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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.
| 91551. |
a. A sample of 0.50g of an organic compound wastreated according to Kjeldahal's method. The ammonia evolved was aboserved in 50 ml of 0.5M H_(2)SO_(4). The residualacid required 60 mL of 0.5M solution of NaOH for neutralisation. Find the percentage composition of hnitrogen in the compound. b. On analysis, 0.2g of a monobasic acid gave 0.505 gm of CO_(2) and 0.0864gm H_(2)O. 0.305 gm of this acid required25 ml of M//10 NaOH solution for neutralisaton. Find the molecular formula of the acid. c. A liquid aromatic organcvi compound (A) conatining carbon (92.3%) and hydrogen (7.7%)decoluidsed KMnO_(4) and on ozonolysis gave methanal and another compound (B). The molecular mass of (A)is 104. On treatmentwith a suitablecatalyisis, (A)gave a high molecularmass solid product (C)havingthe same empirical formula as that of compound(A). Compound (C) is used in making toys and household goods. Identify (A),(B), and (C) and explain the reactions. d. A sample of 0.246 gm of an organic compound gave 0.198 gm fo CO_(2) and 0.104 gm of H_(2)O on completecombusion. .37gm of the compoundgave 0.638 gm of silver bromide in Carius method. What is the molecular formula of the compound if its molecular mass is 109. |
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Answer» Solution :a. TOTAL acid `= 50xx0.5xx2 = 50 mEq`. Excess acid `= 60xx0.5 = 30 mEq`. Acid used to neutralise `NH_(3) = 50 - 30 = 20 mEq`. Perventage of `N = (1.44xx mEq. "of acid")/("Wt fo compd".)` `= (1.4xx20)/(0.5) = 56%` b. Caculate of emprical formula. MASS of sample `= 0.20gm`, mass of `CO_(2)` formed `= 0.505gm` mass of water formed `= 0.864 gm`. Percentage of carbon `= (12xx0.505xx100)/(44xx0.20) = 69%` Percentage of hydrogen `= (2xx0.086xx100)/(18xx0.20) = 4.8%` Percentage of oxygen `= 100-(69+4.8) = 26.20%` Emprical formula `= C_(7)H_(6)O_(2)`  `40 gm` of `NaOH` would meutralise `1 mol` of amononbasic acid `NaOH = 25 xx (1)/(10) = 2.5 mEq. = 2.5xx10^(-3)xx40 = 0.1 gm` `0.1gm` of `NaOH` neuralies `0.305 gm` acid `40gm NaOH` neutralies `(0.305xx40)/(0.1)= 122gm` Molecular mass of acid `= 122 gm mol^(-1)` `n = ("Mol. formula mass")/("EMPIRICAL formula mass") = (122)/(122) = 1` Molecular formula of acid `= (C_(7) H_(6) O_(2))_)1) = C_(7) H_(6) O_(2)` c.  Empirical formula `= CH` Mol mass of `(A) = 2xxV.D. = 2xx52 = 104` `n = (104)/(13) = 8` Molecule formula `= C_(8) H_(8)` Decree of unsaturation `= [(2xx8+2) - 8]//2 = 5^(@)` `5^(@) unsaturation shows that it is can aromatic COMPOUND, `4^(@)` due to BENZENE ring.  `(A)` plymerises as shwon: `n(overset(Ph)overset(|)(CH)= CH_(2)) rarr - overset(Ph)overset(|)(CH) - CH_(2) - overset(Ph)overset(|)(CH) - CH_(2) -`  d. Mass of organic compound`= 0.26 gm` Mass of `CO_(2) = 0.198 gm` Mass of `H_(2)O = 0.1014 gm` Percentage of `C = (12xx0.198xx100)/(44xx0.246) = 21.95%` Percentage of `H = (2xx0.1014xx100)/(18xx0.246) = 4.58%` Percentage of bromic `= (80xx0.638xx100)/(188xx037)` `= 73.37%`  Emprical formula `= C_(2) H_(5) Br`, Molecular mass `= 109` `n = 109//109 = 1` Molecular formula `= (C_(2) H_(5)Br)_(1) = C_(2)H_(5)Br` |
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| 91552. |
(a) A reaction is second order in A and first order in B (i) Write the differential rate equation (ii) How is the rate affected on increasing the concentration of A three times ? (iii) How is the rate affected when the concentration of both A and B are doubled ? (b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t_(1//2) for this reaction. (Given log 1.428=0.1548) |
| Answer» Solution :(i) `-(DA)/(DT)=k[A]^(2)[B]" (ii) RATE increases 8 times (iii) 77.87 min"` | |
| 91553. |
(A): A salt bridge allows the flow of current by completing the electrical circuit (R) : A salt bridge maintains the electrical neutrality of the two half cells |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 91554. |
(A): A reaction cannot become fast by itself unless a catalyst is added. (R): A catalyst always increases the speed of a reaction. |
