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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. |
Law of combining volumes was proposed byA. LavoisierB. Gay LussacC. AvogadroD. Dalton |
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Answer» Correct Answer - B Gay Lussac |
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| 52. |
Which of the following will produce a buffer sollution when mixed in equal volumes ?A. `Co(g)` and `O_(2)(g)` taken in molar ratio 2:1B. 10 ml of `CH_(4)(g)` and `30ml` of `O_(2)(g)`C. `N_(2)(g)` and `H_(2)(g)` taken in a molar ratio 3:1D. `CH_(4)(g)` and `O_(2)(g)` taken in a molar ratio 1:6 |
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Answer» Correct Answer - A::B (A) `{:(CO(g)+,(1)/(2)O_(2)(g)rarr,CO_(2)(g)),(2,1,),(0,0,2):}` % change in mole `=(1)/(3) xx 100` (B) `{:(CH_(4)(g)+,2O_(2)(g)rarrCO_(2)(g)+2H_(2)(l)),(10ml,30ml):}` Volume contraction =20ml % change in volume `=(20)/(40) xx 100` (C) `{:(N_(2)(g)+,3H_(2)(g)rarr,2NH_(3)(g)),(3mol,,1mol),([3=(1)/(4)]mol,L.R,2//3mol):}` % change `=((2)/(3))/(4) xx 100` (D) `{:(CH_(4)(g)+,2O_(2)(g)rarr,CO_(2)(g)+2H_(2)O(l)),(1mol,6mol,),(L.R,4mol,1mol):}` % change in mol `=(2)/(7) xx 100` |
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| 53. |
`NaOH` and `Na_(2)CO_(3)` are dissolvedin 200 ml aqueous solution. In the presence of phenolphthalein indicator. 17.5 ml of `0.1N HCl` are used to titrate this solution. Now methyl orange is added in the same solution titrated and it requires 2.5ml of the same `HCl`. Calculate the normality of `NaOH` and `Na_(2)CO_(3)` and their mass present in the solution. |
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Answer» Milli equivalents (a) of `HCl` used in thepresence of phenolphthalein indicator. `=N xx V`(ml) `= 0.1 xx 17.5 = 1.75` `1.75` (a) = milli eq of `NaOH +1//2` milli eq of `Na_(2)CO_(3)` ..(1) Milli eq (b) of `HCl` used in the presence of methyl orange indicator `= N xx V` (ml) `= 0.1 xx 2.5 = 0.25` `0.25` (b) `= 1//2` milli equivalents of `Na_(2)CO_(3)` .......(2) For `Na_(2)CO_(3)` solution: From equation (2) Milli eq of acid used by `Na(2)CO_(3) = 2b` `= 2 xx 0.25 = 0.5` volume of `Na_(2)CO_(3)` solution =`200ml` Suppose, Normality of `Na_(2)CO_(3) = N` Milli equivalents of `Na_(2)CO_(3) = N xx V` (ml) `= 200` Putting equivalents of acid and `Na_(2)CO_(3)` equal `200N = 0.5` or (Normality of `Na_(2)CO_(3)` solution) `N = (1)/(400)` Mass of `Na_(2)CO_(3) = N xx E xx V` (litre) (E for `Na_(2)CO_(3) = 53) = 0.0265` gram For `NaOH` solution: From equation (1) and (2) Milli eq acid used by `NaOH = a-b = 1.75 - 0.25 = 1.50` Volume of `NaOH` solution `=200ml` Suppose, Normality of `NaOH` solution = N Milli eq of `NaOH = N xx V` (ml) `= 200N` Putting the mili eq of `NaOH` and acid used equal `200 N = 1.5` (Normality of `NaOH` solution) `N = (1.5)/(200)` Mass of `NaOH = N xx E xx` (V litreS) `= (1.5)/(200) xx 40 xx 0.2`(E for `NaOH = 40) = 0.06g` |
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| 54. |
The % of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This proves the law of:A. Constant ProportionB. Reciprocal ProportionC. Multiple ProportionD. Conservation of mass |
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Answer» Correct Answer - A Law of constat proportion |
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| 55. |
100g of a water samples is found to contain 12 mg of `MgSO_(4)` calculate the hardness of water sample. |
