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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. |
The colour of `CuCr_(2)O_(7)` solutioin is water is green becauseA. `Cr_(2)O_(7)^(2-)` ions are greenB. `Cu^(++)` ions are greenC. Both ions are greenD. `Cu^(++)` ions are blue and `Cr_(2)O_(7)^(2-)` ions are yellow |
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Answer» Correct Answer - D `Cu^(++)` ions are blue and `Cr_(2)O_(7)^(-)` ions are yellow, yellow and blue combination gives-green colour. |
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| 52. |
15 ml of 0.2 N alkali is required to complete neutralization of 30 ml acid solution. Concentration of the acid solution isA. 0.1 NB. 0.3 NC. 0.15 ND. 0.4 N |
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Answer» Correct Answer - A `N_(1)V_(1)=N_(2)V_(2),15xx0.2=30xx x` `thereforex=0.1N` |
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| 53. |
`Pb(CH_(3)COO)_(2)` gives . . . Colour with `H_(2)S`A. OrangeB. RedC. BlackD. White |
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Answer» Correct Answer - C `(CH_(3)COO)_(2),Pb+H_(2)S to 2CH_(3)COOH+underset("black ppt.")(PbS) darr` |
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| 54. |
If `Na^(+)` ion and `S^(2-)` ion is larger than `Cl^(-)` ion, which of the following will b least soluble in waterA. `MgS`B. `NaCl`C. `Na_(2)S`D. `MgCl_(2)` |
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Answer» Correct Answer - A `underset("Solubility decreasing order")overset(NaCl gt Mg Cl_(2) gt Na_(2)S gt MgS)to` |
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| 55. |
The equivalent mass of potassium permanganate in alkaline medium isA. `("Molar mass")/(5)`B. `("Molar mass")/(3)`C. `("Molar mass")/(2)`D. Molar mass itself |
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Answer» Correct Answer - B `2KMnO_(4)+H_(2)O overset("alkaline")to2MnO_(2)+2KOH+3[O]` So, equivalent mass`=(Mol." "mass)/(3)` |
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| 56. |
0.45 g an acid (mol wt.=90) required 20 ml of 0.5 N KOH for complete neutralization. Basicity of acid isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B Normality`=N=(W_(B)xx1000)/(Eq." "wtxxV)` `therefore"Eq. Wt"=(0.45xx1000)/(0.5xx20)=45` `therefore"Basicity"=("Molec. Wt")/("Eq. wt")=(90)/(45)=2` |
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| 57. |
`MgSO_(4)` on reaction with `NH_(4)OH` and `Na_(2)HPO_(4)` forms a white crystalline precipitate. What is its formulaA. `Mg(NH_(4))PO_(4)`B. `Mg_(3)(PO_(4))_(2)`C. `MgCl_(2),MgSO_(4)`D. `MgSO_(4)` |
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Answer» Correct Answer - A Test of `Mg^(+2)` ion `Mg^(+2)+NH_(4)OH+Na_(2)HPO_(4)toMg(NH_(4))PO_(4)`. |
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| 58. |
A 0.3 M HCl solution contains the following ions `Hg^(++),Cd^(++),Sr^(++),Fe^(++)`. The addition of `H_(2)S` to above solution will precipitateA. Cd, Cu and HgB. Cd, Fe and SrC. Hg, Cu and FeD. Cu, Sr and Fe |
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Answer» Correct Answer - A The second group radicals will precipitate because their solubility product is very low so sulphates will be precipitated. |
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| 59. |
Addition of solution of oxalate to an aqueous solution of mixture of `Ba^(++),Sr^(++)` and `Ca^(++)` will precipitateA. `Ca^(++)`B. `Ca^(++) and Sr^(++)`C. `Ba^(++) and Sr^(++)`D. All the three |
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Answer» Correct Answer - D Oxalate of these metals are insoluble. |
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| 60. |
Which of the following precipitate does not dissolve even in large excess of `NH_(4)OH`A. AgClB. AgBrC. AgID. None of these |
