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. |
For n moles of a gas, the difference between the molar heat capacity at constant pressure and that at constant volume is equal to "….............". |
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Answer» <P> SOLUTION :`NR(C_(p) -C_(V)=nR)` |
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| 2. |
For N_(2) + 3H_(2) rarr 2NH_(3), enthalpy and internal energy changes respectively are, Delta H " & " Delta U then |
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Answer» `Delta H = O` |
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| 3. |
For metal sulphide K_(sp) of CuS, CdS, ZnS and MnS respectively 1 xx 10^(-44), 1 xx 10^(-28), 1 xx 10^(-22) and 1 xx 10^(-14) one mixture of solution of Cu^(2+), Cd^(2+) and Mn^(2+). In this solution H_2S gas passed than which one precipitate last. |
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Answer» Solution :The salt which has LEAST `K_(sp)` is first form precipitate with `S^(-2)`. So according to this first of all PRECIPITATION CdS take place. The salt which has MAXIMUM `K_(sp)` It form precipitation at LAST. So the precipitation of MnS take plact at last. |
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| 4. |
For MgO_((g)) + C_((g)) hArr Mg_((g)) + CO_((g)) , equilibrium pressure at 2000 K is 100 atm . Now K_(p) for the process is (in atm^(2)) |
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Answer» 2500 |
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| 5. |
For metal M, first ionisation energy is equal to 38 kcal/mole. Which of the following can be a possible value of ionisation enthalpy of M^(+) ion in kcal.mole at 400K. [Given: R = 2 cal//"mole kelvin"] |
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Answer» `38 KCAL//"MOLE"` |
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| 6. |
For Li^(2+), when an electron falls from a higher orbit to nth orbit, all the three types of lines ,i.e., Luyman, balmer and paschen was found int eh spectrum. Here, the value of 'n'will be |
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| 7. |
For isothermal process .......... . |
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Answer» `q=0 and DELTA E=0` |
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| 8. |
For isothermal expansion of an ideal gas, the correct combination of thermodynamic parameters will be |
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Answer» `DeltaU = 0, Q = 0, w ne 0 and DeltaH ne 0` `DeltaU = nC_(V)DeltaT = 0` `DeltaH = nC_(p)DeltaT = 0` ACCORDING to first law of thermodynamics `DeltaU = q + w` Since `DeltaU = 0, q ne w ne 0`. |
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| 9. |
For isomerism……… (i) Geometrical isomers are possible for CHCl=CHCl. (ii) CH_(2)=CHCl has two isomers. (iii) C_(6)H_(4)Cl_(2) has three isomers. (iv) No isomers for chlorobenzene. |
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| 10. |
For isothermal expansion of an ideal gas into vacuum, among the following how many are zero, Pext, g,Delta T, Delta U, Delta H, Delta S, Delta G, Delta S_("surr"), Delta S_("total") |
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Answer» `Delta S_("SYS") = Delta S_("total") + +ve, Delta G_("sys") - ve` |
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| 11. |
For ionic substances which is the proper term to use Formula weight or Molecular weight? |
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| 12. |
For hydrogen atom, Rydberg constant (R_(H)) is xm^(-1). Then for He^(+) ion, the corresponding value of this constant will be |
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Answer» `x cm^(-1)` For H atom, `bar(v) = 0.01 x ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) cm^(-1) " " ( :' Z = 1)` For `He^(+), bar(v) = 0.01 x ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) xx 2^(2) cm^(-1) " " ( :' Z = 2)` `= 0.04 x ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) cm^(-1)` Hence, `R = 0.04 x cm^(-1)` |
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| 13. |
For how many orbitals, the quantum numbers n = 3, l = 2, m = +2 are possible ? |
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Answer» m= + 2 represents ONE particular orbital of this subshell. |
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| 14. |
For helium vander Waals parameter bis 0.024 lit/mol. The diameter of helium is nearly |
