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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. |
What is the normality of `0.3 M H_(3)PO_(4)` when it undergoes the reaction as ? `H_(3)PO_(4) + 2OH^(-) to HPO_(3)^(2-) + 2H_(2)O`A. 0.3 NB. 0.15 NC. 0.60 ND. 0.90 N |
| Answer» Correct Answer - C | |
| 2. |
150 mL of `N//10` HCl is required to react completely with 1.0 g of a sample of limestone. Calculate the percentage purity of calcium carbonate. |
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Answer» `150 mL (N)/(10)HCl-=150 mL (N)/(10)CaCO_(3)` `underset("Mol.mass")(CaCO_(3))+underset(2g eq.)(2HCl) to CaCl_(2)+H_(2)O+CO_(2)` Eq. mass of `CaCO_(3)=(40+12+48)/(2)=(100)/(2)=50` Mass of `CaCO_(3)` present in 150 mL `N//10` solution. `[NxxExx(V)/(1000)]=50xx(1)/(10)xx(150)/(1000)=0.75g` Purity `=(0.75)/(1)xx100=75%` |
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| 3. |
20 mL of a solution containing 0.2 g of impure sample of `H_(2)O_(2)` reacts with 0.316 g of `KMnO_(4)` (acidic). Calculate : (a) Purity of `H_(2)O_(2)` (b) Volume of dry `O_(2)` evolved at `27^(@)C` and 750 mm pressure. |
| Answer» Correct Answer - `% H_(2)O_(2)=85; " Vol" O_(2)=124.79 mL` | |
| 4. |
Which of the following acids is added in the titration of oxalic acid and potassium permanganate ?A. `HNO_(3)`B. HClC. `CH_(3)COOH`D. `H_(2)SO_(4)` |
| Answer» Correct Answer - D | |
| 5. |
1.2 g of a sample of `CaOCl_(2)` were suspended in water made up to 100 mL. 25 mL of this solution was treated with KI and the `I_(2)` liberated corresponded to 10 mL of `N//25` hypo. Calculate the percentage of `Cl_(2)` available in `CaOCl_(2)`. |
| Answer» Correct Answer - `4.7%` | |
| 6. |
1.355 g of pyrolusite sample are added to 50 mL of 1 N oxalic acid solution containing sulphuric acid. After the reaction is completed, the contents are transferred to a measuring flask and the volume made up to 200 mL. 20 mL of this solution is titrated against `KMnO_(4)` solution whose strength is `2 g//L` and 31.6 mL of `KMnO_(4)` solution are required. Calculate the percentage purity in the given sample of pyrolusite. |
| Answer» Correct Answer - 0.9631 | |
| 7. |
1.6 g of pyrolusite was treated with 60 mL of normal oxalic acid and some `H_(2)SO_(4)`. The oxalic acid left undecomposed was made up to 250 mL, 25 mL of this solution required 32 mL of 0.1 N potassium permangante `(KMnO_(4))`. Calcualte the percentage of pure `MnO_(2)` in pyrolusite. |
| Answer» Correct Answer - `76.125%` | |
| 8. |
How many mL of a 0.05 M `KMnO_(4)` solution are required to oxidise 2.0 g of `FeSO_(4)` in a dilute solution (acidic) ? |
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Answer» `underset(10xx151.8)(10 FeSO_(4))+underset(2xx158)(2KMnO_(4))+8H_(2)SO_(4) to K_(2)SO_(4)+2MnSO_(4)+5Fe_(2)(SO_(4))_(3)+8H_(2)O` `10xx151.8g " of" FeSO_(4)` require `KMnO_(4)=2xx158` g 2 g of `FeSO_(4)` will require `KMnO_(4)=(2xx158xx2)/(10xx151.8)g` Suppose, V mL of `KMnO_(4)` solution (0.05 M) is required. Amoount of `KMnO_(4)` in this solution `=(158xx0.05)/(1000)xxV` Thus, `(158xx0.05xx V)/(1000)=(2xx158xx2)/(10xx151.8)` V=52.7 mL |
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| 9. |
Equivalent mass of a substance may be calculated as, Equivalent mass`=("Molecular mass")/("n-factor")=("Atomic mass")/(n-factor")` n-factor= Basicity of acid or acidity of base n-factor= Number of moles of electrons gainted or lost per mole of oxidising or reducing agents n-factor= Total positive or negative valency of a salt n-factor= Valency of an ion. Concept of n-factor is very important for redox as well as for non-redox reactions. `BrO_(3)^(-)` ion reacts with `Br^(-)` to form `Br_(2)`, in acid medium. The equivalent mass of `Br_(2)` in this reaction is :A. `(4 M.w.)/(6)`B. `(3M.w.)/(5)`C. `(5M.w.)/(3)`D. `(5M.w.)/(8)` |