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Answer» Correct Answer - C AgCl and AgBr dissolve in `NH_(4)OH` and form complexes but AgI doesn’t react with `NH_(4)OH`. E.g., `AgCl+2NH_(4)OH to underset("Complex")([Ag(NH_(3))_(2)])Cl+2H_(2)O` |
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| 61. |
To a solution of substance, gradual addition of ammonium hydroxide results in a black precipitate which does not dissolve in excess of `NH_(4)OH`. However, when HCl is added to the original solution, a white precipitate is formed. The solution containedA. Lead saltB. Silver saltC. Mercurous saltD. Copper salt |
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Answer» Correct Answer - C `Hg_(2)Cl_(2)` is white insoluble salt. `Hg_(2)Cl_(2)+2NH_(4)OH to Hg+HGunderset("Black")ubrace((NH_(2))Cl+NH_(4)Cl)+2H_(2)O` |
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| 62. |
Out of `Cr^(2+),Ni^(2+),Co^(2+) and Mn^(2+)` those dissolve in dil. HCl, only one gives a precipitate when `H_(2)S` is passed. Identify the corresponding oneA. `Ni^(2+)`B. `Cu^(2+)`C. `Co^(2+)`D. `Mn^(2+)` |
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Answer» Correct Answer - B Dil. HCl and `H_(2)S` are group-II reagents. `Cu^(2+)` is a group-II ion (in qualitative analysis). |
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| 63. |
A precipitate of calcium oxalate will not dissolve inA. HClB. `HNO_(3)`C. Aqua-regiaD. Acetic acid |
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Answer» Correct Answer - D Calcium oxalate will not dissolve in acetic acid (Weak acid) but dissolve only in strong acid. |
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| 64. |
On performing a borax-bead test with a given inorganic mixture for qualitative analysis, the colour of the bead was found to emerald green both in oxidising and reducing flame. It indicates the possibility of the presence ofA. `Co^(+2)`B. `Ni^(+2)`C. `Cr^(+3)`D. `Cu^(+2)` |
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Answer» Correct Answer - C Chromium ion gives in hot and cold oxidising and reducing gren-colour flame. |
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| 65. |
Which gives violet compourd bead in borax bead testA. `Fe^(2+)`B. `Ni^(2+)`C. `Co^(2+)`D. `Mn^(2+)` |
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Answer» Correct Answer - D As `Mn^(+2)` has all its electrons (5) unpaired in its d-orbital, so its extra stable configuration requires high excitation energy and so it gives violet colour. |
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| 66. |
In borax bead test, which of the following compound is formedA. Meta borateTetra borateB. Double oxideC. Ortho borateD. |
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Answer» Correct Answer - A `Na_(2)B_(4)O_(7).10H_(2)O underset(-10H_(2)O)toNa_(2)B_(4)O_(7) overset(Delta)to2NaBO_(2)+B_(2)O_(3)` `CuO+B_(2)O_(3)to underset(("Copper metaborate blue"))(Cu(BO_(2))_(2)` |
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| 67. |
The metal that does not give the borax bead test isA. CrB. NiC. NaD. Mn |
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Answer» Correct Answer - C Metals which form basic compoounds and have coloured salts give the borax bead test. |
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| 68. |
The metal that does not give the borax-bead test isA. ChromiumB. NickelC. LeadD. manganese |
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Answer» Correct Answer - C Borax bead test is generally given by transition elements. |
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| 69. |
The alkaline earth metal that imparts apple green colour to the bunsen flame when introduced in it ini the form of its chloride isA. BariumB. StrontiumC. CalciumD. Magnesium |
