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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.
| 1. |
The purity of `H_2O_2` in a given sample is `85%.` Calculate the weight of impure sample of `H_2O_2` which requires `10mL` of `M//5 KMnO_4` solution in a titration in acidic mediumA. 2 gB. 0.2 gC. 0.17 gD. 0.15 g |
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Answer» Correct Answer - B `m" Eq of "MnO_(4)^(ɵ)-=m" Eq of "H_(2)O_(2)` `10xx(M)/(5)xx5-=m" Eq of "H_(2)O_(2)` `implies10mEq-=m" Eq of "H_(2)O_(2)` Weight of `H_(2)O_(2)-=10xx10^(-3)xx(34)/(2)=0.17g` Weiht of impure `H_(2)O_(2)=(100)/(85)xx0.17=0.2g` |
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| 2. |
An impure sample of sodium oxalate `(Na_(2)C_(2)O_(4)` weighing 0.20 g is dissolved in aqueous solution of `H_(2)SO_94)` and solution is titrated at `70^(@)C`,requiring 45 mL of 0.02 M `KMnO_(4)` solution. The end point is overrun, and back titration in carried out with 10 mL of 0.1 M oxalic acid solution.Find the purity of `Na_(2)C_(2)O_(4)` in sample:A. 75B. 83.75C. 90.25D. None of these |
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Answer» Correct Answer - B |
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| 3. |
`60ml` of mixture of equal volumes of `Cl_(2)` and an oxide of chloine, i.e., `Cl_(2)O_(n)` was heated and then cooled back to the original temperature. The resulting gas mixture was found to have volume of `75ml`. On treatment with `KOH` solution, the volume contracted to `15ml`. Assume that all measurements are made at the same temperature and pressure. Deduce the value of `n` in `Cl_(2)O_(n)`. The oxide of `Cl_(2)` n heating decomposes quantiatively to `O_(2)` and `Cl_(2)`. |
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Answer» Let the oxide is `Cl_(x).O_(y)`. `Cl_(x)O_(y)to(x)/(2)Cl_(2)+(y)/(2)O_(2)` Volume of `Cl_(x)O_(y)=` volume of `Cl_(2)=3 mL` After cooling, volume `=75mL` This corresponds to volume of `Cl_(2)` initially plus volume of `Cl_(2)` produced and `O_(2)` produced. `thereforeV_(Cl_(2))+V_(Cl_(2))` (produced)`+V_(O_(2))=75mL` `V_(Cl_(2))` (produced) `+V_(O_(2))=75-30=45mL` (Since `V_(Cl_(2))` initiall `=30mL`) NaOH apart from `CO_(2)` also absorbs `Cl_(2)`. So after `NaOH` treatment, the residual volume corresponds to the volume of `O_(2)(=15mL)` `therefore` Volume of `Cl_(2)=45-15=30mL` Volume of `Cl_(2)=30xx(x)/(2)=30impliesx=2` Volume of `O_(2)=30xx(y)/(2)=15impliesy=1` Hence formula of chlorid is `Cl_(2)O`. |
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| 4. |
`12. g` of an impure sample of arsenious oxide was dissolved in water containing `7.5 g` of sodium bicarbonate and the resulting solution was diluted to `250 mL`. `25 mL` of this solution was completely oxidised by `22.4 mL` of a solution of iodine. `25 mL` of this iodine solution reacted with same volume of a solution containing `24.8 g` of sodium thiosulphate `(Na_(2)S_(2)O_(3).5H_(2)O)` in one litre. Calculate teh percentage of arsenious oxide in the sample ( Atomic mass of `As=74`) |
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Answer» `As_(2)O_(3)=12.0g`. It reacts with `NaHCO_(3)` to give `Na_(3)AsO_(3)`. Its reaction with `I_(2)` shows the changes. (a). `undersetunderset(2x=2)(2x-6=0)(overset(+3)(As_(2))O_(3))toundersetunderset(2x=10)(2x-10=0)(overset(+5)(As_(2))O_(5))+4e^(-)(n=4)` (b). `I_(2)+2e^(-)to2I^(ɵ)(n=2)` (c). `2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2e^(-)` (n=(2)/(2)=1)` `m" Eq of "As_(2)O_(3)` in `25mL=m" Eq of "I_(2)=22.4xxN` Aso, `m" Eq of "I_(2)=m" Eq of "hypo (S_(2)O_(3)^(2-))` `=NxxV` `Nxx25=(24.8)/((248)/(1))xx25` `[Ew of Na_(2)S_(2)O_(3).5H_(2)O=(248)/(1)(n=1)]` `therefore_(I_(2))=(1)/(10)` `m" Eq of "As_(2)O_(3)` in `25 m22.4xx(1)/(10)=2.24` or `m" Eq of "As_(2)O_(3)` in `250mL=(2.24xx250)/(25)=22.4` `(W)/(Ew)xx10^(3)=22.4` `W_(As_(2)O_(3))=(22.4xx198)/(4xx10^(3))=1.1088` `% of As_(2)O_(3))=(22.4xx198)/(4xx10^(3))=1.1088` `% of As_(2)O_(3)=(1.1088)/(12)xx100=9.24%` |
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| 5. |
`K_(2)Cr_(2)O_(7)` is obtained in the following steps: `2FeCrO_(4)+2Na_(2)CO_(3)+OtoFe_(2)O_(3)+2Na_(2)CrO_(4)+2CO_(2)` `2Na_(2)CrO_(4)+H_(2)SO_(4)toNa_(2)Cr_(2)O_(7)+H_(2)O+Na_(2)SO_(4)` `Na_(2)Cr_(2)O_(7)+2KCltoK_(2)Cr_(2)+2NaCl` To get 0.25 " mol of "`K_(2)Cr_(2)O_(7)`, " mol of "`50%` pure `FeCrO_(4)` required.A. 1 molB. 0.50 molC. 0.25 molD. 0.125 mol |
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Answer» Correct Answer - A 1 mol `K_(2)Cr_(2)O_(7)` is obtained from 2 " mol of "`FeCrO_(4)` 0.25 mol `K_(2)Cr_(2)O_(7)` is obtained from `-0.50 " mol of "FeCrO_(4) (100%` pure) `=1.00 mol (if 50%` pure) |
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| 6. |