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Answer» Correct Answer - B `2BrO_(3)^(-)+12H^(+)+10e^(-) to Br_(2)+6H_(2)O " " n_(1)=10` `2Br^(-) to Br_(2)+6e^(-) " " n_(2)=2` n-factor`=(n_(1)xxn_(2))/(n_(1)+n_(2))=(10xx2)/(10+2)=(20)/(12)=(5)/(3)` `E.w.=(M.w.)/(5//3)=(3M.w.)/(5)` |
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| 10. |
Equivalent mass of a substance may be calculated as, Equivalent mass`=("Molecular mass")/("n-factor")=("Atomic mass")/(n-factor")` n-factor= Basicity of acid or acidity of base n-factor= Number of moles of electrons gainted or lost per mole of oxidising or reducing agents n-factor= Total positive or negative valency of a salt n-factor= Valency of an ion. Concept of n-factor is very important for redox as well as for non-redox reactions. When `KMnO_(4)` is titrated against ferrous ammonium sulphate in acid medium then equivalent mass of `KMnO_(4)` will be :A. `("Molecular mass")/(10)`B. `("molecular mass")/(5)`C. `("Molecular mass")/(3)`D. `("molecular mass")/(2)` |
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Answer» Correct Answer - B `MnO_(4)^(-)+8H^(+)+5e^(-) to Mn^(2+) +4H_(2)O` n-factor=5 |
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| 11. |
Mixture of 1 mole `BaF_(2)` and 2 mole `H_(2)SO_(4)` can not be neutralised by :A. 2 mole `Ba(OH)_(2)`B. 2 mole `Ca(OH)_(2)`C. 4 mole NaOHD. 2 mole KOH |
| Answer» Correct Answer - D | |
| 12. |
Which one of the following is a primary standard ?A. `KMnO_(4) `B. `CuSO_(4). 5H_(2)O`C. `I_(2)`D. `H_(2)SO_(4)` |
| Answer» Correct Answer - B | |
| 13. |
Which one of the following is not a primary standard :A. Oxalic acidB. Sodium thiosulphateC. Sodium hydroxideD. Potassium dichromate |
| Answer» Correct Answer - C | |
| 14. |
Which one of the following is a standard solution ?A. It contains one gram equivalent mass of the substance in one litre solutionB. Its strength is accurately knownC. Its strength is to be determinedD. A solution which has been prepared from pure substance |
| Answer» Correct Answer - B | |
| 15. |
For the preparation of a litre of `N//10` solution of `H_(2)SO_(4)`, we need :A. 9.8 gB. 4.9 gC. 10 gD. 98 g |
| Answer» Correct Answer - B | |
| 16. |
0.1 N solution of `Na_(2)CO_(3)` is being titrated with 0.1 N HCl, the best indicator to be used is :A. potassium ferricyanide`B. phenolphthaleinC. methyl orangeD. litmus |
| Answer» Correct Answer - C | |
| 17. |
1.0 g of metal carbonate neutralises 200 mL of 0.1 N HCl. The equivalent mass of the metal will be :A. 50B. 40C. 20D. 100 |
| Answer» Correct Answer - A | |
| 18. |
The volume (in mL) of 0.1 `M AgNO_(3)` required to completely precipitat the chloride ions present in 30 mL of 0.01 M of `[Cr(H_(2)O)_(5)Cl]Cl_(2)`, as silver chloride is close to : |
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Answer» Correct Answer - 6 `underset(n_(1)=2)(2AgNO_(3))+[Cr(H_(2)O)_(5)Cl]Cl_(2)to [Cr(H_(2)O)_(5)Cl](NO_(3))_(2)+2AgCl` `(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))` `(0.1xxV)/(2)=(0.01xx30)/(1)` V=6 mL |
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| 19. |
100 mL of 0.2 N HCl solution is added to 100 mL of 0.2 N `AgNO_(3)` solution. The molarity of nitrate ions in the resulting mixture will be :A. 0.05 MB. 0.5 MC. 0.1 MD. 0.2 M |