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Answer» `120 g` of `MgSO_(4) = 100 g` of `CaCO_(3)` `rArr 12 xx 10^(-3) = ?` `rArr W_(CaCO_(3)) = (12 xx 10^(-3)xx 100)/(120) = 10^(-2)g` Hardness `= (W_(CaCO_(3))(g))/(W_(H_(2)O(g))) xx 10^(6) = (10^(-2))/(10^(2)) xx 10^(6) = 100 p p m` |
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| 56. |
Which of the following is displacement reaction :A. `S_((s)) +O_(2(g)) rarr SO_(2(g))`B. `KClO_(3(a)) rarr KCl_((s)) +O_(2(g))`C. `Ca_((s)) +H_(2)O_((l)) rarr Ca(OH)_(2(aq)) +H_(2(g))`D. `F_(2_((g))) +OH_((aq))^(-) rarr F_((aq))^(-) +OF_(2(g)) +H_(2)O_((l))` |
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Answer» Correct Answer - C `Ca(s) +2H_(2)O(l) rarr Ca(OH)_(2) (aq) +H_(2)` is a displacement reaction |
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| 57. |
The number of significant figures in 10500 areA. TwoB. ThreeC. FourD. Five |
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Answer» Correct Answer - A The number of significant figures in 0.0045 are two because zeros to the left of the first non-zero digit are not significant. |
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| 58. |
The number of significant figures in 10500 areA. ThreeB. FourC. FiveD. Can be any of these |
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Answer» Correct Answer - D The number 10500 end in zeros but these zeros are not to the right of a decimal point. These zeros may ot may not be significant. It depends how it is expressed i.e. as `1.5 xx 10^(-4)` or `1.050 xx 10^(4)` or `1.0500 xx 10^(4)` which have significant figures 3,4 and 5 respectively. |
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| 59. |
If 100 ml hydrogen chloride is completely decomposed the volume of `H_(2)` formed will be (P and T are constant).A. 20mlB. 200mlC. 100mlD. 50ml |
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Answer» Correct Answer - D `2HCl rarr H_(2)+Cl_(2)` |
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| 60. |
The volume of `H_(2)`STP required to completely reduce 160 gms of `Fe_(2)O_(3)` isA. `3 xx 22.4L`B. `2 xx 22.4L`C. `22.4L`D. `11.2L` |
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Answer» Correct Answer - A `Fe_(2)O_(3)+3H_(2) rarr 2Fe +3H_(2)O` |
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| 61. |
If one of the atoms in I are completely converted to get compound in II |
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Answer» Correct Answer - `A-p,s,B-q,r,C-p,r,D-p` Apply `POAC` (Principle of atomic conservation) |
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| 62. |
The oxidising agent in the reaction `2MnO_(4)^(-) +16H^(+) +5C_(2)O_(4)^(-2) rarr 2Mn^(+2) +8H_(2)O +10CO_(2)`A. `MnO_(4)^(-)`B. `H^(+)`C. `C_(2)O_(4)^(2-)`D. Both 1 & 2 |
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Answer» Correct Answer - A Oxidising agent undergo reduction |
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| 63. |
`3Cu + 8HNO_(3) rarr 3Cu (NO_(3))_(2) +2NO +4H_(2)O` the wrong statement for the valueA. `Cu` is oxidizedB. `HNO_(3)` is reducedC. `Cu` is reducedD. `Cu` acts as reducing agent |
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Answer» Correct Answer - C O.S of Cu increase From oto `+2` `:.` cu oxidised |
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| 64. |
A metal ion `M^(3+)` loses three electrons , its oxidation number will be |
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Answer» Correct Answer - C `M^(+3) rarr M^(+6) +3e^(-)` |
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| 65. |
In Habers process, the volume at S.T.P. of ammonia relative to the total volume of reactants at STP is:A. One fourthB. One halfC. SameD. Three fourth |