How many moles of `O_(2)` will be liberated by one mole of `CrO_(5)` is the following reaction: `CrO_(5) + H_(2) SO_(4) rarr Cr_(2) (SO_(4))_(3) + H_(2)O + O_(2)`A. `(5)/(2)`B. `(5)/(4)`C. `(9)/(2)`D. none of these |
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Answer» Correct Answer - D Balance the reaction: `2CrO_(5)+3H_(2)SO_(4)toCr_(2)(SO_(4))_(3)+2H_(2)O+(7)/(2)O_(2)` `implies1 mol CrO_(5)` liberated `(7)/(2)` mol `O_(2)` |
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| 7. |
A sample containing 0.4775 of `(NH_(4))_(2)C_(2)O_(4)` and inert material was dissolved in water and made strongly alkaline with KOH which converted `NH_(4)^(o+)` to `NH_(3)` The liberated `NH_(3)` was distilled of `H_(2)SO_(4)` was back titrated with 11.3 " mL of " 0.1214 M NaOH. Calculate (a) `% of (NH_(4))_(2)C_(2)O_(4)=124.10` And atomic weight of `N=14.0078`. |
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Answer» `(NH_(4))_(2)C_(2)O+(4)+2KOHtoK_(2)C_(2)O_(4)+2NH_(3)+2H_(2)O` Total `H_(2)SO_(4)` used `=50xx0.05035xx2NH_(2)SO_(4)` `=5.035m" Eq of "H_(2)SO_(4)` Excess of `H_(2)SO_(4)=11.3xx0.124M NaOH` `=1.372 m" Eq of " NaOH` `=1.372 m" Eq of "H_(2)SO_(4)` `H_(2)SO_(4)` used `=5.035-1.372` `=3.663 m" Eq of "H_(2)SO_(4)` `=3.663 mEq NH_(3)` `=3.663 m" Eq of "(NH_(4))_(2)C_(2)O_(4)` Ew of `(NH_(4))_(2)C_(2)O_(4)=(124.1)/(2)=62.06g` Weight of `(NH_(4))_(2)C_(2)O_(4)=3.663xx10^(-3)xx62.05=0.2273g` `124.1 g of (NH_(4))_(2)C_(2)O_(4)` contains `14.0078 g of N`. `0.2273 g of (NH_(4))_(2)C_(2)O_(4)=(14.0078xx0.2273)/(124.1)=0.02565` g `% of N=(0.02565)/(0.4775)xx100=5.373%` `% of (NH_(4))_(2)C_(2)O_(4)=(0.2273)/(0.4775)xx100=5.373%` |
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| 8. |
2.0 g sample of NaCN is dissolved in 50 " mL of " 0.3 M mild alkaline `KMnO_(4)` and heated strongly to convert all the `CN^(ɵ)` to `OCN^(ɵ)`. The solution after acidification with `H_(2)SO_(4)` requries 500 " mL of " 0.05 M `FeSO_(4)` Calculate the percentage purity of NaCN in the sample. |
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Answer» (a). `undersetunderset(x=2)(x-3=-1)(Coverset(+2)(N^(ɵ)))toundersetunderset(x=4)(x-3-2=-1)(OCoverset(+4)(N^(ɵ)))+2e^(-)(n=2)` (b). `3e^(-)+undersetunderset(x=7)(x-8=-1)(MnO_(4)^(ɵ))toundersetunderset(x=3)(x-4=0)(MnO_(2))` (c). `5e^+MnO_(4)^(ɵ)toMn^(2+)` `(n=5` in acidic medium) (d). `Fe^(2+)toFe^(3+)+2e^(-)(n=1)` `m" Eq of "KMnO_(4)` added in basic medium`=50xx0.3` (n-factor`=3`) `=45.0` `m" Eq of "KMnO_(4)` in acidic medium `(n=5)` left after reaction with `NaCN=m" Eq of "FeSO_(4)` used `=500xx0.5xx1` (n-factor`=1`) `=25.0` `m" Eq of "KMnO_(4)` (n-factor`=3`) left`=(25xx3)/(5)=15` `m" Eq of "NaCN` in sample `=m" Eq of "KMnO_(4)` added `-m" Eq of "KMnO_(4)` left `=45-15=30` `therefore(Weight)/(Ew of NaCN)xx10^(3)=30` `[Ew of NaCN=(49)/(2)` (n factor`=2`)] `(W)/((49)/(2))xx10^(3)=30` `W_(NaCN)=0.735g`. `% of NaCN=(0.735)/(2.0)xx100=36.75%` |
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| 9. |
10 " mL of " `NaHC_(2)O_(4)` is oxidised by 10 " mL of " 0.02 M `MnO_(4)^(ɵ)`. Hence, 10 " mL of " `NaHC_(2)O_(4)` is neutralised byA. 10 " mL of " 0.1 M NaOHB. 10 " mL of " 0.02 M NaOHC. 10 " mL of " 0.1 M `Ca(OH)_(2)`D. 10 " mL of " 0.05 N `Ba(OH)_(2)` |
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Answer» Correct Answer - A In `NaHC_(2)O_(4),C_(2)O_(4)^(2-)` oxidised to `CO_(2)` and `H^(o+)` is neutralised. `Ew_(NaHC_(2)O_(4))(as C_(2)O_(4)^(2-)=(M)/(2)` `(as H^(o+))=(M)/(1)` `10xxN_(1)(NaHC_(2)O_(4))=10xx0.1N MnO_(4)^(ɵ)` `N_(1)=0.1N` |
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| 10. |
`CN^(ɵ)` is oxidised by a strong oxidising agent to `NO_3^(ɵ)` and `CO_2` or `CO_3^(2-)` depending upon the acidity of the reaction mixture. `HNO_3` a strong oxidising agent is reduced by a moderate reducing agent to NO. Write the balanced equation of `HNO_3` with KCN. `CH^(ɵ)toCO_2+NO_3^(ɵ)` `NO_3^(ɵ)NO` If this reaction is carried out, what safety precautions are required? |
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Answer» `(overset(-4)(C)overset(+3)(N))^(-2)orCN^(ɵ)toCO_2+NO_3^(ɵ)` `C^(4-)+2H_2OtoCO_4+8e^(-)+4H^(o+)` `underline(underset(x=3)(N^(3+))+3H_2Otounderset(x=5)(NO_3^(ɵ))+2e^(-)+6H^(o+))` `underline(CN^(ɵ)+5H_2OtoCO_2+10^(o+)+NO_3^(ɵ)+10e^(-))` …. (i) Reduction of `NO_3^(ɵ)` to `NO`: `4H^(o+)+3e^(-)+undersetunderset(x=5)(x-6=-1)(NO_3^(o+))toundersetunderset(x=2)(x-2=0)(NO)+2H_2O+2H_2O` ....(ii) Multiply equation (i) by 3 and equation (ii) by 10 and add them to get the final redox equation: `underline(3CN^(ɵ)+10H^(o+)+7NO_3^(o+)to3CO_2+10NO+5H_2O)` Acid `(H^(o+))` and `CN^(ɵ)` ion react to give `HCN`, a poisonous and deadly gas, unless the oxidising properties of the acid cause the oxidation of `CN^(ɵ)` before the HCN can escape. |