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Answer» `1.7 A^@` `0.04 xx 10^(-3) = 4 xx 6 xx 10^(23) xx 4/3 pi (d/2)^2` `implies d = 2.67 xx 10^(-10) m = 2.67 Å`. |
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| 15. |
For Haber process, N_(2) +3H_(2) harr 2NH_(3)+Q, K_(P) = 10, find correct statements |
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Answer» If PRESSURE is increased by 10 times then new `K_(P)^(')=0.0 XX K_(P)` |
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| 16. |
For the following process H _(2)O (l) (1 bar 373.15K) hArr H_(2)O (g) (1 bar, 373.15 K ) identify the correct set of thermodynamic parameters. |
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Answer» `DELTA G= 0, Delta S= +ve` `Delta S GT 0` because `S_(g) gt S_(L)` |
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| 17. |
For H_(2(g)) + I_(2(g)) hArr 2HI_((g)) , the equilibrium constant is K then which of the following is equilibrium constant of this reaction nH_(2(g))+ nI_(2(g)) hArr 2nHI_((g)) ? |
| Answer» ANSWER :B | |
| 18. |
For H-atom, the energy required for the removal of electron from various sub-shells is given as under: The order of the energies would be : |
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Answer» a. `E_1 GT E_2 gt E_3` |
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| 19. |
For H atom the Bohr radius first orbit is 0.529 Å and the radius of maximum probability for H-atom according to wave mechanical model is also0.529 Å. How do the two approaches differ? |
Answer» Solution :Bohr predicted that the electron will always be found moving around the nucleus in a circular path of radius `0.529 Å`. According to his model, electron cannot be found at distance less than or more than `0.529 Å` . However according to wave MECHANICAL model. electron is most likely to be found at this distance but there is definite probability of finding the electron at DISTANCES both shorter and larger than `0.529 Å`. In other WORDS, according to wave mechanical model, the electron KEEPS on moving towards or away from the nucleus and the maximum probablity of locating it is at a radius of `0.529 Å` from the nucleus. Fig (a) Shows the electron cloud picture of 1s-orbital having RADIS of maximum probability and Fig (b). gives the Bohr picture of orbit in which electron is found only at this distance.
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| 20. |
For given simultaneous reaction: X(s)hArrA(g)+B(s)+C(g) K_(P_(1))=500atm Y(s)hArrD(g)+A(g)+E(s) K_(P_(2))=2000atm If total pressure=x, then write your answer after dividing by 25. |
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| 21. |
For green chemistry...... principle is mentioned. |
| Answer» SOLUTION :Twelve | |
| 22. |
For gaseous state, if most probable speed is denoted by C^(**), average speed by vecCand mean square speed by C, then for a large number of molecules the ratios of these speeds are : |
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Answer» `C^(**):overset(-)C:C=1:1.225:1.128` `:. C^(**):bar(C ):C=sqrt((8)/(pi)):sqrt(3)=1:sqrt((4)/(pi)):sqrt((3)/(2))` `=1:1.128:1.225` |
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| 23. |
For gaseous state, if most probable speed is denoated by C^(**), average speed by C and mean square speed by barC, then for a large number of molecules the ratios of these speeds are: |
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Answer» `C^** :BARC: C = 1.225: 1.128:1` |
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| 24. |
For four elements the IP_2 curve is shown. In the graph the elements represented by A, B, C and D are |
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Answer» Na, Mg, AL, SI |
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| 25. |
for even reaction at a particular temperature, the equilibrium constant has constant value. Is the value of Q also constant? Explain. |
| Answer» Solution :In thechemicalreactionasthe REACTIONPROCEEDS, thereis acontinouschangein the concentrationof REACTANTS andproducts and alsothe Qvalueuntilthe reactionreachesthe equilibrium.SO evenat particularthe equation, Qis notconstant. Evenoncethe EQUILIBRIUMIS achivedthenchangein concentrationof reactantsorproductspressure, volumewillchangethe VALUEOF Q . | |
| 26. |
For estimating ozone in the air, a ceratinvolume of air is passedthrough an acidified or neutralKI solutionwhen oxygen is evolved and iodide is oxidisedto give iodine. When such a solution is acidified, free iodine is evolved which can be be titrated with standard Na_(2)S_(2)O_(3) solution. In an experiment 10 litre of air at1 atmand 27^(@)C were passed through an alkaline KI solution, at the end, the iodine entrapped in a solution on titration as above required 1.5mL of 0.01N Na_(2) S_(2) O_(3) solution. Calculate volume % foO_(3) in sample. |