| Answer» Correct Answer - C | |
| 20. |
When 10 mL of 10 M solution of `H_(2)SO_(4)` and 100 mL of 1 M solution of NaOH are mixed, the resulting solution will be :A. acidicB. neutralC. alkalineD. Cannot be predicted |
| Answer» Correct Answer - A | |
| 21. |
When 20 mL of `(M)/(10)` NaOH are added to 10 mL of `(M)/(10)HCl`, the resulting solution will :A. turn blue litmus redB. turn phenolphthalein solution pinkC. turn methyl orange redD. have no effect on either red or blue litmus |
| Answer» Correct Answer - B | |
| 22. |
The equivalent mass of `MnSO_(4)` is half its molecular mass when it is converted to :A. `Mn_(2)O_(3)`B. `MnO_(2)`C. `MnO_(4)^(-)`D. `MnO_(4)^(2-)` |
| Answer» Correct Answer - B | |
| 23. |
Identify the incorrect statement regarding the volumetric estimation of `FeSO_(4)` :A. `KMnO_(4)` can be used in aqueous HClB. `K_(2)Cr_(2)O_(7)` can be used in aqueous HClC. `KMnO_(4)` can be used in aqueous `H_(2)SO_(4)`D. `K_(2)Cr_(2)O_(7)` can be used in aqueous `H_(2)SO_(4)` |
| Answer» Correct Answer - A | |
| 24. |
In volumetric experiment, it was found that a solution of `KMnO_(4)` is reduced to `MnSO_(4)`. If the normality of solution is 1 N, then molarity of solution will be :A. 0.5 MB. 0.2 MC. 1 MD. 0.4 M |
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Answer» Correct Answer - B Normality=Molarity `xx` Change in oxidation number `overset(+7)(KMnO_(4)) to overset(+2)(MnSO_(4))` Change in oxidation number=5 `therefore N=Mxx5` `1=Mxx5` M=0.2 |
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| 25. |
The number of equivalents of `Na_(2)S_(2)O_(3)` required for the volumetric estimation of one equivalent of `Cu^(2+)` is :A. `1//3`B. `1`C. `3//2`D. `2//3` |
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Answer» Correct Answer - B Number of equivalents of reacting species in a chemical reaction are same. |
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| 26. |
11.2 g carbon reacts completely with 19.63 litre `O_(2)` at NTP. The cooled gases are pased through 2 litre of 2.5 N NaOH solution. Calculate concentration of remaining NaOH and `Na_(2)CO_(3)` in solution. (CO does not react with NaOH under these conditions.) |
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Answer» Let x moles of carbon be converted into CO and y moles of carbon be converted into `CO_(2)`. `underset(x)(C )+(1)/(2)underset(x//2)(O_(2)) to CO` `underset(y)C+underset(y)(O_(2)) to CO_(2)` Total volume of oxygen used `=(x)/(2)xx22.4+yxx22.4` =19.63 11.2x+22.4y=19.63 `x+y=(11.2)/(12)`, i.e., 12x+12y=11.2 Solving eqs. (i) and (ii), we get x=0.11, y=0.82 Number of moles of `CO_(2)` formed =0.82 Number of milliequivalents of NaOH solution through which `CO_(2)` is massed `=NxxV=2.5xx2000=5000` Number of milliequivalents of `CO_(2)` passed`=0.82xx2xx1000` =1640 `2NaOH+CO_(2) to Na_(2)CO_(3)+H_(2)O` Number of milliequivalents of `Na_(2)CO_(3)=1640` `N_(Na_(2)CO_(3))=(1640)/(2000)=0.82` Number of milliequivalents of remaining NaOH =5000-1640=3360 Normality of remaining `NaOH=(3360)/(2000)=1.68` |
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| 27. |
A solution of which substance can best be used as both titrant and its own indicator in an oxidation-reduction titration ?A. `I_(2)`B. NaOClC. `K_(2)Cr_(2)O_(7)`D. `KMnO_(4)` |
| Answer» Correct Answer - D | |
| 28. |
If 100 mL of the acid is neutralised by 100 mL of 4 M NaOH, the purity of concentrated HCl (sp. Gravity =1.2) is :A. 0.12B. 0.98C. 0.73D. 0.43 |
