InterviewSolution
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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.
| 1. |
Phosphoric acid `(H_(3)PO_(4))` is tribasic acid and one of its salt is sodium dihydrogen phosphate `(NaH_(2)PO_(4))`. What volume of 1 M NaOH solution should be added to 12 g of sodium dihydrogen phosphate (mol. Wt. 120) to exactly convert it into trisodium phosphate `Na_(3)PO_(4)`A. 80 mlB. 100 mlC. 200 mlD. 300 ml |
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Answer» Correct Answer - C `1xxV=(12)/(120)xx2=200ml` |
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| 2. |
Weight of `Ca(OH)_(2)` needed to prepare 250 ml of solution with pH=13A. 0.925 gB. 0.0125 gC. 0.25 gD. 1 g |
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Answer» Correct Answer - A We know, `[H^(+)]=10^(-pH)=10^(-13)` `therefore[OH^(-)]=(10^(-14))/(10^(-13))=10^(-1)` hence, normality of solution=0.1 We know, `w=(NEV)/(1000)=(0.1xx74xx250)/(2xx1000)=0.925gm` |
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| 3. |
Which indicator can be used in the titration of strong acid and strong baseA. Only phenoplphthaleinB. Only methyl orangeC. Eitehr of the twoD. Red litmus |
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Answer» Correct Answer - C As the resultant solution will be neutral so either of them can be used. |
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| 4. |
If we use phenolphthalein as an indicator in a titration of `Na_(2)CO_(3)` with HCl, the usual result isA. No visible change is occurB. The indicator reacts with the acidC. The indicator reacts with the baseD. Sodium chloride and carbonic acid will be formed |
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Answer» Correct Answer - A The resulting solution is colourless. |
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| 5. |
How many grams of NaOH are equivalent to 100 ml of 0.1 N oxalic acidA. 0.2B. 2C. 0.4D. 4 |
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Answer» Correct Answer - C `N=(W_(B)xx1000)/("Eq. wt."xxV),W_(B)=(Nxx"Eq. wt"xxV)/(1000)` `=(0.1xx40xx100)/(1000)=0.4` |
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| 6. |
0.53 gm of `Na_(2)CO_(3)` has been dissolved in 100 ml of a sodium carbonate solution. The normality of the solution will beA. `(N)/(5)`B. `(N)/(2)`C. `(N)/(10)`D. N |
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Answer» Correct Answer - C `N=(0.53xx1000)/(53xx100)impliesN=(1)/(10)` So normality of the solution will be `(N)/(10)`. |
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| 7. |
A 0.1 N solution of `Na_(2)CO_(3)` is titrated with 0.1 N HCl solution. The best indicator to be used isA. Potassium ferricyanideB. PhenolphthaleinC. Methyl redD. Litmus Paper |
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Answer» Correct Answer - C As the resulting solution is acidic. |
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| 8. |
25 ml of a solution of `Na_(2)CO_(3)` having a specific gravity of 1.25 required 32.9 ml of a solution of HCl containing 109.5 grams of the acid per litre for complete neutralization. Calculate the volume of 0.84N-`H_(2)SO_(4)` that will be completely neutralized by 125 grams of the `Na_(2)CO_(3)` solutionA. 460 mlB. 540 mlC. 480 mlD. 470 ml |
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Answer» Correct Answer - D `N_(1)V_(1)=N_(2)V_(2)` `Nxx25=(109.25xx32.9)/(36.5)impliesN=(109.5xx32.9)/(36.5xx25)` `N_(3)V_(3)=N_(4)V_(4)" "(V_(3)=(m)/(d),V_(3)=(125)/(1.25))` `(109.5xx32.9)/(36.5xx25)xx100=0.84xxV impliesV=470ml` |
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| 9. |
The number of hydroxide ions, produced by one molecule of sodium carbonae `(Na_(2)CO_(3))` on hydrolysis isA. 2B. 1C. 3D. 4 |
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Answer» Correct Answer - A `Na_(2)CO_(3)+2H_(2)Oto2NaOH+H_(2)CO_(3) and 2NaOH underset(("ionisation"))hArr2Na^(+)+2OH^(-)` Hence, it is clear that `2OH^(-)` ions will be formed on hydrolysis of one molecule of sodium carbonate. |
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| 10. |
