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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. |
C_(2)H_(5)OH can be differentiated from CH_(3)OH by |
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Answer» REACTION with HCl |
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| 2. |
C_(2)H_(5)Cloverset("alc. KOH")rarrAoverset(dil.H_(2)SO_(4))rarrB. Here A and B are |
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Answer» `C_(2)H_(4),C_(2)H_(5)OH` |
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| 3. |
C_(2)H_(5)Cloverset(AlcKOH)rarr Product. This reaction is known as |
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Answer» HYDROHALOGENATION |
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| 4. |
C_2H_5Cl overset(alc.KOH)to A overset(dil.H_2SO_4)to B. Here A and B are |
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Answer» `C_2H_4 , C_2H_5OH` |
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| 5. |
C_2H_5Br + Q_1 to C_2H_5 + Br C_2H_5Br + Q_2 toC_2H_5^((+)) + Br^((-)), then relation between Q_1 and Q_2 is |
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Answer» `Q_1 gt Q_2` |
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| 6. |
C_(2)H_(5)Br + NaOH rarr C_(2)H_(5)OH + NaBr, the above reaction is |
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Answer» FREE RADICAL substitution |
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| 7. |
C_(2)H_(5)Br+2Naoverset(Dry ether)rarrC_(4)H_(10)+2NaBr The above reaction is an example of which of the following |
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Answer» reimer tiemann reqaction |
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| 8. |
C_(2)H_(4) +Cl_(2) rarr C_(2)H_(4)Cl_(2) , Delta H = - 270.6 kJ mol^(-1), Delta S = - 139.0 J K^(-1) mol^(-1) (i)Is the reaction favoured by entropy , enthalpy, both or none ? (ii) Find Delta G if T = 300 K. |
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Answer» Solution :(i) As`DELTA H` is `-` ve , reaction isfavoured by enthalpy. As ` Delta S` is -ve, it is not FAVOURED by entropy. (ii) `Delta G =Delta H - T DELTAS` `= - 270.6 K J mol^(-1) - 300 K ( -139 xx 10^(-3) kJ K^(-1) mol^(-1))` `= -270.6 + 41.7 kJ mol^(-1)` `= - 228.9kJ mol^(-1)` |
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| 9. |
C_(2)H_(4)+H_(2) to C_(2)H_(6) In terms of redox reaction notations A) C_2H_4 is reduced B) H_2 is oxidised |
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Answer» A is TRUE, but B is FALSE |
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| 10. |
C_(2)H_(2)overset(HOCl)underset(excess)rightarrow A overset(Cl_(2))rightarrowB overset(NaOH)underset(aq)rightarrowCoverset(Conc)underset(HNO_(3))rightarrow D Identify A,B,C and D |
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Answer» `CHCl_(3),CHCl_(2)CHO,C Cl_(3)CHO,C Cl_(3)NO_(2)` |
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| 11. |
C_(2)H_(2) underset(HgSO_(4))overset(CH_(3)COOH ("excess"))rarr Z overset("Distil")rarr X + Y X and Y are |
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Answer» ETHANAL, Ethanol |
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| 12. |
C_2H_2 overset("Hot metal tube")(rarr)> A overset("Fuming "H_2SO_4)(rarr)B.Here the product B is |
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Answer» BENZENE |
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| 13. |
C_(2)H_(4)+2HClrarrC_(2)H_(4)Cl_(2) is an example of |
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Answer» ADDITION reaction |
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| 14. |
C^(14) a beta- active nucleus is present in overset(14)(C)H_(4). A sample of (overset(14)(C)H_(4)) is kept in a closed vessel shows increase in pressure with time. It is due to the formation of : |
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Answer» `overset(14)(N)H_(3)` and `H_(2)` |
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| 15. |
C^(14) is |
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Answer» an ARTIFICAL radioactive isotope |
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| 17. |
C-X bond is strongest in |
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Answer» Chloromethane |
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| 18. |
C-X bond is strongest in ……………………… . |
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Answer» CHLOROMETHANE |
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| 19. |
C, Si non metals , germanium , tin and lead are high melting metals. |
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Answer» Solution :False statement C and Si - Non metal GERMANIUM - Semi metal Tin and LEAD - Low MELTING metals |
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| 20. |
C _((s)) + O _(2 (g)) to CO _(2(g)),Delta H =-393.5kJ. This equation can not represent |
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Answer» Heat of transition |
