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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.
| 89651. |
A fruity smell is produced by the reaction of C_2H_5OH with |
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Answer» `PCl_5` |
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| 89652. |
(A) Fructose is the sweetest naturally occuring sugar. (R) Fructose is a functional isomer of glucose. |
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Answer» Both A & R are TRUE and R is the correct EXPLANATION of A |
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| 89653. |
(A) From Nitrogen to phosphorus increase in atomic radii is more when compared to increase in size from P to As (R ) Nitrogen is a gas at room temperature where as P_(4) is a solid atroom temperature |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 89654. |
A fristorderreaction is 40%complete in 50 minutes. Calculatethe value of the constant. In what timewill the reaction be 80 %complete ? (ii)K_(sp) " of" Ag_(2)CrO_(4) " is "1.1 xx 10^(-12) . Whatis the solubility ofAg_(2)CrO_(4) in 0.1MK_(2)CrO_(4) |
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Answer» Solution :(i) For the firstorder reation ` K = (( 2.303)/t) log ( a/(a-x))` When x= ` ( 40/100) a = 0.4 a ` t = 50 m ` thereforek = (( 2.303)/50) log (a/(a-0.4a))` `k = ((2.303)/50) log (1/0.6)` `0.010216min^(-1)` t = ?when x = 0.8a From aboveK = ` 0.010216 min^(-1)` ` thereforet = (2.303/(0.010216)) log (a/(a-0.8a))` ` (2.303/(0.010216)) log (1/0.2) = 157.58 min` The timeat WHICHTHE reaction will be 80%complete is 157.58 min. (ii) ` Ag_(2) CrO_(4)iff_(2x)2Ag^(+) + underset(x)(CrO_(4)^(2-)` x is the solubility of ` Ag_(2)CrO_(4) " in"0.1 M K_(2)CrO_(4)` ` K_(2)CrO_(4) IFF 2K^(+) + CrO_(4)^(2-)` ` [ Ag^(+)]= 2x ` ` [ CrO_(4)^(2-)] = ( x+ 0.1) = 0.1 ` ` K_(SP) = [Ag^(+)]^(2) [ CrO_(4)^(2-)]` `1.1xx 10^(-12) = (2x)^(2) (0.1)` ` 1.1 xx 10^(-12) = 0.4 x^(2)` ` x^(2) = ( 1.1 xx 10^(-12))/(0.4)` ` x = sqrt((1.1 xx 10^(-12))/(0.4))` ` x = sqrt(2.75 xx 10^(-12))` ` x = 1.65 xx 10^(-6) M` |
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| 89655. |
(A)Froath floatation process involves adsorption (R) Pure ore is preferentially wetted by froath in the benefication method. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 89656. |
(A) Fridel-Crafts acylation gives an aromatic ketone from benzene (R) : Acylation reaction is a uncleophilic substitution reaction with respect to benzene . |
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Answer» Both A & R are TRUE, R is the correct explanation of A |
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| 89657. |
A freshly formed precipitate of ferrie hydroxide can be converted to a colloidal sol by shaking it with a small quantity of ferric chloride.Explain. |
| Answer» Solution :When we add `FeCl_3` to a freshly FORMED precipitate of `Fe(OH)_3`, peptisation occurs. The `Fe^(3+)` ions are adsorbed on the SURFACE of the precipitate which ultimately breaks down into SMALLER particles of colloidal SIZE. | |
| 89658. |
A freshly prepared radioactive source of half -life 2 hours emits radiations of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is |
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Answer» 6 hours |
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| 89659. |
A fresh radioactive mixture contians short lived sppecies A and B. Both emitting alpha-particles intially of 8000 alpha particles per minute. 20 minutes later, they emis at the rate of 3500 alpha -particles per minute. If the half-lives of the species A and B are 10 minutes and 500 hours respectively, then the ratio of activities of A: B the intial mixture was: |
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Answer» `4:6` |
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| 89660. |
(A) Frenkel defects are found in silver halides(R) Frenkel defects are commonly found in ionic solids |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 89661. |