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Answer» |
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| 91555. |
Assertion: A pressure cooker reduces cooking time Reason:The boiling point of water inside the cooker is increased |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 91556. |
(A) A pseudo first order recation occurs slowly. (R ) Reactions of higher order can follow kinetics of first order under special conditions. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 91557. |
(a) A liquid mixture of benzene and toluene is composed of 1 mol of benzene and 1 mol of tolune if the pressure over the mixture at 300K is reduced, at what pressure does the first bubble form? (b) What is the composition of the first bubble formed. (c) If the pressure is reduced further, at what pressure does the last trace of liquid disapper? (d) What is the composition of the last drop of liquid? (e) What will be the pressure, composition of the liquid and the composition of vapour, when 1 mol the mixture has been vaporized? Given P_(T^(0))=40mmHg, P_(B^(0))=100mmHg |
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Answer» (b)`Y_(A)=(0.5xx40)/(70)=(2)/(7),Y_(B)=(5)/(7)` LTBRGT (c ) at last TRACE of liquid `Y_(A)=0.5 ,Y_(B)=0.5` `(1)/(P)=(Y_(A))/(P_(A)^(0))+(Y_(B))/(P_(B)^(0))=(0.5)/(40)+(0.5)/(100)` `P=(400)/(7)` (d) `Y_(A)=(Y_(A)P_(A)^(0))/(P) rArr0.5=(X_(A)40)/(400//7) , X_(A)=(5)/(7) , X_(B)=(2)/(7)` (e) LET `X` mole of `B` be present in liquid phase. `{:(,"Mole ofA","Mole of B"),("Liquid",1-x,x),("Vapour",x,1-x):}` `P=40(1-x)+100 x` `(1)/(P)=(Y_(A))/(P_(A)^(0))+(Y_(A))/(P_(B))RARR(1)/(P)=(x)/(40)+(1+x)/(100)=(100x+(1-x)40)/(40xx100)` So `p^(2)=40xx100` `p=20sqrt(10)` `20sqrt(10)=40(1-x)+100x=40+60x` `x=(20sqrt(10)-10)/(60)=(sqrt(10)-2)/(3)` |
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| 91558. |
(a) A liquid mixture of benzene and toluene is composed of 1 mol of benzene and 1 mol of toluene. If the pressure over the mixture at 300K is reduced, at what pressure does the first bubble form? (b) What is the composition of the first bubble formed. (c) If the pressure is reduced further, at what pressure does the last trace of liquid disapper? (d) What is the composition of the last drop of liquid? (e) What will be the pressure, composition of the liquid and the composition of vapour. when 1 mol of the mixture has been vaporized? Given P_(T)^(0)=100mmHg |
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| 91559. |
a) A gaseous mixutre consists ethane and ehylene. How are the individual members recovered? A gaseous mixture consists of ethylene and acetylene. How both are separated from each other? |
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Answer» Solution :Pass mixture through CONC. `H_(2)SO_(4)`. Ethylene is absorbed. The product on heating GIVES again `C_(2)H_(4)` b) Pass mixture through ammonial cuprous chlroide solution. The acetylene forms red precipitate. It is FILTERED and treated with `HNO_(3)` to recover accetylene. `Cu_(2)C_(2)` (red PPT) `+2HNO_(3) to C_(2)H_(2) + Cu_(2)(NO_(3))_(2)` |
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| 91560. |
(A) A gas with higher critical temperature is adsorbed more than a gas with lower criticaltemperature (R ) At higher temperature the gas is more easily liquefiabe |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 91561. |
(A) A first order reaction is always unimolecular (R ) A unimolecular reaction is always a first order reaction |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 91562. |
(a) A first order reaction is 50% complete in 30 minutes at 300K and in 10 minutes at 320k. Calculate activation energy (E_a) for the reaction. (b) Write the two conditions for collisions to be effective reaction. (c ) How order of reaction and molecurity differ towards a complex reaction? |