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Answer» Correct Answer - A `Ba^(2+)` imparts green colour to the flame. |
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| 70. |
To an inorganic mixture dil. `H_(2)SO_(4)` is added in cold, colourless, odourless gas is evolved. The mixture containsA. SulphiteB. AcetateC. NitriteD. Carbonate |
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Answer» Correct Answer - D Sulphide radical will give a burning sulphur odour. Acetate radical will give a sweet, vinegar odour. Nitrite will give a reddish brown gas. Carbonate will give a colourless, odourless gas, i.e., `CO_(2)`. |
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| 71. |
the gas which is absorbed by ferrous sulphate solution giviing blackish brown colour isA. NOB. COC. `N_(2)`D. `NH_(2)` |
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Answer» Correct Answer - A `3FeSO_(4)+NO_(2)+3H_(2)SO_(4)toFe_(2)(SO_(4))_(3)+underset(("Black brown"))(FeSO_(4).NO)+H_(2)O` |
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| 72. |
0.16 gm of a dibasic acid require 25 ml of decinormal NaOH solution for complete neutralization. The moleucular weight of the acidA. 32B. 64C. 128D. 256 |
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Answer» Correct Answer - C `(0.16)/(x)xx(1000)/(25)=(1)/(10)implies(6.4)/(x)=(1)/(10) thereforex=64` Mol. Wt.=`64xx2=128`. |
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| 73. |
Ferric ion forms a prussiann blue coloured ppt. due toA. `K_(4)Fe(CN)_(6)`B. `Fe_(4)[Fe(CN)_(6)]_(3)`C. `KMnO_(4)`D. `Fe(OH)_(3)` |
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Answer» Correct Answer - B `4FeCl_(3)++3K_(4)[Fe(CN)_(6)]tounderset("Prussian Blue colour")(Fe_(4)[Fe(CN)_(6)])+12KCl` |
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| 74. |
When `H_(2)S` gas is passed into a certain solution, it reacts to form a white precipitate. The solution referred to contains ions ofA. LeadB. ZincC. CopperD. Nickel |
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Answer» Correct Answer - B ZnS is white. |
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| 75. |
When `H_(2)S` gas is passed in a metal sulphate solution in presence of `NH_(4)OH`, a white precipitate is produced. The metal is identified asA. ZnB. FeC. PbD. Hg |
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Answer» Correct Answer - A Zinc is that metal, which gives white precipitate when `H_(2)S` gas is passed in the ammoniacal solution of its salt. |
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| 76. |
Which of the following sulphate is insoluble in waterA. `CuSO_(4)`B. `CdSO_(4)`C. `PbSO_(4)`D. `Bi(SO_(4))_(3)` |
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Answer» Correct Answer - D As `Bi(SO_(4))_(3)` is a covalent compound with high lattice energy and hence it is insoluble in water. |
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| 77. |
Which of the following mixture is chromic acidA. `K_(2)Cr_(2)O_(7) and HCl`B. `K_(2)SO_(4) and `conc. `H_(2)SO_(4)`C. `H_(2)SO_(4) and HCl`D. |
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Answer» Correct Answer - C (Conc. `H_(2)SO_(4)+K_(2)Cr_(2)O_(7)`) mixture is called as chromic acid. |
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| 78. |
A mixture of two salts is not water soluble but dissolves completely in dilute hydrochloric acid to form a colourless solution. The mixture could beA. `AgNO_(3) and KBr`B. `BaCO_(3) and ZnS`C. `FeCl_(3) and CaCO_(3)`D. `Mn(NO_(3))_(2) and MgSO_(4)` |
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Answer» Correct Answer - B `BaCO_(3)+2HCl to BaCl_(2)+H_(2)O+CO_(2)` `ZnS+underset((dil.))(2HCl)toZnCl_(2)+H_(2)S` |
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| 79. |
