InterviewSolution
Saved Bookmarks
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. |
40mL gaseous mixture of `CO, CH_(4)` and Ne was exploded with 10mL of oxygen. After treatment with KOH the volume reduced by 9mL and again on treatment with alkaline progallol the volume further reduced by 1.5mL. Percentage of `CH_(4)` in the original mixture is:A. 22.5B. 77.5C. 7.5D. 15 |
|
Answer» Correct Answer - D |
|
| 52. |
Ten millilitre ofa mixture of methane, ethylene, and carbon dioxide was exploded with excess of air: After the explosion, there was contraction of 17 ml and after treatment with `KOH`, there was a further contraction of 14 ml. What was the composition of the mixture? |
|
Answer» Let the volume of `CH_4` be x mL Volume of `(CH_2=CH_2)=ymL` Volume of `CO_2=[10-(x+y)]mL` (a). `CH_4(g)+2O_2(g)toCO_2(g)+2H_2O(l)` (b). `CH_2=CH_2(g)+3O_2(g)to2CO_2(g)+H_2O(l)` Volume of `CO_2` absorbed by `KOH=14mL` `thereforey=4mL` Volume of gaseous reactant `=` volume of gaseous production`+`constraction `x+2x+y+3Y=x+2y+17` `3x+4y-x-2y=17` `2x+2y=17` `2x+2xx4=17` `2x=9` `x=4.5mL` Volume of `CH_4=4.5mL` Volume of `CH_2=CH_2=4mL` Volume of `CO_2=10-(4.5+4)=1.5mL` |
|
| 53. |
20 " mL of " a gasous bydrocarbon (A) was exploded with excess of `O_2` in an eudiometer tube. On cooling, the volume was reduced by 50 mL. On further treatment with KOH, there was further contraction of 40 mL. The molecular formula of the hydrocarbon A is (a). `C_2H_6` (b). `C_2H_4` ` C_3H_6` (d). `C_3H_8` |
|
Answer» `C_(x)H_(y)(g)+(x+(y)/(4))O_2(g)toxCO_2(g)+(y)/(2)H_2O(l)` `CO_2` is absorbed in KOH. `therefore20x=40impliesx=2` `20mL+20(2+(y)/(4))=40+50` `20+40+(20y)/(4)=90` `5y=90-60=30` `y=6` Formula of the compound is `C_2H_6`. |
|
| 54. |
The totak volume of `0.1M KMnO_(4)` solution that are needed to oxidize 100mg each of ferrius oxalate and ferrous sulphate in a mixture in acidic medium is:A. 1.096mLB. 1.32mLC. 5.48mLD. none of these |
|
Answer» Correct Answer - A |
|
| 55. |
What volume of `01` M `KMnO_4` is required to oxidise 100 " mL of " 0.2 M `FeSO_4` in acidic medium ? The reaction involved is |
|
Answer» `5e^(-)MnO_4^(ɵ)toMn^(2+)(n=5)` (reduction or oxidising agent) `Fe^(2+)toFe^(3+)+e^(-)(n=1)` (oxidation or reducing agent) `MnO_4^(ɵ)-=Fe^(2+)` `mEq=mEq` `N_1V_1(mL)-=N_2V_2(mL)` `V_1(mL)xx0.1Mxx5` (n-factor)`-=0.2Mxx1` (n-factor) `xx100` `0.5V_1-=0.2xx100` `V_1-=40mL` |
|
| 56. |
A sample of hydrazine sulphate `(N_(2)H_(6)SO_(4))` was dissolved in `100 mL` water. `10 mL` of this solution was reacted with excess of `FeCl_(3)` solution and warmed to complete the reaction. Ferrous ions formed were estimated and it required `20 mL` of `M//50 KMnO_(4)` solutions. Estimate the amount of hudrazine sulphate in one litre of solution. Given `4Fe^(3+)+N_(2)H_(4)rarrN_(2)+4Fe^(2+)+4H^(+)` `MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarrMn^(2+)+5Fe^(3+)+4H_(2)O` |
|
Answer» 20 " mL of " `(M)/(50)KMnO_4` `=(20)/(50)m" mol of "KMnO_4` `=(2)/(5)xx5m" Eq of "KMnO_4` `=2 m" Eq of "Fe^(2+)=2 m" mol of "Fe^(2+)` `=(2)/(4)m" mol of "N_2H_4` `=0.5xx10^(-3)" mol of "N_2H_6SO_4` `=0.5xx10^(-3)xx130g of N_2H_6SO_4` `=0.065 g i n 10 mL` `=6.5 g i n 1L=6.5gL^(-1)` |
|
| 57. |
A 0.5 g sample of an iron containing mineral mainly in the form of `CuFeS_2` was reduced suitable to convert all the ferric ions into the ferrous form and was obtained as a solution. In the absence of any interfering matter, the solution required 42 " mL of " 0.1 M `K_2Cr_2O_7` solution for titration calculate the percentage of `CuFeS_2` in the mineral `(Cu=63.5,Fe=55.8)` |
|