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Answer» Solution :a) We need to calculate rate constants at 300K and 320 K Using the equations `t_(1//2)=0.693/k or k=0.693/t_(1//2)` `k_(300k)=0.693/(t_(1//2)(300K))=0.693/(30 minutes)=0.0231 minutes^-1` `k_(320 k)=0.693/(t_(1//2) 320 K)=0.693/(10 minutes)=0.0693 minutes^-1` Using ARRHENIUS equation, we have `log""(k_(320k))/(k_(300k))=E_a/(8.314JK^-1 mol^-1)=[(320-300)/(320 times 320)]` SUbstituting the values we have `(0.0693 minutes^-1)/(0.0231 minutes ^-1)=E_A/(8.314J K^-1 mol^-1) times (20K)/(96000 K.K)` `or 3=E_a/(8.314 JK^-1 mol^-1)times1/(4800K)` or `E_A=3 times 8.314 times 4800 J mol^-1` =`119.722 kJ mol^-1` (b) The colliding molecules must possess a certain minimum value of energy (threshold energy). The colliding molecules must have proper orientation to facilitate breaking of bonds and formation of new bonds. (c ) For COMPLEX reaction order is GIVEN by the slowest step and molecularity of the slowest step in same as the order of OVERALL reaction. Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions. |
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| 91563. |
(a) A diatomic moleculae has a dipole moment of 1.2 D. If the bond distance is 1.0Å, What fraction of an electronic charge 'e' exists on each atom ? (b) The dipole moment of LiH is 1.964xx10 ^(-29) C-m and the interatomic distance between Li and H in this molecule is 1.596Å. What is the per cent ionic character in LiH? |
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Answer» Solution :(a) Partical charge `=("DIPOLE moment ")/("BOND dis tance ")` `= (1.2 XX 10 ^(-8)"esu cm")/(1.0 xx 20 ^(-8)cm) = 1. 2 xx 10 ^(-10))=0.25 = 25%`ofe. (B) The dipole moment of `100%` ionic molecule `(Li ^(+)H^(-))` = (1 elecronic charge) (interatomic DISTANCE) ` = (1.602 xx 10 ^(-19) C) ( 1.596 xx 10 ^(-10)m)` `= 2.577 xx 10 ^(-29)Cm` Fractional ionic character `= ("Exp. value of dipole moment")/("Theoretical value of dipole moment") = ( 1. 964 xx 10 ^(-29))/(2.557 xx 10 ^(-29))= 0.768` The bond in LiH is `76.8%` ionic. |
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| 91564. |
(A) A colloidal sol of As_(2)S_(3) is coagulated faster by 0.1 M BaCl_(2) than by 0.1 M NaCl. (R ) BaCl_(2) gives same number of Cl^(-) ions than NaCl |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 91565. |
(a) A current of 1.50 amp was passed through an electrolytic cell containing AgNO_3 solution with inert electrodes. The weight of Ag deposited was 1.50 g. How long did the current flow ? (b) Write the reactions taking place at the anode and cathode in the above cell. (c) Give reactions taking place at the two electrodes if these are made up of Ag. |
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Answer» Solution :(a) `I = 1.50 A, m = 1.50 g` ATOMIC MASS of Ag = 108 Eq. of mass of Ag = 108 108 g silver is deposited by = 96500 C 1.5 g silver is deposited by `=96500/108 xx 1.5 = 1340.28` C Quantity of electricity Q = `l xx t` or `t = Q/l` SUBSTITUTING the values, we get `t = (1340.28)/1.5 = 893.52 s` (b) At cathode: `Ag^(+) + e^(-) to Ag(s)` At anode: `2H_(2)O to 4H^(+) + O_(2) + 4e^(-)` ( c) At anode: `Ag^(+) + e^(-) to Ag(s)` At anode: `Ag to Ag^(+) + e^(-)` |
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| 91566. |
(A): A cis-isomer has a net dipole moment zero. (R): A cis-isomers has two ligands of the same type occupying adjacent positions. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 91567. |
(A): A blue colour is obtained when a copper wire is immersed in AgNO_3 solution (R) : Silver reduces Cu^(2+) to copper |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 91568. |
A:A: Calamine and Dolomite are the carbonate ores. R: Calamine is ZnCO_3 whereas Dolomite is MgCO_3 ZnCO_3 |
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Answer» If both Assertion & REASON are true and the reason is the correct EXPLANATION of the assertion, then mark (1). |
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| 91569. |
A: A catalyst increases the rate of a reaction. R: In presence of a catalyst, the activation energy of the reaction increases. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 91570. |
