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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.
| 50101. |
A group which deactivates the benzene ring towards electrophilic substitution but directs the incoming group towards o - and p- positions is |
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Answer» `–NH_2` |
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| 50102. |
A group 14 elements is to ebe converted into n-type semiconductor by doping it with a sutiable impurity. To which group should this impurity belong ? |
| Answer» SOLUTION :N -type SEMICONDUCTOR means conduction due to presence of exess of negatively charged electrons. Hence, to CONVERT GROUP 14 ELEMENT into n-type semiconductor, it should be doped with group 15 element . | |
| 50103. |
A groups metal (A) which is present in common salt reacts with (B) to give compound in which hydrogen is present in -1 oxidation state. (B) on reaction with a gas (C) to i universal solvent (D). The compound (D) on reacts with (A) to give (E), a strong base Identify A, B, C, D and E. Explain the reactions. |
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Answer» Solution :(i) Isotops of HYDROGEN reacts with which oxygen to give heavy WATER (B) `underset((A))2D_2+O_2tounderset((B))2D_2O` (ii) Deuterium (A) undergoes addition reaction with PROPENE (C ) to give PROPANE deuteride (D) `CH_3underset((C ))CH=CH_2toCH_3-underset((D))CHD-CH_2D` 
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| 50104. |
A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong ? |
| Answer» Solution :n-type semiconductor means conduction due to presence of EXCESS of NEGATIVELY charged electrons. Hence to CONVERT GROUP 14 element into n-type semiconductor, it should be doped with group 15 element | |
| 50105. |
A Grignard reagent (A) and a haloalkene (B) react together to give (C). Compound (C) on heating with KOH yields a mixture of two geometrical isomers, (D) and (E), of which (D) predominates. (C) and (E) have the same molecular formula and (C) gives 1-bromo-3-phenyl propane on reaction with HBr in the presence of a peroxide. Give structures of (A), (B)and (C) and configuration of (D) and (E) with reasons. |
Answer» Solution :Since, (C) GIVES 1-bromo-3-phenylpropane on REACTION with HBr in the presence of a peroxide, therefore (C) is 3-phenylprop-1-ene.  Because (C) and (E) have same MOLECULAR formula, therefore, they are isomers and the reaction is : (C) `overset(KOH)to(D)+(E)` ,it is an isomerisation reaction. So, (D) and (E) are 3-phenylprop-2-ene.  Since, (C) is obtained by the reaction of a Grignard reagent (A) with haloalkene (B), therefore, (A)and (B) are: 
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| 50106. |
A graph of aromic radius versus atomic number is given blow: Using the above graph, how will you account for the varitation of inoisation enthalpy in a period. |
| Answer» Solution :Ionisation enthalpy INCREASE with increase in unclear charge and DECREASE in atomic SIZE. | |
| 50107. |
A graph is plotted between p (atm) vs t^(@)C for 10 mol of an ideal gas as follows: Then slope of curve and volume of container (L) respectively, are: |
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Answer» 0.1,8.21 |
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| 50108. |
A graph is ploted between log V and log T for two moles of gas at constant pressure of 0.0821 atm. [V and T are in litres and kelvin]. Then which of the following is/are correct |
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Answer» The graph is a STRAIGHT line with slope + 1 log v = log T + log 2. |
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| 50109. |
A good quality of water will have |
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Answer» HIGH D.O. |
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| 50110. |
A good dieneophile for a Diels-Alder reaction is? |
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Answer» an alkene |
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| 50111. |
A golfgallhasa massof 40 gand a speedof45 m//sif thespeedcan bemeasuredwithinaccuracyof 2%calculatethe uncertainty inthe position |
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Answer» Solution :Uncertainlty invelocity= VELOCITY`xx %100` Accordingof Heisenberguncertaintyprincipleuncertaintyin position`Delta x` `Delta x = (h)/(4pi m Delta V)`wherem= 40 gm =`40 xx 10^(3) KG` `=(6.626 xx 10^(34) j s)/(4(3.14) (40xx 10^(3)kg) (0.90 MS^(1)))` `=0.01465 xx 10^(31) j s^(2) m^(-1)kg^(-1)` `=1.465 xx 10^(33) m:., 1 J = 1 kgm^(2) s^(2)` thisisnearly`10^(18) ` timessmallerthan thediameterof a typicalatomicnucleus . Asmenthonedearlierfor largeparticle.theuncertaintyprinciplesets so meaningfulimitto theprecisionof measurements . |
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| 50112. |
