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. |
From the following thermochemical equations {:(C_(2)H_(4)+H_(2) to C_(2)H_(6),,DeltaH=-32.7 kcal),(C_(6)H_(6)+3H_(2) to C_(6)H_(12),, DeltaH=-49.2 kcal):} Calculate the resonance energy of benzene. |
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Answer» Solution :`C_(6)H_(6)` has three isolated carbon-carbon DOUBLE bonds. `DeltaH` for hydrogenation of `C_(2)H_(4)`, with one double bond, but the given value is `-49.2 "kcal"` `:.` resonance ENERGY `=3xx(-32.7)-(-49.2)` `=-48.9 "kcal"`. |
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| 2. |
The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of the compound will be : |
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Answer» `C_nh(_2n)O` |
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| 3. |
From the following substances, which carbohydrate has the maximum sweetness ? |
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Answer» Saccharin |
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| 4. |
General formula of an amine is : |
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Answer» `C_nH_(2N+1)N` |
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| 5. |
From the followingstatementregarding H_(2)O_(2) choose the incorrect statement |
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Answer» it can act only as an oxidizing agent |
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| 6. |
General formula of aldehyde is : |
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Answer» `C_nH_(2N)O` |
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| 7. |
Fromthe followingseries of reactions, {:(Cl_(2)+2KOH to KCl+KClO+H_(2)O),("" 2KClO to 2KCl+KClO_(3)),(""4KClO_(3) to 3 KClO_(4) + KCl):} calculate the mass of chlorineneeded to produce100 g of KCiO_(4) |
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Answer» Solution :Wt, of `Cl_(2) = 100 g KCIO_(4) XX ((" 1 mole " KCIO_(4))/( 139 g KCIO_(3)))` `xx (("4 mole " KCIO_(3))/("3mole " KCIO_(4))) xx ((3 "mole " KCIO)/( 1 " mole " KCIO_(3)))` `xx (("1 mole " Cl_(2))/( "1 mole KCIO")) xx ((71 g Cl_(2))/( " 1 mole " Cl_(2)))` 204.5 g |
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| 8. |
General formula of alkyl group is |
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Answer» `c_nH_(2N)` |
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| 9. |
From the followingreaction sequence,{:(CaC_(2) + H_(2) O to CaO+ C_(2)H_(2)),(""C_(2)H_(2)toC_(2)H_(4)),(""nC_(2)H_(4) to (C_(2)H_(4))_(n)):} calculate the mass of polyethylenewhichcan be produced from 10 kg of CaC_(2) |
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Answer» Solution :Wt. of `(C_(2)H_(4))_(n) = (1000 g CaC_(2)) xx ((1 "mole CA"C_(2))/(64 "g Ca"C_(2))) ` `((1 "mole "C_(2)H_(2))/(1"mole Ca" C_(2)))xx ((" 1 mole " C_(2) H_(4))/( " 1 mole "(C_(2)H_(2))))` `(("1 mole " (C_(2)H_(4))_(n))/( "n more " C_(2) H_(4))) xx (( " 28 n g " (C_(2) H_(4))_(n))/("1 mole " (C_(2) H_(4))_(n)))` = 4375 g (note that n cancels) Second Method APPLY POACfor C atoms. |
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| 10. |
General formula for carboxylic acid is |
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Answer» `C_(n) H_(n)O_(2)` |
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| 11. |
Fromthe followingreaction , 2CoF_(2) + F_(2) to 2 CoF_(3)(CH_(2))_(n) + 4 n CoF_(3) to (CF_(2))_(g) _(n) + 2n HF + 4n CoF_(2)calculate how much F_(1)will be consumedto produce 1 kgof (CF_(2))_(n) |
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Answer» |
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| 12. |
General formula for carbohydrate is |
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Answer» `C_(N)H_(2n)O_(2n+2)` |
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| 13. |
From the following reaction sequences 2A = B + C 5B + D = 2E + F E + G = 4H + J calculate moles of H produced by 10 moles of A |
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Answer» 20 |
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| 14. |
General formula for alkenes is |
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Answer» `C_(N)H_(2N)` |
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| 15. |
From the following reaction of H_(2) and X_(2) ,in which reaction catalyst is necessary? |
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Answer» `H_(2)+Br_(2)to2HBr`<BR>`H_(2)+I_(2)to2HI` |
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| 17. |
General electronic configurationof actinoids is (n-2)f^(1-14) ( n-1)d^(0-2) ns^(2) . Which of the followingactinoids have one electron in 6d orbital ? |
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Answer» U ( ATOMIC no. 92) |
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| 18. |
From the following metallurgical flow chart, identify steps A, B and C. "Sulphide" overset((A))rarr "Oxide" overset((B))rarr "Impure metal" overset((C))rarr "Pure metal" |
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Answer» Solution :(A) Roasting (B) Reduction by CARBON or more ELECTROPOSITIVE METAL (C) Electrolysis |
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| 19. |
From the following …………………….. is the example of biopolymer. |