Mark the compound which turns black with `NH_(4)OH`A. Lead chlorideB. Mercurous chlorideC. Mercuric chlorideD. Silver chloride |
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Answer» Correct Answer - B `Hg_(2)Cl_(2)+2NH_(2)OH to NH_(2)-Hg-Cl+underset(("black ppt"))(Hg)` |
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| 11. |
Mark the gas which turns lime water milkyA. `H_(2)S`B. `SO_(2)`C. `Cl_(2)`D. `CO_(2)` |
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Answer» Correct Answer - B::D `CO_(2) and SO_(2)` turns lime water milky, as `Ca(OH)_(2)(aq.)+CO_(2)tounderset(("milky"))(CaCO_(3))darr+H_(2)O` `Ca(OH)_(2)+SO_(2)tounderset(("milky"))(CaSO_(3))darr+H_(2)O` |
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| 12. |
Mark the correct statementA. I group basic radicals precipitate as chloridesB. IV group basic radicals precipitate as sulphidesC. V groups basic radicals precipitate as carbonatesD. All of these statement are correct |
| Answer» Correct Answer - D | |
| 13. |
Flame test is not given byA. `Ba^(2+)`B. `Be^(2+)`C. `Ca^(2+)`D. `Sr^(2+)` |
| Answer» Correct Answer - B | |
| 14. |
In the analysis of basic radicals, the group reagent `H_(2)S` gas is generally used in the groupsA. I and II groupsB. II and III groupsC. III and V groupsD. II and IV groups |
| Answer» Correct Answer - D | |
| 15. |
Sodium carbonate cannot be used in place of ammonium carbonate for the precipitation of the fifth group radicals becauseA. Sodium ions interfere with the detection of fifth group radicalsB. Concentrations of carbonate ions is very lowC. Sodium will react with acidic radicalsD. Magnesium will be precipitated |
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Answer» Correct Answer - D When `(NH_(4))_(2)CO_(3)` is used then the concentration of `CO_(3)^(2-)` is comparatievely low but when `Na_(2)CO_(3)` is added then concentration of `CO_(3)^(2-)` increases so `Mg^(2+)` will be precipitated along with other V group radicals. |
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| 16. |
`Cu^(2+)` ions will be reduced to `Cu^(+)` ions by the addition of an aqueous solution ofA. KFB. KClC. KlD. KOH |
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Answer» Correct Answer - C `underset("Cuprous iodide")(2CuSO_(4)+4KI to 2CuI+2K_(2)SO_(4)+I_(2))` |
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| 17. |
Sometimes yellow turbidity appears o passing `H_(2)S` gas even in the absence of the second group radicals. This happens becauseA. Sulphur is present in the mixture as an impurityB. The fourth group radicals are precipitated as sulphidesC. The `H_(2)S` is oxidized by some acid radicalsD. The third group radicals are precipitated |
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Answer» Correct Answer - B This is due to the precipitation of fourth group radical as sulphides due to high concentration of `S^(2-)` in the solution as a result yellow turbidity is obtained. |
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| 18. |
The strength of a solution (S) in gram/litre, is related to its normality (N) and equivalent weight of solute (E) by the formulaA. `S=(N)/(E)`B. `S=(E)/(N)`C. `S=N.E.`D. All of these |
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Answer» Correct Answer - C Strength`=(W)/(V)=NE` |
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| 19. |
The maximum amount of `BaSO_(4)` precipitated on mixing `BaCl_(2)(0.5M)` with `H_(2)SO_(4)`(1M) will correspond toA. 0.5 MB. 1.0 MC. 1.5 MD. 2.0 M |
| Answer» Correct Answer - A | |
| 20. |
Identify the correct order of solubility of `Na_(2)S,CuS and ZnS` in aqueous mediumA. CuSgtZnSgt`Na_(2)S`B. `ZnS gt Na_(2)S gt CuS`C. `Na_(2)S gt CuS gt ZnS`D. `Na_(2)S gt ZnS gt CuS` |
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Answer» Correct Answer - D Sodium salts are highly soluble. `Cu^(2+)` belongs to the `II^(nd)` group in salt analysis and is precipitated as `CuS`, whereas `Zn^(2+)` belongs to the IV group and is precipitated as ZnS after CuS because of higher `K_(SP)` of `ZnS`. |
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| 21. |