A golf ball has a mass of 40 g and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in position |
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Answer» SOLUTION :UNCERTAINTY in speed = 2% of `40 MS^(-1)`, i.e., `Delta v = (2)/(100) xx 45 xx 0.9 ms^(-1)` Applying uncertainty principle, `Delta x (m xx Delta v) = (h)/(4PI)` or `Delta x = (h)/(4pi m Delta v) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(4 xx 3.14 xx (40 xx 10^(-3) kg) (0.9 ms^(-1))) = 1.46 xx 10^(-33) m` |
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| 50113. |
A golf ball has a mass 40g and a speed of 45 ms^(-1). If the speed can be measured with inaccuracy of 2%, calculate the uncertainty in position. |
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Answer» SOLUTION :The UNCERTAINTY in speed `= 45 XX(2)/(100) = 0.9 ms^(-1)` `"Using the EQUATION, "Deltax=(h)/(4piDeltavxxm)` `Deltax=(6.626xx10^(-34)kgm^(2)s^(-1))/(4xx3.1416xx0.9ms^(-1)xx416xx0.9ms^(-1)xx40xx10^(-3)KG)` `Deltax=1.46xx10^(-33)m` This is nearly 1018 times smaller than the diameter of a typical atomic nucleus. |
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| 50114. |
(A) Gold sol is hydrophobic and multimoleular. (r) Gold sol is prepared by Bredig's arc method. |
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Answer» IF both (A) and (r) are CORRECT and (r) is the correct EXPLANATION for (a). |
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| 50115. |
A glass tube of volume 112ml. containing a gas is partially evacuated till the pressure in it drops to 3.8 xx 10^(-5) torr at 0^@C. The number of molecules of the gas remaining in the tube is |
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Answer» `3 xx 10^17` |
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| 50116. |
(A): Glass possess sharp melting point. (R): Glass is a pseudo solid |
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Answer» Both A and R are TRUE and R is the correct EXPLANATION of A |
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| 50117. |
A glass bulb is connected to an open limb manometre. The level of mercury in both limbs of the manometer was same. The bulb was heated to 57^@C. If the room temperature and the atmospheric pressure were 27^@C and 750mm, the difference of levels in the two limbs now will be |
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Answer» Solution :INITIAL pressure = 75 cm `(P_1)/(T_1) = (P_2)/(T_2) = 75/300 = (P_2)/(320) IMPLIES P_2 = 80 cm`. `:.` diff in LIMB = 5 cm . |
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| 50118. |
A glass bulb is connected to an open limb manometer. The level of mercury in both limbs of the manometer was same. The bulb was heated to 57^(@)C. If the room temperature and the atmospheric pressure were 27^(@)C and 750 mm, the difference of levels in the two limbs now will be |
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Answer» 2.5 CM `:. (750)/(300) = (P_(2))/(330) " or " P_(2) = 825 mm` `:.` Difference of levels `= 825 - 750 = 75 mm` `= 7.5 cm` |
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| 50119. |
A glass bulb contains 2.24 L of H_(2) and 1.12 L of D_(2) at S.T.P. It is connected to a fully evacuated bulb by a stocock with a small opening. The stopcock is opened for some time and then closed. The first bulb now contains 0.10 g of D_(2).Calculate the percentage composition by weight of the gases in the second bulb. |
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Answer» SOLUTION :Weight of 2.24 L of `H_(2)` at S.T.P.=0.2 g 0.1 mol(Mol. Mass of `H_(2)`=2) Weightof 1.12 L of `D_(2)` at S.T.P.=0.2 g =0.05 mol( Mol. Mass of `D_(2)`=4) As NUMBER of moles of the two gases are different but V and T are same, therefore, their partial pressures will be different, i.e., in the ratio of their number of moles. Thus, `(P_(H_(2)))/(P_(D_(2)))=(n_(H_(2)))/(n_(D_(2)))=(0.1)/(0.05)=2` Now, `D_(2)` present in the first bulb=0.1 g (GIVEN) `:.D^(2)` diffused into the second bulb `=0.2-0.1=0.1" g"=0.56" g L"` at S.T.P. Now, `"" (r_(H_(2)))/(r_(D_(2)))=(P_(H_(2)))/(P_(D_(2)))xxsqrt((M_(D_(2)))/(M_(D_(2))))""or ""(v_(H_(2)))/(t)xx(t)/(v_(D_(2)))=(P_(H_(2)))/(P_(D_(2)))xxsqrt((M_(D_(2)))/(M_(H_(2))))` `(v_(H_(2)))/(t)xx(t)/(0.56" L")=2xxsqrt((4)/(2))"or" v_(H_(2))=1.584" L "=014" g"` of `H_(2)` `(`:'` 22.4" L "H_(2)=2" g "H_(2))` `:.` Weight of the gases in 2nd bulb `=0.10 g (D_(2))+0.14 g(H_(2))=0.24" g"` Hence, in the 2nd bulb, % of `D_(2)` by weight `=(0.10)/(0.24)xx100=14.67%` % of `H_(2)` by weight `=100-41.67=58.33%` |
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| 50120. |
(a) Giving reasons arrange the following in the decreasing order of ionic mobility : Li^(+), Na^(+), Rb^(+), K^(+) (b) Explain the various reactions that occur in the Solvay process for the manufacture of Na_2CO_3 (c) Potassium carbonate cannot be prepared by Solvay process, why? |
