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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.
| 51. |
10g sample of bleaching powder was dissolved into water to make the solution one litre. To this solution 35 mL of 1.0 M Mohr salt solution was added containing enough `H_(2)SO_(4)`. After the reaction was complete, the excess Mohr salt required 30 mL of 0.1 M `KMnO_(4)` for oxidation. The % of availabel `Cl_(2)` approximately is (mol wt = 71) |
| Answer» Correct Answer - 0.071 | |
| 52. |
`20ml ofH_(2)O_(2)` after acidification with dilute `H_2SO_(4)`required 30mlof `(N)/(12)KMnO_4` for complete oxidation. The approximate strength of `H_(2)O_(2)` solution (ing/L)is : [Molar mass of `H_2O_2=34`] |
| Answer» Correct Answer - 2.12g/L | |
| 53. |
`x` gram of pure `As_(2)S_(3)` is completely oxididsed to respective highest oxidation states by 50 ml of 0.1 M hot acidified `KMnO_(4)`, then mass of `As_(2)S_(3)` taken is :A. 22.4 gB. 43.92 gC. 64.23 gD. None of these |
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Answer» Correct Answer - D `5As_(2)S_(3)+28KMnO_(4)+H^(+)rarr10H_(3)AsO_(4)+28Mn^(2+)+SO_(4)^(2-)` m.moles of `KMnO_(4)=50xx0.1=5` 28 m.moles of `KMnO_(4)rarr5` m.moles of `As_(2)S_(3)` `:. ` 1 m.mole of `KMnO_(4)rarr` 5/28 m.moles of `As_(2)S_(3)` `:. ` 5 m.mole of `KMnO_(4)rarr(5xx5)/(28)` m.moles of `As_(2)S_(3)` Mass of `As_(2)S_(3)=x=246xx(5xx5)/(28)xx10^(-3)=0.2.2g` |
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| 54. |
25 mL of household bleach solution was mixed with 30 mL of o.50 M Kl and 10 mL of 4N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N `Na_(2)S_(2)O_(3)` was used to reach the end point. The molarity of the household bleach solution is :A. 0.48 MB. 0.96 MC. 0.24 MD. 0.024 M |
| Answer» Correct Answer - C | |
| 55. |
A mixture containing `As_(2)O_(3)` and `As_(2)O_(5)` required 20 mL of 0.05 N iodine solution for titration. The resulting solution is then acidified and excess of Kl was added. The liberated iodine required 1.116 g hypo `(Na_(2)S_(2)O_(3).5H_(2)O)` for complete reaction. Calculate the mass of the mixture. The reactions are : `As_(2)O_(3)+2l_(2)+2H_(2)OrarrAs_(2)O_(5)+4H^(+)+4l^(-)` `As_(2)O_(5)+4H^(+)+4l^(-)rarrAs_(2)O_(3)+2l_(2)+2H_(2)O " " `(Atomic weight : As = 75) |
| Answer» Correct Answer - 0.025075 g | |
| 56. |
0.7g of `(NH_(4))_(2)SO_(4)` sample was boiled with 100mL of 0.2 N NaOH solution was diluted to 250 ml. 25mL of this solution was neutralised using 10mL of a 0.1 N `H_(2)SO_(4)` solution. The percentage purity of the `(NH_(4))_(2)SO_(4)`sample is :A. 94.3B. 50.8C. 47.4D. 79.8 |
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Answer» Correct Answer - A m.eq of `(NH_(4))_(2)SO_(4)+` m.eq of `H_(2)SO_(4)`=m.eq of NaOH `("m.molea"xx2)++(0.1xx10xx(250)/(25))=0.2xx100` `:. " m.mole of " (NH_(4))_(2)SO_(4)=5` wt. of `(NH_(4))_(2)SO_(4)=(5)/(1000)xx132=0.66 g` `:. " % of " (NH_(4))_(2)SO_(4)=(0.66)/(0.7)xx100=94.28% ~~94.3%` |
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| 57. |
Which of the following relations is/are correct for solutions ?A. 3 N `Al_(2)(SO_(4))_(3)=0.5M Al_(2)(SO_(4))_(3)`B. `3 N H_(2)SO_(4)=6 N H_(2)SO_(4)`C. `1 M H_(3)PO_(4)=1//3 N H_(3)PO_(4)`D. `1 M Al_(2)(SO_(4))_(3)=6 N Al(SO_(4))_(3)` |
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Answer» Correct Answer - A::B::D Normality = Molarity `xx` v.f. `" " :. 1M H_(3)PO_(4)=3N H_(3)PO_(4)` |
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| 58. |
Consider the redox reaction `2S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I^(ө)`A. `S_(2)O_(3)^(2-)` gets reduced to `S_(4)O_(6)^(2-)`B. `S_(2)O_(3)^(2-0)` gets oxidised to `S_(4)O_(6)^(2-)`C. `I_(2)` gets reduced to `I^(-)`D. `I_(2)` gets oxidised to `I^(-)` |
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Answer» Correct Answer - B::C S undergoes increase in oxidation number from +2 to +2.5, while I undergoes decrease in oxidation number from 0 to -1. |
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| 59. |
`{:(2NaOH+H_(2)SO_(4)rarrNa_(2)SO_(4)+2H_(2)O),("Base Acid"):}` |
