InterviewSolution
Saved Bookmarks
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. |
What volume at NTP of gaseous ammonia will be required to be passed into `30 cm^(3)` of 1 N `H_(2)SO_(4)` solution to bring down the acid strength of the latter of 0.2 N ? |
| Answer» Correct Answer - 537. 6 mL | |
| 2. |
`50 mL` of `0.1 M` solution of a salt reacted with `25 mL` of `0.1 M` solution of sodium sulphite. The half reaction for the oxidation of sulphite ion is: `SO_(3)^(2-)(aq)+H_(2)O(l) rarr (aq)+2H^(+)(aq)+2e^(-)` If the oxidation number of metal in the salt was `3`, what would be the new oxidation number of metal: |
| Answer» Correct Answer - 2 | |
| 3. |
Volume `V_(1)mL` of 0.1 M `K_(2)Cr_(2)O_(7)` is needed for complete oxidation of 0.678g `N_(2)H_(4)` in acidic medium. The volume of 0.3M `KMnO_(4)` needed for same oxidation in acidic medium will be :A. `(2)/(5)V_(1)`B. `(5)/(2)V_(1)`C. 113`V_(1)`D. can not be determined |
|
Answer» Correct Answer - A Equivalent of `K_(2)Cr_(2)O_(7)` = equivalent of `N_(2)H_(4)` also equivalent of `KMnO_(4)` = equivalent of `N_(2)H_(4)` So, equivalent of `K_(2)Cr_(2)O_(7)` = equivalent of `KMnO_(4)` `0.1xx6xxV_(1)=0.3xx5xxV_(2) " " :. " so" V_(2)=2//5V_(1)` |
|
| 4. |
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. 3B. 3-nC. n-3D. `+n` |
| Answer» Correct Answer - B | |
| 5. |
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 : |
| Answer» Correct Answer - 1 | |
| 6. |
1.2g of carbon is burnt completely in oxygen (limted supply) to produce CO and `CO_(2)` . This mixture of gases is treated with solid `I_(2)O_(5)` (to know the amount of CO produced). The librated iodine required 120 ml of 0.1m hypo solution for complete titration. The percentage carbon converted into CO is :A. 0.6B. 1C. 0.5D. 0.3 |
|
Answer» Correct Answer - D `{:(,C,+,O_(2),rarr,CO,+,CO_(2)," "....(i)),(t=0,0.1,,,,-,,-,),(t=t,0,,,,x,,(0.1-x),),(,5CO,+,I_(2)O_(5),rarr,5CO_(2),+,I_(2),....(ii)),(t=0,x,,,,,,,),(t=t,0,,,,,,x//5,),(,,,,,,,,):}` `I_(2)+2Na_(2)S_(2)O_(3)rarrNa_(2)S_(4)O_(6)+2"NaI"` `:. ` moles of `I_(2)` liberated `=(1)/(2)xx`moles (from of hypo consumed `=(1)/(2)xx120xx10^(-3)xx0.1=60xx10^(-4)` So, x `=5xx60xx10^(-4)=0.03` moles (from reaction (ii) `: 5 xx` mole of `I_(2)`=mole of CO) so, % of C forming CO `=(0.03xx12)/(1.2)xx100=30%` |
|
| 7. |
Hydrogen peroxide acts both as an oxidising and as a reducing agent depending upon the nature of the reacting species. In which of the following cases `H_(2)O_(2)` acts as a reducing agent in acid medium ?A. `MnO_(4)^(-)`B. `Cr_(2)O_(7)^(2-)`C. `SO_(3)^(2-)`D. Kl |
| Answer» Correct Answer - D | |
| 8. |
For standardizing NaOH solution, which of the following is used as a primary standard ?A. Sodium tetraborateB. Ferrous Ammonium SulfateC. Oxalic acidD. dil. HCl |
| Answer» Correct Answer - C | |
| 9. |