| Answer» Correct Answer - A | |
| 29. |
1.0 litre of a solution contains 5.3 g of `Na_(2)CO_(3)` and 8 g of NaOH. 20 mL of this solution are taken and titrated against `N//10` HCl using separately (a) methyl orange as an indicator and (b) phenolphthalein as an indicator. What will be the titre values in these two cases ? |
| Answer» Correct Answer - With phenolphthalein 50 mL, with methyl orange 60 mL | |
| 30. |
Two acids A and B titrated separately each time with 25 mL of N `Na_(2)CO_(3)` solution and require 10 mL and 40 mL respectively for complete neutralisation. What volume of A and B would you mix to produce one litre of normal acid solution ? |
| Answer» Correct Answer - Vol. of A =200 mL, Vol. of B=800 mL | |
| 31. |
Grastric juice contains 3 g of HCl per litre. If a person produces 2.5 litre of gastric juice per day, how many antacid tablets each containing 400 mL of `Al(OH)_(3)` are needed to neutralise all the HCl produced in one day ? |
| Answer» Correct Answer - 14 | |
| 32. |
In the mixture of `(NaHCO_(3) + Na_(2)CO_(3))`, volume of HCl required is x mL with phenolphthalein indicator and y mL with mthyl orange inidicator in same titration. Hence, volume of HCl for complete reaction of `Na_(2)CO_(3)` is :A. 2 xB. yC. `x//2`D. `(y-x)` |
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Answer» Correct Answer - A In pressure of phenolphthalein, `50% Na_(2)CO_(3)` is neutralised whereas `NaHCO_(3)` remains unaffected. In presence of methyl orange, both `Na_(2)CO_(3) " and" NaHCO_(3)` will be 100% neutralised. Let volume of HCl for complete reaction of `Na_(2)CO_(3)=V_(1)` mL and volume of HCl for complete reaction of `NaHCO_(3)=V_(2)` mL . With phenolphthalein, `50% Na_(2)CO_(3)` will be neutralized. `therefore " " (V_(1))/(2)=x, V_(1)=2x` |
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| 33. |
25 mL of a mixture of NaOH and `Na_(2)CO_(3)` when itrated with `N//10` HCl using phenolphthalein indicator required 25 mL HCl. The same volume of mixture when titrated with `N//10 HCl` using methyl orange indicator required 30 mL of HCl. Calculate the amount of `Na_(2)CO_(3)` and NaOH in one litre of this mixture. |
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Answer» When phenolphthalein is the indicator, whole of NaOH has been neutralised and carbonate converted into bicarbonate, i.e., `NaOH+HCl to NaCl+H_(2)O` `Na_(2)CO_(3)+HCl to NaHCO_(3)+NaCl` So, `25 mL (N)/(10)HCl-=NaOH+1//2Na_(2)CO_(3)` present in 25 mL of mixture In another titration when methyl orange is the indicator, whole of NaOH has been neutralised and carbonate converted into carbonic acid, i.e., `Na_(2)CO_(3)+2HCl to 2NaCl +H_(2)CO_(3)` `30 mL(N)/(10) HCl -=NaOH+Na_(2)CO_(3)` present in 25 mL of mixture Hence, `(30-25)mL (N)/(10)HCl-=(1)/(2)Na_(2)CO_(3)` present in 25 mL of mixture Hence, `10 mL (N)/(10)HCl-=Na_(2)CO_(3)` present in 25 mL of mixture `-=10 mL(N)/(10)Na_(2)CO_(3)` solution Amount of `Na_(2)CO_(3)=(53xx10)/(10xx1000)=0.053 g` This amount of `Na_(2)CO_(3)` is present in 25 mL of mixture. The amount present in one litre of mixture `=(0.053)/(25)xx1000=2.12g` `(30-10)mL (N)/(10)HCl-=NaOH` present in 25 mL of mixture `-=20 mL(N)/(10)NaOH` Amount of NaOH in 25 mL of mixture`=(40xx20)/(10xx1000)=0.08 g` The amount present in one litre of mixture`=(0.08)/(25)xx1000=3.20 g` |
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| 34. |
