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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
According to anti Markovnikov, CH_(3)CH=CH_(2) is react with HBr in presence of ……… |
| Answer» SOLUTION :PEROXIDE | |
| 2. |
According to Arrhenius acid-base theory, the strength of acid and base depends on |
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Answer» magnitude of accepting electron. So, As more ionization TAKE place in aqueous solution and more `H^+` or `OH^-` form then it is strong acid/base. |
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| 3. |
According to Andrews isothermals, CO_(2) gas behaves ideally "______________". |
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Answer» at and above `48.1^(@)C` |
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| 4. |
Accetic acid and aq. NH_3are weak monobasic acid and weak monoacidic base respectivelyand K_a of aceticacid is equal to K_bof aq. NH_3 .Which of the following statement are correct? |
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Answer» All proportions of the above mixing would result in an neutral solution having `pH =7 at 25^(@) C ` (b)At HALF neutralisation , buffer ` pOH=pKb + log ""(S)/(B)= pKb` ( c )At half neutralisationbuffer ` pH =pKa+ log ""(S)/(A)= pKa` |
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| 5. |
Boric acid B(OH)_(3)is weak monobasic acid reacts with alkali to form borates. The most common borate of boric acid is borax represented as Na_(2)(B_(4)O_5(OH)_4.8H_2Owhich is made up of two tetrahedral and two triangular units. On dissolution in water, these tetrahedral and triangular units are seperated. Borax is useful primary standard for titra tion against acids The number of B - O - B linkage in borax is/are |
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Answer» All proportions of the above mixing would result in an neutral solution having `PH =7 at 25^(@) C ` (B)At half neutralisation , buffer ` pOH=pKb + log ""(S)/(B)= pKb` ( c )At half neutralisationbuffer ` pH =pKa+ log ""(S)/(A)= pKa` |
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| 6. |
Absorption of hydrogen by palladium is known as |
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Answer» REDUCTION |
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| 7. |
Absolute zero is a temperature that an object can get arbitrarily lose to but ………… will remain unattainable. |
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Answer» absolute zero |
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| 8. |
Absolute zero is, |
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Answer» `-273^(@)C` |
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| 9. |
Absolute ethyl alcohol can be converted to diethyl ether by heating it to 410 K in the presence of |
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Answer» DIL. `H_(2)SO_(4)` |
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| 10. |
Absolute alcohol is |
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Answer» `95%` ETHANOL |
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| 11. |
Absolute alcohol can not be prepared from rectified spirit by simple distillation because |
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Answer» The BOILING points of water and alcohol are very close |
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| 12. |
Absolute alcohol can be prepared from rectified spirit by |
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Answer» Fractional distillation |
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| 13. |
Absolute alcohol can be obtained from rectified spirit by |
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Answer» FRACTIONAL DISTILLATION |
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| 14. |
Absolute alcohol (100% alcohol) is prepared by distilling rectified spirit over |
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Answer» Na |
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| 15. |