In the reaction `H_(2)O_(2)^(18)+O_(3) to ` water + oxygen, radioactivity will be shown by which of the product ?A. waterB. oxygenC. both (1) & (2)D. None of these |
|
Answer» Correct Answer - B `O_(3)` will oxidise `H_(2)O_(2)` into oxygen, hence radioacitve oxygen of `H_(2)O_(2)` will go only in oxygen, not in water. Half reactions : `O_(3)+2H^(+)+2e^(-)rarrO_(2)H_(2)O` , `H_(2)O_(2)rarrO_(2)2H^(-)+2e^(-)` |
|
| 10. |
The volume of `0.1 N` dibasic acid sufficient to neutralize 1g of a base that furnishes `0.04` mole of `OH^-` in aqueous solution is :A. 400 mLB. 600 mLC. 200 mLD. 80 mL |
| Answer» Correct Answer - C | |
| 11. |
1 mole of how many of the following acids neutralize exactly one mol of NaOH, under required favourable conditions ? HCl, `HNO_(3), H_(2)SO_(4),H_(2)SO_(3),H_(3)PO_(4),H_(3)PO_(2),H_(4)P_(2)O_(5),H_(3)BO_(3)`A. 4B. 7C. 8D. 9 |
|
Answer» Correct Answer - A `(HCl,HNO_(3),H_(3)PO_(2),H_(3)BO_(3))H_(2)SO_(4),H_(2)SO_(3),H_(3)PO_(3),H_(4)P_(2)O_(5)` are diprotic. `HCl,HNO_(3),H_(3)PO_(2),H_(3)BO_(3)` are monoprotic. |
|
| 12. |
The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is `Na_(2)CO_(3)(aq.)+2HCl(aq.)rarrNaCl(aq)+CO_(2)(g)H_(2)O(l)` If you had the two solutions of the same concentration, you would have to use double volume of HCl to reach the equivalence point. Indicators change their colours at the end point of the reaction and hence we are able to know the end points (equivalence points of reactions). 25 ml of `Na_(2)CO_(3)` solution requires 100 ml of 0.1 M HCl to reach end point with phenolphthalein indicator. Molarity of `HCO_(3^(-))` ions in the resulting solution isA. 0.008 MB. 0.04MC. 0.16 MD. 0.08M |
|
Answer» Correct Answer - D In presence of phenophthalein indicator valence factor of Hph = 1 Equivalents of HCl = Equivalent of `Na_(2)CO_(3)` `N_(1)V_(1)=N_(2)V_(2)` For HCl `N_(1)=M_(1) ` and `Na_(2)CO_(3)N_(2)=N_(2)M_(1)V_(1)=M_(2)V_(2)` `0.1 xx 100 = N_(2)xx25` `M_(2)=0.4` Reaction is `Na_(2)CO_(3) +HCl rarrNaHCO_(3)+NaCl` in moles of `Na_(2)CO_(3)=MV` `0.4 018xx25xx10^(-3)` =moles of `NaCHO_(3)` molarity of `HCO_(3)^(-)=("Moles of " NaHCO_(3))/("Volume")=(0.4xx25xx10^(-3)xx1000)/(125)=0.08` |
|
| 13. |
The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is `Na_(2)CO_(3)(aq.)+2HCl(aq.)rarrNaCl(aq)+CO_(2)(g)H_(2)O(l)` If you had the two solutions of the same concentration, you would have to use double volume of HCl to reach the equivalence point. Indicators change their colours at the end point of the reaction and hence we are able to know the end points (equivalence points of reactions). How many ml of 1N HCl are required for X milimoles of `Na_(2)CO_(3)` with methyl orange indicatorA. X mlB. 2 X mlC. 3 X mlD. 4 X ml |
|
Answer» Correct Answer - B In presence of MeOH indicator, velence factor of `Na_(2)CO_(3)=2` Equivalent of HCl = Equivalent of `Na_(2)CO_(3)` NV=mole `xx`V.F 1 V `=x xx10^(-3)xx2` `V=2x xx10^(-3)L=2x ml` |