Answer» Equivalent weight of `CuFeS_2=` molecular weight `CuFeS_2toCuSdarr+FeS` `FeS+H_2SO_4toFeSO_4+H_2Suarr` `2FeSO_4+H_2SO_4+OtoFe(SO_4)_3+H_2O` Molecular weight of `CuFeS_2=63.5+55.8+64=183.3g` `42 " mL of " 0.01 M K_2Cr_2O_7=42xx0.01xx6NK_2Cr_2O_7` `=2.52 m" Eq of "K_2Cr_2O_7` `=2.52m" Eq of "Fe^(2+)` `=2.52 m" Eq of "CuFeS_2` `=2.52 m" Eq of "CuFeS_2` `=2.52xx10^(-3)xx18.3g of CuFeS_2` `=0.4619 g of CuFeS_2` `%` of `CuFeS_2=(0.4619xx100)/(0.5)=92.38%` |
|
| 58. |
A sample of `MnSO_4.4H_2O` is reacted in air to give `Mn_3O_4`. The residue `Mn_3O_4` is dissolved in 100 " mL of " `(N)/(12)FeSO_4` containing `H_2SO_4` The solution reacts completely with 50 " mL of " `KMnO_4`. 25 " mL of " this `KMnO_4` requires 30 " mL of " `(N)/(10)FeSO_4` for complete oxidation determine the amount of `MnSO_4.4H_2O` in the sample. |
|
Answer» `Mn_3O_4+3H_2SO_4to3MnSO_4+3H_2O+O` `underline(2FeSO_4+H_2SO_4+OtoFe_2(SO_4)_3+H_2O)` `underline(Mn_3O_4+4H_2SO_4+2FeSO_4to3MnSO_4+4H_2O+Fe_2(SO_4)_3)` `Mn_3O_4+8H^(o+)+2Fe^(2+)to3Mn^(2+)+2F^(3+)+4H_2O` 25 " mL of " `KMnO_4` requires 30 " mL of " 0.1 N `Fe^(2+)` `50mL` of `KMnO_4` requires 60 " mL of " 0.1 N `Fe^(2+)` Volume of `FeSO_4` used by `Mn_3O_4` `=100-60=40" mL of " 0.1 N Fe^(+2)` `=40xx0.1 m" Eq of "FeSO_4` `=4m" mol of "FeSO_4` `6 m" mol of "MnSO_4` `=6xx10^(-3)xx223g of MnSO_4` `=1.338 g of MnSO_4` [2 m" mol of "`FeSO_4=1 m" mol of "Mn_3O_4) `=3 m" mol of "MnSO_4]` |
|
| 59. |
Reduction of the metal centre in aqueous permanganate ion involvesA. 3 electrons in neutral mediumB. 5 electrons in neutral mediumC. 3 electrons in alkaline mediumD. 5 electrons in alkaline medium |
|
Answer» Correct Answer - A::D (a). In neutral medium `:overset(+7)(MnO_(4)^(ɵ))3e^(-)tooverset(+2)(Mn)O_(2)` (c). In alkaline medium: `overset(+7)(Mn)O_(4)^(ɵ)+e^(-)tooverset(+6)(Mn)O_(4)^(2-)` (d). In acidic medium: `overset(+7)(Mn)O_(4)^(ɵ)+5e^(-)toMn^(2+)` |
|
| 60. |
In which of the following reactions, `H_(2)O_(2)` is acting as a reducing agent?A. `SO_(2)+H_(2)O_(2) to H_(2)SO_(4)`B. `2KI+H_(2)O_(2) to 2KOH+I_(2)`C. `PbS+4H_(2)O_(2) to PbSO_(4)+4H_(2)O`D. `Ag_(2)O+H_(2)O_(2) to 2Ag+H_(2)O+O_(2)` |
|
Answer» Correct Answer - D |
|
| 61. |
Three solutions eaach of 100 mL containing 0.4 M `As_(2)S_(3),5M` NaOH and `6MH_(2)O_(2)`, respectively were mixed to form `AsO_(4)^(3-)` and `SO_(4)^(2-)` as products. Q. Percentage strength of the `H_(2)O_(2)` solution left after reaction isA. `0.017%`B. `0.113%`C. `0.51%`D. `0.68%` |
|
Answer» Correct Answer - C `M of H_(2)O_(2)` left`=(2M)/(15)` `N of H_(2)O_(2)` left`=(2xx2)/(15)=(4)/(15)N` `1N of H_(2)O_(2)=(1eq)/(L)H_(2)O_(2)=5.6V` (Volume strength)`=1.7%` `(4)/(15)NH_(2)O_(2)=1.7xx(4)/(15)=0.452%` |
|
| 62. |
Three solutions eaach of 100 mL containing 0.4 M `As_(2)S_(3),5M` NaOH and `6MH_(2)O_(2)`, respectively were mixed to form `AsO_(4)^(3-)` and `SO_(4)^(2-)` as products. Q. When the above solution is allowed to stand for some time what volume of `O_(2)` will be obtained `STP`?A. 0.112 LB. 0.224 LC. 0.224LD. 0.448L |
|
Answer» Correct Answer - C If a `H_(2)O_(2)` solution is allowed to stand, it decomposes to `O_(2)` and `H_(2)O` `underset(40)(H_(2)O_(2)tounderset(0)(H_(2)O)+(1)/(2)underset(0)(O_(2))` `{:(mmol es),(of H_(2)O_(2)),(l eft)]:}-(1)/(2)xx40=20m" mol of "O_(2)` Thus, mmol es of O_(2)formed`=40xx(1)/(2)=20` Volume of `O_(2)` at `STP=20xx10^(-3)xx22.4L` `(because1 " mol of "O_(2) at STP=22.4L)` `=0.448L` |
|
| 63. |
A sample of 28 mL of `H_(2)` `O_(2)` (aq) solution required 10 mL of 0.1 M `KMnO_(4)` (aq) solution for complete reaction in acidic medium. What is the valume strength of `H_(2)O_(2)` ? X |
|
Answer» Correct Answer - 1 |
|
| 64. |
For the redox reaction given, what is the value of `(x)/(z)` ? `xNO_(3)^(-)+yAs_(2)S_(3)+zH_(2)O to -----AsO_(4)^(3-)+-----NO+------SO_(4)^(2-)+-----H^(+)`. |
|
Answer» Correct Answer - 7 |
|
