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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.
| 88901. |
A hypothetical particle 'Xetone' at rest a withmass equivalent to a He nucleus absorbs (n) photons of frequency v_(0). Its de-Broglie wavelength was later found to be (1)/( 8) sqrt((h)/( mv_(0))) ( m = mass of a photon), then value of n is |
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Answer» 4 |
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| 88902. |
(a) List four biological functions of proteins. (b) Name two diseases which are caused by the deficiency of Vitamin A and B. |
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Answer» SOLUTION :(a) Functions of Proteins 1. A transport agent . For example : Oxygen is TRANSPORTED in the body by haemoglobin . 2. For MAINTENANCE of fluid BALANCE in the body. 3. For regulation of metabolism . 4. As structural materials for membranes or CONNECTIVE tissues. (b) The deficiency of Vitamin A causes Xerophthalmia disease and deficiency of Vitamin `B_1` causes Beri-Beri disease. |
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| 88903. |
A hypothetical electrochemical cell shown below overset(" "o+)A|(xM) || B^(+)(yM)|overset(" "o+)B The e.m.f. measured is +0.20" V ". The cell reaction is : |
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Answer» `A^(+)+E^(-) to A , B^(+)+e^(-) to B` `A to A^(+)+e^(-)` `B^(+)+e^(-) to B` `A+B^(+) to A^(+)+B` |
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| 88904. |
A liquid which markedly scatters a beam of light (visible in a dark room) but leaves no residue when passed through a filter paper is best described as |
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Answer» a SUSPENSION |
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| 88905. |
A hypothetical electrochemical cell is shown below ThetaPt(s)|A_(2)(1 atm)|A^(-)(xM)||B^(+)(yM)|B(s)^(o+) The emf measured is 0.30V The cell reaction is :- |
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Answer» `A_(2)+2B^(+)to2A^(-)+2B` |
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| 88906. |
A hypothetical electrochemical cell is shown below overset(o+)(A)|A^(+)(xM)||B^(+)(yM)|overset(Ө)(B) The emf measure is +0.20 V. the cell reaction is |
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Answer» The cell REACTION cannot be predicted |
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| 88907. |
A liquid was mixed with ethanol and a drop of concentrated H_2 SO_4 was added. A compound with a fruity smell was formed. The liquid was: |
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Answer» `HCHO` |
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| 88908. |
A hypothetical electrochemical cell is overset(Theta)(A)|A^(+)(xM)||B^(+)(yM)|overset(o+)(B) The emf measured is +0.20V. The cell reaction is |
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Answer» The cell reaction cannot be PREDICTED |
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| 88909. |
A liquid rises to 1.0 cm in a capillary tube of radius rv How much will it rise if the cross-sectional area of the capillary tube is doubled? |
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Answer» Solution : Let the cross-sectional areas of the TUBES be a and 2a. `therefore a = pir_(1)^(2), r_(1) = sqrt(a/pi)` and, `2a = pir_(2)^(2), r_(2) = sqrt((2a)/pi)` We have, `GAMMA=(hdrg)/2` and `h_(1)=1 CM`. `therefore` for some same liquid, using the above-given equations, we get `h_(2)/h_(1) = r_(1)/r_(2) = 1/sqrt(2), h_(2) = 1/sqrt(2) xx 1 = 0.707 cm` |
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| 88911. |
A hydroxy acid on heating gives a 5-membered lactone. Theacid is : |
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Answer» `CH_(3)CH_(2)CHOHCOOH` |
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| 88912. |
A liquid was mixed with ethanol and a drop of concentrated H_(2)SO_(4) was added. A compound with a fruity smell was formed. The liquid was |
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Answer» `HCHO` |
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| 88913. |
A hydroxyl acid on heating gives a 5 - membered lactone. The acid is |
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Answer» `CH_(2)OHCH_(2)CH_(2)COOH` |
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| 88914. |
A: Hydrometallurgy is used for extraction of Ag and Au. R: Hydrometallurgy is different from pyrometallurgy |