When `Cl_(2)` water is added to a salt solution containing chloroform, chloroform layer turns violet. Salt containsA. `Cl^(-)`B. `I^(-)`C. `NO_(3)^(-)`D. `S^(2-)` |
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Answer» Correct Answer - B As `Cl^(-)` is more electronegative than `Br^(-) and I^(-)` `therefore` it replaces them from their salt as `2I^(-)+Cl_(2) toI_(2)+2Cl^(-)implies`Violet vapour `2Br^(-)+Cl_(2)toBr_(2)+2Cl^(-)implies`Brown vapour |
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| 22. |
A solution which is `10^(-3)M` each in `Mn^(2+),Fe^(2+),Zn^(2+)` and `Hg^(2+)` is treated with `10^(-16)M` sulphide ion. If `K_(SP)` of MnS, FeS, ZnS and HgS are `10^(-15),10^(-23),10^(-20)` and `10^(-54)` respectively, which one will precipitate firstA. FeSB. MgSC. HgSD. ZnS |
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Answer» Correct Answer - C HgS having the lowest `K_(SP)` among the lot will precipitate first. |
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| 23. |
Which BLUE LIQUID is obtained on reacting equimolar amounts of two gases at `-30^(@)C`A. `N_(2)O`B. `N_(2)O_(3)`C. `N_(2)O_(4)`D. `N_(2)O_(5)` |
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Answer» Correct Answer - B `NO(g)+NO_(2)(g) overset(-30^(@)C)to underset("blue")(N_(2)O_(3)(l))` |
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| 24. |
Assertion: Addition of `NH_(4)OH` to an aqueous solution of `BaCl_(2)` in the presence of `NH_(4)Cl` (excess) precipitates `Ba(OH)_(2)` Reason: `Ba(OH)_(2)` in insoluble in water.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 explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - D Additon of `NH_(4)OH` to an aqueous solution of `BaCl_(2)` in presence of `NH_(4)Cl` (excess) does not give a precipitate of `Ba(OH)_(2)` as `(Ba(OH)_(2))` is soluble in water. |
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| 25. |
Precipitate of group IV cations takes place when `H_(2)S` isA. Highly ionisedB. Less ionisedC. Not ionisedD. None of these |
| Answer» Correct Answer - D | |
| 26. |
Statement 1: Basic radical of V group are precipitated as their carbonates in presence of `NH_(4)Cl` Statement 2: `NH_(4)OH` maintains the pH of the solution basic.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 |
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Answer» Correct Answer - D Group reagent for V group is `(NH_(4))_(2)CO_(3)` in presence of `NH_(4)Cl` and because of `NH_(4)OH,pH` of solution is maintained. |
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| 27. |
The only cations present in a slightly acidic solution are `Fe^(3+),Zn^(2+) and Cu^(2+)`. The reagent that when added in excess to this solution would identify the separate `Fe^(3+)` in one step isA. 2M HClB. 6M `NH_(3)`C. 6 NaOHD. `H_(2)S` gas |
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Answer» Correct Answer - B `Fe^(3+)` is a third group radical, whose reagent is `NH_(4)OH` in presence of `NH_(4)Cl`. So if 6M `NH_(3)` is added in the slightly acidic (HCl) solution of ions it will lead to the `Fe^(3+)+Zn^(2+)+Cu^(2+) overset("6 M"NH_(3))to underset("Brown ppt.")(Fe(OH)_(3))+underset("Soluble")([Zn(NH_(3))_(4)]^(2+))+underset("Soluble")([Cu(NH_(3))_(4)]^(2+))` |
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| 28. |
Assertion: White ppt. of AgCl is soluble in `NH_(4)OH` Reason: It is due to the formation of soluble complex.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 explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
| Answer» Correct Answer - A | |
| 29. |
Which compound is soluble in `NH_(4)OH`A. `PbCl_(2)`B. `PbSO_(4)`C. `AgCl`D. `CaCO_(3)` |
| Answer» Correct Answer - C | |
| 30. |
The ratio of amounts of `H_(2)S` needed to precipitate all the metal ions from 100 ml of 1 M `AgNO_(3)` and 100 ml of 1M `CuSO_(4)` isA. `1:2`B. `2:1`C. ZeroD. Infinity |
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Answer» Correct Answer - A In acidic medium 2 molecules of `KMnO_(4)` gives 5 atoms of oxygen `2KMnO_(4)+3H_(2)SO_(4) to K_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5(O)` `2xx158=(316xx8)/(80)=31.6` So, equivalent wt. of `KMnO_(4)` in acidic medium is `=31.6`gm |