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| 11. |
Direct titration of `I_(2)` with a reducing agent is called iodimetry. If `I_(2)` is leberated by the oxidation of `I_(ɵ)` ion by a strong oxidising agent in neutral or acidic medium, the liberated `I_(2)` is then titrated with reducing agent. Iodometry is used to estimate the strngth of the oxidising agent. For example, in the estimation of `Cu^(2+)` with `S_(2)O_(3)^(2-)` `Cu^(2+)+I^(ɵ)toCuI_(2)+I_(2)` (iodometry) `I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(ɵ)` (iodimetry) Strach is used as an indicator at the end point, which forms bluecoloured complex with `I_(3)^(ɵ)` Disappearance of blue colourindicates the end point whe free `I_(2)` in not present. Q. The volume of KI solution used for `CuSO_(4)` is:A. 1LB. 2LC. 4LD. 5L |
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Answer» Correct Answer - C `1 mol CuSO_(4)-=1 mol S_(2)O_(3)^(2-)` `-=(1)/(2)mol I_(2)-=2 mol KI` `-=0.4xx2=0.8mol` `therefore0.2MxxV=0.8` `V=4L` |
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| 12. |
Hydrogen peroxide solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified with dilute of `H_(2)SO_(4)`. The same volume of the `KMnO_(4)` solution is just decolourised by `10 mL` of `MnSO_(4)` in neutral medium simultaneously forming a dark brown precipitate of hydrated `MnO_2`. The brown precipitate is dissolved in `10 mL` of `0.2 M` sodium oxalate under boiling condition in the presence of dilute `H_(2)SO_(4)`. Write the balanced equations involved in the reactions and calculate the molarity of `H_(2)O_(2)`. |
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Answer» `2KMnO_(4)+3H_(2)SO_(4)+5H_(2)O_(2)toK_(2)SO_(4)+2MnSO_(4)+8H_(2)O` `2KMnO_(4)+3MnSO_(4)+2H_(2)Oto5MnO_(2)+2H_(2)SO_(4)+K_(2)SO_(4)` `MnO_(2)+Na_(2)C_(2)O_(4)+2H_(2)SO_(4)toMnSO_(4)+Na_(2)SO_(4)+2H_(2)O+2CO_(2)` millimoles of `Na_(2)C_(2)O_(4)=10xx0.2=2` `m" Eq of "Na_(2)C_(2)O_(4)=4` `m" Eq of "MnO_(2)=4` `m" Eq of "KKMnO_(4)=4` `m" Eq of "H_(2)O_(2)=4` millimoles of `H_(2)O_(2)=2` millimoles of `H_(2)O_(2)=` Molarity`xxV_(mL)` or `2=` molarity `xx20` or molarity `=0.1` |
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| 13. |
Find the molality of 1.0 L solution of 90% `H_2SO_4` by weight/volume. The density of the solution is 1.47 `gc c^(-1)` |
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Answer» (a). `90% H_2SO_4 (weight)/(volume)implies(90g)/(100mL)` `implies(Mass)/(L)=900g` `("Mass of solution")/(L)=1000xx1.47=1470g` `impliesMass of H_2O=1470-900=570g` `m=((900)/(98))/(570)xx1000=16.11` |
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| 14. |
Borax in water gives the following: `B_4O_7^(2-)+7H_2Oto4H_2BO_3+2overset(ɵ)(O)H` How many grams of borax `(Na_2B_4O_7.10H_2O)` are required to (a). Prepare 50 " mL of " 0.2 M solutionl. (b). Neutralise 25 " mL of " 0.2 M `H_2SO_4` `(Mw` of borax`=382`) |
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Answer» `B_4O_7^(2-)+7H_2Oto4H_3BO_3+2overset(o+)(O)H` So, borax is a diacidic bese. (a). Millimoles `=(Weight)/(Mw)xx1000` `implies50xx0.2=(weight)/(382)xx1000impliesweight=3.82g` (b). `mEq=(weight)/(E)xx1000implies2xx0.2xx25` `=(weight)/((383)/(2))xx1000` Weight`=1.91g` |
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| 15. |
How many grams of borax `(Na_2B_4O_7.10H_2O)` are required to neutralise 25 " mL of " 0.2 M HCl and `H_2SO_4` separately? |
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Answer» Borax in water gives `underset((1 mol))(B_4O_7^(2-))+7H_2Oto4H_3BO_3+underset(2 mol)(2overset(ɵ)(O)H)` 1 " mol of "borax `-=2 mol overset(ɵ)(O)Himplies` acidity `=2` (a). m" Eq of "borax `=m" Eq of "HCl` `("weight")/((382)/(2))xx1000=(0.2xx2)xx25` [Since HCl is a monobasic acid] Weight `=0.955g=` Mass of borax (b). m" Eq of "borax `=m" Eq of "H_2SO_4` `("weight")/((382)/(2))xx1000=(0.2xx2)xx25` [Since `H_2SO_4` is a dibasic acid]` Weight `=1.91g=` mass of borax |
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| 16. |
When a mixture of NaBr and NaCl is separately digested with `H_2SO_4` all the halogens are expelled and `Na_2SO_4` is formed quantitatively with a particular mixture, it was found that the weight of `Na_2SO_4` obtained was precisely the same as the weight of `NaBr+NaCl` mixture taken. Calculate the ratio of the weight of NaCl and NaBr in the mixture. |
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Answer» Let the weight of `NaCl` be x g and the weight of NaBr be y g. ,brgt We have to calculate `(x)/(y)` `2NaCl-=Na_2SO_4` `2xx58.5g of NaCl-=142g of Na_2SO_4` `x g of NaCl-=(142)/(2xx58.4)xx x g of Na_2SO_4` `2NaBr-=Na_2SO_4` `2xx103g of NaBr-=142 g of Na_2SO_4` `y g of NaBr-=(142)/(2xx103)xx x g of Na_2SO_4` `therefore(142x)/(2xx58.5)+(142x)/(2xx103)=x+y` `1.21x-x=y-0.69y` `0.21x=0.31y` `(x)/(y)=(0.31)/(0.21)=1.475` The ratio of weights of `NaCl:NaBr` is `1.476:1` |