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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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| 88915. |
A liquid mixture of benzene and toluene is composed of 1 mol of benzene and 1 mol of toluence. a.If the pressure over the mixture at 300 K is reduced, at what pressure does the first vapour form? b.What is the composition of the first trace of vapour formed? c. If the pressure is reduced further, at what pressure does the last trace of liquid disappear? d.What is the composition of the last trace of liquid? e. What will be the pressure, the composition of the liquid, and the composition of the vapour, when 1 mol of the mixture is vapourized? , Given:p_(T)@ = 32.05 mm Hg , p_(B)@ = 103 mm Hg |
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Answer» SOLUTION :a. The first vapour will be formed when the external PRESSURE becomes equal to the vapour pressure of the SYSTEM. `P=chi_(T)p_(T).^(@)+p_(B).^(@)chi_(B)` `P = 1/2 (32.05) + 1/2 (103) = 67.52 mm Hg` b. Composition of the first trace of vapour formed `chi_(T)= (P_(T).^(@)chi_(T))/(P)= (0.5 xx 32.05)/(67.52)=0.24` `chi_(B) = 1 - 0.24 = 0.76` c. The last trace of liquid will disappear when the composition of the vapour phase become `chi_(B) = 0.5` and `chi_(T) = 0.5`. The pressure at which this OCCURS can be calculated as `1/P = (chi_(T))/(P_(T)@)+(chi_(B))/(P_(B)@)=(0.5)/(32.05)+(0.5)/(103)` `P=49.01 mm Hg` d. Composition of the last trace of the liquid will be `chi_(B)=(P_(B)@chi_(B))/(P)` `0.5 =(32.05 chi_(B))/(49.01)` `chi_(B) = 0.76` and `chi_(T) = 0.24` e. `chi_(T) = 0.642`, `chi_(B) = 0.358` and `chi_(B) = 0.642`,`chi_(T) = 0.358`, `P = 57.46 mm Hg` |
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| 88916. |
(A) Hydrolysis of ethyl acetate in presence of acid is a reaction of first order whereas in presence of alkali, it is a reaction of second order. (R ) Acid only acts as a catalyst whereas alkali acts as one of the reactants. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 88917. |
(A) Hydrolysis of BrF_(3) gives HF and HBrO_(2) (R ) In the hydrolysis of interhalogen compounds ,lower halogen forms halide & higher halogen form oxyhalide |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 88918. |
A liquid mixture of A and B is placed in a cylinder-and-piston arrangement. Thepiston is slowly pulled out isothermally so that the volume of the liquid decreases and that of the vapour increases. At the instant when the quantity of the liquid still remaining is negligibly small, the mole fraction of A in the vapour is 0.4. P_A^@=0.4 atm, P_B^@=1.2 atm at the temperature in question. Calculate the total pressure at which the liquid has almost evaporated. Assume ideal behaviour. |
| Answer» SOLUTION :0.667 ATM | |
| 88919. |
A: Hydrolysis of an ester is a slow reaction R: Reactions between covalent species involve breaking and making of bonds. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 88920. |
A liquid is stirred in thermally insulated container , for about 2 hrs. Which of the following si correct ? |
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Answer» `w LT 0, Q = 0 , DeltaU lt 0 ` |
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| 88921. |
A hydrogenation reaction is carried out at 500K. If the same reaction is carried out in presenceof catalyst at the same rate, the temperature required is 400K. Calculate the activation energy of the reaction if the catalyst lowers the activation energy barrier by 20kJ/mol. |
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Answer» SOLUTION :Let `E_(a)` and `E._(a)` be the energy of activation in absence and presence of CATALYST for hydrogenation reaction, as `k=Ae^(-E_(a)//RT)` `k_(1)=Ae^(-E_(a)//Rxx500)` (In absence of catalyst) `k_(2)=Ae^(-E._(a)//Rxx400)` (In presence of catalyst) Given , `r_(1)=r_(2)`, Hence, `k_(1)=k_(2)` `e^(-E_(a)//Rxx500)=e^(-E._(a)//Rxx400)` `IMPLIES(E_(a))/(Rxx500)=(E._(a))/(Rxx400)` or `(E_(a))/(500)=(E_(a)-20)/(400)` (As `E_(a)-E._(a)=20`) `:.E_(a)=100kJ//mol^(-1)` |
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| 88922. |
A hydrogenation reaction is carried out at 500 K. If the same reaction is carriedout in the presence of a catalyst at the same rate, the temperature required is 400 K. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by 20 kJ "mol"^(-1) |
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Answer» SOLUTION :From Arrhenius.s equation, under the GIVEN condition, we have, `- (E)/(RT_1) = (E - 20)/(RT_2)` `100KJ "mol"^(-1)` |
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| 88923. |
