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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.
| 51. |
Normality of 2 % `H_(2)SO_(4)` solution by volume is nearly :A. 2B. 4C. 0.2D. 0.4 |
| Answer» Correct Answer - D | |
| 52. |
Assertion : 5 M HCl solution is diluted 10 times, its molarity becomes 50. Reason : On dilution, molarity of the solution decreases.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 - D | |
| 53. |
The molecular mass of `KMnO_(4)` is M. Its equivalent mass in acidic medium will be :A. MB. `M//2`C. `M//5`D. `M//4` |
| Answer» Correct Answer - C | |
| 54. |
Molecular mass of a tribasic acid is M. Its equivalent mass will be :A. `M//3`B. `3M`C. `M//2`D. `2M` |
| Answer» Correct Answer - A | |
| 55. |
Assertion : Equivalent weight of a base `=("Molecular weight")/("Acidity")` Reason : Acidity is the number of replaceable hydrogen atoms in one molecule of the base.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 - C | |
| 56. |
For the reaction, `N_(2)(g)+3H_(2)(g)to 2NH_(3)(g)`, if molecular masses of `NH_(3) " and " N_(2) " and " M_(1) " and" M_(2)`, their equivalent masses are `E_(1) " and" E_(2)`, then `(E_(1)-E_(2))` is :A. `(2M_(1)-M_(2))/(6)`B. `M_(1)-M_(2)`C. `3M_(1)-M_(2)`D. `M_(1)-3M_(2)` |
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Answer» Correct Answer - A Equivalent mass of `N_(2), i.e., E_(2)=(M_(2))/(6)` Equivalent mass of `NH_(3), i.e., E_(1)=(M_(1))/(3)` Then, `" " E_(1)-E_(2)=(M_(1))/(3)-(M_(2))/(6)=(2M_(1)-M_(2))/(6)` |
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| 57. |
The reagent commonly used to determine harness of water titrimetrically is :A. Oxalic acidB. disodium salt EDTAC. sodium citrateD. sodium thiosulphate |
| Answer» Correct Answer - B | |
| 58. |
How many moles of acidified `FeSO_(4)` can be completely oxidised by one mole of `KMnO_(4)~ ?A. 10B. 5C. 6D. 2 |
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Answer» Correct Answer - B `2KMnO_(4)+3H_(2)SO_(4) to K_(2)SO_(4) +2MnSO_(4)+3H_(2)O +5[O]` `([2FeSO_(4)+H_(2)SO_(4)+[O]toFe_(2)(SO_(4))_(3)+H_(2)O]xx5)/ul(2KMnO_(4)+10FeSO_(4)+8H_(2)SO_(4)to K_(2)SO_(4)+2KnSO_(4)+5Fe_(2)(SO_(4))_(3)+8H_(2)O)` `2" mole" KMnO_(4)-=10 " mole" FeSO_(4)` `1 "mole" KMnO_(4)-=5 " mole" FeSO_(4)` |
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| 59. |
3.0 g of pyrolusite ore were treated with 20 g of pure ferrous ammonium sulphate (Mol.mass `=392 " g mol"^(-1))` and dilute `H_(2)SO_(4)`. After the reaction, the solution was diluted to 500 mL. 50 mL of diluted solution required 10 mL of 0.1 N `K_(2)Cr_(2)O_(7)` solution. Calculate the % of pure `MnO_(2)` in pyrolusite. |
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Answer» Correct Answer - `59.4%` `MnO_(2)` present in pyrolusite oxidises ferrous ammonium sulphate into ferric ammonium sulphate, i.e., `Fe^(2+) to Fe^(3+)`. Unreacted ferrous ammonium sulphate is estimated by potassium dichromate solution. `MnO_(2)+2Fe^(2+)+4H^(+) to Mn^(2+) +2H_(2)O+2Fe^(3+)` `Cr_(2)O_(7)^(2-)+6Fe^(2+) +14H^(+) to 2Cr^(3+)+7H_(2)O+6Fe^(3+)` 50 mL diluted ferrous ammonium sulphate solution `=10 mL " of" 0.1 N K_(2)Cr_(2)O_(7)` 500 mL diluted ferrous ammonium sulphate solution `=10xx10 mL " of" 0.1 N K_(2)Cr_(2)O_(7)` =100 mL of 0.1 N FeAS `=(0.1xx392)/(1000)xx100=3.92 g` Used FeAS=(20-3.92)=16.08 g `MnO_(2)` present in pyrolusite `=(87)/(392xx2)xx16.08=1.784 g` Percentage of pure `MnO_(2)=(1.784)/(3.0)xx100=59.4 %` |