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Answer» Correct Answer - B `{:(N_(2)(g)+,3H_(2)(g)rarr,2NH_(3)(g)),(1L,3L,2L):}` |
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| 66. |
Match the following |
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Answer» Correct Answer - A::B::C::D Follow definitions of molarity, molarity and more frations. |
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| 67. |
Match the following columns |
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Answer» Correct Answer - A::B::C::D `overset(-3)(NH_(3))rarr overset(+5)(NO_(3)^(-))" "overset(+2)(Fe)overset(+3)(C_(2))O_(4) rarr Fe^(3+)+overset(+4)(C)O_(2)` `H_(2)overset(+6)(S)O_(5)rarr S_(8)" "Koverset(+7)(Mn)O_(4) rarr Mn^(2+)` |
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| 68. |
Match the following columns |
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Answer» Correct Answer - A::B::C::D `overset(+3)(Cr)overset(-1)(I_(3)) rarr overset(+6)(Cr_(2)) O_(7)^(2-) +overset(+7)(I)O_(4)^(-)` `overset(+2)(Fe)(SCN)_(2) rarr Fe^(3+) +SO_(4)^(-2) +CO_(3)^(2-) +NO_(3)^(-)` `NH_(4)SCN rarr SO_(4)^(-2) +CO_(3)^(2-) +NO_(3)^(-)` `AS_(2)S_(3) rarr AsO_(4)^(-3) +SO_(4)^(-2)` |
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| 69. |
An oxide of nitrogen contains `36.8%` by weight of nitrogen. The formula of the compound isA. `N_(2)O`B. `N_(2)O_(3)`C. `NO`D. `NO_(2)` |
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Answer» Correct Answer - B Calculate relative No: of atoms of Nitrogen and oxygen `N: O = (26.8)/(14): (100 - 36.8)/(18)` |
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| 70. |
An organic compound contains `40%C, 6.6%H`. The empirical formula of the compound isA. `CH_(2)`B. `CH_(2)O`C. `CHO`D. `CHO_(2)` |
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Answer» Correct Answer - B `(%C)/(12) (%H)/(1) (%O)/(16)` |
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| 71. |
`3.0gms` of an organic compound on combustion give `8.8 gm` of `CO_(2)` and `5.4gm` of water. The empirical formula of the compound isA. `CH_(3)`B. `C_(2)H_(4)`C. `C_(2)H_(2)`D. `C_(2)H_(6)` |
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Answer» Correct Answer - A `C_(x)H_(y)+ (x+(y)/(4)) O_(2) rarr xCO_(2) +(y)/(2) H_(2)O` |
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| 72. |
The empirical formula of a compound is `CH_(2)`. IF one mole of the compound has a mass of `42 g`, its molecular formula isA. `CH_(2)`B. `C_(2)H_(2)`C. `C_(3)H_(6)`D. `C_(3)H_(8)` |
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Answer» Correct Answer - C `n = (M. Wt)/(E.F Wt)` |
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| 73. |
Equivalent weight of `NH_(4)OH` isA. 35B. 17.5C. 42D. 72 |
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Answer» Correct Answer - A `NH_(4)OH rarr NH_(4)^(+) +OH^(-)` `:. GEW = GMW//` No. of `OH^(-)` ions per molecule |
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| 74. |
The emperical formula of Acetic acid isA. `CH_(3)-COOH`B. `C_(2)H_(4)O`C. `CH_(2)O`D. `CHO` |
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Answer» Correct Answer - C Acetic acid `M.F = C_(2)H_(4)O_(2) rArr E.F = CH_(2)O` `M.F. = n xx E.F` |
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| 75. |
Equivalent weight of `CaCI_(2)` isA. `("Formula weight")/(2)`B. `("Formula weight")/(1)`C. `("Formula weight")/(3)`D. `("Formula weight")/(4)` |
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Answer» Correct Answer - A For `CaCl_(2)`, n-factor =2 |
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| 76. |