A freshly cut piece of wood gives 16100 counts of beta - ray emission per minute per kg and old wooden bowl gives 13200 counts per minute per kg. The age of wooden bowl is ( half-life of ""^(14) C is 5568 years ) [Given : log 1.219 = 0.0857 ] |
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Answer» 576 years |
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| 89663. |
A fractional column is used in : |
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Answer» SUBLIMATION |
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| 89664. |
A forms hcp lattice and B are occupying 1/3rd of tetrahedral voids, then the formula of compound is- |
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Answer» AB `therefore` Number of tetrahedral VOIDS = 2n `therefore` Number of B atoms = `1/3xx2n=2/3n` Ratio of A : B = 3 : 2 Then the formula of the compound is `A_(3)B_(2)`. |
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| 89665. |
A forms hcp lattice and B are occupying (1/3)^(rd) of tetrahedral voids. The formula of compound is ……. |
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Answer» `AB` Total number of tetrahedral VOIDS = 12 Total number of B-atoms `= 12xx 1/3 = 4` RATIO of A-atoms to B-atoms `= A:B = 6:4 = 3:2` |
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| 89666. |
(A) Formic acid reduces Tollen's reagent. (R) Compounds containing free-CHO group reduce Tollen's reagent. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 89667. |
(A) : Formic acid is stronger acid than acetic acid . (R) : +1 groups decrease acidic nature of carboxylic acids. |
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Answer» Both A & R are TRUE , R is the CORRECT EXPLANATION of A |
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| 89668. |
(A) Formic acid reduces mercuric chloride. (R) Formic acid has reducing aldehydic group |
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Answer» If both (A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A). |
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| 89669. |
(A) Formic acid is made poisonous with the presence of sulphuric acid (R ) Sulphuric acid acts as dehydrating agent liberating carbon monoxide from formic acid |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 89670. |
(a) Formaldehyde gives Cannizzar's reaction while acetaldehydedoes not. Explain (b) complete the follows : (i) C_(6)H_(5)COCH_(3)overset(Zn//Hg//HCl)rarr (ii) C_(6)H_(5)CHO overset(NaOH)rarr |
| Answer» Solution :(b) (i) `C_(6)H_(5)CH_(2)CH_(3)` , (ii) `C_(6)H_(5)CH_(2)OH + C_(6)H_(5)COONA` | |
| 89671. |
(A) Formation of ether from alcohol is an example of bimolecular nucleophilic substitution reaction (SN^2)(R) Alkyl groups of alcohol are preferably 1^@ alkyl groups |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 89672. |
(A) For the reduction of esters in to alcohols Ni//H_2 is preferred than LiAlH_4 (R) LiAIH_4 is more expensive than Ni//H_2 |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 89673. |
a) For the electrochemical cell represented as: Cu_((s))|Cu_((aq))^(2+)||Ag_((aq))^(+)|Ag_((s)), write the half cell reaction that occurs at (i) anode (ii) cathode |
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Answer» Solution :Given CELL representation is `Cu|Cu^(2+)||Ag^(+)|Ag` The half cell reaction that occurs at i) anode `=cu hArr cu^(2+)+2E^(-)` ii) CATHODE `=2AG^(+)+2e^(-)rarr2Ag` |
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| 89674. |
(A) For iron extraction usually Fe_(2) O_(3)is used but not FeS_(2) (R) Fe_(2) O_(3)does not produce polluting gas like SO_(2)but FeS_(2) |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 89675. |
(a) For a reaction A+BrarrP, the rate is given by Rate =K[A][B]^(2) (i) How is the rate of reaction affected if the concentration of B is doubled ? (ii) What is the overall order of reaction if A is present in large excess ? (b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. |