AB(s)hArrA(g)+B(g)K_(p)=4,DeltaH=+ve In a container,A(g) "and" B(g) are filled to partial pressure of 1 atm each. Now AB(s) is added (in excess quantity). Which of the following is CORRECT? (No other gas is present in container): |
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Answer» At equilibrium, the total pressure in the container is `4` ATM. |
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| 16. |
Above the Boyle point, for real gases shows _______. |
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Answer» POSITIVE deviation |
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| 17. |
Above conversion can be achieved by : |
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Answer» WOLF - Kishner - reduction |
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| 18. |
Above conversion can be achieved by: |
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Answer» Wolf - Kishner - reduction 
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| 19. |
About NaOH some statements are given below i) It is used for mercerising cotton ii) It can be stored in moist air iii) It can be used for the purification of Baxuite |
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Answer» I and II are correct |
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| 20. |
About meso tartaric acid which of the following statements is/are not correct |
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Answer» It is optically inactive due to INTERNAL compensation B) It can be separated into two optically active isomers D) It is optically inactive due to external compensation |
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| 21. |
About 80% of the mass of the atmosphere is present in ........... |
| Answer» SOLUTION :TROPOSPHERE | |
| 22. |
About 393K, when Plaster of Paris is heated, it forms _________ |
| Answer» SOLUTION :DEAD BURNT PLAST | |
| 23. |
Abnonnal boiling point of a compound is due lo |
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Answer» VAN DER Waal.s forces |
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| 24. |
Abnormal boiling point of a compound is due to |
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Answer» VAN der Waal's FORCES |
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| 25. |
A,B,C and D are four different gases with critical temperatures 304.1, 154.3 , 405.5 and 126.0 K respectively. While cooling, the gas which gets liquefied first is "_____________". |
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Answer» B |
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| 26. |
A,B,C and D are four 100-mlL samples of water. Fluoride ion concentrations in these samples are 2, 5, 8 and 12 ppm respectively. One drop of alizarine - S dye is added to each of these samples. The order of intesity of pink colour in these water samples is |
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Answer» `A GT B gt C gt D` |
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| 27. |
(a)Based on the nature of intermolecular forces , classify the following solids : Sodium sulphate, Hydrogen (b)What happen when AgCl is doped with CdCl_2 (c )Why do ferrimagnetic substances show better magnetism than antiferromagnetic substances ? |
| Answer» Solution :(a)`Na_2SO_4`=IONIC solid, `H_2`=Non-polar molecular solidand (B)ELECTRICAL CONDUCTIVITY increases | |
| 28. |
(a)Based on the nature of intermolecular forces , classify the following solids : Silicon carbide, Argon (b)ZnO turns yellow on heating. Why ? (c )What is meant by groups 12-16 compounds . Give an example. |
| Answer» SOLUTION :(a)SiC=covalent/network SOLID , Ar=Non POLAR MOLECULAR solid | |
| 29. |
AB,A_(2) and B_(2) are diatomic molecules. If the bond enthlpies of A_(2),AB and B_(2) are in the ratio 1 : 1 : 0.5 and enthalopy of formation of AB from A_(2) to B_(2) is -100kJ//mol^(-1). What is the bound enthalpy of A_(2)? |
| Answer» Answer :A | |
| 30. |
(a)Based on the nature of intermolecular forces, classify the following solids : Benzene , Silver (b)AgCl shows Frenkel defect while NaCl does not . Give reason. (c )What type of semiconductor is formed when Ge is doped with Al ? |