An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with `H_(2)S` in a dilute mineral acid medium. however, it gave a precipitate (R) with `H_(2)S` in an ammonical medium. the precipitate R gave a coloured solution (S), when treated with `H_(2)O_(2)` in an aqueous NaOH medium. Q. The precipitate P containsA. `Pb^(2+)`B. `Hg_(2)^(2+)`C. `Ag^(+)`D. `Hg^(2+)` |
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Answer» Correct Answer - A `Pb^(2+)+2Cl^(-) to PbCl_2`(white ppt) |
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| 80. |
The aqueous solution of the following salts will be coloured in the case ofA. `Zn(NO_(3))_(2)`B. `LiNO_(3)`C. `CrCl_(3)`D. Potash alum |
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Answer» Correct Answer - C `CrCl_(3)`, as `Cr^(+3) implies3d^(3)`, has unpaired electron in d-orbital as a result it will show paramagnetism and thus forms coloured complexes. |
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| 81. |
Which reagent is used to remove `SO_(4)^(2-)` and `Cl^(-)`A. `BaSO_(4)`B. `NaOH`C. `Pb(NO_(3))_(2)`D. `KOH` |
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Answer» Correct Answer - C `PbSO_(4)` and `PbCl_(2)` are insoluble in cold water hence the reagent `Pb(NO_(3))_(2)` is used to remove `SO_(4)^(2-) and Cl^(-)`. |
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| 82. |
Te reagents `NH_(4)Cl` and aqueous `NH_(3)` will precipitateA. `Ca^(2+)`B. `Al^(+3)`C. `Mg^(2+)`D. `Zn^(2+)` |
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Answer» Correct Answer - B `NH_(3)+H_(2)OtoNH_(4)OH` due to common ion effect `NH_(4)^(+)` ion concentration increases which leads to the precipitation of `Al(OH)_(3)`. |
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| 83. |
What product is formed by mixing the solution of `K_(4)[Fe(CN)_(6)]` with the solution of `FeCl_(3)`A. Ferro-ferricyanideB. Ferri-ferrocyanideC. Ferri-ferricyanideD. None of these |
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Answer» Correct Answer - B `Fe^(3+)+K_(4)[Fe(CN)_(6)]toKunderset("Prussian blue")([Fe[Fe(CN)_(6)]])+3K^(+)` |
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| 84. |
The reagent, `NH_(4)Cl` and aqueous `NH_(3)` will precipitateA. `Ca^(2+)`B. `Al^(3+)`C. `Bi^(3+)`D. `Mg^(2+)` |
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Answer» Correct Answer - A::B `Al^(3+)` (thrid group radical) and `Ca^(2+)` (fifth grou radical) precipitate out as their hydroxide with `NH_(4)Cl` and aq. `NH_(3)(NH_(4)OH)` which are the group reagents. |
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| 85. |
Which of the following change the colour of the aqueous solution of `FeCl_(3)`A. `K_(4)[Fe(CN)_(6)]`B. `H_(2)S`C. `NH_(4)CNS`D. `KCNS` |
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Answer» Correct Answer - A::B::C::D `FeCl_(3)+K_(4)[Fe(CN)_(6)]to underset("Ferri ferrocyanide (blue)")(Fe_(4)[Fe(CN)_(6)]_(3))` `2FeCl_(3)+3H_(2)S to Fe_(2)S_(3)+6HCl` `3NH_(3)CNS+FeCl_(3)to underset(("blood red"))(Fe(CNS)_(3))+3NH_(4)Cl` `FeCl_(3)+3KCNSto underset(("Blood red"))(Fe(CNS)_(3))+3KCl` |
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| 86. |
A black sulphide is formed by the action of `H_(2)S` onA. Cupric chlorideB. Cadmium chlorideC. Zinc chlorideD. Sodium chloride |
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Answer» Correct Answer - A `CuCl_(2)+H_(2)S to underset(("black ppt."))(Cus)+2HCl` |
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| 87. |
Which of the following statement (s) is (are) correct when a mixture of NaCl and `K_(2)Cr_(2)O_(7)` is gently warmed with conc. `H_(2)SO_(4)`A. A deep red vapour is evolvedB. The vapours when passed into NaOH solution gives a yellow solution of `Na_(2)CrO_(4)`C. Chlorinegas is evolvedD. Chromyl chloride is formed |
