(7)^(2-) (aq) + 3SO_(3)^(2-) (aq) + 8 H^(+) to 2 Cr^(3+) (aq) + 3 SO_(4)^(2-)(aq)` Step : 5 Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., 4`H_(2)O`) on the right to achieve balanced redox change. `Cr_(2)O_(7)^(2-) (aq) + 3SO_(3)^(2-) (aq) + 8H^(+) (aq) to 2 Cr^(3+) (aq) + 3 SO_(4)^(2-) (aq) + 4H_(2)O (1)` |
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| 342. |
Which is not true about the oxidation state of the following elements ?A. Sulphur +6 to -2B. Carbon +4 to -4C. Chlorine +7 to -1D. Nitrogen +3 to -1 |
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Answer» Correct Answer - D Oxidation state of N varies from +5 to -3 . |
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| 343. |
For the reaction : `Cl_(2)(aq) +2Br^(-) (aq) to Br_(2) (aq) +2Cl^(-)(aq)` Which of the following could be used to monitor the rate ? (P) pH meter (Q) SpectrophotometerA. P onlyB. Q onlyC. Either P or QD. Neither P nor Q |
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Answer» Correct Answer - b |
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| 344. |
Identify the compounds which are reduced and oxidised in the following reaction: `3N_(2)H_(4)+2BrO_(3)^(-)rarr3N_(2)+2Br^(-)+6H_(2)O`A. `N_(2)H_(4)` is oxidised and `BrO_(3)^(-)` is reduced.B. `BrO_(3)^(-)` is oxidised and `N_(2)H_(3)` is reduced.C. `BrO_(3)^(-)` is both reduced and oxidised.D. This is not a redox reaction. |
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Answer» Correct Answer - A `3overset(-2)N_(2)H_(4)+2overset(+5)BrO_(3)^(-)rarr3overset(0)N_(2)+overset(-1)(2B)r^(-)+6H_(2)O` |
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| 345. |
Drinking is injurious of health. But for revenue purpose government has fixed some permissible value for alcohol . The permissible value for the alcohol content in the blood is 1 % mass . On analysis of blood sample of a driver of being drunk over than the pernissible value , it was obtained that 60 gm sample reacted with 30 mL of `8 M K_(2)Cr_(2)O_(7)`(Acidic solution) . Reaction : `2K_(2)Cr_(2)O_(7) + 8 H_(2)SO_(4) + C_(2)H_(5)OH rarr 2Cr_(2) (SO_(4))_(3) + 11 H_(2)O + 2K_(2)SO_(4) + 2CO_(2)` Assume `K_(2)Cr_(2) O_(7)` reacts only with the alcohol present in blood . What is the percentage of alcohol in the blood sample?A. 0.092B. 0.088C. 0.008D. 0.072 |
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Answer» Correct Answer - a |
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| 346. |
Drinking is injurious of health. But for revenue purpose government has fixed some permissible value for alcohol . The permissible value for the alcohol content in the blood is 1 % mass . On analysis of blood sample of a driver of being drunk over than the pernissible value , it was obtained that 60 gm sample reacted with 30 mL of `8 M K_(2)Cr_(2)O_(7)`(Acidic solution) . Reaction : `2K_(2)Cr_(2)O_(7) + 8 H_(2)SO_(4) + C_(2)H_(5)OH rarr 2Cr_(2) (SO_(4))_(3) + 11 H_(2)O + 2K_(2)SO_(4) + 2CO_(2)` Assume `K_(2)Cr_(2) O_(7)` reacts only with the alcohol present in blood . Will the driver be prosecuted for drunken driving?A. yesB. noC. may or may notD. Date insufficient |
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Answer» Correct Answer - a |
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| 347. |
`CuS` is dissolved in dil. `HNO_(3)`. Balanced equation with correct products isA. `Cus+2H^(+)+3NO_(3)^(-) rarr Cu(NO_(3))_(2)+H_(2)S+H_(2)O+NO_(2)`B. `3Cus+8H^(+)+8NO_(3)^(-) rarr 3Cu(NO_(3))_(2)+3S+4H_(2)O+2NO`C. `Cus+4NO_(3)^(-) rarr Cu(NO_(3))_(2)+H_(2)S+H_(2)O`D. None of the above in correct |
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Answer» Correct Answer - B `CuS` (black ppt.) on boiling with `HNO_(3)` results in the formation of colloidal `S` and blue coloured solution of `Cu(NO_(3))_(2)`. |
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| 348. |
Calculate the oxidation numbers of `Cr` in `K_(3)CrO_(8)`: |
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Answer» Correct Answer - 5 `K_(3)CrO_(3)` is `K_(3)[Cr(O_(2))_(4)]`, i.e., `4(O-O)` bond `:.` Oxidation no. of `Cr=+5` |
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| 349. |
The reaction `5H_(2)O_(2)+XClO_(2)+2OH^(-) rarr XCl^(-)+YO_(2)+6H_(2)O` is balanced ifA. X=5,Y=2B. X=2,Y=5C. X=4,Y=10D. X=5,Y=5 |
| Answer» Correct Answer - B | |
| 350. |
The standard electrode potential corresponding to the reaction ` Au^(3+) (aq) + 3e^(-) rarr Au (s)` is `1. 42 V`. This implies that . (i) gold dissolves in ` 1M HCl` (ii) metallic gole will be precipitated on passing hydrogen gas through gold salt solution (iii) gold does not dissolve in ` 1M HCl` solution (iv) metallic gold will not be precipitated on passing hydrogen gas through gold salt solution.A. (i), (ii)B. (i), (iv)C. (ii), (iv)D. (ii), (iii) |
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Answer» Correct Answer - D Gold will dissolve in `1 M HCl` if it gets oxidized and ` H+` ions fo `HCl` are reduced : ` Au(s) + 3H^(+) (aq.) rarr Au^(3+) (aq.) + 3/2 H_2(g) ` But this is not possible because ` E_(Au//Au^(3+))^(Ө) (-1.42 V)` is less than ` E_(H^+//H_(2))^Ө (0.00V)`. In other words, the standard reduction potential of gold is greater than zero . Therefore, the reduced from of gold, i.e., Au (s) is mor stable than the reduced from of hydrogen (i.e., `H_2`) i.e., gold cannot be oxidized by ` H^(+)` ions to give ` H_2`. Hence, gold does not dissolve in ` 1M HCl` solution. Since ` E_(Au^(3+) //Au)^(Ө) (1.42 V)` is greater than ` E_(H^+ //H_2)^Ө` (0.00 V), hydrogen gas passed through the salt solution of gold will reduce auric ions and metallic gold will be percipitated. |
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