| 65. |
Following reaction describes the rusting of iron `4Fe+3O_(2) rarr 4Fe^(3+)+6O^(2-)` Which one of the following statements is incorrect?A. This is an example of a redox reactionB. Metallic iron is reduced to `Fe^(2+)`C. `Fe^(3+)` is an oxidising agentD. Metallic iron is a redoxing agent |
|
Answer» Correct Answer - B |
|
| 66. |
0.71 g of a sample of bleaching powder `(CaOCl_2)` is dissolved in 100 " mL of " water. 50 " mL of " this solution is titrated with KI solution. The `I_2` so liberated required 10 mL 0.1 M `Na_2S_2O_3` (hypo) solution in acidic medium for complete neutralisation. Calculate the percentage of available `Cl_2` from the sample of bleaching power. |
|
Answer» `CaOCl_2-=Cl_2-=I^(ɵ)-=S_2O_3^(2-)` `mEq-=mEq-=mEq-=mEq` `mEq-=mEq-=mEq-=mEq` `--=--=--=10mLxx0.1xx1` (n-factor) `--=1mEq-=1mEq` Thus m" Eq of "`Cl_2` in `50 mL` of solution `=1mEq` m" Eq of "`Cl_2` in 100 " mL of " solution `=1xx2=2mEq` Weight of `Cl_2=mEqxx10^(-3)xxEq(Cl_2)` `=2xx10^(-3)xx(71)/(2)` `{:((n-fact or fo r Cl_2=2),(Cl_2+2e^(-)to2Cl^(ɵ)))` `=0.071 g` `%` of available `Cl_2=("Weight of" Cl_2)/("Weight of" CaOCl_2)xx100` `=(0.071xx100)/(0.71)=10%` |
|
| 67. |
0.5 g sample of copper ore is converted into `CuSO_4` solution. The resulting solution is acidified with dilute `CH_3COOH` (acetic acid) and excess KI added. The liberated `I_2` required 0.248 g `Na_2S_2O_3.5H_2O` for complete reaction. Calculate the percantage of Cu in the ore. |
|
Answer» `Mw(Na_2S_2O_3.5H_2O)=248g` `Ew=(248)/(1)(n=1)` `Ew(CuSO_4 or Cu)=(M)/(1)` `Cu-=CuSO_4-=KI-=I_2-=Na_2S_O_3` `-=(0.248)/(248)=10^(-3)Eq` " Eq of "`Cu=10^(-3)Eq` "Weight of" of `Cu=EqxxEq=10^(-3)xx(63.5)/(1)g(n=1)` `%` of `Cu=(63.5xx10^(-3)gxx100)/(0.5g)=(635xx100)/(1000xx5)=12.7%` |
|
| 68. |
The oxidation state of oxygen of `H_(2)O_(2)` in the final products when it reacts with `ClO_(3)^(ɵ)` is |
|
Answer» Correct Answer - A `H_(2)O_(2)+ClO_(3)^(ɵ)to[Cl_(2) or Cl^(ɵ)+underset(2x=0)(O_(2)^(0)` |
|
| 69. |
Which of the following is redox reaction ?A. `H_(2)SO_(4)` reach with NaOHB. In atmoshere, `O_(3)` is formed from `O_(2)` by lightningC. Evaporation of `H_(2)O`D. Oxides of nitrogen are formed form nitrogen & oxygen by lightning |
|
Answer» Correct Answer - D |
|
| 70. |
0.5 gmixture of `K_(2)Cr_(2)O_(7)` and `KMnO_(4)` was treated with excess of KI in acidic medium. Iodine liberated required `150 cm^(3)` of 0.10 N solution of thiosulphate solution for titration. Find trhe percentage of `K_(2)Cr_(2)O_(7)` in the mixture :A. 14.64B. 34.2C. 65.69D. 50 |
|
Answer» Correct Answer - A |
|
| 71. |
10 g mixture of `K_(2)Cr(2)O_(7)` and `KMnO_(4)` was treated with excess of KI in acidic medium. Iodine liberated `100 cm^(3)` of 2.2 N sodium thiosulphate solution for titration. If the mass percent of `KMnO_(4)`in the mixture Z, then what is the value of 2Z/5 ? |
|
Answer» Correct Answer - 6 |
|
| 72. |
A mixture contaning 0.05 moleof `K_(2)Cr_(2)O_(7) and 0.02 "mole of" KMnO_(4)` was treated eoith excess of KI in acidic medium. The liberated iodine required `1.0L "of" Na_(2)S_(2)O_(3)` solution for titration. Concentration of `Na_(2)S_(2)O_(3)` solution was:A. `0.4 molL^(-1)`B. `0.20 molL^(-1)`C. `0.25 molL^(-1)`D. `0.30 molL^(-1)` |
|
Answer» Correct Answer - A |
|
| 73. |
When `BrO_(3)^(-)` ion reacts with `Br^(-)` ion in acidic medium, `Br_(2)` is liberated. Calculate the ratio of molecular mass and equivalent mass of `KBrO_(3)` |
|
Answer» Correct Answer - 5 |
|
| 74. |
18 " mL of " 1.0 M `Br_(2)` solution undergoes complete disproportionation in basic medium to `Br^(The hardness of water in terms of )` and `BrO_(3)^(ɵ)`. Then the resulting solution required 45 " mL of " `As^(3+)` solution to reduce `BrO_(3)^(ɵ)` to `Br^(ɵ)`. `As^(3+)` is oxidised to `As^(5+)` which statements are correct?A. `Ew(Br_(2))=(M)/(10)`B. `Ew(Br_(2))=(5M)/(3)`C. Molarity of `As^(+3)=0.4M`D. Molarity of `As^(3+)=0.2M` |