|
| 14. |
The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is `Na_(2)CO_(3)(aq.)+2HCl(aq.)rarrNaCl(aq)+CO_(2)(g)H_(2)O(l)` If you had the two solutions of the same concentration, you would have to use double volume of HCl to reach the equivalence point. Indicators change their colours at the end point of the reaction and hence we are able to know the end points (equivalence points of reactions). How many ml of 1N HCl are required for X milimoles of NaOH + Y milimoles of `Na_(2)CO_(3)+Z` milimoles of `NaHCO_(3)`with methyl orange indicatorA. (2X+Y+Z)mlB. (X+2Y+2Z)mlC. (X+2Y+3Z)mlD. (X+2y+Z) ml |
|
Answer» Correct Answer - D Equivalent of HCl = Equivalent of `(NaOH + Na_(2)CO_(3)+NOHCO_(3)]` `1xxV=[x xx10^(-3)xx1]+[Yxx10^(-3)xx2]+[Zxx10^(-3)xx1]` `V=[x +2Y+Z]xx10^(-3)L` `V=[X+2Y+Z]ml` |
|
| 15. |
One litre of a solution contains 18.9 g of `HNO_(3)` and one litre of another solution contains 3.2 g of NaOH. In what volume ratio must these solution be mixed to obtain a neutral solution ?A. `3:8`B. `8:3`C. `15:4`D. `4:15` |
| Answer» Correct Answer - D | |
| 16. |
22.7 mL of (N/10) `Na_(2)CO_(3)` solution neutralises 10.2 mL of a dilute `H_(2)SO_(4)` solution. The volume of water that must be added to 400 mL of this `H_(2)SO_(4)` solution in order to make it exactly N/10.A. 490.2 mLB. 890.2 mLC. 90.2 mLD. 290.2 mL |
|
Answer» Correct Answer - A meq of `Na_(2)CO_(3)` = meq of `H_(2)SO_(4)` `(1)/(10)xx22.7=Nxx10.2` Normality = 0.2225 N `0.2225xx400=(1)/(10)xxV_(f)` or `V_(f)=890.2` mL `:. ` Volume of `H_(2)O` mixed = 890.2-400=490.2 mL |
|
| 17. |
`Na_(2)CO_(3), " " Fe_(2)(SO_(4))_(3), " " FeSO_(4).7H_(2)O` |
| Answer» `{:("V.f.",2" ",2xx3=6" "," 2",),("Eq. wt.","M/2 "," M/6","M/2",):}` | |
| 18. |
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. 3.54B. 7.08C. 1.77D. None of these |
| Answer» Correct Answer - A | |
| 19. |
`25ml` of a `0.1(M)` solution of a stable cation of transition metal `z` reacts exactly with `25 ml` of `0.04(M)` acidified `KMnO_(4)` solution. Which of the following is most likely to represent the change in oxidation state of `Z` correctly?A. `Z^(+)rarrZ^(2+)`B. `Z^(2+)rarrZ^(3+)`C. `Z^(3+)rarrZ^(4+)`D. `Z^(2+)rarrZ^(4+)` |
| Answer» Correct Answer - D | |
| 20. |
How many millilitres of `0.1 NH_(2)SO_(4)` solution will be required for complete reaction with a solution containing 0.125 g of pure `Na_(2)CO_(3)`?A. 23.6 mLB. 25.6 mLC. 26.3 mLD. 32.6 mL |
| Answer» Correct Answer - A | |
| 21. |
Calculate the normality of a solution containing 15.8 g of `KMnO_(4)` in 50 mL acidic solution. |
|
Answer» Normality(N) `=(Wxx1000)/(ExxV)` Here W=15.8g, V=50mL E = `("molar mass of KMnO"_(4))/("Valency factor")=158//5=31.6` So, normality = 10 N |
|
| 22. |
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. 1 NB. 0.5 NC. 1.8 ND. 0.8 N |
| Answer» Correct Answer - D | |