Equivalent weight of hydrated oxialic acid isA. 44B. 45C. 63D. 126 |
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Answer» Correct Answer - C For `H_(2)C_(2)O_(4), 2H_(2)O, n-` factor = 2 `:. GEW = (GMW)/(2) = (126)/(2) = 63g` |
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| 77. |
What is the mass per cent of carbon in carbon dioxide?A. `0.034%`B. `27.27%`C. `3.4%`D. `28.7%` |
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Answer» Correct Answer - B `% C = (12)/(44) xx 100 = 27.27%` |
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| 78. |
Equivalent weight of sulphate ion and bicarbonate ion : |
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Answer» sulphate ion `(SO_(4)^(-2)) = (96)/(2) = 48` bicarbonate ion `(HCO_(3)^(-)) = (61)/(1) = 61` |
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| 79. |
A `2.0 g` sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of `CO_(2)` ceases. The volume of `CO_(2)` at `750mm Hg` pressure and at `298 K` is measured to be `123.9 mL`. A `1.5 g` of the same sample requires `150 mL` of `(M//10) HCl` for complete neutralisation. Calculate the percentage composition of the components of the mixture.A. `10%`B. `71%`C. `31.5%`D. `50%` |
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Answer» Correct Answer - C `2NaHCO_(3) rarr Na_(2)CO_(3)+H_(2)O+CO_(2)` No. of moles of `NaHCO_(3)` in the mixture `=(112)/(22400) xx 2 = 0.01` No. of equivalents of `NaHCO_(3) = 0.01` No. of eqts of `(NaHCO_(3)+Na_(2)CO_(3)) =` No. of eqts of HCI reacted `= 0.2 xx 0.1 = 0.01` weight of `(Na_(2)CO_(3) +NaHCO_(3)) = (0.01 xx 53) +(0.01 xx 84) = 1.37gm` % of `Na_(2)SO_(4) = ((2-1.37)/(2)) xx 100 = 31.5%` |
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| 80. |
If an ore sample containing `Mn`, is treated with `50 mL` of `0.2750M Na_(2)C_(2)O_(4)` and the unreacted `Na_(2)C_(2)O_(4)` required `18.28 mL` of `0.1232 M KMnO_(4)` in acidic medium,then the number of moles of `Mn` in the ore isA. `1.38 xx 10^(-2)`B. `1.49 xx 10^(-3)`C. `1.15 xx 10^(-2)`D. `8.35 xx 10^(-3)` |
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Answer» Correct Answer - D `MnO_(2) +2e^(-) Mn^(2+)` (n-factor =2) m.eqts of unreacted `Na_(2)C_(2)O_(4) =` `(18 xx 0.12 xx 5) =`m .eqts of `KMnO_(4) = 10.8` Total m.equts of `Na_(2)C_(2)O_(4) = 50 xx 0.275 xx 2 = 27.5` m.eqts of `Na_(2)C_(2)O_(4)` reacted with ore =`27.5 - 10.8 = 16.7 = 16.7 xx 10^(-3)` eqts of `MnO_(2)` `:.` moles of `MnO_(2) = (16.7 xx 10^(-3))/(2) = 8.35 xx 10^(-3)` |
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| 81. |
The percentage of Carbon in `CO_(2)` isA. `27.27%`B. `29.27%`C. `30.27%`D. `26.97%` |
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Answer» Correct Answer - A `%C = ("No. of C -atoms" xx "AT. Wt. of C")/(GMW CO_(2)) xx 100` |
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| 82. |
A compound has `20%` of nitrogen by weight. If one molecule of the compound contains two nitrogen atoms, the molecular weight of the compound isA. 35B. 70C. 140D. 280 |
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Answer» Correct Answer - C `100 rarr 200 ? rarr 28` |
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| 83. |
Which of the following is a penta-atomic moleculeA. PhosphorusB. SulphurC. EthaneD. Methane |
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Answer» Correct Answer - D Atomicity of `CH_(4)` is 5 |
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| 84. |
Which of the following is an octa-atomic moleculeA. PhosphorusB. SulphurC. MethaneD. Oxygen |
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Answer» Correct Answer - B Rhombic SUlphur exists has `S_(8)` rings |