|
Answer» Correct Answer - B::C `[2e^(-)+Br_(2)to2Br^(ɵ)]xx5` `underline(Br_(2)to2BrO_(3)^(ɵ)+10e^(-))` `underline(6Br_(2)to5Br^(ɵ)+2BrO_(3)^(ɵ))` Ew of `Br_(2)=(M)/(2)+(M)/(10)=(10M)/(6)=(5M)/(3)` mmoles of `Br_(2)=18xx1=18` So mmoles of `BrO_(3)^(ɵ)=(18)/(6)xx2=6` `m" Eq of "underset((n=2))(As^(+3))tomEq . underset((n=6))(of BrO_(3)^(ɵ))` `6e^(-)+BrO_(3)^(ɵ)toBr^(ɵ)` `m" Eq of "As^(3+)-=m" Eq of "BrO_(3)^(ɵ)` `45xxMxx2` ("n factor")`-=6xx6` `thereforeM_(As^(3+))=(36)/(90)=0.4`M |
|
| 75. |
To a 10mL, 1M aqueous solution of `Br_(2)`, excess of NaOH is added so that all `Br_(2)` is disproportionated to `Br^(-)` and `BrO_(3)^(-)`. The resulting solution is free from `Br^(-)`, by extraction and excess of `OH^(-)` neutralised by acidifying the solution. The resulting solution is suffcient to react with 2 g of impure `CaC_(2)O_(4)`(M= 128g/mol) sample. The % purity of oxalate sample is :A. 0.853B. 0.125C. 0.9D. 0.64 |
|
Answer» Correct Answer - B |
|
| 76. |
10 " mL of " a gaseous organic compound containing C, H and O only was mixed with 100 " mL of " `O_(2)` and exploded under condition which allowed the `H_(2)O` formed to condense. The volume of the gas after explosion was 90 mL. On treatment with KOH solution, a further contraction of 20 mL in volume was observed. The vapour density of the compound is 23. All volume measurements were made under the same condition. Q.The molecular formula of the compound isA. `C_(3)H_(8)O`B. `C_(3)H_(6)O`C. `C_(2)H_(6)O`D. `C_(2)H_(4)O` |
|
Answer» Correct Answer - C Volume of `O_(2)` reacted with 10 " mL of " compound `=100-70=30mL` Let the formula of the compound be `C_(x)H_(y)O_(z)`. `undersetunderset(10 mol)(10mL)(C_(x)H_(y)O_(z))toundersetunderset(30mL)(30mL)(O_(2))toundersetunderset(20 mol)(20 mL)(xCO_(2))+(y)/(2)H_(2)O` Applying POAC (principle of atom conservation) for C atoms. `x xx` moles of `C_(x)H_(y)O_(z)=1xx` " mol of "`CO_(2)` `x xx10=1xx20` `x=2` Again applying POAC for H and O atoms. `yxx` moles of `C_(x)H_(y)O_(z)=2xx` moles of `H_(2)O` `yxx10=2xx` moles of `H_(2)O` `zxx` moles of `C_(x)H_(y)O_(z)+2xx` moles of `O_(2)` `10z+2xx30=2xx20+` moles of `H_(2)O` ...(ii) Eliminating mols of `H_(2)O` from equation (i) and (ii) we get `y-2z=4` Now, molecular weight of the compound`=2xx23` `=46` `thereforeC_(x)H_(y)O_(z)=2xx12+yxx1+zxx16=46` `y+16z=22` Solve for y and z from equation (iii) and (iv) we get `y=6,z=1` Formula of the compound`=C_(2)H_(6)O` |
|
| 77. |
10 " mL of " a gaseous organic compound containing C, H and O only was mixed with 100 " mL of " `O_(2)` and exploded under condition which allowed the `H_(2)O` formed to condense. The volume of the gas after explosion was 90 mL. On treatment with KOH solution, a further contraction of 20 mL in volume was observed. The vapour density of the compound is 23. All volume measurements were made under the same condition. Q.The volume of unreacted `O_(2)` isA. 20 mLB. 50 mLC. 70 mLD. 90 mL |
|
Answer» Correct Answer - C Volume of unreacted `O_(2)=90-20=70mL` |
|
| 78. |
10 " mL of " a gaseous organic compound containing C, H and O only was mixed with 100 " mL of " `O_(2)` and exploded under condition which allowed the `H_(2)O` formed to condense. The volume of the gas after explosion was 90 mL. On treatment with KOH solution, a further contraction of 20 mL in volume was observed. The vapour density of the compound is 23. All volume measurements were made under the same condition. Q.The volume of `CO_(2)` isA. 20 mLB. 50 mLC. 70 mLD. 90 mL |
|