(a) A blackish brown coloured solid 'A' when fused with alkali metal hydroxide in presence of air, produces a dark green coloured compoun 'B', which on electrolytic oxidation in alkaline mediumgives a dark purplecoloured compound C. Identify A,B and C and write the reaction involved. (b) Whathappens when an acidic solution of the green compound (B) is allowed to stand for some time ? Give the equation involved. What is this type of reaction called ? |
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Answer» Solution :(a) `underset("(A)BLACKISH brown")underset("Pyrolusite")(MnO_(2)) + 4KOH + O_(2) overset("Fuse")(rarr) underset("(B) - Green coloured")underset("Potassium manganate ")( 2K_(2)MnO_(4))+ 2H_(2)O` `underset((B))(2K_(2)MnO_(4)) + H_(2)O+(O) underset("medium")overset("Alkaline")(rarr) underset("(C ) -Purple coloured")underset("Potassium permanganate")(2KMnO_(4)) + 2KOH` or `MnO_(4)^(2-) rarrMnO_(4)^(-) + E^(-)` (b) When acidic solution of green compound (B) , i.e., potassium manganate is allowed to stand for some time, it disproportionates to GIVE permanganate as follows `:` `3MnO_(4)^(2-) + 4H^(+) rarr 2MnO_(4)^(-) + MnO_(2) + 2H_(2)O` This reaction is called disproportionation reaction. |
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| 91571. |
(a) A 5% solution (by mass) of cane - sugar in water has freezing point of 271 K. Calculate the frttzing point of 5% solution (by mass) of glucose in water if the freezing point of pure water is 273.15 K. [Molecular masses : Glucose C_(6)H_(12)O_(6) : 180 amu, Cane - sugar C_(12)H_(12)O_(11) : 342 amu] (b) State Henry's law and mention two of its important applications. |
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Answer» Solution :5% by mass cane-sugar = 5 g cane-sugar (MW = 342 g/litre) `n_(B) = (5)/(342) ` and `w_(A) = 95 g` `DeltaT_(F)= 273.15-271= 2.15 K` `K_(f)= (DeltaT_(f)w_(A))/(n_(B)xx1000)=(2.15xx95)/(5/(342)xx1000)` = 13.97 K kg `mol^(-1)` 5% by mass glucose = 5 g glucose (mw = 18/mol) in 100 g solution `DeltaT_(f)=K_(f)xx(n_(B)1000)/(w_(A))=13.97xx(5)/(180)xx(1000)/(95)` =4.085 K `T_(f)= 273.15-4.085` Freezing point of glucose solution (b) According to Henry.s law, the partial pressure of a GAS in vapour phase (p) is directly proportional to the mole fraction (x) of the gas in the solution. Henry.s law is applied in the production of carbonated beverages and for the deep-sea-divers. |
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| 91572. |
A 95 mass precent solution of ethanol is further diluted with water. If the mole fraction of ethanol in diluted solution is 0.25, what is the mole fraction of waer in the diluted solution. Is it still an ideal solution |
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Answer» The mole fraction of WATER = 1-0.25=0.75 In this solution. Water is solute and alcohol is solvent. Upon DILUTION, the mass PRESENT of water BECOME more than 5% This means that the solution is no longer dilute or ideal. |
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| 91573. |
A 8.2 L cylinder of nitrogen gas at 5.0 atm pressure and 27^(@)C developed a leakage. When the leakage was repaired , 3.5 atm of nitrogen remained in the cylinder still at that temperature . How many moles of gas escaped ? |
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Answer» 0.75 mol Moles after LEAKAGE is repaired (n) `= ((3.5atm) xx( 8.2L))/( ( 0.082) xx ( 300K))` Moles of gas escaped `=n-n.` `= ( 8.2)/( 0.082 xx ( 3.0))[5.0-3.5]` = 0.5 mol |
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| 91574. |
A 8. 0 g sample contained Fe_(3)O_(4) , Fe2O3 and inert materials . It was treated with an excess of aqueous Kl Solution in acidic medium,which reduced all the iron to Fe^(+2) ions. The resulting solution was diluted to 50.0 cm^(3) and a 10. 0 cm^(3) of it was taken. The liberated iodine in this solution required 7. 2 cm^(3) of 1. 0 M Na_(2) S_(2)O_(3) for reduction to iodide. The iodine from another 25. 0 cm^(3) sample was extracted , after which the Fe^(+2) ions was titrated against 1. 0 M MnO_(4)^(-)in acidic medium. The volume of KMnO_(4)solution used was found to be 4. 2 cm^(3) . Calculate the mass percentages of Fe_(3)O_(4)and of Fe_(2)O_(3) in the original mixture |