A mixture of KOH and `Na_(2)CO_(3)` solution required 15 mL of `N//20` HCl using phenolphthalein as indicator. The same amount of alkali mixture when titrated using methyl orange as indicator required 25 mL of same acid. Calculate amount of KOH and `Na_(2)CO_(3)` present in solution. |
| Answer» Correct Answer - `Na_(2)CO_(3)=0.053 g; KOH=0.014 g` | |
| 35. |
Assertion : In the titration of HCl against NaOH, phenolphthalein is used as suitable indicator. Reason : Phenolphthalein is pink coloured in basic medium.A. If both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.C. If Assertion is correct, but Reason is incorrect.D. If Assertion is incorrect, but Reason is correct. |
| Answer» Correct Answer - B | |
| 36. |
Assertion : Starch is used as absorption indicator in iodometric and iodimetric titrations. Reason : Starch forms iodostarch complex with iodine, which is blue coloured.A. If both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.C. If Assertion is correct, but Reason is incorrect.D. If Assertion is incorrect, but Reason is correct. |
| Answer» Correct Answer - A | |
| 37. |
The M mass of NaOH is 40.50 mL of a solution containing 2 g of NaOH in 500 mL will require for complete neutralisation :A. 10 mL decinormal HClB. 20 mL decinormal HClC. 50 mL decinormal HClD. 25 mL decinormal HCl |
| Answer» Correct Answer - C | |
| 38. |
20 mL of `0.1 M H_(3)BO_(3)` solution on complete neutralisation requires x mL of 0.05 M NaOH solution. The value of x will be :A. 20 mLB. 40 mLC. 120 mLD. 80 mL |
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Answer» Correct Answer - B Boric acid is monobasic acid. `H_(3)BO_(3)+NaOH to Na[B(OH)_(4)]` `(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))` `(0.1xx20)/(1)=(0.05xx x)/(1)` x=40 mL |
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| 39. |
Assertion : Equivalent mass of `KMnO_(4)` is equal to one-fifth of its molecular mass when it acts as an oxidising agent in acidic medium. Reason : Oxidation number of Mn in `KMnO_(4)` is `+7`.A. If both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.C. If Assertion is correct, but Reason is incorrect.D. If Assertion is incorrect, but Reason is correct. |
| Answer» Correct Answer - B | |
| 40. |
In order to prepare one litre nomal solution of `KMnO_(4)`, how many grams of `KMnO_(4)` are required if the solution is to be used in acid medium for oxidation ?A. 158 gB. 31.6 gC. 62 gD. 790 g |
| Answer» Correct Answer - B | |
| 41. |
Some amount of `NH_(4)Cl` was boiled with 50 mL of 0.75N NaOH solution till the reaction was complete. After the completion of the reaction, 10 mL of `0.75 N H_(2)SO_(4)` were required for the neutralisation of the remaining NaOH. Calculate the amount of `NH_(4)Cl` taken. |
| Answer» Correct Answer - 1.605 g | |
| 42. |
`28 NO_(3)^(-) + 3As_(2)S_(3) + 4H_(2)O to 6AsO_(4)^(3-)+28 NO+ 9SO_(4)^(2-) + 8H^(+)`. What will be the equivalent mass of `As_(2)S_(3)` in above reaction ?A. `(M. wt.)/(2)`B. `(M. wt.)/(4)`C. `(M. wt.)/(24)`D. `(M. wt.)/(28)` |
| Answer» Correct Answer - D | |
| 43. |
An equal volume of reducing agent is titrated separately with `1 M KMnO_(4)` in acid, neutral and alkaline medium. The volumes of `KMnO_(4)` required are `20 mL, 33.3 mL` and `100 mL` in acid, neutral and alkaline medium respectively. Find out oxidation state of `Mn` in each reaction product. Give balance equation. Find the volume of `1 M K_(2)Cr_(2)O_(7)` consumed if same volume of reductant is titrated in acid medium. |