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Answer» teflon NATURAL polymers LIKE polysaccharide, protein, nuclic ACID is necessary for human life is called biopolymer substance. Two TYPES of Nuclic acid (1) DNA & (2) RNA |
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| 20. |
General empirical formula of silicone is ………………. . |
| Answer» Answer :A | |
| 22. |
General electronic configuration of transition elements is |
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Answer» `(n-1)d^(1-10)NS^(1-2)` |
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| 23. |
From the following half cells a galvanic cell is made, A^(2+)+2e=A" "..." "E_(1)^(0)=0.8V B=B^(3+)+3e" "..." "E_(2)^(0)=-0.3V E^(0) cell is |
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Answer» `E_(1)^(0)-E_(2)^(0)` |
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| 24. |
General electronic configuration of lanthanide is : |
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Answer» `(n-2)F^(1-14), (n-1) s^2p^6d^(0-1) , ns^2` |
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| 25. |
From the following ions magnetic moment is maximum for |
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Answer» `Mn^(2+)` |
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| 26. |
General electronic configuration of actionoids is (n-2) f^(1-14) (n-1) d^(0-2) ns^(2). Which of the following actinoids have one electron in 6d orbital? |
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Answer» U(atomic no92) `""_(93)NP: [Rn]5f^(4) 6d^(1) 7s^(2)` |
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| 27. |
From the following graph, identify order of reaction and mention the unit of its rate constant. |
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Answer» SOLUTION :FIRST ORDER reaction Unit - `s^(-1)` |
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| 28. |
General electronic configuration of d-block elements is |
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Answer» `(n-1)d^(1-10) NS^(1-2)` |
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| 29. |
From the following graph , identify order of reaction and mention the unit of its rate constant. |
Answer» Solution :FIRST ORDER Reaction , Unit `RARR s^(-1)`
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| 30. |
General electronic configuration of group 16 elements is |
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Answer» `ns^(2)NP^(3)` |
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| 31. |
From the following equations, calculate the standard molar heat of formation of AgCl at 25^(@)C. Ag_(2)O(s)+2HCl (g) to 2AgCl(s)+H_(2)O(l), DeltaH^(0)=-77.61 kcal 2Ag(s)+(1)/(2)O_(2)(g) to Ag_(2)O(s), DeltaH^(0)=-73.1 kcal (1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g) to HCl (g), DeltaH^(0)=-22.06 kcal H_(2)(g)+(1)/(2)O(g) to H_(2)O(l),DeltaH^(0)=-68.32 kcal |
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Answer» |
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| 32. |
From the following E^(o) value of helf cells: (i) A+e^(-)rarr A ^(-) , E^(o) = -0.824 V (ii) B^(-)+e^(-)rarr B ^(2-) , E^(o) = +1.25V (iii) C^(-)+2e^(-)rarr C ^(3-) , E^(o) = -1.25V (iv) D^(-)+2e^(-)rarr D ^(2-) , E^(o) = +0.68V What combination of two half cells would result is a cell with the largest cell potential? |
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Answer» (II)&(III) |
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| 33. |
General electronic configuration of element of Group 16 is : |
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Answer» `ns^2np^6` |
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| 34. |
From the following equations pick up the possible fusion reactions : |
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Answer» `""_(6)^(13) C + ""_(1)^(1) H + ""_(0)^(1) e + 4.3 MeV` |
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| 35. |
Gemstone is the impure form of which compound? |
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Answer» `Al_2O_3` |
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| 36. |
From the following data, the activation energy for the reaction is (cal/mol):H_2 + I_2---->2HI |
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Answer» `4xx10^4` |
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| 37. |
From the following data , show that the decomposition of hydrogen peroxide is a reaction of the first order : {:(t(min),0,10,20),(V(ml),46.1,29.8,19.3):} where t is the time in minutes and V is the volume of standard KMnO_4 solution required for titrating the same volume of the reaction mixture. |
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Answer» Solution :Volume of `KMnO_4` , solution used `prop ` AMOUNT of `H_2O_2` present . Hence if the given reaction is of the first ORDER , it must obey the equation. `k=(2.303)/tlog.(a)/((a-x))` `k = (2.303)/tlog.V_o/V_t` In this case, `V_o=46.1` ml The value of k at each INSTNAT can be calculated as follows : THUS, the value of k comes out to be nearly constant . Hence it is a reaction of the first order. |
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| 38. |
Gelatine is mostly used in making ice creams in order to |
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Answer» prevent forming the COLLOIDAL sol |
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| 39. |
From the following data, the activation energy for the reaction (cal/mol): |