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Answer» SOLUTION :`Rb^(2+) gt K^(+) gt NA^(+) gt Li^+` because of lesser hydration of ion as size increases. In other words, hydrated radii of ions decrease as `Li^(+),Na^(+),K^(+),Rb^(+)`. (c) Potassium carbonate cannot be prepared by SOLVAY process because potassium bicarbonate being more soluble than sodium bicarbonate does not get precipitated when `CO_2` is passed through a CONCENTRATED solution of KCl SATURATED with `NH_3` |
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| 50121. |
A given p-orbitals is_______fold degenerated |
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Answer» Two |
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| 50122. |
A given orbital is labelled by the magnetic quantum number, m = -1. This can not be |
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Answer» s - orbital |
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| 50123. |
A given orbital is labelled by the magnetic quantum number , m = -1. This can not be |
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Answer» s-orbital |
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| 50124. |
A given metal crystllizes out with a cubic structure having edge length of 361 pm. If there are four metal stoms in one unit cell. What is the radius of one atom ? |
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Answer» Solution : Z = 4 means structure is fcc, for fcc. ` r = a/(2SQRT2) = ( 361)/( 2 XX 1.414) = 121 . 65 "pm"= 127 "pm" ` |
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| 50125. |
A given metal crystallizes out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom ? |
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Answer» 80 pm |
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| 50126. |
A given mass of gas expands from the state A to the state B by three paths 1,2 and 3 as shown in the figure. If w_(1), w_(2) and w_(3) respectively be the work done by the gas along three paths then: |
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Answer» `w_(1) gt w_(2) gt w_(3)` |
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| 50127. |
A given mass of a gas occuples 919.0 mL in dry state at STP. The same main when collected over water at 15^(@)C and 750 mm pressure occupies one litre volume. Calculate the vapour pressure of water at 15^(@)C. |
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Answer» Solution :Step 1. To calculate the pressure of the dry gas at `15^(@)C` and 750 mm pressure `{:("Given Conditions at STP","Final Conditions"),(V_(1)=919 mL,V_(2)=1000 mL),(P_(1)=760 mm,P_(2)=?("Dry state")),(T_(1)=273 K",",T_(2)=273+15=288 K):}` By applying gas EQUATION, we have `(760xx919)/(273)=(P_(2)xx1000)/(288)"or" P_(2)=(760xx919xx288)/(1000xx273)=736.7mm` Step 2. To Calculate the vapour pressure of water at `15^(@)C`. Vapour pressure of water =Pressure of the moist gas-Pressure of the dry gas, =750-736.7=13.3 mm Alternatively, if p is the vapour pressure of water at `15^(@)C`, then take `P_(2)`=(750-p)mm.Substituting in the equation, `(P_(1)V_(1))/(T_(2))=(P_(2)V_(2))/(T_(2))`, we get `(760xx919)/(273)=((750-p)xx1000)/(288)`. Solve for p. |
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| 50128. |
A given compound always contatins exactly the same proportion of elements by weight’. Write the name of the scientist who proposed it |
| Answer» SOLUTION :JOSEPH PROUST | |
| 50129. |
A given compound always contatins exactly the same proportion of elements by weight'. Name the law |
| Answer» SOLUTION :LAW of Definite PROPORTIONS Or Law of Definite COMPOSITION | |
| 50130. |
A given compound always contatins exactly the same proportion of elements by weight’. Calculate the number of molecules in each of the following:1g CO_2 (Given that N_A is 6.02xx10^23, molecular mass of N_2 is 28 and CO_2 is 44). |
| Answer» SOLUTION :NUMBER of MOLECULES PRESENT in ` 1gCO_2="MASS"/"Grammolecular mass"xxN_A=1/44xx6.02xx10^23=1.37xx10^22` | |
| 50131. |
A given compound always contatins exactly the same proportion of elements by weight’. Calculate the number of molecules in each of the following: 1g N_2(Given that N_A is 6.02xx10^23, molecular mass of N_2 is 28 and CO_2 is 44). |
| Answer» SOLUTION :NUMBER of MOLECULES PRESENT in `1gN_2="MASS"/"Grammolecular mass"xxN_4=1/28xx6.02xx10^23=2.15xx10^22` | |
| 50132. |
(a) Give uses of chromatography (b) Give meaning of chromatography word (c ) Write first use of chromatography |