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| 17. |
Find the amound of iron pyrites `(FeS_2)` which is sufficient to produce enough `SO_2` on roasting (heating in excess of `O_2`) such that is `(SO_2)` completely decolourise a 1 L solution of `KMnO_4` containing 15.8 g `L^(-1)` of it. The equation are `FeS_2+O_2toFe_2O_3+SO_2` `KMnO_4+SO_2toMnSO_4+H_2SO_4+KHSO_4` |
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Answer» First calculate the amount of `SO_2` required to decolourise `15.8gL^(-1)` of `KMnO_4` solution. For this, balance the following chemical reaction. The balanced equation is as: `KMnO_4+SO_2toMnSO_4+H_2SO_4+KHSO_4` `2KMnO_4+5SO_2+2H_2Oto2MnSO_4+H_2SO_4+2KHSO_4` `2" mol of "KMnO_4-=5 " mol of "SO_2` Calculate moles in `15.8g L^(-1) of KMnO_4` Using strength `(gL^(-1))=(M)/(Mw)` `implies1.0L of KMnO_4` contains 0.1 mol Hence, moles of `SO_4` required `=(5)/(2)(0.1)=0.25` To calculate the amount of pyrites, we have to balance the following reaction. `FeS_2+O_2toFe_2O_3+SO_2` Balancing the reaction, we have `4FeS_2+11O_2to2Fe_2O_3+8SO_2` From stoichiometry of roasting, we have: `8 " mol of "SO_2-=4" mol of "FeS_2` `0.25 " mol of "SO_2-=(4)/(8)(0.25)" mol of "FeS_2` `=0.125 " mol of "FeS_2` Mass of `FeS_2=0.125xx120=15gL^(-1)` |
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| 18. |
One litre of an acidified solution of `KMnO_4` containing 15 g of `KMnO_4` is decolourised by passing sufficient amount of `SO_2`. If `SO_2`. If `SO_2` is produced by roasting of iron pyrites `(FeS_2)`. What will be the amount of pyrites required to produce the necessary amount of `SO_2` |
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Answer» `underset(4xx120)(4FeSO_(2))+11O_(2)to2Fe_(2)O_(3)+underset(8xx64)(8SO_(2))` Equivalents of `KMnO_(4)=(15.0)/(31.6)xx1` litre `=0.5eq` `=0.5" Eq of "SO_(2)` `=0.5xx(64)/(2)g of SO_(2)` `=16g of SO_(2)` `8xx64g of SO_(2)` is produced from `2xx120g of FeS_(2)` `therefore16 g of SO_(2)` is produced from `(4xx120)/(8xx64)xx16=15g` |
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| 19. |
20 " mL of " `(M)/(60)` `KBrO_3` was reacted with a sample of `SeO_(3)^(2-)` 20 The `Br_(2)` thus evolved was removed and the excess of `NaAsO_(2)` The reaction involved are `SeO_(3)^(2-)+BrO_(3)^(ɵ)+H^(o+)toSeO_(4)^(2-)+Br_(2)+H_(2)O` .. (i) `BrO_(3)^(ɵ)+ASO_(2)^(ɵ)+H_(2)OtoBr^(ɵ)+AsO_(4)^(3-)+H^(o+)` ..(ii) `[Mw(SeO_(3)^(2-))=79+48=127gmol^(-1)]` Q. n-factors of `BrO_(3)^(ɵ)` ion in equations (i) and (ii) respectively areA. 10,6B. 5,6C. 6,10D. 6,5 |
| Answer» Correct Answer - B | |
| 20. |
20 " mL of " `(M)/(60)` `KBrO_3` was reacted with a sample of `SeO_(3)^(2-)` 20 The `Br_(2)` thus evolved was removed and the excess of `NaAsO_(2)` The reaction involved are `SeO_(3)^(2-)+BrO_(3)^(ɵ)+H^(o+)toSeO_(4)^(2-)+Br_(2)+H_(2)O` .. (i) `BrO_(3)^(ɵ)+ASO_(2)^(ɵ)+H_(2)OtoBr^(ɵ)+AsO_(4)^(3-)+H^(o+)` ..(ii) `[Mw(SeO_(3)^(2-))=79+48=127gmol^(-1)]` Q. Amount of `SeO_(3)^(2-)` in mg isA. 19.4 mgB. 194 mgC. 970 mgD. 97 mg |
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Answer» Correct Answer - D `m" Eq of "SeO_(3)^(2-)=(55)/(36)` Weight of `SeO_(3)^(2-)=mEqxx10^(-3)xxEw` `(Ew of SeO_(3)^(2-)=(127)/(2))(n=2)` `=(55)/(36)xx10^(-3)xx(127)/(2)g` `=0.097g=97g` |
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| 21. |
20 " mL of " `(M)/(60)` `KBrO_3` was added to a sample of `SeO_3^(2-)`, The bromine evolved was removed and the excess of `KBrO_3` was titrated with 5.1 " mL of " `(M)/(25)` solution of `NaAsO_2`. Calculate the amound of `SeO_3^(2-)` and balance the equation. `SeO_3^(2-)+BrO_3^(ɵ)+H^(o+)toSeO_4^(2-)Br_2+H_2O` `BrO_3^(ɵ)+AsO_2^(ɵ)+H_2OtoBr^(ɵ)+AsO_4^(3-)+H^(o+)` `(Br=80, K=39, As=75, Se=79)` |
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Answer» This question cannot be solved by the equivalent method, since the numbers of electrons in the two reactions are different. The n-factors for `BrO_3^(ɵ)` to `Br_2` and `BrO_3^(ɵ)` to `Br^(ɵ)` are different. So it has to be solved by mol method by balancing the two equations. First Reation: `{:[(undersetunderset(x=4)(x-6=2)(SeO_3^(2-))toundersetunderset(x=6)(x-8)(SeO_4^(2-))+2e^(-)),(undersetunderset(2x=10)(2x-12=-2)(10e^(-)+2BrO_3^(-))tounderset(2x=10)(Br_2)):}]_((n=(10)/(2)=5))^((n-2))` Second reaction: `{:[(undersetunderset(x=5)(x-6=-1)(6e^(-)+BrO_3^(ɵ)tounderset(x=-1)(Br^(ɵ)))),(undersetunderset(x=3)(x-4=-1)(AsO_2^(ɵ))toundersetunderset(x=5)(x-8=-3)(AsO_4^(3-)+2e^(-))):}]_((n=2))^((n=6))` First