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| 85. |
When oxalic acid is heated with concentrated `H_(2) SO_(4)` it producesA. `CO`B. `CH_(4)`C. `CO_(2)`D. `C_(2)H_(4)` |
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Answer» Correct Answer - A `H_(2)C_(2)O_(4) overset(H_(2)SO_(4))underset(-H_(2)O)rarr CO+CO_(2)` `CO +CO_(2) overset(c)underset(Delta)rarr 3CO` |
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| 86. |
In an organic compound of molar mass greater than `100` containing only `C`, `H` and `N`, the percentage of `C` is `6` times the percentage of `H` while the sum of the percentage of `C` and H is `1.5` times the percentage of `N`. What is the least molar mass :A. 175B. 140C. 105D. 210 |
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Answer» Correct Answer - B Mass ratio of `C: H : N = 18: 3 :14` Empirical formula `=C_(3)H_(6)N(2)` E.f. Wt `=70` |
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| 87. |
What will be the normally of a solution obtained by mixiing 0.45N and 0.60N NaOH in the ration 2:1 by volume?A. `0.5g`B. `25g`C. 20gD. 5g |
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Answer» Correct Answer - C `N = (V_(1)N_(1)+V_(2)N_(2))/(V), N = (w)/(GEW) xx (1)/(V("in lits"))` |
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| 88. |
`3HOCl rarr 2HCl +HClO_(3)` is an example ofA. Disproportionation reactionB. Displacement reactionC. Chemical combination reactionD. Decomposition reaction |
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Answer» Correct Answer - A disproportionation reaction |
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| 89. |
`Na_(2)SO_(3)+H_(2)O_(2) rarr Na_(2)SO_(4) +H_(2)O`, in reactionA. `H_(2)O_(2)` is bleachedB. `H_(2)O_(2)` is reducedC. `H_(2)O_(2)` is dehydratedD. `H_(2)O_(2)` is neither oxidised nor reduced |
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Answer» Correct Answer - B `H_(2)O_(2)` is reduced |
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| 90. |
The oxidation state of the most electronegative atom in each of the product is `H_(2) +O_(2) rarr H_(2)O_(2) +H_(2)O`A. `-2,-2`B. `+1,-2`C. `+2,-1`D. `-1,-2` |
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Answer» Correct Answer - D `H_(2)+O_(2) rarr H_(2) overset(-1)(O_(2)) +H_(2) overset(-2)(O)` |
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| 91. |
Which of the following reactions involves neither oxidation nor reduction ?A. `CrO_(4)^(-2) rarr Cr_(2)O_(7)^(2-)`B. `Cr rarr CrCl_(3)`C. `Na rarr Na^(+)`D. `2S_(2)O_(3)^(2-) rarr S_(4)O_(6)^(2-)` |
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Answer» Correct Answer - A `overset(+6)(Cr)O_(4)^(-2) rarr overset(+6)(Cr_(2))O_(7)^(-2)` |
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| 92. |
The oxidation state of oxygen is maximum inA. Bleaching powder `(CaOCI_(2))`B. Oxygen difluoride `(OF_(2))`C. Dioxygen difluoride `(O_(2)F_(2))`D. Hydrogen peroxide `(H_(2)O_(2))` |
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Answer» Correct Answer - B `overset(+2)(OF_(2)) gt overset(+1)(O_(2)F_(2)) gt H_(2)overset(-1)O_(2) gt Caoverset(-2)(OCl_(2))` |
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| 93. |
Study the following table: Which two compounds have least weight of oxygen (molecular weights of compounds are given in brackets) A. II & IVB. I & IIIC. I & IID. III & IV |
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Answer» Correct Answer - A find out the weight of oxygen as per the data given by the formula weight of oxygen =no. of moles `xx32` |