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Answer» Solution :(a) The use of chromatography technique is as under. (i) In purification of compounds (ii) To test the purity of compounds (III) To SEPARATE mixture into their compounds (B) The name chromatography is BASED on the Greek word chroma, for colour. On the base of colour so it is called chromatography. (c ) The chromatography was first used for the separation of coloured substances found in plants. Procedure: (i) The mixture of coloured substances obtain from the plant is applied on to a stationary phase, which may be a solid or a liquid. (ii) A pure solvent, a mixture of solvent or a gas is allowed to move slowly over the stationary phase. (iii) The components of absorbate mixture get GRADUALLY separated from one another. (iv) The moving phase in stationary phase, liquid or gas is called mobile phase. |
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| 50133. |
(a) Give relation of value of K and amount of products Reactants. (b) Give relation of value of Deltan_((g))and value of K_p and K_c. |
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Answer» Solution :(a)K=1 , so, [Products] = [Reactants] `K GT 1`, So, [Products] `gt` [Reactants] `K lt 1` , So, [Products ] `lt` [Reactants] (b) `Deltan_((g))=0` So, `K_p=K_c` `Deltan_((g)) gt 0` So, `K_p gt K_c` `Deltan_((g)) lt 0`So, `K_p lt K_c` |
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| 50134. |
(a) Give reactivity of alkaline earth metals towards air, water, dihydrogen and acids.(b) Explain reducing nature of alkaline earth metals. (c) Give chemical reaction of alkaline earth metals with liquid ammonia. |
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Answer» Solution :(i) REACTIVITY towards air and WATER : Beryllium and magnesium are kinetically inert to oxygen and water because of the formation of an oxide film on their surface. However, powdered beryllium burns brilliantly on ignition in air to give Beo and `Be_(3)N_(2)`. Magnesium is more electropositive and burns with dazzling brilliance in air to give Mgo and `Mg_(3)N_(2)`. Calcium, strontium and barium are readily attacked by air to form the oxide and nitride. They also react with water with increasing vigour even in cold to form hydroxides. (ii) Reactivity towards the halogens : All the alkaline earth metals combine with HALOGEN at elevated temperatures forming their halides. `M+X_(2) to MX_(2)(X=F, Cl, Br, I)` Thermal decomposition of `(NH_(4))_(2), BeF_(4)`is the best route for the preparation of `BeF_(2),and BeCl_(2)` is conveniently made from the oxide. `BeO+C+Cl_(2)overset(600-800K)hArr BeCl_(2) +CO` (iii) Reactivity towards hydrogen : All the elements except beryllium combine with hydrogen upon heating to form their hydrides, `MH_(2). BeH_(2)`,however, can be prepared by the reaction of `BeCl_(2)`with `LiAlH_(4)`. `2BeCl_(2) + LiAlH_(4) to 2BeH_(2) + LiCl + AlCl_(3)` (iv) Reactivity towards acids : The alkaline earth metals readily react with acids liberating dihydrogen. `M + 2HCl toMCl_(2) + H_(2)` (v) Reducing nature : Like alkali metals, the alkaline earth metals are strong reducing agents. This is indicated by large NEGATIVE values of their reduction potentials. However their reducing power is less than those of their corresponding alkali metals. Beryllium has less negative value compared to other alkaline earth metals. However, its reducing nature is due to large hydration energy associated with the small size of`Be^(2+)`ion and RELATIVELY large value of the atomization enthalpy of the metal. (vi) Solutions in liquid ammonia: Like alkali metals, the alkaline earth metals dissolve in liquid ammonia to give deep blue black solutions forming ammoniated ions. From these solutions, the ammoniates, `[M(NH_(3))_(6)]^(2+)`can be recovered. `M_((s))+(x+y) NH_(3) to [M(NH_(3))_(x)]^(2+)+2[e(NH_(3))_(y)]^(-)` |
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| 50135. |
(a) Give principal of Adsorption chromatography (b) Adsorbent (c ) Mobile phose and (d) What is |
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Answer» Solution :(a) Principle: This method is based upon the differential adsorption of the various components of a mixture on a suitable adsorbent to DIFFERENT degree. (b) Adsorbent: The compound adsorbed on which is CALLED adsorbent. E.g. silica gel and alumina Adsorbent is a stable so it is called stationary phase. (C ) Mobile phase : In adsorbed (stationary phase) which is liquid SOLVENT pass through it which one is known as mobile phase. (d) Adsorbate: The compound or the mixture of compound which is kept in adsorbent is called adsorbate. It REMAIN seen suitable solvent then migrate and separate in stationary phase |
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| 50136. |
(a) Give a detailed account on the different mechanisms followed in elimination reaction. |