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| 60. |
Pyrolusite, `MnO_(2)`, is the main ore from which manganese is produced. The manganese content of the ore may be determined by reducing the `MnO_(2)` under acetic conditions to `Mn^(2+)` with the oxalate ion, `C_(2)O_(4)^(2-)`, the oxalate ion being oxidised to carbon dioxide during the reaction. The analytical determination is carried out by adding a known excess volume of oxalate solution to a suspension of the pyrolusite and digesting the mixture on a hot water bath until all the `MnO_(2)` has been reduced. The excess, unreacted oxalate solution is then titrated with standardised potassium permanganate, `KMnO_(4)` solution after which the manganese content of the ore can be calculated. A student prepared a standard solution of sodium oxalate by weighing 3.2 g of the dry anhydrous salt, dissolving it in distilled water and making the solution up to 500 mL. 25 mL of the oxalate solution required 24.76 mL of `KMnO_(4)` solution. Molarity of the sodium oxalate solution is ..A. 0.04776B. 0.07446C. 0.06447D. 0.07644 |
| Answer» Correct Answer - A | |
| 61. |
Pyrolusite, `MnO_(2)`, is the main ore from which manganese is produced. The manganese content of the ore may be determined by reducing the `MnO_(2)` under acetic conditions to `Mn^(2+)` with the oxalate ion, `C_(2)O_(4)^(2-)`, the oxalate ion being oxidised to carbon dioxide during the reaction. The analytical determination is carried out by adding a known excess volume of oxalate solution to a suspension of the pyrolusite and digesting the mixture on a hot water bath until all the `MnO_(2)` has been reduced. The excess, unreacted oxalate solution is then titrated with standardised potassium permanganate, `KMnO_(4)` solution after which the manganese content of the ore can be calculated. A student prepared a standard solution of sodium oxalate by weighing 3.2 g of the dry anhydrous salt, dissolving it in distilled water and making the solution up to 500 mL. 25 mL of the oxalate solution required 24.76 mL of `KMnO_(4)` solution. What is the molarity of `KMnO_(4)` solution ?A. 0.04776B. 0.01929C. 0.038D. 0.028 |
| Answer» Correct Answer - B | |
| 62. |
25 ml of `(N)/(10)` caustic soda solution exactly neuralised 20ml of an acid solution contaning `7.875gm` of acid per litre. What will be the equivalent mass of acid? |
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Answer» `underset((NaOH))(N_(1)V_(1))=underset(("Acid"))(N_(2)V_(2))` `(1)/(10)xx25=N_(2)xx20` `N_(2)=(25)/(10xx20)=0.125` Strength =Normality `xx` Eq. mass Eq. mass of the acid`=(7.875)/(0.125)=63.00` |
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| 63. |
(a) What is the normality of a 96 per cent solution of `H_(2)SO_(4)` of specific gravity 1.84 ? (b) How many mL of 96 per cent sulphuric acid solution is necessary to prepare one litre 0.1 N `H_(2)SO_(4)` ? ( c) To what volume should 10 mL of 96 per cent `H_(2)SO_(4)` be diluted to prepure 2 N solution ? |