Answer» Correct Answer - A Since `H_(2)O` vapours are condensed, Volme of unreacted `O_(2)+` volume of `CO_(2)=90mL` Volume of `CO_(2)` (absorbed by KOH)`=20mL` |
|
| 79. |
245 g of iodine and 142 g of chlorine are made to react completely to given a mixture of `ICl` and `ICl_(3)` .How many moles of each are formed ?A. 0.1 mole of IC l and 0.1 mole of `ICI_(3)`B. 1 mole of I Cl and 1 mole of `ICL_(3)`C. 0.5 mole of IC l and 0.1 mole of `ICl_(3)`D. 0.5 mole of IC l and 0.1 mole of IC l mole of `ICl_(3)` |
|
Answer» Correct Answer - B Both reasctants are completely consumed, hence, both are limiting. `I_(2) + 2Cl_(2) to ICl+ ICl_(3)` 254g or 1 mole `I_(2)` and 142 g or `Cl_(2)` will react to given 1 mole Icl and 1 mole `IC l_(3)`.] |
|
| 80. |
(a). Take 1 L of a mixture of CO and `CO_(2)` Pass this mixture through a tube containing red hot charcoal. The volume now becomes 1.6 L. The volumes are measured under the same conditions. Find the composition of the mixture by volume. (b). A compound contains 28 percent of nitrogen 72 percent of a metal by weight. Three atoms of the metal combine with two atoms of N. find the atomic weight of the metal. |
|
Answer» (a). Let the volume of CO in the mixture be V. The volume of `CO_(2)` in the mixture is `(1-V)` `CO_(2)+Cto2CO` `(1-V)LCO_(2)` forms `2(1-V)LCO`. Total CO, `V+2(1-V)=1.6` or `V=0.4` `CO=0.4L` `CO_(2)=1-V=0.6L` (b). 28 g of nitrogen combines with 72 g of metal. three atoms of metal. Three atoms of metal combine with two atoms of nitrogen. This means the valency of metal is 2. " Eq of "`N=(28)/((14)/(3))=(1)/(6)` " Eq of "metal `=(72)/((M)/(2))=(144)/(M)` `therefore(1)/(6)=(144)/(M)impliesM=(144)/(6)=24g` |
|
| 81. |
0.4 g of polybasic acid HnA (all the hydrogens are acidic) requries 0.5 g of NaOH for complete neutralisation. The number of replaceable hydrogen atoms and the molecular weight of A would be (Mw of `acid=96)`A. `2,94`B. `1,95`C. `3,93`D. `4,92` |
|
Answer» Correct Answer - D `mEq Acid=mEq base` `implies(0.4)/((Mw)/(n))xx1000=(0.5)/((40)/(1))xx1000` `impliesn=3 and H_(n)A-=96impliesA=96-3xx1=93` |
|
| 82. |
0.108 g of finely divided copper was treated with an excess of ferric sulphate solution untill Cu was completely dissolved. The solution after the addition of excess dil `H_2SO_4`, required `33.7mL` of 0.1 N `KMnO_4` for complete oxidation. Find the equation which represents the reaction between metallic copper and ferric sulphate solution. `(Cu=63.7,Fe=56)` |
|
Answer» 33.7 " mL of " `0.1 N KMnO_4=3.37 m" Eq of "KMnO_4` `=0.00337 " Eq of "KMnO_4` `[Fe^(2+)` ions are formed on reduction of `Fe_2(SO_4)_3` by Cu] " Eq of "`Cu(0.108)/(63.5)=0.0017` 0.0017 " Eq of "Cu produces 0.00337 " Eq of "`Fe^(2+)` 1 Eg of Cu produces `=(0.00337)/(0.0017)approx2" Eq of "Fe^(2+)` `approx2 " mol of "Fe^(2+)` `approx2 " mol of "FeSO_4` Hence, the equation is `Cu+Fe_2(SO_4)_3toCuSO_4+FeSO_4` |
|
| 83. |
Hydrogen and oxygen combine to from `H_(2) O_(2) ` and `H_(2)O` containing `5.93%` and `11.2%` hydrogen respectively . The data illustrates :A. law of conseravtion of massB. law of constant proportionC. law of reciparocal proporationD. law of multiple proporetion |
|
Answer» Correct Answer - 4 |
|
| 84. |
`H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` behave as acids as well as reducing agents. Which of the following are correct statements?A. Equivalents weights of `H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` are equal to their molecular weights when acting as reducing agents.B. Equivalents weight of `H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` are equal to half their molecular weights when acting as reducing agents.C. 100 " mL of " 1 M solution of each is neutralised by equal volumes of 1 Nca (OH)_(2)`D. 100 " mL of " 1 M solution of each is oxidised by equal volumes of 1 M `KMnO_(4)`. |