A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in the presence of a catalyst at the same rate with same frequency factor, the temperature required is 400 K. What is the activation energy of the eaction. If the catalyst lowers the activation energy barrier by 16 KJ/mol? |
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Answer» 100kJ/mol `(epsilon_(a_(1)))/(T_(1))=(epsilon_(a_(2)))/(T_(2)),(epsilon_(a_(1)))/600=((epsilon_(a_(1))-16))/400,4epsilon_(a_(1))=5(epsilon_(a_(1))-16),4epsilon_(a_(1))=5epsilon_(a_(1))-80,epsilon_(a_(1))=80` |
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| 88924. |
A hydrogenation reaction is carried at 500K. If the same reaction is carried out in presence of the catalyst at the same rate, the temperature required is 400K. If the threshold energy of the reactant is 120 kJ mol^(-1) and the normal energy of the reactant is 20kJ mol^(-1) , calculate the magnitude of activation barried ( in kJ mol^(-1) ) that is lowered by the catalyst. |
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Answer» Solution :Let the amount of activation BARRIER LOWERED is x `kJ mol^(-1)` is presenve of catalyst. `:. K = A .e^(-) ( Ea)/( R T)` For uncatalysed reaction, `K_(1) = A. e^(-) ( Ea_(1))/( R T_(1))` For catalysed reaction `K_(2) = A. e^(-) ( Ea_(2))/( R T_(2))` From equation, `K_(1) = K_(2)` `e^(-) ( Ea_(1))/( RT_(1)) = e^(-)( Ea_(2))/( RT_(2)) RARR( Ea_(1))/( T_(1)) = ( Ea_(2))/( T_(2))` `(Ea_(1))/( T_(1)) = ( Ea_(1) - x)/( T_(2)) [ :. Ea_(2) = Ea_(1) -x]` `:. ( 100)/( 500)= ( 100 - x)/( 400)rArr x = 20 kJ mol^(-1)` |
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| 88925. |
A hydrogenation reaction is carrid out at 500 K. If the same reaction is carried out in the presence of a catalyst at the same rate, the termperature required is 400 K. Calculatethe activation energy of the reaction if the catalyst lowers the activation barrier by 20 kj mol^(-1). |
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Answer» Solution :RATE of hydrogenation at `500^@K` for a REACTION is equal to the rateof reaction in presense of catalyst at `400^@ K` Rate = k `xx` conc of REACTANT If rate of reaction are equal therefore their rate constant will also form Arrhenius equation `k = AE^(-E_a//RT)` `k_500 =Ae^(-(E-a//RT)`...(i)`k_400 = Ae^(-E_a//RT_2)`...(ii) When `k_500 = k_400` `:.((E_0)_1)/(RT_1) = ((E_a)_2)/(RT_2)`(From eqs (i) and (ii))or `((E_a)_1)/((E_a)_2) = T_1/T_2` `E_(a_1)/E_(a_2)=T_1/T_2 = 500/400 = 5/4`... (i) In presence of catalyst`E_a`decreases by 20 kJ `mol^(-1)` So,`E_a_1 = 100kJ//mol^(-1)` `E_a_2 = 80 KJ//mol^(-1)` Hence, `E_a` or reaction OS `100kJ mol^(-1)` |
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| 88926. |
A hydrogen sample is prepared in a particular excited state. Photons of energy 2.55 eV get absorbed into the sample to take some of the electrons to a further excited state B. Find orbit numbers of the states A and B. Given the allowed energies of hydrogen atom: E_(1)=-13.6 eV, E_(2)=-3.4 eV, E_(3)=-1.5 eV, E_(4)=-0.85 eV. E_(6)=-0.54 eV |
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Answer» |
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| 88927. |
A liquid is in equlibrium with its vapour at its boiling point. On the average, the molecules in the two phases have equal |
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Answer» Inter-molecular forces |
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| 88928. |
A hydrogen on oxidative ozonlysis produces Oxalic acid and Butanedioic acid. Its structrure is |
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Answer»
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| 88929. |
A liquid is kept in a closed vessel. If a glass plate (negligible mass) with a small hole is kept on top of the liquid surface, then the vapour pressure of the liquid in the vessel is: |
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Answer» More than what WOULD be if the GLASS PLATE were removed |
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| 88930. |
A hydrogen like specie is in a spherical symmetrical orbitalS_(1) having 3 radial node. It gets dexcited to another level S_(2) having no radial node. Energy of S_(2) orbital is 2.25 times energy of 1st bohr orbit of hydrogen atom. Based on this information answer the questions that follow. What is the combined total number of nodes (radial+angular) in S_(1) and S_(2) ? |
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Answer» 4 |
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| 88931. |