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Answer» Mass of 1 litre of `H_(2)SO_(4)` solution `="Vol."xx"Density"` `=1000xx1.84=1840 g` Mass of `H_(2)SO_(4)` present in one litre 96% `H_(2)SO_(4)` solution `= (96)/(100)xx1840=1766.4 g` Strength of `H_(2)SO_(4)` solution `=1766.4 g//L` (a) Normality `=("Strength")/("Eq. mass")=(1766.4)/(49)=36.05 N` (b) Let the volume taken be `V_(1)` mL Applying `" " N_(1)V_(1)=N_(2)V_(2)` `N_(1)=36.05 N, V_(1) = ?, N_(2)=(N)/(10), V_(2)=1000` mL `36.05xxV_(1)xx1000` So, `V_(1)=(1000)/(36.05xx10)=2.77` mL i.e., 2.77 mL of `H_(2)SO_(4)` is diluted to one litre. (c ) ` underset("Before dilution")(N_(B)V_(B))= underset("After dilution ")(N_(A)V_(A))` `10xx36.05=V_(A)xx2` `V_(A)=180.25` mL i.e., 10 mL of given `H_(2)SO_(4)` is diluted to 180.25 mL. |
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| 64. |
500 mL of 2 M HCl, 100 mL of 2 M `H_(2)SO_(4)` and one gram equivalent of monoacidic alkali are mixed together. 30 mL of this solution required 20 mL of `Na_(2)CO_(3). xH_(2)O` solution obtained by dissolving 143 g `Na_(2)CO_(3).xH_(2)O` in one litre solution. Calculate the water of crystallisation of `Na_(2)CO_(3) .xH_(2)O`. |
| Answer» Correct Answer - `10 H_(2)O` | |
| 65. |
A 0.5 g sample containing `MnO_(2)` is treated with HCl liberating `Cl_(2)`. The `Cl_(2)` is passed into a solution of KI and `30.0 cm^(3)` of 0.1 M `Na_(2)S_(2)O_(3)` are required to tirate the liberated iodine. Calculate the percentage of `MnO_(2)` in the sample. |
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Answer» `30.0 mL 0.1 M Na_(2)S_(2)O_(3)-=30.0 mL 0.1 N Na_(2)S_(2)O_(3)` `-=30.0 mL 0.1 N I_(2)` `-=30.0 mL 0.1 N Cl_(2)` `-=30.0 mL 0.1 N MnO_(2)` Amount of `MnO_(2)` present `=(NxxExxV)/(1000)` `=(1)/(10)xx(87)/(2)xx(30)/(1000)` `% MnO_(2)=(87xx30xx100)/(10xx2xx1000xx0.5)=26.1` |
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| 66. |
One litre sample of water contains 0.9 mg `CaCl_(2)` and 0.9 mg of `MgCl_(2)`. Find the total hardness in terms of parts permillion of `CaCO_(3)`. |
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Answer» `111 g CaCl_(2)-=100 g CaCO_(3)` `therefore " " 0.9 mg CaCl_(2) -=(100)/(111)xx0.9 mg CaCO_(3)` `-=0.81 mg " of" CaCO_(3)` ` 95 g MgCl -=100 g CaCO_(3)` ` therefore " " 0.9 mg MgCl_(2) -=(100)/(95)xx0.9 mg CaCO_(3)` `-=0.94 mg " of " CaCO_(3)` Thus, 1 litre water to `10^(6)`. mg water contains (0.81+0.94 = 1.75 mg) i.e., hardness of water is 1.75 ppm. |
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| 67. |
A solution of `H_(2)O_(2)` is titrated against a solution of `KMnO_(4)`. The reaction is : `2MnO_(4)^(-)+5H_(2)O_(2)+6H^(+) to 2Mn^(2+)+5O_(2)+8H_(2)O` If it requires 46.9 mL of 0.145 M `KMnO_(4)` to oxidise 20 g of `H_(2)O_(2)`, the mass percentage of `H_(2)O_(2)` in this solution is :A. 2.9B. 29C. 21D. 4.9 |
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Answer» Correct Answer - A Number of moles of `KMnO_(4)=(MV)/(1000)=(0.145xx46.9)/(1000)` `=6.8xx10^(-3)` Number of moles of `H_(2)O_(2)=6.8xx10^(-3)xx2.5=0.017` Mass of `H_(2)O_(2)=0.017xx34=0.578` Mass % of `H_(2)O_(2)=(0.578)/(20)xx100=2.9` |
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| 68. |
One mole of a mixture of CO and `CO_(2)` requires exactly 20 g of NaOH to convert all the `CO_(2) " into " Na_(2)CO_(3)`. How many more grams of NaOH would it require for conversion into `Na_(2)CO_(3)` if the mixture (one mole) is completely oxidised to `CO_(2)` ?A. 60 gB. 80 gC. 40 gD. 20 g |
| Answer» Correct Answer - A | |
| 69. |