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Answer» Let `N_(1), N_(2) " and" N_(3)` be the normalities of `1M KMnO_(4)` solution in acid, neutral and alkaline mediums, repectively. `20 mL N_(1)-=33.4 mL N_(2)-=100 mL N_(3)` In acidic medium, the half reaction is : `MnO_(4)^(-) +8H^(+)+5e^(-)=Mn^(2+)+4H_(2)O` `1 M KMnO_(4)=5N KMnO_(4)` Thus, from above relation, `N_(2)=(20)/(33.4)xxN_(1)=(20)/(33.4)xx5N =3N` and `N_(3)=(20)/(100)xx N_(1)=(20)/(100)xx5N =1N` The equations in the three media are : `MnO_(4)^(-)+5e^(-) overset("Acid")to Mn^(2+)` `MnO_(4)^(-)+3e^(-) overset("Neutral")to Mn^(4+)` `MnO_(4)^(-) +e^(-) overset("Alkaline") to Mn^(6+)` The balanced equations are : `MnO_(4)^(-)+8H^(+)+5e^(-) overset("Acid")to Mn^(2+)+4H_(2)O` `MnO_(4)^(-)+2H_(2)O+3e^(-) overset("Neutral")to MnO_(2)+4OH^(-)` `MnO_(4)&(-)+e^(-) overset("Alkaline") to MnO_(4)^(2-)` The balanced equation in the case of acidified `K_(2)Cr_(2)O_(7)` solution can be written as : `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 3Cr^(3+)+7H_(2)O` `1M K_(2)Cr_(2)O_(7)=6 N K_(2)Cr_(2)O_(7)` The volume required for the titration of the same volume of reducing agent with acidified `K_(2)Cr_(2)O_(7)` solution as follows : `20 mL 5 N KMnO_(4)-= V 6 N K_(2)Cr_(2)O_(7)` `V=(20xx5)/(6)=16.66` mL |
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| 44. |
In a sample of sodium carbonate some sodium sulphate is also mixed. 1.25 g of this sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralises 20 mL of `(N)/(10)` sulphuric acid. Calculate the percentage of sodium carbonate in the sample. |
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Answer» 25 mL of sample solution neutralises `20 mL(N)/(10)H_(2)SO_(4)` 250 mL of sample solution will neutralise `=200 mL (N)/(10)H_(2)SO_(4)` `200 mL (N)/(10)H_(2)SO_(4)-=200 mL(N)/(10)Na_(2)CO_(3)` solution Amount of `Na_(2)CO_(3)` present `=(ExxNxxV)/(1000)` `=(53xx200)/(10xx1000)=1.06` % of `Na_(2)CO_(3)` in the sample `=(1.06)/(1.25)xx100=84.8` |
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| 45. |
In a sample of sodium carbonate, some sodium sulphate is mixed. 2.50 g of this sample is dissolved and the volume made up to 500 mL . 25 mL of this solution neutralises 20 mL of `N//10` in the sample. |
| Answer» Correct Answer - `Na_(2)CO_(3)=84.8 %` | |
| 46. |
The volume of water to be added to 400 mL of `N//8` HCl to make it exactly `N//12`, is :A. 400 mLB. 300 mLC. 200 mLD. 100 mL |
| Answer» Correct Answer - C | |
| 47. |
250 mL of x M solution and 500 mL of y M solution of a solute A are mixed and diluted to 2 litre to produce a final concentration of 1.6 M. |
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Answer» We Know that, `M_(1)V_(1)+M_(2)V_(2)=M_(R )(V_(1)+V_(2))` `x xx 250+yxx500=1.6(2000)` `x+2y=1.6xx8` x+2y = 12.8 `(x)/(y)+2=(12.8)/(y)` `(5)/(4)+2=(12.8)/(y)` `(13)/(4)=(12.8)/(y)` `y=(12.8xx4)/(13)=3.94` Similary, `" " ` x=4.92 |
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| 48. |
One mole of acidic `KMnO_(4)` reacts with :A. `(5)/(3) " mol of " FeC_(2)O_(4)`B. `(5)/(2) " mol of " SO_(2)`C. 4 mol of FeSD. 1 mol of `H_(2)SO_(4)` |
| Answer» Correct Answer - A::B | |
| 49. |
0.5 g of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas, the gas is absorbed in 150 mL of `N//5 H_(2)SO_(4)` solution. Excess sulphuric acid required 20 mL of 1N NaOH for complete neutralization. The percentage of `NH_(3)` in the ammonium chloride is :A. 0.68B. 0.34C. 0.48D. 0.17 |
| Answer» Correct Answer - B | |
| 50. |
The normality of a 26 % `"mass"//"volume"` solution of ammonia (density `0.885 g//mL`) is approximately :A. 1.5B. 4C. 0.4D. 15.3 |
| Answer» Correct Answer - D | |