|
Answer» Correct Answer - B::D `H_(2)C_(2)O_(4)`: As acid `n=2` As reducing agent `n=2` `NaHC_(2)O_(4)`: As acid `n=1` As reducing agent, `n=2` `E_(H_(2)CO_(2)O_(4))]_("as reducing agent")=(M)/(2)`: `E_(NaHC_(2)O_(4))]_("as reducing agent")=(M)/(2)` On reaction with `KMnO_(4)` `H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)`: both are acting as reducing agents with same n factor of 2. `m" Eq of "KMnO_(4)-=m" Eq of "H_(2)C_(2)O_(4)` `m" Eq of "KMnO_(4)-=m" Eq of "NaHC_(2)O_(4)` |
|
| 85. |
100 mL sample of hard water is titrated with 500 " mL of " 0.001 M EDTA solution at `pH=10`, using eriochrome black-T indicator to reach equivalence point. An equal another amount of hard water sample is boiled for 30 min. After filtration and cooling, the same sample is titrated with 200 " mL of " 0.011 M EDTA solution at `pH=10` using Mg-EDTA complex solution and erichrome black-T indicator to reach equivalence point. (i). Calculate the total hardness of water sample (temporary`+`permanent) in ppm of `CaCO_3`. (ii). Calculate the permanent hardness of water sample in ppm of `CaCO_3`. (iii). Calculate the temporary hardness of water sample in ppm of `CaCO_3` |
|
Answer» (i). Total hardness (First titration) `M_1V_1(EDTA)=M_2V_2` (Total `Ca^(2+)` and `Mg^(2+)` in Temporary and permanent hardness) `500xx0.001=M_2xx100` `M_2=0.005` (Total `Ca^(2+) and Mg^(2+)`) Total hardness in ppm of `CaCO_3=(0.005xx100xx10^(6))/(10^(3))` `=500ppm` (ii). Permanent hardness (second titration): `M_1V_1(EDTA)=M_2V_2` (total `Ca^(2+)` and `Mg^(2+)` due to permanent hardness) `200xx0.001=M_2xx100` `M_2=0.002` (Total `Ca^(2+)` and `Mg^(2+)` in permanent hardness) Permanent hardness in ppm of `CaCO_3`) `=(0.002xx100xx10^(6))/(10^(3))=200ppm` (iii). Temporary hardness in ppm of `CaCO_3=500-200=300ppm` |
|
| 86. |
Two acids A and B are titrated separately each time with 25 " mL of " `N-Na_(2)CO_(3)` solution to requrie 10 mL and 40 mL respectively, of their solution for complete neutralisation. What volume of A and B would you mix to produce 1 L of N-acid solution? |
|
Answer» `N_(A)=(25)/(10)=2.5N` `N_(B)=(25)/(40)=0.625N` Let V is the volume for solution of `N_(A)` and for the solution B volume is `(1-V)L`. `therefore2.5V+0.625(1-V)=1` equivalent of acid `thereforeV=0.2L` `V_(A)=200mL,V_(B)=800mL` |
|
| 87. |
If a g is the mass of `NaHC_(2)C_(2)O_(4)` required to neutralize 100mL of 0.2M NaOH and b g that required to reduce 100mL of `0.02mL `KMnO_(4)` in acidic medium then:A. a=bB. 2a=bC. a=2bD. none of these |
|
Answer» Correct Answer - D |
|
| 88. |
A mixture of `n_(1)` moles of `Na_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` is titrated separately with `H_(2)O_(2)` and `KOH`, to reach at equivalence point. Which of the following statement is/are correct?A. Moles of `H_(2)O_(2)` and `KOH` are `n_(1)+n_(2)` and `n_(2)`B. Moles of `H_(2)O_(2)` and `KOH` is `n_(1)+(n_(2))/(2)` and `n_(1)`C. n-factors of `NaHC_(2)O_(4)` with KOH and `H_(2)O_(2)`, respectively, are 1 and 2.D. n-factors of `Na_(2)C_(2)O_(4)` with `H_(2)O_(2)` and KOH, respectively, are 2 and 1. |
|
Answer» Correct Answer - A::C `Na_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` both reacts with `H_(2)O_(2)` as reducing agent only. (n factor for both `=2`) With `H_(2)O_(2)`: " Eq of "`H_(2)O_(2)=" Eq of "Na_(2)C_(2)O_(4)+" Eq of "NaHC_(2)O_(4)` `2xx` moles of `H_(2)O_(2)=n_(1)xx2+n_(2)xx2` Moles of `H_(2)O_(2)=(2(n_(1)+n_(2))/(2)=(n_(1)+n_(2))` With KOH: Only `NaHC_(2)O_(4)` reacts with KOH as acid base titration n factor `=1(one H^(o+)` ion) " Eq of "`KOH=" Eq of "NaHCO_(3)` `1xx` Moles of `KOH=n_(2)xx1` Moles of KOH`=n_(2)` `therefore` Moles of `H_(2)O_(2)` and KOH are: `(n_(1)+n_(2))` and `n_(2)`. n-factor of `NaHC_(2)O_(4)` with KOH and `H_(2)O_(2)=1 and 2` n-factor of `Na_(2)C_(2)O_(4)` with `H_(2)O_(2)` and `KOH=2 and 2`. |