A molal solution is one that contains one mole of the solute in :A. 1000 g of the solventB. one litre of the solventC. one litre of the solutionD. 22.4 litre of the solvent |
| Answer» Correct Answer - A | |
| 70. |
50 mL of `10 N H_(2)SO_(4)`, 25 mL of 12 N HCl and 40 mL of 5 N `HNO_(3)` are mixed and the volume of the mixture is made 1000 mL by adding water. The normality of the resulting solution will be :A. 1 NB. 2 NC. 3 ND. 4 N |
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Answer» Correct Answer - A `N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=N_(R )xxV_(R )` `10xx50+12xx25+5xx40=N_(R )=xx1000` `500+300+200=N_(R )=1000` `N_(R )=1` (Resulatnt normality) |
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| 71. |
22.6 g of an ammonium salt were treated with 100 mL of normal NaOH solution and boiled till no more of ammonia gas was given off. The excess of NaOH solution left over required 60 mL normal sulphuric acid. Calculate the percentage of ammonia in the salt. |
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Answer» 60 mL normal `H_(2)SO_(4)-=60 mL` normal NaOH Thus, (100-60) mL normal NaOH were consumed by ammonium salt. So, 40 mL normal `NaOH -= 40 mL` normal `NH_(3)` Amount of `NH_(3)` in 40 mL normal `NH_(3)` `=("Eq. mass of " NH_(3)xx40)/(1000)` `=(17xx40)/(1000)=0.68` SO, % of ammonia in the ammonium salt `=(0.68)/(2.26)xx100` =30.09 |
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| 72. |
A quantity of ammonium chloride was heated with 100 mL of 0.8 N NaOH solution till the reaction was complete. The excess of NaOH was neutralised with 12.5 mL of `0.75N H_(2)SO_(4)`. Calculate the quantity of ammonium chloride. |
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Answer» 12.5 mL of `0.75 N H_(2)SO_(4)-=12.5 mL " of" 0.75 N NaOH` 12.5 mL of 0.75 N NaOH `-= 11.72 mL " of" 0.8 N NaOH` NaOH solution used by `NH_(4)Cl` `=(100-11.72)mL ` of 0.8N NaOH =88.28 mL of 0.8 N NaOH `-=88.28 mL " of " 0.8 N NH_(4)Cl` Mass of `NH_(4)Cl` present in 88.28 mL of `0.8N NH_(4)Cl` solution `=(NxxExxV)/(1000)=(0.8xx53.5xx88.28)/(1000)=3.7783 g` [Eq. mass of `NH_(4)Cl=53.5]` |
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| 73. |
A solution containing 4.2 g of KOH and `Ca(OH)_(2)` is neutralised by an acid. If it consumes 0.1 g equivalents of the acid, calculate the composition of the sample. |
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Answer» Let mass of KOH be present in mixture=a g and Mass of `Ca(OH)_(2)=(4.2-a)g` Eq. mass of KOH=56, Eq. mass of `Ca(OH)_(2)=(74)/(2)=37` g equivalent of KOH + g equivalent of `Ca(OH)_(2)` = g equivalent of the acid `(a)/(56)+((4.2-a))/(37)=0.1` or `37a-56a=0.1xx56xx37-4.2xx56` or 19a=28 `a=(28)/(19)=1.47` Mass of KOH in the sample =1.47 g Percentage of KOH = 35 and Percentage of `Ca(OH)_(2)=100-35=65` |
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| 74. |
200 mL of 3 N HCl were mixed with 200 mL of 6 N `H_(2)SO_(4)` solution. The final normality of `H_(2)SO_(4)` in the resultant solution will be:A. 9 NB. 3 NC. 6 ND. 2 N |
| Answer» Correct Answer - B | |
| 75. |
3.45 g of a metallic carbonate were mixed with 240 mL of `N//4` HCl. The excess acid was neutralised by 50 mL of `N//5` KOH solution. Calculate the equivalent mass of the metal. |
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Answer» Equivalent mass of metal carbonate `=(3.45xx4xx1000)/(200)=69` Equivalent mass of metal =69- Eq. mass of carbonate =(69-30)=39 |