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| 21. |
C+O_(2)rarrCO_(2)the reaction is |
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Answer» CHEMICAL combination |
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| 22. |
[C] Match the solutions in Column-I with their naturein Column-II : {:(,"Column-I",,"Column-II"),("(a)","n-hexane + n heptane","(p)","Can be perfectly separated by distillation"),("(b)","Acetone + chloroform","(q)","Maximum boiling azeotrope"),("(c)","Acetone + aniline","(r)","Cannot be perfectly separated by distillation"),("(d)","Ethanol + water","(s)","Nearly ideal"):} |
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Answer» |
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| 23. |
C is isomers are less stable than transisomers |
Answer» SOLUTION : Amongcisand TRANSISOMERS .Cisisomeris lessstablethantransisomer. In theisomermakethe MOLECULE muchmore unstable.Similargroupare diagonallyisomersis lessstablethantransisomer. |
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| 24. |
(C) is |
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Answer» CALCIUM hydroxide `CaO+H_(2)OrarrCa(OH)_(2)` |
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| 25. |
C-II bond is polar but CH_(4) is non polar. |
| Answer» Solution :METHANE has symmetrical tetrahedral sturctur hence DIPOLE moment is ZERO. Therefore it is non POLAR. | |
| 26. |
(c) H_(2)O(l) to H_(2)O(g), DeltaH = 40.7 kJ Delta_(r)H^(@) is the heat of .......of water |
| Answer» SOLUTION :C) VAPORISATION | |
| 27. |
C_(("graphite")) + H_2O_((g)) to CO_((g)) + H_(2(g)), DeltaH = +131.4 kJ. (Assume reaction occures at STP) How much energy is absorbed when one litre of hydrogen gas is formed at STP ? |
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Answer» Solution :The heat of REACTION, `Delta H = + 131.4kJ` It indicates that formation of one mole of hydrogen INVOLVE absorption of 131.4 kJ of heat. One mole gas at STP has a VOLUME of 22.4 L. Formation of 22.4 litre of hydrogen gas absorbs 131.4 kJ. Formation of one litre of hydrogen gas absorbs 5.87 kJ |
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| 28. |
C_("graphite") + O_(2(g)) : Delta H = -393.5 kJ. Delta H of the above reaction cannot be |
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Answer» formation of `CO _(2)` |
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| 29. |
C-F is the strongest among C-F, C-Cl, C-Br, C-I because |
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Answer» the strongest bonds are formed by the overlap of orbitals of the same PRINCIPLE quantum number. In C-F the overlap involves orbitals of the same principle quantum number (SECOND) |
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| 30. |
C ("diamond") rarr C ("graphite") , DeltaH = -ve , this indicates that |
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Answer» GRAPHITE is more STABLE than diamond |
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| 31. |
C Cl_(4) gt CHCl_(3) gt CH_(2) Cl_(2) gt CH_(3) Cl is the decreasing order of boiling point in haloalkanes . Give reason . |
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Answer» Solution : The boiling point of chloro, bromo and iodoalkanes INCREASES with INCREASE in the number of halogen ATOMS. So the correct DECREASING order of boiling point of haloalkanes is: `C Cl_(4) gt CHCl_(3) gt CH_(2) Cl_(2) gt CH_(3) CL` . |
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| 32. |
C Cl_3-CH=CH_2overset(HOCl)toP(Major product ), C Cl_3-CH=CH_2overset(HOCl)to 'P' is : |
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Answer» `undersetunderset(OH)(|)(C Cl_3CHCH_2Cl)` |
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| 33. |
C Cl_(4) does not act as Lewis acid while SiCl_(4) and SnCl_(4) act as Lewis acids. Why? |
| Answer» SOLUTION :`SiCl_(4) and SnCl_(4)` act as Lewis acids. This is because they can EXTEND their coordination NUMBER beyond FOUR due to the presence of vacant d-orbitals in the valance shell | |
| 34. |
C-C sigma bond enthalpy of ethane is 397 KJ mol^(-1) and C-C pi-bond enthalpy in ethene is 284 kJ mol^(-1). So what is the bond enthalpy of double bond in kJ mol^(-1) in ethane ? |
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Answer» 384 `(397 + 284) = 681 kJ mol^(-1)` |
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| 36. |
C-C bond length in Diamond is |
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Answer» `1.33A^(@)` |
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| 37. |
C and Si are always tetravalent but Ge, Sn , Pb show divalency . Why ? |
| Answer» SOLUTION :GE, SN, Pb show divalency due to the INERT pair effect . `Pb^(2+)` is more stable than `Pb^(4+)` . | |
| 38. |
C and Si are almost alwaystetravalent but Ge, Sn and Pb show divalency. Why ? |