reaction: `5SeO_3^(2-)+5H_2Oto5SeO_4^(2-)+10H^(o+)+10e^(-)` `underline(2BrO_3^(ɵ)+12H^(o+)+10e^(-)toBr_2+6H_2O)` `underline(5SeO_3^(2-)+2BrO_3^(ɵ)+2H^(o+)to5SeO_4^(2-)+Br_2+H_2O)` Second `BrO_3^(ɵ)+6H^(o+)+6e^(-)toBr^(ɵ)+2H_2O` `underline(3AsO_2^(ɵ)+6^(ɵ)+12H^(o+))` `underline(BrO_3^(ɵ)+3AsO_2^(ɵ)+3H_2OtoBr^(ɵ)+3AsO_4^(3-)+6H^(o+))` Excess of `KBrO_3=5.1xx(1)/(25)=0.204 ` mol `=0.204` m" mol of "`NaAsO_2` `3 mmol` of `AsO_2^(ɵ)=1 mmol` of `KBrO_3` `0.204 mmol` of `AsO_2^(ɵ)=0.068 m` " mol of "`KBrO_3` Total `KBrO_3=20xx(1)/(60)=0.333` mmol `KbrO_3` used `=0.333-0.068=0.265 mmol` `2mmol` of `BrO_3^(ɵ)=(5)/(2)xx0.265` `=0.6625 mmol` of `SeO_3^(2-)` (molecular weight of `SeO_3^(2-)` is 127) `implies0.6625xx10^(-3)xx127g of SeO_3^(2-)` `=0.8143 g of SeO_3^(2-)` |
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| 22. |
The molarity (m) of `KMnO_(4)` in the acidic medium is (densityh of `KMnO_(4)` solution`=1.58gmL^(-1)Mw(KMnO_(4))=158gmol^(-1))`A. 0.025B. 0.25C. 0.12D. 0.012 |
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Answer» Correct Answer - D `d_(sol)=M((Mw_(2))/(1000)+(1)/(m))` `M=0.02` `1.58=0.02((158)/(1000)+(1)/(m))` Solve for m: `m=0.012` |
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| 23. |
20 " mL of " `(M)/(60)` `KBrO_3` was reacted with a sample of `SeO_(3)^(2-)` 20 The `Br_(2)` thus evolved was removed and the excess of `NaAsO_(2)` The reaction involved are `SeO_(3)^(2-)+BrO_(3)^(ɵ)+H^(o+)toSeO_(4)^(2-)+Br_(2)+H_(2)O` .. (i) `BrO_(3)^(ɵ)+ASO_(2)^(ɵ)+H_(2)OtoBr^(ɵ)+AsO_(4)^(3-)+H^(o+)` ..(ii) `[Mw(SeO_(3)^(2-))=79+48=127gmol^(-1)]` Q. m" Eq of "`SeO_(3)^(2-)` isA. `(55)/(72)`B. `(55)/(36)`C. `(11)/(36)`D. `(11)/(6)` |
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Answer» Correct Answer - B m" mol of "`BrO_(3)^(ɵ)` reacted with `SeO_(3)^(2-)` in equation (i) `=(11)/(36) m" mol of "BrO_(3)^(ɵ)` `=(11)/(36)xx5` (n-factor `BrO_(3)^(ɵ)` in equation (i)) `=(55)/(36)m" Eq of "BrO_(3)^(ɵ)-=m" Eq of "SeO_(3)^(2-)` reacrted |
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| 24. |
(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): In the titration of strong acid and strong base, phenolphthalein is used as suitable indocator. Reason (R): IN the titration of strong acid and strong base, the equivalence points lies is the pH range of `(3.0-10.5)` and phenolphthalein have pH range of `(8.0-9.8)`. |
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Answer» Correct Answer - A Both (A) and (R) are correct and (R) is correct explanation of (A). |
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| 25. |
(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): Concentration of `H_(2)O_(2)` is expressed in volume Reason (R): Volume strength `=` normality `xx5.6` |
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Answer» Correct Answer - B Both (A) and (R) are correct but (R) is not the correct explanation of (A). |
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| 26. |
(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): The amount CO in a gas sample can be determined by using the reaction`I_(2)O_(5)+5COtoI_(2)+5CO_(2)` IF gas sample liberates `127.g of I_(2)`. Then 70 g of CO were present in the sample. Reason (R): CO gas is absorbed in ammonical CuCl solution. |
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Answer» Correct Answer - B Moles of `I_(2)=(127)/(254)=(1)/(2)mol` Moles of `CO=(5)/(2)` Weight of `CO=(5)/(2)xx28=70.0` Both (A) and (R) are correct but (R) is not the correct explanation of (A). |
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| 27. |
(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): Estimation of reducing substance by the use of standard `I_(2)` is called iodometry. Reason (R): in the reaction `I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2I^(ɵ)` The n factor of `S_(2)O_(3)^(2-)` is one. |
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Answer» Correct Answer - D (A) is false, because it is called iodimatry. (R) is true `("n-factor"=(2)/(2)=1)` |
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| 28. |
1 M NaOH solution was slowly added in to 1000 mL of 183.75 g impure `H_(2)SO_(4)` solution and the following plot was obtained. The percentage purity of `H_(2)SO_(4)` sample and slope of the curve respectively are: A. 75%, `-(1)/(3)`B. 80%, `-(1)/(2)`C. 80%, -1D. None of these |
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Answer» Correct Answer - C |
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| 29. |