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Answer» Solution :This PROBLEM can be done by two methods. In the first method, we break up `Fe_(3) O_(4)` as an equimolar mixture of `FeO` and `Fe_(2) O_(3)` Method ( 1) `Fe_(3)O_(4) ` is` FeO. Fe_(2)O_(3)` Equivalents of `Na_(2)S_(2) O_(3) = ( 1 xx 7.2 ) /( 1000)= 7.2 xx 10^(-3)` Equivalents of `I_(2)` in 10cc `= 7.2 xx 10^(-3)` Equivalents of `I_(2)` in 50 CC `= 7.2 xx 10^(-3) xx 5= 3.6 xx 10^(-2)` Since equivalents of `I_(2)` is equal to that of KI which in turn is equal to the total equivalents of `Fe_(2) O_(3) ( Fe_(2) O_(3)` in the free state and `Fe_(3) O_(3)` COMBINED with `FeO)` `:.` equivalents of `KMnO_(4)` solution ` = ( 4.2 xx 1 xx 5)/( 1000) = 2.1 xx 10^(-2)` Since `KMnO_(4)` reacts with the total `Fe^(2+) ( Fe^(2+)` in FeO and `Fe^(2+)` that was produced by the action of Kl on `Fe_(2) O_(3))` `:.` equivalents of total `Fe^(2-)` in 50ml `= 2.1 xx 10^(-2) xx 2 = 4.2 xx 10^(-2)` Since equivalent of `Fe^(2+)` produced from `Fe_(2)O_(3)` is equal to that of equivalents of `Fe_(2) O_(3)` `:.` Equivalents of `FeO` `= 4.2 xx 10^(-2) - 3.6 xx 10^(-2) = 6 xx 10^(-3)` `:.` moles of `FeO = 6 xx 10^(-2)` moles of `Fe_(2)O_(3)` combined with FeO `= 6 xx 10^(-3)` total moles of `Fe_(2) O_(3) = ( 3.6 xx 10^(-2))/( 2)` ( because when `Fe_(2) O_(3) gt Fe^(2+) ` 'n' factor is 2 ) ` = 1.8 xx 10^(-2)` moles of `Fe_(2) O_(3)` in the free state `= 1.8 xx 10^(_2) - 6 xx 10^(-3) = 1.2 xx 10^(-2)` mass of `Fe_(3) O_(4) = 6 xx 10^(-3) xx 232 = 1.392g ` mass of `Fe_(3) O_(4) = 1.2 xx 10^(-2) xx 160 = 1.92 g ` percentage `Fe_(2) O_(4) = 17.4%` percentage of `Fe_(2) O_(3) = 23.75 %` Method2`:` Here we take `Fe_(3) O_(4)` as a single entity. Equivalents of `Na_(2) S_(2) O_(3) =( 7.2 xx 1)/( 1000) = 7.2 xx 10^(-3) ` Equivalents of `I_(2) ` in 50 cc `= 7.2 xx 10^(-3) xx 5 = 3.6 xx 10^(_2)` `:.` Equivalents of `Fe_(3) O_(4) +Fe_(2) O_(3) = 3.6 xx 10^(_2)` Let us assume that the moles of `Fe_(3) O_(4)` in xg and that of `Fe_(2) O_(3)` is y g . Since on reacting with KI both `Fe_(3) O_(4)` and `Fe_(2) O_(3)` give `Fe^(2+)` 'n' factor for both is two . `:. 2x + 2y = 3.6 xx 10^(-3) `........(1) Equivalent of `KMnO_(4)= ( 4.2 xx 1 xx 5 ) /( 1000) = 2.1 xx 10^(-2)` moles of `Fe^(2+)` in 50 mL `= 4.2 xx 10^(-2)` Since the moles of `Fe_(3) O_(4)` are x, moles of `Fe^(2+)` produced from `Fe_(3) O_(4)` will be 3x and that produced from `Fe_(2) O_(3) `will be 2y `:. 3x + 2y = 4.2 xx 10.2`.....(2) `( 2) - ( 1) ` gives `x = 6 xx 10^(-3)` `:. y = 1.2 xx 10^(-2)` Solving this percentage of `Fe_(3) O_(4)` is 17.4 % and `Fe_(2) O_(3)` is 23.75% |
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| 91575. |
a. 7- Bromocycloheptraiene (tropylium bromide) completely dissociates in water and givesa yellow precipiate of AgBr with AgNO_(3). Why? (b) Why is cyclopentadiene (K_(a) = 10^(-15)) much more acidic than 1,3- cyclohexadience? C. Explain the following 1,3,5-cycloheptatrience through a cycle, planner triene with six pi vec(e)'s is not armoatic, whereas tropolonebehaves like a phenol. |
Answer» Solution : The cycloheptatrienyl carbocation formed by the loss of `Br^(o-)` is STABLE due to AROMATIC character. b.  C. Troplylim is not aromatic, while tropolone is aromatic. SEE Illustration `11.6(13)` and `(16)` |
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| 91576. |
A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of the KOH solution. |
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Answer» Solution :Molarity and density are RELATED as under : Molarity =(% by mass x d x 10 )/(Molar mass) `6.90= (30 XX d xx 10)/(56) ` ` d = (6.90 xx 56)/(30 xx 10) = 1.288 g CM^(-3)` |
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| 91577. |
A 6.90 M Solution of KOH in water contains 30% by mass of KOH. Calculate the density of the KOH solution. (Molar mass of KOH = "56 g mol"^(-1)) |