| Answer» SOLUTION :(a)`C_6H_6`=NON polar molecular SOLID, AG =METALLIC solid | |
| 31. |
AB_(3) molecule has trigonal pyramidal shape state no. of bonding and non bonding electrons ? |
| Answer» Answer :C | |
| 32. |
AB_(2(g)) dissociates as AB_(2(g)) harr AB_(2(g))+B_((g)). The initial pressure of AB_2 is 600 mm Hg and equilibrium pressure of the mixture of gases is 800 mm Hg. Then |
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Answer» `alpha = 33.3%`  `P_(EQN)=600+P=800 implies P=200 prop (200)/(600)=1//3` |
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| 33. |
AB_(2) dissociates as AB_(2(g)) Leftrightarrow AB_((g))+B_((g)). Whwn the initial pressure of AB_(2) is 600mm of Hg, the total equilibrium pressure is 800mm of Hg. Calculate K_(p) for the reaction, assuming that the volume of the system remains unchanged |
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Answer» 50 |
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| 34. |
AB_(2) " dissociates as " : AB_(2) (g) hArr AB (g) + B (g). If the intial pressure is 500 mm of Hg and the total pressure at equilibriumis 700 mm 0f Hg, calculate K_(p)for the reaction. |
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Answer» Solution :After dissociation , suppose the decrease in the pressure of `AB_(2)` at equilibrium is p MM. Then ` {:(,AB_(2) (G),hArr,AB (g),+,B (g)),(" Intial pressure",500 mm,,0,,0),(" Pressures at eqm.",(500-)p " mm", ,p" mm",,p " mm"):} ` `:. " Total pressure at equilibrium " = 500 - p + p+ p = 500 + p" mm"` `500 + p = 700" (Given) " or p= 200 " mm"` Hence , at equilibrium `p _(AB_(2)) = 500 - 200 = 300 " mm" , 200 " mm ", p_(B) = 200 mm ` `:. K_(p) = (p_(AB) XX p_(B))/p_(AB_(2)) = (200 xx 200)/300 = 133*3MM` Note. With RESPECT to standard state pressure of 1 bar , i.e., 0*987 atm , i.e., 750 mm, `K_(p) = (133*3)/750=0*178.` |
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| 35. |
AB_(2) dissociates as : AB_(2 (g)) hArr AB_((g)) + B_((g))When the intial pressure of AB_(2) is 500 mm Hg , the total equilibrium pressure is 700 mm Hg . Calculate equilibrium constant for the reaction , assuming that the volume of the system remains unchanged . |
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Answer» 100 MM HG |
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| 36. |
AB is predominantly ionic as A^+ B^- if |
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Answer» `(IP)_A lt (IP)_B` |
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| 37. |
AB is an ionic solid. The ionic radii of A^+and B^(-) are respectively r_cand r_a. Lattice energy of AB is proportional to |
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Answer» `(r_(C))/(r_(a))` |
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| 38. |
AB crystallizes in a body centredcubic lattice with edge length 'a' equal to 387 pm. The distancebetween two oppositely charged ions in the lattices is |
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Answer» 250 pm `d=(sqrt3a)/2 = (sqrt3times387)/2"pm"=335.15"pm"` |
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| 40. |
AB AB ____________type of packing is called ________whereas ABCABC __________type of packing is called ___________ |
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Answer» |
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| 41. |
AB , A_(2) and B_(2) are diatomic molecules, if the bond enthalpies of A_(2) , AB and B_(2) are in the ratio of 1:1:0.5 and the enthaly of formation of AB from A_(2) gives AO and DeltaH_(c)=-1200KJ mol^(-1) . Bond energy of (O=O) bond is 500KJ mol^(-1) . What is the bond enthalpy of (A-O) bond ? |
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Answer» `400KJmol-1` `(1)/(2)A_(2)+(1)/(2)B_(2)rarrAB-100KJ//mol^(-1)` ...(i) `(1)/(2)A_(2)+(1)/(2)O_(2)rarrAO-1200KJ//mol` (given) `A_(2)+O_(2)rarr2AO-2400KJ//mol` ...(ii) `O=Orarr500KJ//mol` From eq. (i) `(1)/(2)B.E._(A_(2))+(1)/(2)B.E._(B_(2))-B.E._(AB)=-100` `(1)/(2)x+(x)/(4)-x=-100` `-(x)/(4)=-100` `x=400(kJ)/(mol)` Now from eq (ii) `B.E._(A_2)+B.E._(O_2)-2B.E._(AO)=-2400` `x+(+500)-2B.E._(AO)=-2400` `400+500-2B.E._(AO)=-2400` `-B.E._(AO)=-3300` `B.E._(A-O)=1650(kJ)//(mol)` |