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Answer» Valency factor of base = 1 Here, two molecule of NaOH replaced `2H^(+)` ion from the `H_(2) SO_(4)`. Therefore, each molecule of NaOH replaced only one `H^(+)` ion of acid, so v.f = 1. |
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| 60. |
An ion is reduced to the element when it absords `6xx10^(20)` electrons. The number of equivalents of the ion is:A. 0.1B. 0.01C. 0.001D. 0.0001 |
| Answer» Correct Answer - C | |
| 61. |
Find the valency factor (n) for `NH_(2)OH` in given reaction : `Fe^(3+)+NH_(2)OHrarrFe^(2+)+N_(2)O+H^(+)+H_(2)O` |
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Answer» Correct Answer - 2 `{:(NH_(2)OHrarrN_(2)O),(" (-1) (+1) oxidation number of nitrogen"):}` `:. ` Vf=change in oxidation number of nitrogen = 2. |
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| 62. |
Write the balanced redox reaction and calculate the equivalent weight of oxidising agent and reducing agent for titration of `K_(2)Cr_(2)O_(7)` Vs Ferrous ammonium sulphate. |
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Answer» The reaction : `6[FeSO_(4)(NH_(4))_(2)SO_(4).6H_(2)O]+K_(2)Cr_(2)O_(7)+7H_(2)SO_(4)rarr3Fe_(2)(SO_(4))_(3)+K_(2)SO_(4)+6(NH_(4))_(2)SO_(4)+43H_(2)O` Redox changes : `(E_(FeSO_(4))=(M)/(6)) , (E_(K_(2)Cr_(2)O_(7))=(M)/(6))` |
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| 63. |
What volume of 0.05 M `Ca(OH)_(2)` solution is needed for complete conversion of 10mL of 0.1 M `H_(3)PO_(4)` into `Ca(H_(2)PO_(4))_(2)` ? |
| Answer» Correct Answer - 10 ML | |
| 64. |
When `N_(2)` is converted into `NH_(3)`, the equivalent weight of nitrogen will be:A. 1.67B. 2.67C. 3.67D. 4.67 |
| Answer» Correct Answer - D | |
| 65. |
1 mole of `OH^(-)` ions is obtained from 85 g of hydroxide of a metal. What is the equivalent weight of the metal ? |
| Answer» Correct Answer - 68 | |
| 66. |
When `HNO_(3)` is converted into `NH_(3)`, the equivalent weight of `HNO_(3)` will be:A. M/2B. M/1C. M/6D. M/8 |
| Answer» Correct Answer - D | |
| 67. |
the hardness of a water sample (in terms of Equivalents of `CaCO_(3)`) containing `10^(-3)MCaSO_(4)` Is : (Molar mass of `CaSO_(4)=1365h mol^(-1)`)A. 10 ppmB. 50 ppmC. 90 ppmD. 100 ppm |
| Answer» Correct Answer - D | |
| 68. |
The amount of lime, `Ca(OH)_(2)` required to remove the hardness in 60 L of pond water containing 1.62 mg of calcium bicarbonate per 100 ml of water, will be :A. `4.44g`B. `0.222g`C. `2.22 g`D. `0.444 g` |
| Answer» Correct Answer - D | |
| 69. |
0.00012% `MgSO_(4)` and `0.000111% CaCl_(2)` is present in water. What is the measured hardness of water and millimoles of washing soda required to purigy water 1000 L water ? |
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Answer» Basis of calculation = 100 g hard water `MgSO_(4) = 0.00012g = (0.00012)/(120) ` mole `CaCl_(2)=0.00012 g = (0.000111)/(111) ` mole `:. ` equivalent moles of `CaCO_(3)=((0.00012)/(120)+(0.000111)/(111))` mole `:.` mass of `CaCO_(3)=((0.00012)/(120)+(0.000111)/(111))xx100=2xx10^(-4)g` Hardness (in terms of ppm of `CaCO_(3)`)=`(2xx10^(-4))/(100)xx10^(6)=2`ppm `CaCl_(2)+Na_(2)CO_(3)rarrCaCO_(3)+2NaCl` `NaSO_(4)+Na_(2)CO_(3)rarrMgCO_(3)+Na_(2)SO_(4)` `:.` Required `Na_(2)CO_(3)` for 100g of water `=((0.00012)/(120)+(0.000111)/(111))` mole `=2xx10^(-6)` mole `:.` Required `Na_(2)CO_(3)` for 1000 litre water `=(2xx10^(-6))/(100)xx(2)/(100)` mole (`:.` d=1g/mL) `=(20)/(1000)` mole = 20 m mole |
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| 70. |
In the conversion `NH_(2)OH rarr N_(2)O`, the equivalent weight of `NH_(2)OH` will be:A. M/4B. M/2C. M/5D. M/1 |
| Answer» Correct Answer - B | |
| 71. |
Which of the following is not a redox reaction ?A. `KCN+Fe(CN)_(2)rarrK_(4)[Fe(CN)_(6)]`B. `Rb+H_(2)OrarrRbOH+H_(2)`C. `H_(2)O_(2)rarrH_(2)O+O`D. `CuI_(2)rarrCuI+I_(2)` |
| Answer» Correct Answer - A | |
| 72. |
Number of moles of CaO required to remove hardness from 1000 litre water having 324 ppm of calcuim bicarbonate and 74.5 ppm of potassium chloride is :A. 8B. 4C. 3D. 2 |
| Answer» Correct Answer - D | |