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Answer» Solution :6.90 M KOH solution means 6.90 moles of KOH are present in 1 litre of the solution. But 6.90 moles of KOH `=6.90 xx56 g=386.4g` Thus, volume of solution containing 386.4 g KOH= 1000 mL `30%` by mass of KOH solution means that 30 g of KOH are present in 100 g of the solution. `THEREFORE"Mass of solution containing 386.4 g KOH"=(100)/(30)xx386.4g=1288g` `therefore"Density of solution "=("Mass")/("Volume")=("1288 g")/("1000 mL")="1.288 g mL"^(-1)` |
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| 91578. |
A 6.4 gm sample of methanol (CH_(3) OH) was placed in an otherwise empty 1 litre flask and heated to 227^(@)C to varpoise the methanol. Methanol vapour decomposes by following gasesous compound to effuse out of flask. Measurement shows it contains 32 times as much as H_(2) (g) as CH_(3)OH (g). Then Total pressure of mixture at equilibrium. |
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Answer» `2.08` atm `n_("total") = a_(0) (1 + 2 alpha) = 0.2 (1 + (8)/(5)) = (2.6)/(5)` `P_("total") = (n_(t))/(V) RT = (2.6)/(5) xx 0.08 xx 500` `P_("total") = 20.8` atm |
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| 91579. |
A 6.4 gm sample of methanol (CH_(3) OH) was placed in an otherwise empty 1 litre flask and heated to 227^(@)C to varpoise the methanol. Methanol vapour decomposes by following gasesous compound to effuse out of flask. Measurement shows it contains 32 times as much as H_(2) (g) as CH_(3)OH (g). Then Value of K_(C) for this reaction is : |
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Answer» `(16)/(25)` `sum n_("eqm") = a_(0) (1 + 2 alpha)` `(r_(H_(2)))/(r_(CH_(3)OH)) = (p_(H_(2)))/(P_(CH_(3)OH)) sqrt(M_(CH_(3)OH)/(M_(H_(2)))) = (2 a_(0) alpha)/(a_(0) (1 - alpha)) sqrt((32)/(2))` `(32)/(1) = (Deltan_(H_(2)))/(Deltan_(CH_(3)OH)) = (2 alpha)/(1 - alpha) xx 4` `4 = (alpha)/(1 - alpha) rArr 4 - 4 alpha = alpha` `alpha = (4)/(5)` `K_(C) = (a_(0) alpha (2a_(0) alpha)^(2))/(a_(0) (1 - alpha)) xx (1)/(V^(2))` `K_(C) = (4 alpha^(3) xx a_(0)^(2))/(1 - alpha) xx (1)/(V^(2)) = 4 xx ((4)/(5))^(3) xx ((0.2)^(2))/(1 - (4)/(5)) = ((16)/(25))^(2)` |
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| 91580. |
A 6.00g sample contained Fe_(3) O_(4), Fe_(2) O_(3) and inert materials. It was treated with an excess of aqueous Kl in acidic medium which reduced all the iron to Fe^(2+). The resulting solution was dilution to 50ml.and a 10ml.sample of it was taken. The liberated iodine reacts with 5.5ml. of 1M Na_(2) S_(2) O_(3) solution, yielding S_(4) O_(6)^(2-). The iodine from another 25ml. sample was extracted, after which the Fe^(2+) was titrated with 3.2 ml. of 1M MnO_(4)^(-) in H_(2)SO_(4) solution. Calculate the percentage of Fe_(3) O_(4) and Fe_(2)O_(3) in the original mixture. |
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Answer» Solution :Equivalents of `Na_(2)S_(2) O_(3)` reacting with the liberated `I_(2) = ( 5.5 xx 1)/( 1000) = 5.5 xx 10^(-3)` `:.` Equivalent of `I_(2)` in 10ml `= 5.5 xx 10^(-3)` Equivalent of `Fe^(3+)` in 50ml = `5.5 xx 10^(-3) xx 5 = 0.0275`[ lt is `Fe^(3+)` that reacts with KL to give `I_(2) `] Total moles of `Fe_(2) O_(3) = ( 0.0275)/( 2) = .01375` ( free `+` combined ) Equivalengt of `MnO_(4)^(-) = ( 1 xx 5 xx 3.2 )/( 1000) = 0.016` `:.` Eq of `Fe^(2+)` in 25 ml= 0.016 Eq of `Fe^(2+)` in 50ml = `0.016 xx 2 = 0.0 32 ` Eq. of FeO `= 0.032 - 0.0275 = 4.5 xx 10^(-3)` Moles of `FeO = 4.5 xx 10^(-3)` `:.` Moles of `Fe_(3) O_(4) = 4.5 xx 10^(-3)` Mass of `Fe_(3) O_(4) = 4.5 xx 10^(-3) xx 232= 1.044g m`. Moles of `Fe_(2) O_(3) ` present FREELY `= 0.01375 - 4.5 xx 10^(-3) = 9.25 xx 10^(-3)` Mass of `Fe_(2) O_(3) ` present freely `= 9.25 xx 10^(-3) xx 160 = 1.48 g ` % of `Fe_(3) O_(4) = ( 1.044)/(6) xx 100 = 17.4 % ` %`Fe_(2) O_(3) = 24.67%` |
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| 91581. |
A 62 g quantity of white phosphorous was burned in an excess of Oxygen and the product was dissolved in water to form of an acid. Calculate the number of moles of acid obtained. |
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Answer» 124 g pf phosphorous PRODUCES 4 moles of `H_(3)PO_(4)` 62g produces 2 moles |