| 23. |
Calculate the normality of a solution containing 13.4 g fo sodium oxalate in 100mL solution. |
|
Answer» Normality = `("wt. in g/eq. wt")/("vol. of solution in litre")` Here, eq. wt. of `Na_(2)C_(2)O_(4)=134//2=67` so `N=(13.4//67)/(100//1000)=2N` |
|
| 24. |
The number of moles of ferrus oxalate oxidised by one mole of `KMnO_(4)` in acidic medium is:A. `5//2`B. `2//5`C. `3//5`D. `5//3` |
|
Answer» Correct Answer - D Eq. of `1FeC_(2)O_(4)`=Eq. of `KMnO_(4)` moles of `FeC_(2)O_(4)xx3` = moles of `KMnO_(4)xx5` so, moles of `FeC_(2)O_(4)=5//3` |
|
| 25. |
By which reason temporary and permanent hardness occur ? |
|
Answer» Correct Answer - Temporary hardness - due to bicarbonates of Ca & Mg Permanent hardness - due to chlorides & sulphates of Ca & Mg. |
|
| 26. |
A sample of water from a river was analyzed for the presence of metal ions and the observations were recorded as given below `{:("Reagent added","Observation"),("dil. HCl","No change"),("aq. " Na_(2)CO_(3),"White precipitate"),("Aq. " Na_(2)SO_(4),"No change"):}` The water sample is likely to containA. `Ba^(2+)`B. `Cu^(2+)`C. `Li^(+)`D. `Mg^(2+)` |
| Answer» Correct Answer - D | |
| 27. |
The volume of water which must be added to `0.4 dm^(3)` of 0.25 N oxalic acid in order to make it exactly decinormal is :A. `0.2 dm^(3)`B. `0.4 dm^(3)`C. `0.6 dm^(3)`D. `0.8 dm^(3)` |
| Answer» Correct Answer - C | |
| 28. |
You are given a solution of an alkali. In order to estimate its concentration in terms of normality , you need to knowA. the volume of the solution, the volume of the alkali present in it and its formula weightB. the mass of the solution, the mass of the alkali present in it and its equivalent weight.C. the volume of the solution, the solution, the mass of the alkali present in it and its equivalent weight.D. the mass of the solution, the volume of the alkali present in it and its equivalent weight. |
| Answer» Correct Answer - C | |
| 29. |
The number of moles of oxalate ions oxidised by one mole of `MnO_(4)^(-)` ion in acidic medium is : |
|
Answer» Equivalents of `C_(2)O_(4)^(2-)`=equivalents of `MnO_(4)^(-)` x(mole)`xx2=1xx5` `(.:. "v.f. of " C_(2)O_(4)^(2-)=2(4-3)=2 " and v.f. of " MnO_(4)^(-)=1(7-2)=5)`. `x=(5)/(2)` mole of `C_(2)O_(4)^(2-)` ions. |
|
| 30. |
Define two method by which we can soften the water sample. |
|
Answer» Correct Answer - There are some method by which we can soften the water sample . `{:("(a)","By boiling :",2HCO_(3)^(-)rarrH_(2)O+CO_(2)CO_(3)^(2-),),("or","By Slaked lime :",Ca(HCO_(3))_(2)+Ca(OH)_(2)rarrCaCO_(3)+2H_(2)O,),(,,Ca^(2+)CO_(3)^(2-)rarrCaCO_(3),),("(b)","By Washing Soda :",CaCl_(2)+Na_(2)CO_(3)rarrCaCO_(3)+2NaCl,),("(c)"," ionexchange resins",Na_(2)R+Ca^(2+)rarrCaR+2Na^(+),),("(d)","By adding chelating agents like",(PO_(3)^(-))_(3)"etc.",):}` |
|
| 31. |