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Answer» SOLUTION :(a) (i) `A+BrarrP" Rate "=k[A] [B]^(2)` Differential rate equation of REACTION is `(dn)/(dt)=k[A]'[B]^(2)=k[A][B]^(2)` When concentration of B is doubled it means concentration of B becomes `[2xxB]" "therefore` New rate of reaction, `(dn)/(dt)=k[A] [2B]^(2)=4k[A] [B]^(2)=4[(dn)/(dt)]` i.e., rate of reaction will become 4 times. (ii) A useful protocol for determining the order of reaction with respect to a particular components is to measure the concentration dependence of rate when all other REACTANTS are in great EXCESS. `v=K[A]_("excess")^(a)[B]^(2)` `v=K'[B]^(2)["Over all order "=2]` `k'=K[A]_("excess")^(a)` |
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| 89676. |
(a) For a reaction A+B to P, the rate is given by Rate =k[A][B]^(2) (i) How is the rate of reaction affected if the concentration of B is doubled ? (ii) What is the overall order of reaction if A is present in large excess ? (b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90%. Completion of this reaction (log2=0.310). |
| Answer» SOLUTION :(a) (i) 4 times (ii) 2 (b) `k=(0.693)/(30)MIN^(-1),t_(90%)=(2.303)/(k)log""(100)/(100-90)=99.69" min"` | |
| 89677. |
(A) Food preservatives prevent the growth of microorganisms. (R) Antioxdants preserve food by retarding the action of oxygen on food. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 89678. |
Which set of d-orbitals of a metal atom/ion experience more repulsion inoctahedral field created by the ligands? |
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Answer» SOLUTION :(a) `Co (NH_(3))_(5) NO_(2)Cl_(2)` Pentaamminenitrito-N-cobalt(in) chloride. (B)Eg set of orbitals. |
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| 89679. |
A follows parallel path, first order reaction giving B and C as If initial concentration of A is 0.25M, calculate the concentration of C after 5 hours of reaction. [Given : k_(1)=1.5xx10^(-5)s^(-1), k_(2)=5xx10^(-6)s^(-1) |
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Answer» |
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| 89680. |
A follows first order reaction. (A) rarr Product The concentration of A changes form 0.1 M to 0.025 M in 40 min. Find the rate of reaction of A when the concentration of A is 0.01 M. |
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Answer» `3.47xx10^(-4)M"MIN"^(-1)` |
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| 89681. |
(a) Following reactions occur at cathode during the electrolysis of aqueous copper (II) chloride solution : 1Cu^(2+) (aq) + 2e^( -) toCu (s),E° = + 0.34 V H^(+)+ (aq) + e^(-) to1/2 H_(2) (g) On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why? (b) State Kohlrausch law of independent migration of ions. Write its one application. |
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Answer» Solution :(a) That reaction will take PLACE which has greater value of electrode potential. Thus the following reaction takes place `Cu^(2+) (aq) + 2E^(-) toCu (s)` (b) Limiting molar conductivity of an electrolyte can be represented as the sum of the individualcontributions of the anion and cation of the electrolyte. Application : Using Kohlrausch law of independent migration of ions, it is POSSIBLE to calculate `Lambda_(m)^(@)` for any electrolyte from the `LAMBDA^(@)`values of individual ions. |
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| 89682. |
(a) Following are the transition metal ions of 3d series : Ti^(4+),V^(2+),Mn^(3+),Cr^(3+) (Atomic numbers : Ti = 22, V = 23, Mn = 25, Cr = 24) Answer the following : (i) Which ion is most stable in an aqueous solution and why ? (ii) Which ion is a strong oxidising agent and why ? (iii) Which ion is colourless and why ? (b) Complete the following equations : (i) 2MnO_(4)^(-)+16H^(+)+5S^(2-)to (ii) KMnO_(4)overset("Heat")to |
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Answer» Solution :(a) (i) `Cr^(3+)` ion is most stable in an aqueous solution because it has stable half-filled `(t_(2g))` configuration. (ii) `Mn^(3+)` is a strong oxidising AGENT because `Mn^(2+)` formed as a result of reduction has half-filled `(d^(5))` stable configuration. (iii) `TI^(4+)` is colourless. There is no possibility of d-d transition and HENCE no absorption of LIGHT. (b) (i) `2MnO_(4)^(-)+16H^(+)+5S^(2-)to2Mn^(2+)+8H_(2)O+5S` (ii) `2KMnO_(4)overset("Heat")toK_(2)MnO_(4)+MnO_(2)+O_(2)` |