Answer» Solution :(a) Elimination reactions MAY proceed through two different mechanisms namely `E_(1) and E_(2)`  (i) The rat of `E_(2)` reaction depends on the concentration of alkyl halide and base Rate =K [alkyl halide] [base] (ii) It is therefore , a SECOND order reaction. Generally primary alkyl halide UNDERGOES this reaction in the presence of alcoholic KOH. It is a one step process in which the abstraction of the proton from the `beta` carbon and expulsion of halide from the `alpha` carbon occur simultaneously. The mechanism is shown below.  (iii) Generally, TERTIARY alkyl halide which undergoes elimination reaction by this mechanism in the presence of alcoholic KOH. It follows first order kinetics . Let us consider the following elimination. Step-1 : Heterolytic FISSION to yield a carbocation  Step-2 : Elimination of a proton from the `beta` - carbon to produce an alkene. 
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| 50137. |
A general mechanism for aromatic electrophilic substitution reaction is. (K_(1) and K_(2) are rate constant for the forward reaction) also C-D bond in harder to break than a C-H bond, and consequently reaction in which C-D bond broken proceed more slowly than the reaction in which C-H bond are broken. However experimetns reveal that nitration of C_(6)H_(6) and C_(6)D_(6) proceeds at equal rates while the same is not true for sulphonation of C_(6)H_(6) and C_(6)D_(6). Q. What can be inferred regarding mechanism of sulphonation of C_(6)H_(6) and C_(6)D_(6) |
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Answer» Activation energy for the first step in `C_(6)H_(6)` is GREATER than that of `C_(6)D_(6)` |
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| 50138. |
A general mechanism for aromatic electrophilic substitution reaction is. (K_(1) and K_(2) are rate constant for the forward reaction) also C-D bond in harder to break than a C-H bond, and consequently reaction in which C-D bond broken proceed more slowly than the reaction in which C-H bond are broken. However experimetns reveal that nitration of C_(6)H_(6) and C_(6)D_(6) proceeds at equal rates while the same is not true for sulphonation of C_(6)H_(6) and C_(6)D_(6). Q. When one of the carbons of benzene is labelled (C^(14)), in which case we expect greater yield of the product obtained at labelled carbon. |
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Answer» Nitration |
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| 50139. |
A general mechanism for aromatic electrophilic substitution reaction is. (K_(1) and K_(2) are rate constant for the forward reaction) also C-D bond in harder to break than a C-H bond, and consequently reaction in which C-D bond broken proceed more slowly than the reaction in which C-H bond are broken. However experimetns reveal that nitration of C_(6)H_(6) and C_(6)D_(6) proceeds at equal rates while the same is not true for sulphonation of C_(6)H_(6) and C_(6)D_(6). Q. In the nitration reaction |
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Answer» `k_(1)=k_(2)` |
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| 50140. |
(A): GeF_(4) and SiCl_(4) act as Lewis bases. (R) : Ge and Si have d-orbitals to accept electrons |
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Answer» Both (A) and (R) are CORRECT, (R) is not the correct EXPLANATION of (A) |
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| 50141. |
A gasesous mixture contains 56 gN_(2),44 g CO_(2) and 16 g CH_(4). The total pressure of the mixture is 720 mm Hg. What is the partial pressure of CH_(4) ? |
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Answer» |
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| 50142. |
A gaseous system is initially characterised by 500 mL volume and 1 atm pressure at 298 K. This system is allowed to do work as In isobaric conditions it expands to 800 mL resulting a decrease in pressure and temperature to 0.6 atm and 273 K respectively. (ii) In adiabatic conditions it is allowed to expand upto 800 mL and results a decrease in pressure and temperature to 0.6 atm and 273 K respectively. If Gibbs energy change in (i) is DeltaG_(a) and in (ii) is DeltaG_(b), then what will be the ration of (DeltaG_(a))/(DeltaG_(b))? |
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Answer» 0 |
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| 50143. |
A gaseous system changes from state A(P_(1), V_(1), T_(1)) " to " B(P_(2), V_(2), T_(2)), B " to" C(P_(3), V_(3), T_(3)) and finally from C to A. The whole process may be called |
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Answer» REVERSIBLE process |
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| 50144. |
A gaseous substance dissolves in water giving a pale blue solution which decolourises KMnO_(4) and oxidises KI to I_(2). Gaseous substance is |
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Answer» `N_(2)O_(5)` |
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| 50145. |