The quantity of electricity requried to reduce 0.05 mol of `MnO_(4)^(-)` to `Mn^(2+)` in acidic medium would beA. 0.01 FB. 0.05 FC. 0.15 FD. 0.25 F |
| Answer» Correct Answer - D | |
| 32. |
X gm of metal gave Y gm of its oxide, so equivalent mass of metal is :A. `(y-x)/(x)xx8`B. `(x)/((y-x))xx8`C. `(x)/(y)xx8`D. `(x+y)/(x)xx8` |
| Answer» Correct Answer - B | |
| 33. |
What volume of water is requried to make `0.20 N` solution from `1600 mL` of `0.2050 N` solution?A. 40 mLB. 50 mL of 0.1 M `H_(3)AsO_(3)` to be oxidized to `H_(3)AsO_(4)`C. 100 mLD. 20 mL |
| Answer» Correct Answer - A | |
| 34. |
If the equivalent mass of a metal is double that of oxygen then the weight of its oxide is ___ times greater than the weight of the metal.A. 1.5B. 2C. 0.5D. 3 |
| Answer» Correct Answer - A | |
| 35. |
`5H_(2)C_(2)O_(4)(aq)+2MnO_(4)(aq)+6H^(+)(aq) to 2Mn^(2+)(aq)+10CO_(2)(g)+8H_(2)O(l)` Oxalic acid, `H_(2)C_(2)O_(2)` , reacts with permanganate ion accroding to the balanced equation above. How many mL of 0.0154 M `KMnO_(4)` solution are required to react with 25.0mL of 0.0208 M `H_(2)C_(2)O_(4)` solution?A. 13.5 mLB. 18.5 mLC. 33.8 mLD. 84.4 mL |
| Answer» Correct Answer - A | |
| 36. |
Oxalic acid, `H_(2)C_(2)O_(4)`, reacts with paramagnet ion according to the balanced equation `5H_(2)C_(2)O_(4)(aq)+2MnO_(4)^(-)hArr2Mn^(2+)(aq)+10CO_(2)(g)+8H_(2)O(l)`. The volume in mL of 0.0162 M `KMnO_(4)` solution required to react with 25.0 mL of 0.022 M `H_(2)C_(2)O_(4)` solution is :A. 13.6B. 18.5C. 33.8D. 84.4 |
| Answer» Correct Answer - A | |
| 37. |
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. 4.5 gB. 6.3 gC. 0.63 gD. 0.45 g |
|
Answer» Correct Answer - C `H_(2)C_(2)O_(4).2H_(2)O=2+24+64+36=126` and Equivalent wt. `=[(126)/(2)]` `0.2=(Wxx1000)/(((126)/(2))xx50) " " :. W=0.63 g` |
|
| 38. |
If 1 mL of a `KMnO_(4)` solution react with 0.140g `Fe^(2+)` and if 1 mL of `KHV_(2)O_(4)`. `H_(2C_(2)O_(4)` solution react with o.1 mL of previous `KMnO_(4)` solution, how many millilitres of 0.20 M NaOH will react with 1 mL of previous `KHC_(2)O_(4)`. `H_(2)C_(2)O_(4)` solution in which all the protons `(H^(+))` are ionisable ?A. 15/6 mLB. 13/16C. `11/14`D. None of these |
| Answer» Correct Answer - A | |
| 39. |
A mixture of CuS (molecular weight = `M_(1)`) and `Cu_(2)S` (molecular weight `=M_(2)`) is oxidised by `KMnO_(4)` (molecular weight `=M_(3)`) in acidic medium, where the product obtained are `Cu^(2+),Mn^(2+)` and `SO_(2)`. Find the equivalent weight of CuS, `Cu_(2)S` and `KMnO_(4)` respectively. |
| Answer» Correct Answer - `(M_(1))/(6), (M_(2))/(8), (M_(3))/(5)` | |
| 40. |
25 mL of a solution of `Fe^(2+)` ions was titrated with a solution of the oxidizing agent `Cr_(2)O_(7)^(2-)` . 50 mL of 0.01 M `K_(2)Cr_(2)O_(7)` solution was required. What is the molarity of the `Fe^(2+)` solution ? |
| Answer» Correct Answer - 0.12 M. | |
| 41. |