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| 89683. |
(a) Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution :Ag+ (aq) + e^(-) toAg (s), E° = + 0.80 V H^(+) (aq) + e^(-) to 1/2H_(2)(g) ,E^(@) = 0.00 V On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why? (b) Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration ? |
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Answer» Solution :(a) The reaction with a higher value of electrode potential is preferred. Hence `Ag^(+)` will be deposited in preference to `H^(+)`. `Ag^(+) (AQ) + e^(-) toAg (s) ` (b) When the concentration of the solution APPROACHES zero, molar conductivity is called limiting molar conductivity. Conductivity of an electrolyte solution decreases with decrease in concentration. Conductivity of a solution at any concentration is the conductance of one unit volume of the solution kept between platinum electrodes with unit area of cross section and at a DISTANCE of unit LENGTH. With DILUTION, the number of ions per unit volume decreases. Therefore conductivity of the solution decreases with decrease in concentration. |
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| 89684. |
(A) : Fluorine is the best oxidising agent (R ) SRP is highest for Fluorine |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 89685. |
(A) Fluorine forms only one oxyacid HOF (R ) Fluorine has small size and high electronegativity |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 89686. |
(a) Fluorine is a stronger oxidizing agent than chlorine. Why?(b) Write one use of chlorine gas(c) Complete the following equation CaF_(2)+H_(2)SO_(4)rarr |
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Answer» Solution :(a) It is due to (i) low enthalpy of dissociation of F-F bond (ii) HIGH hydration enthalpy of ` F^(–)` . (b) (i) for bleaching WOOD pulp (required for manufacture of paper and rayon), cotton and textiles. (ii) In the metallurgy (extraction) of gold and platinum. (III) In the manufacture of dyes, drugs and organic compounds such as ` CHCl_(3), C Cl_(4)`, DDT, refrigerants (`C Cl_(2)F_(2)`, freon), and bleaching powder. (iv) In the preparation of poisonous gases such as phosgene ` (COCl_(2))` , tear gas `(C Cl_(3)NO_(2))` , mustard gas `(ClCH_(2)CH_(2)SCH_(2)CH_(2)Cl)` , etc. Mustard gas was used by Germany in World War I. (v) In sterilizing drinking water. `(c) CaF_(2)+H_(2)SO_(4) rarr CaSO_(4)+2HF` |
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| 89687. |
(A): Fluorine does not form oxyacids (R): Electronegativity of fluorine is higher than that of oxygen. |
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Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) |
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| 89688. |
(A) Fluorine exhibits only -l oxidation state (R ) Fluorine is called super halogen |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 89689. |
(A) Fluoride is oxidised by managanese dioxide (R ) Manganese dioxide is reduced to mana |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 89690. |
A flask of volume 1L contains 1 mol of an ideal gas. The flask is connected to an evacuated flask, and as a result the volume of the gas becomes 10 L. the change in entropy (J*K^(-1)) of the gas is this process is- |
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Answer» 9.56 |
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| 89691. |
A flask of volume 1 litre contains vapour of CH_(3)OH at a pressure of 1 atm and 25^(@)C. The flask was then evacuated till the final pressure dropped to 10^(-1) mm. Find the number of molecules of methyl alcohol left in the flask. |
| Answer» SOLUTION :`3.2 XX 10^(15)` | |
| 89692. |