Nitric acid canbe produced from `NH_(3)`in three steps process given below (I)`4NH_(3)(g)+5O_(2)(g)to4NO(g)+6H_(2)O(g)` (II)`2NO(g)+O_(2)(g)to 2NO_(3)(g)` `3NO_(2)(g)+H_(2)O(l)to 2HNO_(3)(aq)+NO(g)` percent yield of `1^(st)` ,`2^(nd)` and `3^(rd)` steps are respectively 50%,60% and 80% respectivley then what volume of `NH_(3)`(g) at 1 atm and `0^(@)`required to produced1575 g of `HNO_(3)` .A. 156.25B. 350 LC. 3500 LD. None of these |
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Answer» Correct Answer - C |
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| 30. |
`KMnO_4` reacts with `Na_2S_2O_3` in acidic, strongly basic and aqueous (neutral) media. `100mL` of `LMnO_4` reacts with 100 mL of 0.1 M `Na_2S_2O_3` in acidic, basic and neutral media. Q. The molarity `(M)` of `KMnO_4` solution in the acidic medium isA. 0.2 MB. 0.02 MC. 0.4 MD. 0.04 M |
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Answer» Correct Answer - B `MnO_(4)^(ɵ)-=S_(2)O_(3)^(2-)` `(2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2e^(-))` `100mLxxN-=100mLxx0.1xx1` (n-factor) `N=0.1` `M=(0.1)/("n factor")=(0.1)/(5)=0.02M` |
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| 31. |
`KMnO_4` reacts with `Na_2S_2O_3` in acidic, strongly basic and aqueous (neutral) media. `100mL` of `LMnO_4` reacts with 100 mL of 0.1 M `Na_2S_2O_3` in acidic, basic and neutral media. Q. The molarity (M) of `KMnO_4` in aqueous medium isA. 0.8 MB. 0.08 MC. 0.26 MD. 0.026 M |
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Answer» Correct Answer - C `MnO_(4)^(ɵ)-=S_(2)O_(3)^(2-)(S_(2)O_(3)^(2-)to2HSO_(4)^(ɵ)+8e^(-))(n=8)` `100mLxxN-=100mLxx0.1xx8` `N=0.8` `M=(0.8)/(3)=0.26` |
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| 32. |
`KMnO_4` reacts with `Na_2S_2O_3` in acidic, strongly basic and aqueous (neutral) media. `100mL` of `LMnO_4` reacts with 100 mL of 0.1 M `Na_2S_2O_3` in acidic, basic and neutral media. Q. The molarity (M) of `KMnO_4` solution in basic medium is:A. 0.8 MB. 0.08 MC. 0.26 MD. 0.026 M |
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Answer» Correct Answer - A `MnO^(4)^(ɵ)-=S_(2)O_(3)^(2-)(S_(2)O_(3)^(2-)to2SO_(4)^(2-)+8e^(-))(n=8)` `100 mLxxN-=100mLxx0.1xx8` `N=0.8` `M=(N)/("n factor")=0.8M` |
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| 33. |
100 mL solution of ferric alum `[Fe_(2)(SO_(4))_(3).(NH_(4))_(2)SO_(4).24H_(2)O` `(Mw=964g mol^(-1))` containing `2.41 g ` salt was boiled with Fe when the reaction `Fe+Fe_(2)(SO_(4))_(3)to3FeSO_(4)` Takes place. The unreacted iron was filtered off and the solution was titrated with `(M)/(60)K_(2)Cr_(2)O_(7)` in acidic medium. Q. Moles of `FeSO_(4)` formed when Fe reacts with `Fe(2)(SO_(4))_(3)` isA. 0.0075B. 0.005C. 0.001D. 0.002 |
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Answer» Correct Answer - A `Fe+Fe_(2)(SO_(4))_(3)tounderset(3 mol)(3FeSO_(4))` `1 " mol of "Fe_(2)(SO_(4))_(3).(NH_(4))_(2)SO_(4).24H_(2)O-=1 mol Fe_(2)(SO_(4))_(3)` `(2.41)/(964)` " mol of "ferric alum `-=(2.41)/(964) " mol of "Fe_(2)(SO_(4))_(3)`. `-=(3xx2.41)/(964)` `=0.0075 " mol of "FeSO_(4)` |
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| 34. |
A solution of `Na_(2)S_(2)O_(3)` is standardized iodometrically against 0.167g of `KBrO_(3)`. The process requires 50mL of the `Na_(2)S_(2)O_(4)` solution. What is the normality of the `Na_(2)S_(3)O_(3)?`A. 0.2NB. 0.12NC. 0.72ND. 0.02N |
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Answer» Correct Answer - B |
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| 35. |
1.5 g of chalk was treated with 10 " mL of " 4 N HCl. The chalk was dissolved and the solution was made to 100 mL. 25 " mL of " this solution required 18.75 " mL of " 0.2 N NaOH solution for comjplete neutralisation. Calculate the percentage of pure `CaCO_3` in the sample of chalk. |
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Answer» Excess of HCl in `25 mL=18.75xx0.2mEq` Excess of HCl in 100 mL`=18.75xx0.2xx4=15mEq` Total HCl used `=10xx4=40mEq` HCl used `=40-15=25mEq=25mEq` of `CaCO_3` `=25xx10^(-3)xx50g of CaCO_3=1.25g` `% of CaCO_3=(1.25xx100)/(1.5)=83.35%` |
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| 36. |
Oxygen is prepared by catalytic decomposition of potassium chlorine `(KClO_(3))`. Decomposition of potassium, chloride gives potassium chloride (KCl) and oxygen `(O_(2))`. How many moles and how many grams of `KClO_(3)` are required to produce 2.4 mole `O_(2)`? |
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Answer» Decomposition of `KClO_(3)` takes places as, `2KClO_(3)(s) to 2KCl(s)+3O_(2)(g)` `2"mole" of KClO_(3)-=3"mole of"O_(2)` `therefore "3 mole "O_(2)"formed by 2 mole" KClO_(3)` `therefore 2,4 "mole" O_(2) "will be formed by" ((2)/(3)xx2.4)"mole"KClO_(3)` `=1.6"mole of" KClO_(3)` Mass of `KClO_(3)`= Number of moles `xx` Molar mass `=1.6xx122.5=196g` |
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| 37. |