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| 91582. |
A 6.00 g sample contained Fe_(3)O_(4), Fe_(2)O_(3) , and inert materials. It was treated with an excess of aqueous Kl in acidic medium which reduced all the iron to Fe^(2+). The resulting solution was diluted to50 ml . and a 10 mlsample of it was taken . The liberated iodine reacts with 5.5 ml. of 1 M Na_(2)S_(2)O_(3)solution, yielding S_(4)O_(6)^(2-) . The iodine from another 25 ml . sample was extracted , after which the Fe^(2+) was titrated with 3.2ml of 1 M MnO_(4) " in " MnO_(4)^(-) solution . Calculate the percentage of Fe_(3)O_(4)" and " Fe_(2)O_(3) in the original mixture. |
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Answer» |
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| 91583. |
A 6.0 gm sample contained Fe_3O_4, Fe_2O_3 and inert materials. It was treated with an excess of KI in acid, which reduced all iron to Fe^(2+). The resulting solution was diluted to 50 ml, and a 10 ml sample of it was taken. The liberated iodine in the small sample was titrated with 5.5 ml of 1M Na_2S_2O_3 solution yielding S_4O_6^(2–). Theiodine from another 25 ml sample was extracted, after which the Fe^(2+) was titrated with 3.2 ml of 1.0 M MnO_4^– in H_2SO_4 solution. Calculate the percentage of Fe_3O_4 and Fe_2O_3, in the original mixture. |
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Answer» |
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| 91584. |
A 6% solution of urea is isotonic with |
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Answer» 0.05 M solution of glucose `=(6xx1000)/(60xx100)=1M` The SOLUTIONS having same CONCENTRATIONS are ISOTONIC. Therefore, 6% solution of urea is isotonic with 1M solution of glucose. |
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| 91585. |
A+5HIO_4 to 4 HCOOH+ 2HCHO The probable structure of A is |
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Answer» `CH_(2)OH(CHOH)_(3)CHO` |
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| 91586. |
A 5g sample containing FeO_4 (FeO+Fe_2O_3) and an inert impurity is treated with excess of KI solution in the presence of dilute H_2SO_4.The entire Iron converted to Ferrous ion along with liberation of Iodine.The resulting solution is diluted to 100 ml. 20 ml of the diluted solution requires 10 ml of 0.5M Na_2S_2O_3 solution to reduce the Iodine present.Amongs the following select correct statements. |
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Answer» % of `Fe_2O_3` in sample is 40% eq of `Fe_2O_3` (in `Fe_3O_4`) =eq of KI = eq of `I_2` =eq of `Na_2S_2O_3` or 2x=0.025 or x=0.0125 mol of `Fe_2O_3` So mass of `Fe_2O_3=0.0125xx160=2g` % of `Fe_2O_3=2/5xx100=40%` |
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| 91587. |
A 5.8% (wt/vol.)NaCl solution will exert an osmotic pressure closest to which one of the following: |
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Answer» `5.8%` (wt/vot.) SUCROSE SOLUTION |
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| 91588. |
A 5.25% solutions of a substance is isotonic with 1.5% solution of urea in the same solvent. If the densities of both the solutions are assumed to be equal to lg cm the molar mass of the substance will be |
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Answer» 115 g `"MOL"^(-1)` |
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| 91589. |
A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60g mol^(-1)) in the same solvent. If the densities of both the solutionsare assumed to be equal to 1.0 g cm^(-3), molar mass of the substance will be: |
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Answer» 210.0 G `mol^(-1)` `pi_(1)=pi_(2)orC_(1)RT=C_(2)RT` `W_(("urea"))/M_(("urea"))=W_("SUBSTANCE")/M_("substance")` `((1.5g))/((60G mol^(-1)))=((5.25g))/M` `M=((5.25 g)XX(60 g mol^(-1)))/((1.5g))` `=210.0 g `mol^(-1)` |
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| 91590. |
A 5.25 % solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g.mol^(1)) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm^(-3), then molar mass of the substance will be |