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| 31. |
3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042N. The amount of acetic acid absorbed (per gram of charcoal) isA. 18 mgB. 36 mgC. 42 mgD. 54 mg |
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Answer» Correct Answer - A Initial m moles of `CH_(3)COOH=0.06xx50` Final m moles of `CH_(3)COOH=0.043xx50` Hence, mass of `CH_(3)COOH` adsorbed per gram of charcoal`=((0.06-0.042)xx50xx10^(-3)xx60xx10^(3))/(3)=18mg` |
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| 32. |
An aqueous solution of `FeSO_(4),Al_(2)(SO_(4))_(3)` and chrome alum is heated with excess of `Na_(2)O_(2)` and filtered. The materials obtained asA. A colourless filtrate and a green residueB. A yellow filtrate and a green residueC. A yellow filtrate and a brown residueD. A green filtrate and a brown residue |
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Answer» Correct Answer - B Yellow filtrate is due to chromate ions `(CrO_(4)^(- -))` and green residue to due to `Fe(OH)_(2)`. |
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| 33. |
Amongst the following, the total number of compound soluble in concentrated `NH_(3)` solution is (a) `Ag_(2)CrO_(4)` (b) `Cu(OH)_(2),CuSO_(4)` (c) `PbSO_(3)` ltBrgt (d) `Al(OH)_(3)` (e) `Ni(OH)_(2)` (f) `Zn_(3)(PO_(4))_(2)` (g) `BaSO_(4)` (h) `Bi(OH)_(2)NO_(3)` (i) `Mn(OH)_(2)`. |
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Answer» Correct Answer - D (a),(b) and (c) are soluble in sodium hydroxide not in `NH_(3)` (d) Soluble in sodium hydroxide not in `NH_(3)`. (e) `Ni(OH)_(2)darr(green)+6NH_(3)to[Zn(NH_(3))_(4)]^(2+)("Deep blue colouration"))+2OH^(-)` (f) `Zn_(3)(PO_(4))_(2)darr(white)+12NH_(3) to 3[Zn(NH_(3))_(4)]^(2+)+2PO_(4)^(3-)` (g) Insoluble in `NH_(3)`, soluble appreaciable in boiling concentrated `H_(2)SO_(4)`. (h) `Bi(OH)_(2)NO_(3)` is insoluble in `NH_(3)`. (i) Insoluble in `NH_(3)` but soluble in ammonium salts liberating `NH_(3)`. |
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| 34. |
Which one of the following metals will give blue ash when its salt is heated with `Na_(2)CO_(3)` solid and `Co(NO_(3))_(2)` on a charcoal pieceA. `Cu`B. `Mg`C. `Al`D. `Zn` |
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Answer» Correct Answer - C `CoAlO_(2)` is formed which is blue. Follow cobalt nitrate-charcoal test. |
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| 35. |
Which of the following metals, Fe,Zn,Pb,Ag and Pt, do not give a metal nitrate on treatment with concentrated `HNO_(3)`A. Fe and ZnB. Fe and PtC. Pb, Ag and PtD. Fe, Zn and Pt |
| Answer» Correct Answer - D | |
| 36. |
The equivalent weight of `Zn(OH)_(2)` is the following reaction is equal to its, `[Zn(OH)_(2)to(NO_(3))toZn(OH)(NO_(3))+H_(2)O]`A. `("Formula wt.")/(2)`B. `("Formula wt.")/(1)`C. `3xx`formulaD. `2xx`formula |
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Answer» Correct Answer - B Equivalent weight of `Zn(OH)_(2)=("Molecular weight")/("Acidicty")=(M)/(1)` |
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| 37. |
Which one of the following anions is not easily removed from aqueous solutions by precipitationA. `Cl^(-)`B. `NO_(3)^(-)`C. `CO_(3)^(-2)`D. `SO_(4)^(-2)` |
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Answer» Correct Answer - B As most of the nitrates are soluble in water. |
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| 38. |
`S^(2-)` and `SO_(3)^(2-)` can be distinguihsed by usingA. `(CH_(3)COO)_(2)Pb`B. `Na_(2)[Fe(CN)_(5)NO]`C. both (a) and (b)D. None of these |
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Answer» Correct Answer - C `S^(2-)+(CH_(3)COO)_(2)Pb underset("black ppt.")(PbS darr) +2CH_(3)COO^(-)` `S^(2-)+Na_(2)[Fe(CN)_(5)NO]tounderset("violet ppt.")([Fe(CN)_(5)NOS]^(4-))+2Na^(+)` Hence, they are distinguished by both the reagents as only `S^(2-)` gives precipitate with these reagents. |
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| 39. |
A brown ring appears in the test forA. NitrateB. NitriteC. BromideD. Iron |