A bottle of `H_(3)PO_(4)` solution contains 70% acid. If the density of the solution is `1.54 g cm^(-3)` , the volume of the `H_(3)PO_(4)` solution required to prepare 1L of IN solution is .A. 90 mLB. 45mLC. 30mLD. 23mL |
| Answer» Correct Answer - C | |
| 42. |
A substance which participates readily in both acid-base and oxidation-reduction reactions is:A. `Na_(2) CO_(3)`B. KOHC. `KMnO_(4)`D. `H_(2)C_(2)O_(4)` |
| Answer» Correct Answer - D | |
| 43. |
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. 79.75B. 159.5C. 329D. None of these |
|
Answer» Correct Answer - B `2CuSO_(4)+4KIrarrCu_(2)I_(2)+I_(2)+2K_(2)SO_(4)`. `Cu^(2+)+1e^(-)rarrCu^(+)`. `E_(cu)=?. V.f.=1`. `E_(cuSO_(4))=(159.5)/(1)=159.5` |
|
| 44. |
One mole of acidified `K_(2)Cr_(2)O_(7)` on reaction with excess of KCl will liberate….., moles of `I_(2)`.A. 6B. 1C. 7D. 3 |
| Answer» Correct Answer - D | |
| 45. |
If `10g` of `V_(2)O_(5)` is dissolved in acid and is reduced to `V^(2+)` by zinc metal, how many mole `I_(2)` could be reduced by the resulting solution if it is further oxidised to `VO^(2+)` ions? [Assume no change in state of `Zn^(2+)` ions] (`V=51`, `O=16`, `I=127`)A. 0.11B. 0.22C. 0.055D. 0.44 |
| Answer» Correct Answer - A | |
| 46. |
The percentage of available chlorine in a commercial sampleof bleaching powder isA. `52.9%`B. `55.9%`C. `58%`D. `60%` |
| Answer» Correct Answer - B | |
| 47. |
In the redox reaction, `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) to 2Mn^(2+)+10CO_(2)+8H_(2)O` 20mL of 0.1 M `KMnO_(4)` reacts quantitatively with :A. 20 mL of 0.1 M oxalateB. 40 mL of 0.1 M oxalateC. 50 mL of 0.25 M oxalateD. 50 mL of 0.1 M oxalate |
| Answer» Correct Answer - D | |
| 48. |
Some amount of "20V" `H_(2)O_(2)` is mixed with excess of acidified solution of Kl. The iodine so liberated required 200 mL of 0.1 N `Na_(2)S_(2)O_(3)` for titration. The volume of `H_(2)O_(2)` solution is :A. 11.2 mLB. 37.2 mLC. 5.6 mLD. 22.4 mL |
| Answer» Correct Answer - C | |
| 49. |
To a 25 mL `H_(2)O_(2)` solution excess acidified solution of Kl was added. The iodine liberated 20 ml of 0.3 N sodium thiosulphate solution. Use these data to choose the correct statements from the following :A. The weight of `H_(2)O_(2)` present in 25 ml solution is 0.102 gB. The molarity of `H_(2)O_(2)` solution is 0.12 MC. The weight of `H_(2)O_(2)` present in 1 L of the solution is 0.816 gD. The volume strength of `H_(2)O_(2)` is 1.344 L |
| Answer» Correct Answer - A::B::D | |
| 50. |
A 7.1 g sample of bleaching powder suspended in `H_(2)O` was treated with enough acetic acid and KI solution. Iodine thus liberated required 80 mL of 0.2 N hypo solution for titration. Calcutale the % of available chlorine : |
|
Answer» Correct Answer - 8 molesof iodine = moles of chlorine `=(80xx0.2)/(2)xx10^(-3)=8xx10^(-3)` so required % `=(8xx71xx10^(-3))/(7.1)xx100%=8%` |
|