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Answer» `90 G mol^(-1)` |
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| 91591. |
A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol^(-1)) in the same solvent. If the densities of both solutions are assumed to be equal to 1.0 g cm^(-3), molar mass of the substance will be |
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Answer» `90.0 g mol^(-1)` Let the MOLAR MASS of the substances be M `pi = C_(1)RT=C_(2)RT=pi_(2)` So, `C_(1)=C_(2)` As DENSITY of the solutions are same So, `(5.25)/(M)=(15)/(60)RARR M =(5.25xx60)/(1.5)=210` |
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| 91592. |
A 5.2 molal aqueous solution of methyl alcohol, CH_(3)OH, is supplied. What is the mole fraction of methyl alcohol in the solution ? |
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Answer» 0.086 Moles of water, `n_("water")=(1000)/(18)=55.55` `x_(CH_(3)OH)=(5.2)/(5.2+55.55)=0.086` |
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| 91593. |
A 5.2 molal aqueous solution of methyl alcohol CH_3OHis supplied. What is the mole fractionof methyl alcohol in the solution ? |
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Answer» 0.1 |
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| 91594. |
A 5.2 molal aqueous of methyl alcohol, CH_(3)OH, is supplied. What is the molefraction of methyl alcohol in the solution ? |
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Answer» 0.1 `=(5.2)/(5.2+1000/1B)=(5.2)/(5.2+55.5)=0.086` |
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| 91595. |
A 5.2 molal aqueous solution of methyl alcohol, CH_(3)OH, is supplied. What is mole fraction of methyl alcohol in the solution? |
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Answer» 0.19 `THEREFORE "MOLE fraction of "CH_(3)OH` `=(5.2)/(5.2+55.55)=0.086` |
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| 91596. |
A 500 mL glass flask is filled at 298 K and 1 atm, pressure with three diatomic gases, X, Y and Z . The initial volume ratio of the gases before mixing was 5:3:1 .The density of the heaviest gas in the mixture is not more than 25 times that of the lighest gas. When the mixture was heated, vigorous reactions take place between X and Y and X and Z in which all the three gases were completely used up. The gases X,Y, Z respectively are |
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Answer» `H_2, O_2 , N_2` |
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| 91597. |
A 500g tooth paste sample has 0.2g fluoride concentration. The concentration of F^(-) ion in terms of ppm level is 100 xx X. What is the value of X ? |
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Answer» <P> Since to find the level of `F^(-) `ion in PPM we NEED to convert the conc. In Mg AS Conc of `F^(-)`ion in `Mg //Kg = 0.4 xx 100 mg // g m = 400 mg // kg ` Hence level of `F^(-)`ion in ppm `= ( " weight of solute")/("weigth of ion") = 400 mg //kg = 400 p p m = 100 xx 4 `p p m |
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| 91598. |
A 500 g tooth paste sample has 0.2 g fluoride concentration. What is the concentration of F^- in ppm: |
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Answer» 250 |
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| 91599. |
A 50.0 cm^3 portion of a mixture of H_2SO_4 and H_2C_2O_4 required 48.9 cm3 of 0.15 M NaOH solution for titration. Another 50 cm^3 required 38.9 cm^3 of 0.10 N KMnO_4 solution for titration. Calculate the masses of H_2SO_4 and H_2C_2O_4 present per dm3 of the solution. |
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Answer» |
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| 91600. |
A 500-cc bubl weighs 38.734 grams when evacuated and 39.3135 grams when filled with air at 1 atm pressure and 24^(@)C. Assuming that air behaves as an ideal gas at this pressure, calculate effective mass of 1 mole of air. |
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Answer» |
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