1 " mol of "`IO_(3)^(ɵ)` ions is heated with excess of `I^(ɵ)` ions in the presence of acidic conditions as per the following equation `IO_(3)^(ɵ)+I^(ɵ)toI_(2)` How many moles of acidified bypo solution will be required to react completely with `I_(2)` thus produced?A. 1B. 3C. 5D. 6 |
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Answer» Correct Answer - D (i). `10e^(-)+2IO_(3)^(ɵ)toI_(2)` `underline(2I^(ɵ)toI_(2)+2e^(-)]xx5)` `underline(2IO_(3)^(ɵ)+10I^(ɵ)to6I_(2))` (ii). `I_(2)+2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2I^(ɵ)` `1 " mol of "IO_(3)^(ɵ)=5 mol I^(ɵ)-=3 mol of` `I_(2)=6 " mol of "S_(2)O_(3)^(2-)` |
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| 38. |
12 g of impure cyanogen undergoes hydrolysis by two different paths. (i). `(CN)_(2)+4H_(2)Oto(NH_(4))_(2)C_(2)O_(4)` (ii). `(CN)_(2)+2H_(2)OtoNH_(2)CONH_(2)` When 11.52 g of pure ammonium carbonate `[(NH_(4))_(2)CO_(3)]` was heated, the exact amount of urea was obtained. 20 " mL of " 1.6 M acidic `KMnO_(4)` is required to completely oxidise `(NH_(4))_(2)C_(2)O_(4)`. Q. The percentage purity of cyanogenA. `86.67%`B. `76.67%`C. `66.67%`D. `56.67%` |
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Answer» Correct Answer - A `underset(1mol)(NH_(4))_(2)CO_(3)toH_(2)N-overset(O)overset(||)underset(1mol)C-NH_(2)+2H_(2)O` (Mw of `(NH_(4))_(2)CO_(3)=96)` Moles of urea`=(11.56)/(96)=0.12` `MnO_(4)^(ɵ)-=(NH_(4))_(2)C_(2)O_(4)(n=2)` `[C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)]` `20xx1.6xx5mEq-=mEq]` `160mEq-=160mEq` `-=(160)/(2)mmol=80mmol=0.08mol` Therefore, moles of `(NH_(4))_(2)C_(2)O_(4)=0.08` Let a and b " mol of "`(CN)_(2)` react in reactions (i) and (ii), respectively. (i). `underset(a=0.08 mol)((CN)_(2)+4H_(2)O)tounderset(a=0.08mol)((NH_(4))_(2)CONH_(2))` `thereforea=0.08,b=0.12` Total moles of `(CN)_(2)=0.08+0.12=0.2` Weight of `(CN)_(2)=0.2xx52=10.4` `%` purity of `(CN)_(2)=(10.4)/(12)xx100=86.67%` |
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| 39. |
Find out % of oxalate ion ina given sample of an alkali metal oxalate salt, 0.30g of it is dissolve in 100mL water and its required 90mL OF Centimolar `KMnO_(4)` solution in aicdic medium:A. 0.66B. 0.55C. 0.44D. 0.066 |
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Answer» Correct Answer - A |
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| 40. |
5 g sample contain only `Na_(2)CO_(3)` and `Na_(2)SO_(4)` . This sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralizes 20 mL of 0.1 M `H_(2)SO_(4)`. Calcalute the % of `Na_(2)SO_(4)` in the sample .A. 42.4B. 57.6C. 36.2D. None of these |
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Answer» Correct Answer - B |
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| 41. |
320g mg of a sample of magnessium having a coasting of its oxide required 20mL of 0.1M hydrochloric acid for the complete neutralisation of the latter. The composition of the sample is:A. 87% Mg and 12.5%MgOB. 12.5% Mg and 87.5 %MgOC. 80% Mg and 20%MgOD. 20% Mg and 80%MgO |
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Answer» Correct Answer - C |
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| 42. |
Calculate the weight of `N_2H_4` (hydrazine) oxidised to `N_2` by `19.4 g K_2CrO_4`, which is reduced to `Cr(OH)_4^(ɵ)` in basic medium. |
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Answer» Mw of `K_2CrO_4=2xx39+52+4xx16=194g` Mw of `N_2H_4=2xx14+4=32g` `3N_2H_4+4H_2O+4CrO_4^(2-)to3N_2+4Cr(OH)_4^(ɵ)+4overset(ɵ)(O)H` `4xx194g of K_2CrO_4-=3xx32 g of N_2H_4` `19.4 g K_2CrO_4=(3xx32xx19.4)/(4xx194)=2.4g` |
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| 43. |
When 100 " mL of " 0.06 M Fe `(NO_3)_3,50mL ` of `0.2M FeCl_3` and `100mL` of `0.26M` Mg `(NO_3)_2`, are mixed In the final solution….. `[Fe^(3+)]=` …. `[NO_3^(ɵ)]=`….. `[Cl^(ɵ)]=`…… `[Mg^(2+)]=`…… |
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Answer» (a). 100 " mL of " 0.06 M `Fe(NO_3)_3-=6` " mmol of " `Fe^(3+)+6xx3` " mmol of " `NO_3^(ɵ)` (b). 50 " mL of " 0.26 `M FeCl_3-=100"m mol of "Fe^(3+)+10xx3m" mol of "Cl^(ɵ)` (c). `100mL` of 0.26 M `Mg(NO_3)_2-=26"m mol of "mg^(2+)+26xx2"m mol of "NO_3^(ɵ)` (d). Total`-=16m" mol of "Fe^(3+)+70" m mol of "NO_3^(ɵ)+30"m mol of "Cl^(ɵ)+26"m mol of "Mg^(2+)` Total volume `=100+50+100=250mL` `[Fe^(3+)]=(16)/(250)=0.064M` `[NO_3^(ɵ)]=(70)/(250)=0.28M` `[Cl^(ɵ)]=(30)/(250)=0.12M` `[Mg^(2+)]=(26)/(250)=0.104M` |
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| 44. |
0.75 " mol of "solid `X_(4)` and 2 " mol of "gaseous `O_(2)` are heated to react completely in sealed vessel to produce only one gaseous compound Y. After the compound is formed the vessel is brought to the initial temeprature, the pressure is found to half the inital pressure. Calculate the molecular formula of compound? |