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| 76. |
(a) 2 g of metal carbonate were dissolved in 50 mL of N HCl. 100 mL of 0.1 N NaOH were required to neutralise the resultant solution. Calculate the equivalent mass of the metal carbonate. (b) How much water should be added to 75 mL of 3 N HCl ot make it a normal solution ? |
| Answer» Correct Answer - (a) 50, (b) 150 mL | |
| 77. |
An aqueous solution of 6.3 g of oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralise 10 mL of this solution is :A. 40 mLB. 20 mLC. 10 mLD. 4 mL |
| Answer» Correct Answer - A | |
| 78. |
How many mL of `1 M H_(2)SO_(4)` acid solution is required to neutralise 10 mL of 1 M NaOH ?A. 5 mLB. 2.5 mLC. 10 mLD. 20 mL |
| Answer» Correct Answer - A | |
| 79. |
30 mL of `N//10` HCl are required to neutralise 50 mL of a sodium carbonate solution. How many mL of water must be added to 30 mL of this solution so that the solution obtained may have a concentration equal to `N//50` ? |
| Answer» Correct Answer - 60 mL | |
| 80. |
To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorus acid `(H_(3)PO_(3))`, the volume of 0.1 M aqueous KOH solution required is :A. 10 mLB. 20 mLC. 40 mLD. 60 mL |
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Answer» Correct Answer - C `H_(3)PO_(3)+2KOH to K_(2)HPO_(3)+2H_(2)O` `(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))` `(0.1xx20)/(1)=(0.1xx V_(2))/(2)` `V_(2)= 40 mL` |
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| 81. |
If equal volume of `1M KMnO_(4) " and" 1M K_(2)Cr_(2)O_(7)` solutions are allowed to oxidise `Fe^(2+) " to " Fe^(3+)` in acidic medium, then `Fe^(2+)` will be oxidised :A. more by `KMnO_(4)`B. more by `K_(2)Cr_(2)O_(7)`C. equal in both casesD. cannot be determined |
| Answer» Correct Answer - A | |
| 82. |
5.0 g of `H_(2)O_(2)` is present in 100 mL of the solution. The molecular mass of `H_(2)O_(2)` is 34. The molarity of the solution is :A. 1.5 MB. 0.15 MC. 3.0 MD. 50 M |
| Answer» Correct Answer - A | |
| 83. |
The molarity of pure water is :A. 18 MB. 50.0 MC. 55.6 MD. 100 M |
| Answer» Correct Answer - C | |
| 84. |
For a series of indicators the following colours and pH range over which colour change takes place are as follows : Indicator V could be used to find the equivalence point for 0.1 M acetic acid and 0.1 M ammonium hydroxide solution : (a) True (b) False |
| Answer» Correct Answer - B | |
| 85. |
For the half cell reaction, `2BrO_(3)^(-) + 12H^(+)+10e to Br_(26H_(2)O` the equivalent mass of sodium bromate is :A. equal to its mol. MassB. `1//3` of its mol. MassC. `1//6` of its mol. MassD. `1//5` of its mol. Mass |
| Answer» Correct Answer - D | |
| 86. |
The equivalent mass of phosphoric acid `(H_(3)PO_(4))` is 49. It behaves as . . . . Acid.A. monobasicB. dibasicC. tribasicD. tetrabasic |
| Answer» Correct Answer - B | |
| 87. |
Statement-1: Equivalent mass of `H_(3)PO_(2)` is equal to its molecular mass. Statement-2: `H_(3)PO_(2)` is a monobasic acid.A. Statement-1 is true, statement -2 is true, statement-2 is a correct explanation for statement-1.B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1.C. Statement-1 is true, statement-2 is false.D. Statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - A | |
| 88. |