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| 42. |
AB, A_(2) and B_(2) are diatomic molecule. Enthalpies of AB A_(2) and B_(2) are in the ratio of 1:1:0.5 the value. Enthalpy of formation of AB, Delta_(f) H- 100 KJ mol-1. Find the dissociation enthalpy of A_2 ? Reaction : (1)/(2) A_(2) + (1)/(2) B_(2) to AB |
| Answer» SOLUTION :400 kJ/mol | |
| 43. |
(a)An element crystallises in BCC structure. The edge length of its unit cell is 288 pm. If the density of the crystals is 7.2 "g cm"^(-3), what is the atomic mass of the element ? (b)How many atoms of this element are present in 100 g ? |
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Answer» |
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| 44. |
aA+bB hArr cC + dD here A,B, C , D are in gaseous phase. Derive the relation between K_p and K_c. |
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Answer» SOLUTION :General reaction, `aA_((g)) + bB_((g)) hArr cC_((g)) +dD_((g))` `K_c=([C]^C [D]^d)/([A]^a[B]^b)` Here , concentration of component in [ ] is in mol `L^(-1)` . If partial pressure of GAS A,B,C,D is `p_A, p_B,p_C` and `p_D` at equilibrium , so Equilibrium constant = `K_p` and `K_p =((p_C^c)(p_D^d))/((p_A^a )(p_B^b))` but p=cRT So, `p_C^c=([C]RT)^c =[C]^c (RT)^c` `p_D^d = ([D]RT)^d =[D]^c (RT)^d` `p_A^a =([A]RT)^a = [A]^c (RT)^a` `p_B^b=([B]RT)^b=[B]^b(RT)^b` `THEREFORE K_p=([C]^c (RT)^c [D]^d (RT)^d)/([A]^a (RT)^a [B]^b (RT)^b)` `=([C]^c [D]^d)/([A]^a[B])(RT)^((c+d)-(a+b))` `therefore K_p = K_c(RT)^(DELTAN)`....(Eq.-i) where , `Deltan`=(c+d)-(a-b) `Deltan`=(Addition of coefficient of gases mole of products )- (Addition of coefficient of gases mole of reactant ) For equilibrium equation , `(n_p-n_r)=(c+d)-(a+b)` |
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| 45. |
A5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculatge the freezing point of 5% glucose in water if freezing point of pure water is 273.15K |
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Answer» Solution :Molar MASS of cane sugar `C_(12) H _(22) O _(11) = 342 g MOL ^(-1)` Molality of sugar `= (5 xx 100)/( 342 xx 100) = 0.146` `Delta T_(F)` for sugar solution `=273.15 - 271=2.15^(@)` `Delta T_(f) = K_(f) xxm` `K _(f) = (2.15)/(0.146)` Molality of glucose solution `= (5)/(180) xx (1000)/(100) = 0.278 =4.09^(@)K` `therefore` Freezing point of glucose solution `=273.15-4.09 =269.06K` |
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| 46. |
A_(2) and B_(2) are two diatomic molecules with bond energies of A-A and B-B bonds as x and y respectively. If the bond energy of the molecule, A-B formed up from A_(2) and B_(2) is z. then, the responance energy of molecules A-B will be |
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Answer» `(DELTAE)_(A-B) = Z - SQRT (XY)` |
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| 47. |
A1 atomloseselectronssuccessively to form A1^(+) ,A1^(2+) andA1^(3+)ions. Whichstep will havehighestionizationenthalpy ? |
| Answer» SOLUTION :`A1^(2+) toA1^(3+) + e^(-)` (largerthe + vecharge on the cationfrom whichthe ELECTRONIS losthighestis the ionizationenthalpy ). | |
| 48. |
A 0.25 M glucose solution at 370 . 28 K has approxmately the pressure as blood does what is the osmotic pressure of blood? |
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Answer» Solution :`C = 90.25 M` `T=370.28K` `(pi) _("glucose") = CRT` `(pi) =0.25 MOL L ^(-1) xx 0.082 L ATM K^(-1) mol ^(-1) xx 370. 28K = 7.59 ` atm |
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| 49. |
(A): Zn is not a transition element. (R) : Elements with incompletely filled d-orbital are called transition elements |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 50. |
(A): Zinc is not considered as a transition element (R): Zn or Zn^(+2) does not contain unpaired d-electon |
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Answer» Both A and R are TRUE and R is the CORRECT explanation of A |
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