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Answer» `4xx10^4` |
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| 40. |
Gemstone is an impure form of .......... |
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Answer» `Cu_2O` |
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| 41. |
From the following data, show that the decomposition of hydrogen peroxide is a reaction of the order : {:("t (min)",0,10,20),("V (ml)",46.1,29.8,19.3):} Where t is the time in minutes and V is the volume of standard KMnO_(4) solution required for titrating the same volume of the reaction mixture. |
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Answer» SOLUTION :`k=(2.303)/(t)"log"([A_(0)])/([A])` `k=((2.303)/(t))log((V_(0))/(V_(t)))` In the PRESENT CASE, V0 = 46.1 ml. The value of k at each instant can be calculated as follows : ![]() Thus, the value of k comes out to be NEARLY constant. HENCE it is a reaction of the first order. |
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| 42. |
Gem dihalides on hydrolysis give : |
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Answer» Acetone |
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| 43. |
From the following data of DeltaH of the following reactions, C(s)+(1)/(2)O_(2)(g) to CO(g),DeltaH=-110 kJ and C(s)+H_(2)O(g) to CO(g)+H_(2)(g),DeltaH=132 kJ Calculate the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, keeping the reaction temperature constant. |
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Answer» Solution :From the QUESTION we see that the first reaction is exothermic and the second ONE is endothermic. Thus, if a mixture of oxygen and steam `(H_(2)O)` is passed over coke and at the same time, the temperature does not change, the composition should be such that `DeltaH` of both the reactions are numerically EQUAL. In the first reaction, consumption of `1//2` "mole" of `O_(2)` evolves 110 kJ of energy, while in the second reaction, for 1 mole of steam `(H_(2)O)`, 132 kJ of energy is absorbed. `:. "mole"` of `O_(2)` needed to EVOLVE `132kJ=(0.5)/(110)xx132=0.6` `:. "mole"` ratio of `O_(2)` and steam `(H_(2)O)=0.6:1`. |
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| 44. |
Gelatin which is a peptide is added in ice-creams. What can be its role ? |
| Answer» Solution : Ice-creams are EMULSIONS which get STABILISED by emulsifying agents like GELATIN. | |
| 45. |
From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order : {:(t,,,0,,,10,,,20),(x,,,46.1,,,29.8,,,19.3):} where t is the time in minutes and x is the volume of standard KMnO_(4) solution in "cm"^(3) required for titrating the same volume of the reaction mixture. |
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Answer» Solution :Volume of `"KMnO"_(4)` solution used `prop`amount of `H_(2)O_(2)` present. Hence, ifthe given reaction is of the first order, it must obey the equation : `k=(2.303)/(t)log""(a)/(a-x)=(2.303)/(t)log""(V_(0))/(V_(t))` (Note that the SYMBOL x in the numerical PROBLEM isin FACT `a-x, i.e., V_(t)`) In the present CASE, `V_(0)=46.1" cm"^(3)` The value of k at each instant can be calculated as follows: `{:("t (min)",,,V_(t),,,""k=(2.303)/(t)log""(V_(0))/(V_(t))),(10,,,29.8,,,k=(2.303)/(10min)log""(46.1)/(29.8)=0.0436min^(-1)),(20,,,19.3,,,k=(2.303)/(20min)log""(46.1)/(19.3)=0.0435min^(-1)):}` Thus, the value of k comesout to be nearly constant. Hence, it is a reaction of the first order. |
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| 46. |
From the following data show that the decomposition of H_(2)O_(2) is a reaction of first order. Also calculate the value of the rate constant. |
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Answer» Solution :Let us substitute the values in the rate EQUATION for FIRST order reaction and calculate the value of rate constant(k) `k=2.303/tlog(a/(a-x)) = (2.303)/tlogV_(0)/V_(t)` CASE I. `V_(0) = 22.8 mL, V_(t) = 13.8 mL, t=10 min` `k=(2.303)/(10 min) log(22.8)/(13.8) = (2.303)/(10 min) (log 22.8 - log13.8)` `=(2.303)/(10 min) (1.358-1.140) = (2.303 XX 0.128)/(10 min) = 0.0502 min^(-1)` Case II. `V_(0) =22.8 mL, V_(t)=8.3, t=20min` `therefore k=(2.303)/(20 min)(1.388 - 0.919) = (2.303 xx 0.439)/(20 min) = 0.0505 min^(-1)` Since, the value of k comes out to be almost constant, the reaction is of first order. |
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| 47. |
Gelatin which is a peptide is added is ice-creams. What can be its role ? |
| Answer» SOLUTION :Ice-creams are emulsions. These are STABILIZED by emulsifying agents LIKE GELATIN. | |
| 48. |
From the following data show that decomposition of H_(2)O_(2) in aqueous solution is first order. Time (in minutes) 0 10 20 Volume (in c.c. of KMnO_(4)) 22.8 13.3 8.25 |
| Answer» SOLUTION :FIRST ORDER | |
| 49. |
Gelatin which is a peptide is added in ice creams. What can be its role? |
| Answer» SOLUTION :GELATIN ACTS as an emulsifying AGENT. It STABILIZES the ice-cream. | |
| 50. |
From the following data of DeltaH of the following reactions C (s) + (1)/(2) O_(2) (g) , Delta H = -110 KJ and C(s) + H_(2) O (s) to CO_(g) + H_(2) (g) , Delta H = 132 KJ Calculate the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, keeping the reaction temperature constant. |
| Answer» SOLUTION :MOLE % `O_(2) (G) = 37.5 , H_(2) O (g) = 62.5` | |