|
| 89. |
A mixture of `NaHC_(2)O_(4) and KHC_(2)O_(4) .H_(2)C_(2)O_(4)` required equal volumess of `0.2 N KMnO_(4) and 0.12N NaOH ` separtely. What is the molar ration `NaHC_(2)O_(4) and KHC_(2)O_(4) .H_(2)O_(4)` in the mixture?A. `6:1`B. `1:6`C. `1:3`D. `3:1` |
|
Answer» Correct Answer - D |
|
| 90. |
STATEMENT-1: `0.1 MH_(3)PO_(3)`(aq) solution has normality equal to 0.3 N when completely reacted with NaOH. STATEMENT-2 : `H_(3)PO_(3)` is a dibasic acid.A. If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT-1B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT- is 1 TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
|
Answer» Correct Answer - D |
|
| 91. |
`100 cm^(3)` of a solution of an acid (Molar mass =98) containing 29.4 g of the acid per litre were completely neutrazed by `90.0cm^(3)` of aq. NaOH cotaining 20 g of NaOH per`500 cm^(3)`. The basicity of the acid isA. 3B. 2C. 1D. data insufficient |
|
Answer» Correct Answer - A |
|
| 92. |
`30 mL` of `CH_(3)OH (d = 0.8 g//cm^(3))` is mixed with `60 mL` of `C_(2)H_(5)OH(d = 0.92 g//cm^(2))` at `25^(@)C` to form a solution of density `0.88 g//cm^(3)`. Select the correct option(s) :A. Molarity and molality of resulting solution are 6.33 and 13.59 respectivelyB. The mole fraction of solute and molality are 0.385 and 13.59 respectivelyC. Molarity and % change in volume are 0.615 and zero respectivelyD. Mole fraction of solvent and molality are 0.615 and 13.59 respectively |
|
Answer» Correct Answer - B::C |
|
| 93. |
The strength of `H_(2)O_(2)` is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of `H_(2)O_(2)` on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of `H_(2)O_(2)`gives 10 litre of `O_(2)` at 1 atm and 273 K The decomposition of `H_(2)O_(2)` is shown as under : `H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g)` `H_(2)O_(2)` can acts as oxidising as well as reducing agent. As oxidizing agent `H_(2)O_(2)` is converted into `H_(2)O` and as reducing agent `H_(2)O_(2)` is converted into `O_(2)`. For both cases its n-factor is 2. ` :. "Normality " "of" H_(2)O_(2)` " solution " `=2xx "molarity of" H_(2)O_(2)` solution 40 g `Ba(MnO_(4))_(2)` (mol.mass=375) sample containing some inert impurities in acidic medium completely reacts with 125 mL of "33.6 V" of `H_(2)O_(2)`. What is the percentage purity of the sample ?A. 0.2812B. 0.7031C. 0.85D. None of these |
|
Answer» Correct Answer - B |
|
| 94. |
In the complex with formula `MCl_(3).4H_(2)O` the co-ordination number of the metal M is six. And there is a no molecule of hydration in it. The volume of 0.1 M `AgNO_(3)` solution needed to precitate the free chloride ions in 200 ml of 0.01 M solution of the complex isA. 40mLB. 20mLC. 60mLD. 80mL |
|
Answer» Correct Answer - B The complex will be `M[(H_(2)O)_(4) Cl_(2)[Cl_(2)` the complex will have free chloride ion concentration of (0.01 M). `1 Cl^(-) + 1AgNO_(3) to AgCl + NO_(3)^(-)` ` (M_(1)V_(1))/(n_(1)) = (M_(2)V_(2))/(n_(2)) or (0.01 xx 200)/(1) = (0.1 xx V_(2))/(1)` `V_(2) = mL ]` |
|
| 95. |
Sulphur forms the chlorides `S_(2)Cl_(2)` and `SCl_(2)`. The equivalent mass of sulphur in `SCl_(2)` isA. 8g/molB. 16g/molC. 64.8g/molD. 3g/mol |
|
Answer» Correct Answer - B |
|
| 96. |
`C_(7)H_(6)O_(3) + C_(4)H_(6)O_(3) rightarrow C_(9)H_(8)O_(4)O_(2)` What is the percent yield if 0.85g of aspirin is formed in the reaction of 1.00g of salicylic acid with excess acetic anydride ? `{:("Substance", "Molar Mass"), (C_(7)H_(6)O_(3), 138.12g.mol^(-1)), (C_(4)H_(6)O_(3), 102.09g.mol^(-1)), (C_(9)H_(8)O_(4), 180.15g. mol^(-1)), (C_(2)H_(4)O_(2), 60.05g.mol^(-1)):}`A. 0.65B. 0.77C. 0.85D. 0.91 |