The molecular mass of `H_(3)PO_(3)` is 82. Its equivalent mass, if it is completely neutralised, is:A. 82B. 27.3C. 41D. 246 |
| Answer» Correct Answer - C | |
| 89. |
Equivalent mass of `H_(3)PO_(2)` when it disproportionates into `PH_(3) " and" H_(3)PO_(3)` is (Molecular mass=M) :A. MB. `(M)/(2)`C. `(M)/(4)`D. `(3M)/(4)` |
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Answer» Correct Answer - D `overset(+1)(H_(3)PO_(2))to overset(+3)(H_(3)PO_(3)), " Eq. wt. "=(M)/(2)` (Change in oxidation number=2) `overset(+1)(H_(3)PO_(2) to overset(-3)(PH_(3)), " Eq. wt." = (M)/(4)` (Change in oxidation number=4) The equivalent mass of `H_(3)PO_(2)` in the process of disproportionation`=(M)/(2)+(M)/(4)=(3M)/(4)` |
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| 90. |
A solution contains `Na_(2)CO_(3) " and " NaHCO_(3)`. 10 mL of the solution required 2.5 mL of 0.1 M`H_(2)SO_(4)` for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 mL of `0.2 M H_(2)SO_(4)` was required. Calculate the amount of `Na_(2)CO_(3) " and" NaHCO_(3)` in one litre of the solution. |
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Answer» 2.5 mL of `0.1 M H_(2)SO_(4)=2.5 mL " of" 0.2 N H_(2)SO_(4)` `=(1)/(2)Na_(2)CO_(3)` present in 10 mL of mixture So, `5 mL " of" 0.2 N H_(2)SO_(4)=Na_(2)CO_(3)` present in 10 mL of mixture `-=5 mL " of" 0.2 N Na_(2)CO_(3)` `-=(0.2xx53)/(1000)xx5=0.053 g` Amount of `Na_(2)CO_(3)=(0.053)/(10)xx1000=5.3 g//L` of mixture Between first and second end points, =2.5 mL of `0.2 M H_(2)SO_(4)` used =2.5 mL of `0.4N H_(2)SO_(4)` used =5 mL of `0.2 N H_(2)SO_(4)` used `-=(1)/(2)Na_(2)CO_(3)+NaHCO_(3)` present in 10 mL of mixture `(5-2.5)mL 0.2 N H_(2)SO_(4)` `-=NaHCO_(3)` present in 10 mL of mixture `-=2.5 mL 0.2 N NaHCO_(3)` `-=(0.2xx84)/(1000)xx2.5=0.042 g` Amount of `NaHCO_(3)=(0.042)(10)xx1000=4.20 g//L` of mixture. |
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| 91. |
Hypo is the common name of sodium thiosulphate, with molecular formula `Na_(2)S_(2)O_(3)`. It is used as intermediate in iodometric as well as in iodimetric titrations. Iodine and chlorine react with hypo in different ways as follows : `2na_(2)S_(2)O_(3)+I_(2) to 2NaI+Na_(2)S_(4)O_(6)` `Cl_(2)(g)+S_(2)O_(3)^(2-) to SO_(4)^(2-) + Cl^(-) +S` Suppose, 50 mL of `0.01 M Na_(2)S_(2)O_(3)` solution and `5xx10^(-4)` mol of `Cl_(2)` are allowed to react according to the above equation. Hypo is also used in photography to dissolve AgBr, forming a complex compound. `2Na_(2)S_(2)O_(3) +AgBr to Na_(3)[Ag(S_(2)O_(3))_(2)]+NaBr` The balanced chemical reaction with `Cl_(2)` is :A. `Cl_(2)+2Na_(2)S_(2)O_(3) to 2NaCl+Na_(2)S_(4)O_(6)`B. `Cl_(2)+H_(2)O+Na_(2)S_(2)O_(3) to Na_(2)SO_(4)+2HCl+S`C. `Cl_(2)(g)+S_(2)O_(3)^(2-) to SO_(4)^(2-)+S+Cl^(-)`D. none of the above |
| Answer» Correct Answer - B | |
| 92. |
Hypo is the common name of sodium thiosulphate, with molecular formula `Na_(2)S_(2)O_(3)`. It is used as intermediate in iodometric as well as in iodimetric titrations. Iodine and chlorine react with hypo in different ways as follows : `2na_(2)S_(2)O_(3)+I_(2) to 2NaI+Na_(2)S_(4)O_(6)` `Cl_(2)(g)+S_(2)O_(3)^(2-) to SO_(4)^(2-) + Cl^(-) +S` Suppose, 50 mL of `0.01 M Na_(2)S_(2)O_(3)` solution and `5xx10^(-4)` mol of `Cl_(2)` are allowed to react according to the above equation. Hypo is also used in photography to dissolve AgBr, forming a complex compound. `2Na_(2)S_(2)O_(3) +AgBr to Na_(3)[Ag(S_(2)O_(3))_(2)]+NaBr` Number of moles of `S_(2)O_(3)^(2-)` present in the sample is :A. 0.0005B. 0.01C. 0.0025D. 0.02 |