A hydrogen like specie is in a spherical symmetrical orbitalS_(1) having 3 radial node. It gets dexcited to another level S_(2) having no radial node. Energy of S_(2) orbital is 2.25 times energy of 1st bohr orbit of hydrogen atom. Based on this information answer the questions that follow. What is the orbital angular momentum quantum number of S_(2) ? |
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Answer» 1 `2p to 2-1-1=0` radial node |
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| 88932. |
A hydrogen like specie is in a spherical symmetrical orbitalS_(1) having 3 radial node. It gets dexcited to another level S_(2) having no radial node. Energy of S_(2) orbital is 2.25 times energy of 1st bohr orbit of hydrogen atom. Based on this information answer the questions that follow. Identify the specie involved |
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Answer» `He^(+)` `13.6xx(Z^(2))/(n^(2)) =9/4 XX 13.6` `Z=3 rArr Li^(2+)` |
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| 88933. |
A hydrogen like, atom ( atomic number =Z) is in a higher excited state of quantum number n. This excited atom can make a transition to first excited state by successively emitting two photons of energies 10.20eV and 17.0 eV respectively. Alternatively, the atom from the same excited state can make a transition to second excited state by successively emitting two phons of energy 4.25 eV and 5.95eV respectively. Determine the values of n and Z. |
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Answer» Solution :The energy emitted during transition of ELECTRON from `n^(th)` shell to FIRST excited state ( i.e. `2^(nd)` shell ) `10.20 + 17.0 = 27.2 eV = 27.2 XX 1.6 = 10^(-19) J` `Delta E = R_(H) Z^(2) hc [ (1)/( 2^(2)) - ( 1)/( n ^(2)) ]` `27.2 xx 1.6 xx 10^(-19)` `= 1.09 xx 10^(7) xx Z^(2) xx 6.6 xx 10^(-34) xx 3 xx 10^(8) [ ( 1)/( 2^(2)) - ( 1)/( n^(2)) ]`....(1) Similarly total energy liberated during transition of electron from NTH shell to second excited state ( i.e. `3^(rd)` shell ) `= 4.25 xx 5.95= 10.20 eV = 10.20 xx 1.6 xx 10^(-19) J ` `:. 10.20 xx 1.6 xx 10^(-19)` `= 1.09 xx 10^(7) xx Z^(2) xx 6.6 xx 10^(-34) xx 3 xx 10^(8) [ ( 1)/( 3^(2)) - ( 1)/( n ^(2)) ] `......(2) Dividing equation (1) by equation (2) n = 2 On substituting the value of n in equation (1) or (2) Z = 3 |
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| 88934. |
A liquid is in equilibrium with its vapour at a certain temperature. If DeltaH_(vap) = 60.24 kJ mol^-1 and DeltaS_(vap) = 150.6 J mol^-1 then the boiling point of the liquid will be |
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Answer» 275 K |
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| 88935. |
(A): Hydrogen iodide is most stable among hydrogen halides (R): Iodide is most powerful reductant among halides. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 88936. |
A hydrogen gas electrode is mae by dipping platinum wire in a solution of HCl of pH=10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be |
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Answer» 1.81 V `E_(O.P.)=E_(O.P.)^(o)-(0.059)/(n)log([H^(+)])/((P_(H_(2)))^(1//2))` `E_(O.P.)=0-(0.059)/(1)"log"(10^(-10))/((1)^(1//2))(pH=10,[H^(+)]=10^(-10)M)` `E_(O.P.)=0.59V` |
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| 88937. |
A liquid is immiscible in water was steam distilled at 95.2^(@)C at a pressure of 0.983 atm. What is the mass of the liquid present per gram of water in the distullate. Molar mass of the liquid is 134.3 g/mol and the vapour pressure of water is 0.84 atm. Also, Vapour pressure of pure liquid is 0.143 atm. |
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Answer» `1 G` |
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| 88938. |
A hydrogen gas electrode is made by dipping platinum wire is a solution of HCl of pH=10 and by passing hydrogen gas around the platinum wire at one atom prssure. The oxidation potential of the electrode would be |
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Answer» 0.059 V `H^(+)+e^(-)to(1)/(2)H_(2)` `E_(H^(+)//(1)/(2)H_(2))^(@)=E_(H^(+)//(1)/(2)H_(2))^(@)-(0.0591)/(n)"LOG"(p_(H_(2))^(1//2))/([H^(+)])` `=0-(0.0591)/(n)"log"(1)/(10^(-10))=-0.591` `(pH=10" MEANS "[H^(+)]=10^(-10)M)` `therefore`OXIDATION potential`=0.591V` |
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| 88939. |
A hydrogen gas electrode is made by dipping platinum wire in a solution having pH =4 and by passing hydrogen gas around the platinum wire at one atm. Pressure. The oxidation potential of electrode would be |