A flask of methane (CH_4) was weighed Methane was then pushed out and the flask again weighed when filled with oxygen at the same temperature and pressure. The mass of oxygen would be : |
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Answer» The same as the METHANE |
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| 89693. |
A flask of 8.21 litre contains CH_(4) gas at a pressure of 2 atm. Find moles of CH_(4) gas at 400 K? |
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Answer» 0.5 MOLE |
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| 89694. |
A flask is partially evecuated to 400 torr pressure of air. A small amount of benzene is introduced into the flask in order that some liwquid will remain after equilibrium has been established. The vapour pressure of at 25^(@)C is 200 torr. What is the total pressure in the flask at equilibrium at 25^(@)C ? |
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Answer» 120 TORR |
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| 89695. |
A flask contains a mixture of two compounds AB and XY. Both of these decompose on heating by first order reaction. If the half-life of AB is 30 min and that of XY is 10 min, how long will it take for the concentration of AB to be four times that of XY ? Assume that the initial concentration of both AB and XY to be same. |
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Answer» Solution :SUPPOSE time taken for [AB] to be EQUAL to 4 [XY] is t min. Half-lives of AB in t min `=(t)/(30)` Amount of AB LEFT after `t//30` half-lives `=((1)/(2))^(t//30)[AB]_(0)` Half-lives of XY in t min `=(t)/(10)` Amount of XY left after `t//10` half-lives `=((1)/(2))^(t//10)[XY]_(0)` But `[AB]=4[XY]:.((1)/(2))^(t//30)[AB]_(0)=4XX((1)/(2))^(t//10)[XY]_(0)` But `[AB]_(0)=[XY]_(0)("Given")` `:.((1)/(2))^(t//30)=4xx((1)/(2))^(t//10)" or "((1)/(2))^(t//30-t//10)=4" or "2^((t//10-t//30))=2^(2)" or "(t)/(10)-(t)/(30)=2` or `20t=600"or"t=30" min"` |
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| 89696. |
A flask contains 16 gm helium (""_(2)^(4)He) gas (Gram atomic mass of He=4). Find (i) Moles of He (ii) Moles of proton (iii) Total number of neutrons |
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Answer» (i) 4 (II) 8 (iii) `8N_(A)` |
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| 89697. |
A flask containing air ( open to atmosphere ) is heated from 300K to 500K . The fraction of air escaped to the atmosphere is about. |
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Answer» 0.166 `( V_(1))/( T_(1))= ( V_(2))/( T_(2)) ` or `V_(2) = V_(1) xx ( T_(2))/( T_(1)) = V xx( 500)/( 300) = ( 5)/( 3) V ` Increase in volume `= ( 5)/( 3) V - V = ( 2)/( 3) V ` % of AIR escaped `= ( 2)/( 3) V xx ( 100)/( V) = 66.7%` |
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| 89698. |
A first order reaction which is 30% complete in 30 minutes has a half-life period of…… |
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Answer» 24.2 minutes |
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| 89699. |
A first order reaction was started with a decimolar solution of the reactant , 8 minutes and 20 seconds later its concentration was found to be M/100 . So the rate of the reaction is |
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Answer» `2 . 303 xx 10^(-5) sec^(-1)` GIVEN : `a = (1)/(10) = 0.1` m , a -x = `(1)/(100) = 0.01` m , t = 500 sec `thereforek = (2.303)/(500) "log" (0.10)/(0.01)= (2.303)/(500) ` log 10 = `(2.303)/(500) = 0.004606 = 4.6 xx 10^(-3) sec^(-1)`. |
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| 89700. |
Define an expression for the rate constant of a 1st order reaction. Define half life period. A first order reaction takes 69.3 minutes for 50% completion. How much time will be needed for 80% completion? |
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Answer» SOLUTION :Given that `t_1/2=69.3`MIN. `K=0.693/(69.3min)=10^-2min^-1`APPLYING the RELATION:`t=2.303/K"log"100/100-80)` (for`80%` completion ,when` a =100,K=80) t=2.303/(10^-2S^-1"log5"=(230.3xx0.6989)`minute=160.96minutes: |
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