| Answer» Correct Answer - A | |
| 93. |
The molar concentration of the chloride ion in the solution obtained by mixing 300 mL of 3.0 M NaCl and 200 mL of 4.0 M solution of `BaCl_(2)` is :A. 1.6 MB. 1.8 MC. 5.0 MD. 0.5 M |
| Answer» Correct Answer - C | |
| 94. |
On dissolving 1 mole of each of the following acids in one litre water, the acid which does not give a solution of strength 1 N is :A. HClB. `HClO_(4)`C. `HNO_(3)`D. `H_(3)PO_(4)` |
| Answer» Correct Answer - D | |
| 95. |
Potassium permanganate acts as an oxidising agent in acidic, alkaline as well as neutral media. Which among the following statements is incorrect ?A. `N=M//5` (in acid medium )B. `N=M//3` (in alkaline medium)C. `N=M//3` ( in neutral medium)D. `N=M` ( in alkaline medium) |
| Answer» Correct Answer - B | |
| 96. |
Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is :A. `Cl_(2)O`B. `Cl_(2)O_(7)`C. `ClO_(2)`D. `Cl_(2)O_(6)` |
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Answer» Correct Answer - A Bleaching powder` to CaOCl_(2)` `to Ca^(2+)(Ocl^(-))(Cl^(-))` It is salt of hypochlorous acid HOCl, Anhydride of HOCl is `Cl_(2)O` `2HOCl to Cl_(2)O+H_(2)O` |
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| 97. |
Calculate the percentage of available chlorine in a given sample of bleaching powder from the following data : 3.55 g of bleaching powder when treated with acetic acid and excess of KI liberated iodine which required 60 mL of 0.5 N sodium thiosulphate solution. |
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Answer» `60 mL 0.5 N Na_(2)S_(2)O_(3)-=60 mL 0.5 N I_(2)` `-=60 mL 0.5 N Cl_(2)` Amount of chlorine `=(35.5xx0.5)/(1000)xx60=1.065 g` % of available chlorine `=(1.065)/(3.55)xx100=30.0` |
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| 98. |
0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. `H_(2)SO_(4), I_(2)` was liberated which required 50 mL of `N//10` hypo `(Na_(2)S_(2)O_(3). 5H_(2)O)` in presence of starch. The reactions involved are : I. `CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2)` II. `2KI+Cl_(2) to 2KCl+I_(2)` III. `2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI` Starch forms iodo-starch complex in the given titration. The colour of the complex will be :A. greenB. blueC. pale yellowD. milky white |
| Answer» Correct Answer - B | |
| 99. |
(a) 50 mL of `0.2 N KMnO_(4)` is required for complete oxidation of 0.45 g of anhydrous oxalic acid. Calculate the normality of oxalic acid solution. (b) In the titration of `Fe^(2+)` ions with `KMnO_(4)` in acid medium, why is dilute `H_(2)SO_(4)` used and not dilute HCl ? |
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Answer» Correct Answer - (a) 0.2 N The reaction involved in the titration is : `10FeSO_(4)+2KMnO_(4)+3H_(2)SO_(4) to 5Fe_(2)(SO_(4))_(3)+K_(2)SO_(4)+2MnSO_(4)+3H_(2)O` If HCl is taken in place of `H_(2)SO_(4)`, then HCl will be oxidised to `Cl_(2)` |
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| 100. |
0.25 g of an oxalate salt was dissolved in 100 mL of water. 10 mL of this solution required 8 mL of `N//20 KMnO_(4)` for its oxidation. Calculate the percentage of oxalate in the salt. |
| Answer» Correct Answer - `70.4%` | |