A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion ( in the intermediate stages of expansion/ compression the states of gases are not defined). The work done can be calculated using dw=-P_("ext")dV while in case of reversible process the work done can be calculated using dw=-PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process, since P=(nRT)/(V),so, w=intdw=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT ""In((V_(f))/(V_(i))) Since,dw=PdV, so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. Two samples (initially under same states) of an idea gas are first allowed to expand to doubletheir volume using irreversible isothermal expansion against constant external pressure, then samples are turned back to their original volume first by reversible process having equation PV^(2)= constant then: |
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Answer» final TEMPERATURE of both samples will be equal |
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| 50146. |
A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion ( in the intermediate stages of expansion/ compression the states of gases are not defined). The work done can be calculated using dw=-P_("ext")dV while in case of reversible process the work done can be calculated using dw=-PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process, since P=(nRT)/(V),so, w=intdw=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT ""In((V_(f))/(V_(i))) Since,dw=PdV, so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. In the above problem: |
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Answer» work DONE by GAS Ist SAMPLE `gt` work done by gas in Iind sample |
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| 50147. |
A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion ( in the intermediate stages of expansion/ compression the states of gases are not defined). The work done can be calculated using dw=-P_("ext")dV while in case of reversible process the work done can be calculated using dw=-PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process, since P=(nRT)/(V),so, w=intdw=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT ""In((V_(f))/(V_(i))) Since,dw=PdV, so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. If four identical samples of an ideal gas initially at similar state (P_(0),V_(0),T_(0)) are allowed to expand to double their volumes by four different processes. (P)By isothermal irreversible process (Q) By reversible process having equation P^(2)V= constant (R) By reversible adiabatic process (S)By irreversible adiabatic expansion against constant external pressure. Then in the graph shown the final state is represented by four different points then, the correct match can be : |
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Answer» `1-P,2-Q,3-R,4-S` |
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| 50148. |
A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion ( in the intermediate stages of expansion/ compression the states of gases are not defined). The work done can be calculated using dw=-P_("ext")dV while in case of reversible process the work done can be calculated using dw=-PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process, since P=(nRT)/(V),so, w=intdw=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT ""In((V_(f))/(V_(i))) Since,dw=PdV, so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. There are two sample of same gas initially at same initial state. Gases of both the sample are expanded. Ist sample using reversible isothermal process and IInd sample using reversible adiabatic process till final pressures of both the samples become half of the initial pressure , then : |
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Answer» Final VOLUME of IST SAMPLE `LT` final volume of Iind sample |
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| 50149. |
A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion ( in the intermediate stages of expansion/ compression the states of gases are not defined). The work done can be calculated using dw=-P_("ext")dV while in case of reversible process the work done can be calculated using dw=-PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process, since P=(nRT)/(V),so, w=intdw=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT ""In((V_(f))/(V_(i))) Since,dw=PdV, so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. An ideal gaseous sample at time state i(P_(0)V_(0)T_(0)) is allowed to expand to volume 2V_(0) using two different processes, in the first process the equation of process is PV^(2)=K_(1) and in second process the equation of the process is PV=K_(2). Then : |
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Answer» WORK done in the first PROCESS will be greater than work in second process ( magnitude WISE) |
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| 50150. |
A gaseous paraffin requires five times its volume of oxygen for complete combustion. How many carbon atoms are present in a molecule of that paraffin? |
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Answer» |
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