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Answer» `underset(`{:(i nitial mol),(Fi nal mol):}](0.75)(2)(0) `)(X_(4)+O_(2)toY)` Initial pressure is due to `O_(2)` (2 mol), because `X_(4)` is solid and final pressure is due to Y only (a mol). Temperature remains constant. Now, `Pprop" mol of "O_(2)prop2` `(P)/(2)prop" mol of "Ypropa` In other words, only 1 " mol of "Y is formed. The mol ratio of `X_(4):O_(2)` in `Y=0.75:2` Ratio of atoms of X and O in `Y=4xx0.75:2xx2=3:4` Thus, Y is `X_(3)O_(4)` |
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| 45. |
(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): If x " mL of " 0.2 M `H_(2)SO_(4)` solution requires 10 " mL of " 0.24 M KOH solution then x " mL of " 0.1 M `H_(2)SO_(3)` would require 20 " mL of " 0.01 M acidified `K_(2)Cr_(2)O_(7)` solution. Reason (R): `H_(2)SO_(3)` is dibasic acid. |
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Answer» Correct Answer - B `H_(2)SO_(4)` vs `KOH` (acid base titration) `m" Eq of "H_(2)SO_(4)=m" Eq of "KOH` `x xx0.2xx2=10xx0.24xx1` `x=6mL` `underset((n=2))(H_(2)SO_(3))vsunderset((n=6))((K_(2)Cr_(2)O_(7))/(H^(o+))` `(SO_(3)^(2-)toSO_(4)^(2-)+2e^(-))(6e^(-)+Cr_(2)O_(7)^(2-)to2Cr^(3+))` `m" Eq of "H_(2)SO_(3)-=m" Eq of "Cr_(2)O_(7)^(2-)` `x xx0.1xx2-=20xx0.01xx6` `x=6 mL` (A) is correct (R) is the correct explanation of (A). |
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| 46. |
11.6 g of an organic compound having formula `(C_(n)H_(2n+2))` is burnt in excess of `O_(2)(g)` initially taken in a 22.41 litre steel vessel. Reaction the gaseous mixture was at 273 K with pressure reading 2 atm. After complete complete combustion and loss of considerable amount of heat, the mixture of product and excess of `O_(2)` had a temperature of 546 K and 4.6 atm pressure. The formula of organic compound is :A. `C_(6)H_(6)`B. `C_(3)H_(8)`C. `C_(5)H_(12)`D. `C_(4)h_(14)` |
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Answer» Correct Answer - D |
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| 47. |
Number of ions present in `2.0 "litre"` of a solution of `0.8 M K_(4)Fe(CN)_(6)` is: |
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Answer» `K_4[Fe(CN)_6]to4K^(o+)+[Fe(CN)_6]^(4-)` 5 ions (mol)`=5xx6xx10^(23)` ions Number of ions `=5xx6xx10^(23)xx0.8xx2=48xx10^(24)` ions |
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| 48. |
Alcohol level in blood is determined by the reaction with `K_2Cr_2O_7` solution in acidic medium. Calculate the blood level in mass percent if 10 " mL of " 0.05 M solution of `K_2Cr_2O_7` is required for the reaction of a 10.0 g sample of blood. |
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Answer» Mole method: Balance the redox reaction of `C_2H_5OH` with `Cr_2O_7^(2-)` in acidic medium (Mw of `C_2H_5OH=46g)` `cancel6e^(-)+Cr_2O_7^(2-)to2Cr^(3+)]xx2` `underline(C_2H_5OHto2CO_2+cancel(12e^(-))` `underline(C_2H_5OH+2Cr_2O_7^(2-)to4Cr^(3+)+2CO_2)` For 2 " mol of "`Cr_2O_7^(2-)-=1 " mol of "C_2H_5OH` is required `10xx0.05xx10^(-3) " mol of "Cr_2O_7^(2-)` `-=(10xx0.05xx10^(-3))/(2) " mol of "C_2H_5OH` `-=25xx10^(-5)` " mol of "`C_2H_5OH` `-=25xx10^(-5)xx46g of C_2H_5OH` Percent of alcohol `-=(25xx10^(-5)xx46)/(10)xx100=0.115%` Equivalent method: `Cr_2O_7^(2-)-=C_2H_5OH` `1mEq-=1mEq` `10xx0.05xx6m" Eq of "Cr_2O_7^(2-)=3 mEq-=3 m" Eq of "C_2H_5OH` `3 mEq-=3 mEq` `3m" Eq of "C_2H_5OH-=3xx10^(-3)eq-=3xx10^(-3)xx(46)/(12)g` Percent of alcohol `-=(3xx10^(-3)xx46xx100)/(12xx10)-=0.115%` |
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| 49. |
After 20 " mL of " 0.1 M `Ba(OH))_(2)` is mixed with 10 " mL of " 0.2 M `HClO_(4)`, the concentration of `overset(ɵ)(O)H` ions isA. `2xx10^(-3)M`B. `10^(-3)`MC. 0.066 MD. `overset(ɵ)(O)H` ions are completely neutralised. |
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Answer» Correct Answer - C `m" Eq of "overset(ɵ)(O)H=20xx0.1xx2=4mEq` m" Eq of "`HClO_(4)(x=1)=10xx0.2xx1=2mEq` m" Eq of "`overset(ɵ)(O)H` left`=4-2=2mEq` Concentration or `N of overset(ɵ)(O)H=(2mEq)/(30mL)` `=0.066N of overset(ɵ)(O)H` `=0.066 M of overset(ɵ)(O)H` |
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`20mL` of a mixture of `CO` and `H_(2)` were mixed excess of `O_(2)` and exploded & cooled. There was a volume contraction of 23mL. All volume measurements corresponds to room temperature `(27^(@)C)` and one atmospheric pressure. Determine the volume ratio `(V_(1):V_(2) "of" Co` anf `H_(2)` in the original mixture .A. `1:2`B. `3:2`C. `2:3`D. `4:1` |
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Answer» Correct Answer - B |
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