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Answer» Correct Answer - A::B::D The reactions are `4NaCl+K_(2)Cr_(7)O_(7)=6H_(2)SO_(4)to2CrO_(2)Cl_(2)+4Naunderset(("Red vapours"))(HSO_(4)+2KH)SO_(4)+3H_(2)O,underset("Chromyl chloride")(CrO_(2)Cl_(2))+4NaOH to underset("yellow solution")(Na_(2)CrO_(4))+2NaCl+2H_(2)O` |
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| 88. |
The best explanation for the solubility of MnSin dil. HCl is thatA. Solubility product of `MnCl_(2)` is less than that of MnSB. Concentration of `Mn^(2+)` is lowered by the formation of complex ions with chloride ionsC. Concentration of sulphide ions is lowered by oxidation to free sulphurD. Concentration of sulphide ions is lowered by formation of the weak acid `H_(2)S` |
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Answer» Correct Answer - D It is a characteristic property of IV group, concentration of sulphide ions is lowered by formation of the weak acid `H_(2)S` . |
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| 89. |
10 ml of concentrated HCl were diluted to 1 litre. 20 ml of this diluted solution required 25 ml of 0.1 N sodium hydroxide solution for complete neutralization, the normality of the concentrated acid will beA. 8B. 9.5C. 12.5D. 15 |
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Answer» Correct Answer - C `{:(HCl,NaOH),(N_(1)V_(1)=,N_(2)V_(2)):}` `N_(1)xx20ml=0.1xx25impliesN_(1)=(0.1xx25)/(20)=0.125` If one litre HCl present in 0.125 Therefore in 10 ml `=(0.125)/(1000)xx100=12.5` The normality of conc. `HCl` is 12.5 N. |
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| 90. |
Which mixture is separated by conc. Aqueous solution of sodium hydroxide:-A. `Al^(3+) and Sn^(2+)`B. `Al^(3+) and Fe^(3+)`C. `Al^(3+) and Zn^(2+)`D. `Zn^(2+) and Pb^(2+)` |
| Answer» Correct Answer - B | |
| 91. |
Excess of carbon dioxide is passed through 50 ml of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate was completely neutralised with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At. mass of calcium=40)A. `200cm^(3)`B. `500cm^(3)`C. `400cm^(3)`D. `300cm^(3)` |
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Answer» Correct Answer - B According to the question, the reaction occurs as `underset(74(=1mol))(Ca(OH)_(2))+underset(44(=1mol))(CO_(2))to underset(100(=1mol))(CaCO_(3))+underset(18(=1mol))(H_(2)O)` Given, 50 ml of 0.5 M `Ca(OH)_(2)` reacts with excess of `CO_(2)` `therefore` No. of millimoles of `Ca(OH)_(2)` reacts with excess of `CO_(2)` `therefore` No. of millimoles of `Ca(HO)_(2)` reacted=25 `therefore` 1 mole of `Ca(OH)_(2)` gives 1 mole of `CaCO_(3)`, `therefore`No. of millieq. `=("Weight in mg")/("eq. Wt")=(25xx100)/(50)=50` `implies`No. of milliequivalents of `CaCO_(3)=50` As, volume of `CaCO_(3)` solution=50 ml So, normality of `CaCO_(3)` solution=1N (millieq.=`NxxV` in ml) Normality of HCl=0.1N (given) Volume of HCl=? `N_(HCl)xxV_(HCl)=N_(CaCO_(3))xxV_(CaCO_(3))` `implies0.1xxV_(HCl)=1xx50impliesV_(HCl)=(50)/(0.1)=500cm^(3)`. |
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| 92. |
When 100 ml 1N-NaOH solution and 10 ml of 10 N sulphuric acid solution are mixed together, the resulting solution will beA. AlkaliB. Weakly acidicC. Strongly acidicD. Neutral |
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Answer» Correct Answer - D If `N_(1)V_(1)=N_(2)V_(2)` then the solution will be neutral `therefore1xx100=10xx10` `100=100implies` solution in neutral. |
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| 93. |
100 `cm^(3)` of 0.1 N HCl solution is mixed with 100 `cm^(3)` of 0.2 N NaOH solution. The resulting solution isA. 0.1 N and the solution is basicB. 0.05 N and the solution is basicC. 0.1 N and the solution is acidicD. 0.05 N ad the solution is acidic |
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Answer» Correct Answer - B Acid=0.1`xx100=10` base`=0.2xx100=20` `therefore` Solution will be basic. 0.1 N of HCl is neutralised by 0.1 N of NaOH and the remaining 0.1 N of NaOH is in 200 `cm^(3)` of solution. `therefore` Resulting normality`=0.5`N, basic. |