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| 94. |
A sample of `H_(2)O_(2)` is x% by mass x ml of `KMnO_(4)` are required to oxidize one gram of this `H_(2)O_(2)` sample. Calculate the normality of `KMnO_(4)` solution. |
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Answer» Suppose, Mass of `H_(2)O_(2)` solution `= 100g` Mass of `H_(2)O_(2)` present =x gram Mass of `H_(2)O_(2)` solution taken = 1 gram Mass of `H_(2)O_(2)` present in 1 gram solution `=(x)/(100)` Equivalents of `H_(2)O_(2) = (w)/(E) = (X)/(100 xx 17)` ....(1) (E for `H_(2)O_(2)) = 17` equivalents of `KMnO_(4) = N xx V` (litre) `= N xx X xx 10^(-3)` Putting equivalents of `H_(2)O_(2)` and `KMnO_(4)` equal `(X)/(100 xx 17) = N xx x xx 10^(-3)` `N = 0.59` (Normality of `KMnO_(4))` |
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| 95. |
Equivalent weight of `N_(2)H_(4)` isA. 32B. 16C. 48D. 24 |
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Answer» Correct Answer - B Pressure of two L.P its acidity is two |
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| 96. |
The isotopes of chlorine with mass numbers 35 and 37 exist in the ratio ofA. `1:1`B. `2:1`C. `3:1`D. `3:2` |
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Answer» Correct Answer - C Average atomic mass `= (p xx m_(1)+ (100-p)m_(2))/(100)` |
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| 97. |
12 gm of an alkaline earth metal gives 14.8 g of its nitride. The atomic mass of metal isA. 12B. 24C. 20D. 40 |
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Answer» Correct Answer - D 2.8 gms of nitrogen combians with 12 gm of metal `14//3` gms of nitrogen combins with 20 gm of metal (eqwt. Atomic weight = eqwtx valency `=20 xx 2 = 40 gms`. |
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| 98. |
A mixture of `HCOOH` and `H_(2)C_(2)O_(4)` is heated with conc. `H_(2)SO_(4)`. The gas produced is collected and on treating with `KOH` solution the volume of the gas decreases by `1//6th`. Calculate molar ratio of two acids in original mixure.A. `4:1`B. `1:4`C. `3:1`D. `2:1` |
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Answer» Correct Answer - A `underset("a moles")(HCOOH) overset(H_(2)SO_(4))underset(("Conc"))rarr H_(2)O +underset("a moles")(CO)` `underset("a moles")(H_(2)C_(2)O_(4)) overset(H_(2)SO_(4))underset(("Conc"))rarr H_(2)O +underset("b moles")(CO)+underset("b moles")(CO_(2))` Total no. of moles released as gases `a +b +b = a +2b` `CO_(2)` is absorbed by KOH `b = (1)/(6) (a+2b)` (moles `oo` volume) `(a)/(b) = 4` or `a:b :: 4:1` |
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| 99. |
An acid solution of 0.2 " mol of "`KReO_(4)` was reduced with Zn and then titrated with 1.6 " Eq of "acidic `KMnO_(4)` solution for the reoxidation of the ehenium `(Re)` to the perrhenate ion `(ReO_(4)^(ɵ))`. Assuming that rhenium was the only elements reduced, what is the oxidation state to which rhenium was reduced by Zn?A. 1B. 2C. `-1`D. `-2` |
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Answer» Correct Answer - C Eq of `K Re O_(4) -=` Eq of `MnO_(4)^(-)` `0.2 xx n = 1.6` Eq, `n = (1.6)/(0.2) = 8` Hence, ther is an `8e^(-)` reduction. `8e^(-) +overset(+7)(ReO_(4)^(-)) rarr underset(x =- 1)(Re^(-1))` oxidation state of `Re =- 1` |
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| 100. |
Which of the following is not a homgeneous mixture.A. AirB. BrassC. Solution of sugar in waterD. Smoke |
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Answer» Correct Answer - D Smoke is hetrogeneous mixture |
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