| Answer» Solution :Among the elements of group 14, CARBON does not have d- or f-electrons . Therefore , it does not SHOW inert pair effect. CONSEQUENTLY, it shows an oxidationstate of`+4`due to the presenceof two electrons in the s- andtwo electronsin the p-orbitalof thevalenceshell. In contrast , all other elements from GE to Pb containeither dor both d- and f-electron increases, the inertpair effect becomesmore and more prominent. In other words,as we move dow the group from Ge to Pb, the stability of +2 oxidation stateincreases whilethat of `+4` oxidationstate decreases.Therefore , the tendencyof Ge, Sn and Pb to EXHIBIT +2 oxidation state increases with increasing atomic number in group 14. | |
| 39. |
C _(6) H _(6) to C _(6) H _(12), Delta H =-204 KJ.Heat of hydrogenation of each C=C (assumed to be) in benzene is [in KJ] |
| Answer» ANSWER :A | |
| 40. |
Bywhich instrument atomic mass is measured ? |
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Answer»
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| 41. |
Bya adding which of the following the permanent hardeness of water can be removed? |
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Answer» SODA lime |
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| 42. |
By X-ray studies, the packing of atoms in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly identical . The density of gold is found to be 19.4 g cm^(-3) and its atomic mass is 197 a.m.u. Assuming gold atom to be spherical , its radius will be |
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Answer» 203.5 PM =0.3535 x 407 pm =143.9 pm |
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| 43. |
By X-ray studies, the packing of atoms in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly identical . The density of gold is found to be 19.4 g cm^(-3) and its atomic mass is 197 a.m.u. The approximate number of unit cells present in 1 g of gold is |
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Answer» `3.06xx10^21` |
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| 44. |
By X-ray studies, the packing of atoms in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly identical . The density of gold is found to be 19.4 g cm^(-3) and its atomic mass is 197 a.m.u. The length of the edge of the unit cell will be |
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Answer» 407 pm or `a^3=(ZxxM)/(rhoxxN_0xx10^(-30))` For ccp, i.e., f.c.c., Z=4. HENCE,`a^3=(4xx197)/(19.4xx6.02xx10^23xx10^(-30))` `=6.747xx10^7` =`67.47xx10^6` or `a=(67.47)^(1//3)xx10^2` `=4.07xx10^2` pm =407 pm |
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| 45. |
By X-ray studies, the packing of atoms in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly identical . The density of gold is found to be 19.4 g cm^(-3) and its atomic mass is 197 a.m.u. The fraction occupied by gold atoms in the crystal is |
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Answer» 0.52 |
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| 46. |
By X-ray studies, the packing of atoms in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly identical . The density of gold is found to be 19.4 g cm^(-3) and its atomic mass is 197 a.m.u. The coordination number of gold atom in the crystal is |
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Answer» 4 |
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| 47. |
By X-ray studies, the packing atoms, in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly idenctical. The density of gold is found to be19.4 g cm^(-3) and its atomic mass is 197 a.m.u. Assuming gold atom to be spherical , its radius will be |
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Answer» 203.5 "pm"^(2)` ` = 0.3535 XX 407 "pm" = 143.9" pm"` |
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| 48. |
By X-ray studies, the packing atoms, in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly idenctical. The density of gold is found to be19.4 g cm^(-3) and its atomic mass is 197 a.m.u. The length of the edge of the unit cell will be |
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Answer» <P>407pm For CCP , i.e, f.c.c , Z= 4 Hence , ` a^(3) = ( 4xx 197)/( 19.4 xx 6.02 xx 10^(23) xx 10^(-30))` ` = 6.747 xx 10^(7) = 67.47 xx 10^(6)` ` a = (67.47)^(1//3) xx 10^(2)` ` 4.07 xx 10^(2) "pm" = 407 "pm"` |
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| 49. |
By X-ray studies, the packing atoms, in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly idenctical. The density of gold is found to be19.4 g cm^(-3) and its atomic mass is 197 a.m.u. The fraction occupied by gold atoms in the crystal is |
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Answer» 0.52 |
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| 50. |
By X-ray studies, the packing atoms, in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly idenctical. The density of gold is found to be19.4 g cm^(-3) and its atomic mass is 197 a.m.u. The coordination number of gold atom in the crystal is |
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Answer» 4 |
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