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Answer» 0.236V |
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| 88940. |
(A): Hydrogen gas always evolved only at cathode during electrolysis (R) : H^(+) ions undergo reduction by gaining electrons The correct answer is |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 88941. |
A hydrogen electrode placed in a buffer solution of CH_(3)COONa and CH_(3)COOH in the ratios of x:y and y:x has electrode potential values E_(1) volts and E_(2) volts, respectively at 25^(@)C The pK_(a) values of acetic acid is (E_(1) and E_(2) are oxidation potential) |
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Answer» `(E_(1)+E_(2))/(0.118)` or `pH_(1)=E_(1)//0.059=pK_(a)+log((X)/(y))` `pH_(2)=E_(2)//0.059=pK_(a)+log((y)/(x))` or `(E_(1)+E_(2))/(0.059)=2pK_(a)` or `pK_(a)=(E_(1)+E_(2))/(0.118)` |
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| 88942. |
A liquid is found to scatter a beam of light but leaves no residue when passed through filter paper. The liquid can be described as |
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Answer» a SUSPENSION |
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| 88943. |
A hydrogen electrode placed inbuffer solution of CH_(3)COONa and acetic acid in the ratios x:y and y:x and electrode potential values E_(1) and E_(2) respectively at 25^(@)C. Calculate the pK_(a) value of acetic acid in terms of E_(1) and E_(2) (E_(1) and E_(2) are oxidation potentials). |
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Answer» Solution :For oxidation REACTION, `(1)/(2)H_(2)TOH^(+)+e^-` `E_(1)=E^(@)-(0.0591)/(1)LOG[H^(+)]_(1)` `E_(2)=E^(@)-(0.0591)/(1)log[H^(+)]_(2)` As `E^(@)=0`, adding EQNS, (i) and (ii) `E_(1)+E_(2)=-(0.0591)/(1){log[H^(+)]_(1)+log[H^(+)]_(2)}` For `CH_(3)COOHhArrCH_(3)COO^(-)+H^(+)` `K_(1)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])` or `[H^(+)]=(K_(a)[CH_(3)COOH])/([CH_(3)COO^(-)])` `therefore[H^(+)]_(1)=K_(1)(y)/(x) and [H^(+)]_(2)=K_(a)xx(x)/(y)` `thereforeE_(1)+E_(2)=-(0.0591)/(1)["log"(K_(a)y)/(x)+"log"(K_(a)x)/(y)]=-0.0591[2logK_(a)]` `thereforelogK_(a)=(E_(1)+E_(2))/(2(-0.0591))=-(E_(1)+E_(2))/(0.118)` or `pK_(a)=-logK_(a)=(E_(1)+E_(2))/(0.118)` |
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| 88944. |
A liquid is in equilibrium with its vapours at its boiling point. On the average the molecules in thetwo phases have equal: |
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Answer» INTERMOLECULAR forces |
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| 88945. |
A hydrogen electrode placed in a buffer solution of CH_3COONa and acetic acid in the ratio.sx:y and y:x has electrode potential values E_1 volt and E_2 volt respetively at 25^@C thepKa values of acetic acid is (E_1 and _2 are oxidation potential ): |
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Answer» `(E_1+E_2)//(0.118)` |
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| 88946. |
A liquid is found to scatter a beam of light but leaves no residance when passed through the filter paper. The liquid can be described as |
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Answer» A suspension |
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| 88947. |
A hydrogen electrode placed in a buffec solution of CH_3COONA and acetic acld in the ratio's x: y and y : x has electrode potential values E_1 volt and E_2 volts respectively at 25^@C. The pK_a values of acetic acid is E_1 and E_2 are ox-potential) |
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Answer» `(E_(1)+E_(2))/(0.118)` |
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| 88949. |
A hydrogen atom is in its ground state absorbs a photon. The maximum energy of such a photon is: |
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Answer» 1.5 eV |
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| 88950. |
A hydrocation (A) (C_(8) H_(10)) (E) is a stream volatile compound and on nitrationgives two mononitro derivatives. (A) gives the followingreactions. Degree fo unsaturation (DU) in (H) is: |
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Answer» `7` `4 D.U`. and `C:H = 1:1` suggst that `(A)` contains benzene ring with two extra `C` atom [i.e., two `(Me)` groups]. Since COMPOUND`(A)` is steam volatie and on nitration gives two nitro-derivaties, so `(A)` is ortho-xylene.  `D.U` in `H` = Two benzene ring `+` cycloburtane ring `= (2xx4+1) = 9` |
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