|
Answer» Correct Answer - A Mass of aspirin (theoretical) formed from 1g salicylic acid=`(180.15)/(138.12)=13.04g` `"% yield"=("Actual amount of product")/("Theoretical amount of product")xx100` `=(0.85)/(1.304)xx100=65%` |
|
| 97. |
Magnesium hydroxide, `Mg(OH)_(2)` is the white milky substance in milk of magnesia. What mass of `Mg(OH)_(2)` is formed when 15mL of 0.18M of NaOH combined with 12mL of 0.14M `MgCl_(2)`? The molar mass of `Mg(OH)_(2)` is 58.3`g mol^(-1)`A. 0.079gB. 0.097C. 0.16gD. 0.31g |
|
Answer» Correct Answer - A `n_(MgCl_(2))=(MV)/(1000)=(12xx0.14)/(1000)=1.68xx10^(-3)` `n_(NaOH)=(15xx0.18)/(1000)=2.7xx10^(-3)` NaOH will be limiting reagent because on complete consumption NaOH give least amount of `Mg(OH)_(2)`. `"Mass of "Mg(OH)_(2)=(1)/(2)xx2.7xx10^(-3)xx58.3=0.079g` |
|
| 98. |
One gram of charcoal adsorbs 400 " mL of " 0.5 M acetic acid to form a mono layer and the molarity of acetic acid reduced to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. The surface area of charcoal is `3.01xx10^(2)m^(2)g^(-1)`. |
|
Answer» Initial millomoles of `CH_(3)COOH=100xx0.5=50` Final millimoles of `CH_(3)COOH=100xx0.49=49` millimoles adsorbed `=50-49=1` moles adsobed `=(1)/(1000)` Molecules adsorbed`=(1)/(1000)xx6.023xx10^(23)` `=6.023xx10^(20)` Area per molecule`=("Total area")/("Number of molecules")` `=(3.01xx10^(2))/(6.023xx10^(20))=5xx10^(19)m^(2)` |
|
| 99. |
When 1.82 g a of mixture of Al and an unkown metal. Arranged in the series of the standard reduction electrode potential below hydrogen, was dissolved in `HCl`, 0.672 L of `H_2`, measured at STP was liberated. To oxidise this mixture, 0.56 L of `O_2` measured at STP was needed which unknown metal was taken? Determine the mass percentage of metals in the mixture. |
|
Answer» (a). Only Al gives `H_2`, not the other metal below `H_2`. `2Al+6"HCl"to2AlCl_3+3H_2` Ew of `H_2=" Eq of "Al (xg)` `(0.672)/(11.2)=((x)/(27))/(3)` `(1 " mol of "H_2=2 " Eq of "H_2=2.4L)` `(1 " Eq of "H_2=11.2L)` `thereforex=0.54 of Al` The weight of unknown metal `(M)=1.82-0.54=1.28g` (b). `4Al+3O_2to2Al_2O_3` `{:(4 mol Al, 3 " mol of "O_2),(4xx27g,3xx32g O_2):}` `4xx27g of Al implies3xx32g O_2` `0.54 of "Al "implies(3xx32)/(4xx27)xx0.54 g of O_2` `(1 " mol of "O_2=32g=22.4L)` Volume of `O_2=(22.4)/(32)xx(3)/(4)xx(32)/(27)xx0.54` `=0.336 L of O_2` `O_2` used for metal `M=(0.56-0.336)L=0.224L` Let M be monovalent. `1 " mol of "M= 1 " mol of "O_2` `implies(1.28)/("Atomic weight")=(0.224)/(22.4)" mol of "O_2` `therefore` Atomic weight`=(1.28)/(0.224)xx22.4` `=128` which is not possible So the metal is divelent `2 mol=1 " mol of "O_2` `therefore` Atomic weight `=(128)/(2)=64` i.e., metal is copper. |
|
| 100. |
Mercuric iodate `[Hg_5(IO_6)_2]` reacts with a mixutre of KI and HCl according to the following equation. `Hg_5(IO_6)_2+34KI+24"HCl"to5K_2HgI_4+8I_2+24KCI+12H_2O` or `Hg_5(IO_6)_2+34I^(ɵ)+24H^(o+)to5HgI_4^(2-)+8I_2+12H_2O` The liberated `I_2` is titrated with `Na_2S_2O_3` solution. 1.0 " mL of " `Na_2S_2O_3` is equivalent to 3.992 g of `CuSO_4.5H_2O`. What volume (in mL) of the `Na_2S_2O_3` solution will be required to react with liberated `I_2` from `14.485 g Hg_5(IO_6)_2`.`[Hg=200.5,Cu=63.5,I=127]`? `Mw(Hg_5(IO_6)_2)=1448.5g` `Mw(CuSO_4.5H_2O)=249.5g` |
|
Answer» 1 " mol of "`Hg_5(IO_6)_2-=8I_20=16" mol of "CuSO_4.5H_2O`. (Since 1 " mol of "`I_2` reacts with 2 " mol of "`CuSO_4.5H_2O` in iodometric titration) `1448.5g of Hg_5(IO_6)_2-=8I_2-=16xx249.5g of CuSO_4.5H_2O` `3.992 g CuSO_4.5H_2O-=1" mL of " Na_2S_2O_3` `therefore39.92 g CuSO_4.5H_2O-=10" mL of " Na_2S_2O_3` `therefore` Volume of `Na_2S_2O_3=10mL` |
|