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Answer» Correct Answer - A At the junction of salt solution and `FeSO_(4)` solution with conc. `H_(2)SO_(4)` a brown ring is obtained `underset("Brown ring")((FeSO_(4).NO))` |
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| 40. |
Which of the following combines with Fe (II) ions to form a brown complexA. `N_(2)O`B. `NO`C. `N_(2)O_(3)`D. `N_(2)O_(5)` |
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Answer» Correct Answer - B Brown ring test with `(FeSO_(4).NO)` |
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| 41. |
If 30 ml of `H_(2) and 20ml` of `O_(2)` reacts to form water, what is left at the end of the reactionA. 10 ml of `H_(2)`B. 5 ml of `H_(2)`C. 10 ml of `O_(2)`D. 5 ml of `O_(2)` |
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Answer» Correct Answer - D `H_(2)+(1)/(2)O_(2)toH_(2)O` 1 mole `(1)/(2)` mole 1 mole 1 volume `(1)/(2)` volume 1 ml `H_(2)` reacts with `(1)/(2)`ml `O_(2)` 30 ml of `H_(2)` reacts with `=(1)/(2)xx30=15ml" "O_(2)` (20-15)=50ml of `O_(2)` will left at the end of the reaction. |
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| 42. |
The volume of `(N)/(10)NaOH` require to neutralise 100 ml of `(N)/(25)` HCl isA. 30 mlB. 100 mlC. 40 mlD. 25 ml |
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Answer» Correct Answer - C For complete neutralisation, Milli equivalent of base=milli equivalent of acid `N_(1)V_(1)=N_(2)V_(2) implies(1)/(10) xxV_(1)=(1)/(25)xx100,V_(1)=40ml` |
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| 43. |
haemoglobin is a complex ofA. `Fe^(3+)`B. `Fe^(2+)`C. `Fe^(4+)`D. `Cu^(2+)` |
| Answer» Correct Answer - B | |
| 44. |
For preparing one litre N/10 solution of `H_(2)SO_(4)`, we need `H_(2)SO_(4)`A. 98 gmsB. 10 gmsC. 100 gmsD. 4.9 gms |
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Answer» Correct Answer - D `(1)/(10)=(Wxx1000)/(eq." "wtxxvol.)=(Wxx1000)/(49xx1000)impliesW=4.9gm` |
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| 45. |
The volume of 0.6 M NaOH required to neutralise 30 `cm^(3)` of 0.4 M HCl isA. `40cm^(3)`B. `30cm^(3)`C. `20cm^(3)`D. `10cm^(3)` |
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Answer» Correct Answer - C Normality=molarity`xx`basicity or acidity (for HCl) `N_(2)=0.4xx1=0.4N` basicity=1 (for NaoH acidity=1) `N_(1)=0.6xx1=0.6N,V_(1)=?,V_(2)=30cm^(3)` From the equation, `N_(1)V_(1)=N_(2)V_(2)` `0.6xxV_(1)=0.4xx30` `V_(1)=(0.4xx30)/(0.6)=20cm^(3)` |
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| 46. |
Which one of the following sulphides is yellowA. Zinc sulphideB. Cadmium sulphideC. Nickel sulphideD. Led sulphide |
| Answer» Correct Answer - B | |
| 47. |
When dilute aqueous solution of `AgNO_(3)` (excess) is added to KI solution, positively charged solution particles of AgI are formed due to adsorption of ionA. `NO_(3)^(-)`B. `O_(2)^(-)`C. `Ag^(+)`D. `K^(+)` |
| Answer» Correct Answer - C | |
| 48. |
The ion that can be precipitated by HCl as well as `H_(2)S` isA. `Pb^(2+)`B. `Fe^(3+)`C. `Zn^(2+)`D. `Cu^(2+)` |
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Answer» Correct Answer - A `Pb^(2+)` as it precipitated as chloride and sulphide in `I^(st)` and `II^(nd)` group respectively. |
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| 49. |
Sodium nitrophrusside when added to an alkaline solution of sulphide ions produce asA. Red colourationB. Blue colourationC. Purple colourationD. Brown colouration |
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Answer» Correct Answer - C `Na_(2)S+underset("Sodium nitroprusside")(Na_(2)[Fe(NO)(CN)_(5)])to underset("Violet coloured complex")(Na_(3)[Fe(ONSNa)(CN)_(5)])` or `underset("Purple colour complex")(Na_(4)[Fe(CN)_(5)NOS])` |
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| 50. |
The ion that cannot be precipitated by both HCl and `H_(2)S` isA. `Pb^(2+)`B. `Cu^(+)`C. `Ag^(+)`D. `Sn^(2+)` |
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Answer» Correct Answer - D `Sn^(2+)` can be precipitated by `H_(2)S` but not HCl. |
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