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| 94. |
If 20 ml of 0.25N strong acid and 30 ml of 0.2 N of strong base are mixed, then the resulting solution isA. 0.25 N basicB. 0.2 N acidicC. 0.25 N acidicD. 0.2 N basic |
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Answer» Correct Answer - D `V_(1)=20ml,N_(1)=0.25N,V_(2)=30ml,N_(2)=0.2N` `thereforeN=(N_(1)V_(1)+N_(2)V_(2))/(V_(1)+V_(2))=(20(0.25)+30(0.2))/(20+30)` `=(5+6)/(50)=(11)/(50)=0.2N` basic. |
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| 95. |
What is the molarity of `H_(2)SO_(4)` solution if 25 ml of exactly neutralised with 32.63 ml of 0.164M, NaoHA. 0.107 MB. 0.126 MC. 0.214 MD. `-0.428M` |
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Answer» Correct Answer - A 0.164 M NaOH`-=`0.164 N NaOH We know, `N_(1)V_(1)=N_(2)V_(2),N_(1)xx25=0.164xx32.63` 0.214N `H_(2)SO_(4)-=(0.214)/(2)"M "H_(2)SO_(4)` (`because"basicity of "H_(2)SO_(4)` is 2) `-=0.107" M "H_(2)SO_(4)` |
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| 96. |
50 ml 10N-`H_(2)SO_(4),25ml12N-HCl and 40 ml" "5N-HNO_(3)` were mixed together and the volume of the mixture wasmade 1000 ml by adding water. The normality of the resultant solution will beA. 1NB. 2NC. 3ND. 4N |
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Answer» Correct Answer - A `H_(2)SO_(4)" HCl "HNO_(3)` Total volume `N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=Nxx1000ml` `N=(N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3))/(1000)` `=(50xx10+25xx12+40xx5)/(1000)` `N=(500+300+200)/(1000)=(1000)/(1000)=1N` |
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| 97. |
Ozone when reacts with potassium iodide solution liberates certain product, which turns starch paper blue. The liberated substance is . . . .A. OxygenB. IodineC. Hydrogen ioidideD. Potassium hydroxide |
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Answer» Correct Answer - B Ozone react with KI solution to liberate `I_(2)` gas. |
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| 98. |
To a 25 ml of `H_(2)O_(2)` solution, excess of acidified solution of KI was mixed. The liberated `I_(2)` require 20 ml of 0.3 M hypo solution for neutralization. The volume strength of `H_(2)O_(2)` will beA. 1.34 mlB. 1.44 mlC. 1.60 mlD. 2.42 ml |
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Answer» Correct Answer - A 20 ml of 0.3N `Na_(2)S_(2)O_(3)` =20ml of 0.3 N `I_(2)` solution =20ml of 0.3N `H_(2)O_(2)` solution =25ml of 0.08N `H_(2)O_(2)` solution Mass of 100 ml `H_(2)O_(2)` solution Mass of 100 ml `H_(2)O_(2)` evolve oxygen at N.T.P.=22400 ml 0.00136 gm `H_(2)O_(2)` evolve oxygen at N.T.P. `=(22400)/(68)xx0.00136=0.448` For 0.1N, the solution is of 0.448 volume. `therefore3N`, volume =0.448`xx3=1.344ml`. |
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| 99. |
`MnO_(2) and H_(2)SO_(4)` added to NaCl, the greenish yellow gas liberated isA. `Cl_(2)`B. `NH_(3)`C. `N_(2)`D. `H_(2)` |
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Answer» Correct Answer - A Yellowish-green gas (chlorine) with suffocating odour is evolved when the sodium chloride mixed with manganese dioxide is heated with concentrated `H_(2)SO_(4)`. `NaCl+H_(2)SO_(4)toNaHSO_(4)+HCl` `MnO_(2)+4HCl toMnCl_(2)+2H_(2)O+Cl_(2)` |
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| 100. |
When `H_(2)S` gas is passed through the HCl containing aqueous solutions of `CuCl_(2),HgCl_(2),BiCl_(3) and CoCl_(2)`, which does not precipiate outA. `CuS`B. `HgS`C. `Bi_(2)S_(3)`D. `CoS` |
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Answer» Correct Answer - D Cobalt sulphide does not ppt. out second group. |
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