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| 27. |
For equilibrium mixture PCl_(3(g)) + Cl_(2(g)) hArr PCl_(5(g)).The value of K_c at 250^@ C is 26. The value of K_p at this temperature is ____ |
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Answer» `0.20` |
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| 28. |
For electron affinity of halogen which of the following is correct? |
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Answer» `BR GT F` |
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| 29. |
For each of the following,give the level designation,the allowable m values and the number of orbitals.(i)n=4,l=2,(ii)n=5,l=3(iii)n=7,l=0 |
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Answer» Solution :(i)n=4,L=2 If l =2,.m. VALUE are-2,-1,0,+1,+2 so,5 orbitals such as `d_(xy),d_(yz),d_(XZ),d_(X^(2)-y^(2))` and `d_(Z)^(2)` (ii)n=5,l=3 if l=3,.m. values are--3,-2,-1,0,+1,+2,+3, so,7 orbitals such as `f_(z)^(3),f_(xz)^(2),f_(yz)^(2),f_(xyz),f_(z(x^(2)-y^(2)),f_(x(x^2-3y^(2))),f_(y(3x^2-3z^(2)))` (iii)n=7,l=0 if l=0,.m. values are 0.only one value. So,1 ORBITAL such as 7s orbital. |
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| 30. |
For each pair, indicate which bond will be stronger a) C-H , Li-F b) Li-F, Mg-Oc) Li - F , Cs - 1 |
| Answer» SOLUTION :a) `Li_(F)` (B) MgO (c) LiF | |
| 31. |
For each of the following substances, identify the intermolecular force or forces that predominate. Using your knowledge of the relative strength of the vaious forces, rank the substance in order of their normal boiling points. Al_(2)O_(3), F_(2), H_(2)O, Br_(2), IC l, NaCl |
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Answer» `F_(2) LT Br_(2) lt IC l` ICl - DIPOLE dipole `H_(2)O` - dipole - dipole (H - bonding) NaCl - ionic `F_(2) lt Br_(2) lt ICl lt H_(2)O lt NaCl lt Al_(2)O_(3)` |
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| 32. |
Foreach of the followingpairspredictwhich one haslowerfirstionization enthalpy ? (i) N or O (ii) Na or Na^(+)(iii) Ba^(+)"or" Mg^(2+)(iv)" or" I^(-) |
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Answer» (ii)Na haslower `Delta_(i) H_(1)` than `Na^(+)` becauseof the followingtwo reasons : (a)In caseof `Na^(+)`an electron hasto belostfrom astableinert gasconfigurationbut incase ofNa , lossof anelectrongivesstableinertgasconfiguration(B) Na is neutralbut `Na^(+)` is + vely charged (III) `Be^(+)` haslower`Delta_(i)H_1)` than `Mg^(2+)` becausein caseof `Be^(+)` the lossof oneelectrongivesa stableinertgasconfigurationbut in caseof `Mg^(2+)` the electron has to belost fromthe stable INERTGAS configuration . (iv)I haslower `Delta_(i) H_(1)` than `I^(-)` becausein caseof `I^(-)` an electronhas to belostfroma stableinertgas configuration |
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| 33. |
For each of the following pair of hydrogen orbitals, indicate which is higher in energy ? (a) 1s, 2s (b) 2p, 3p (c) 3d_(xy), 3d_(yz) (d) 3s, 3d (e) 4f, 5s |
| Answer» Solution :(a) `2s GT 1S` (b) `3p gt 2p` (c) `3d_(xy) = 3d_(YZ)` (e) `5S gt 4f` | |
| 34. |
For each of the following, give the sub level designation, the allowable m values and the number of orbitals (ii) n = 5, l = 3. |
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Answer» SOLUTION :`n = 5, l = 3` If l = 3, .m. values are -3, -2, -1, 0, +1, +2, +3 So, 7 orbitals such as `f_(X^3), d_(xz^2), f_(yz^2), f_(xyz), f_(z)(x^2 - y^2), f_(x)(x^2 - 3y^2), f_y(3x^2 - z^2)` . |
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| 35. |
For each of the following, give the sub level designation, the allowable m values and the number of orbitals (iii) n = 7, l = 0 |
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Answer» Solution :`N = 7, l =0` If `l = 0,` .m. values are 0. only ONE VALUE So, 1 ORBITAL such 7 s orbital . |
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| 36. |
For each of the following, give the sub level designation, the allowable m values and number of orbitals. |
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Answer» SOLUTION :(i) N=4,l=2, (ii) n=5,l=3 (iii) n= 7,l=0 (i) n=4,l=2 If l=2, .m. values are -2, -1, 0, +1, +2. So, 5 orbitals such as `d_(xy),d_(yz),d_(xz),d_(x^(2)-y^(2))` and `d_(1)^(2)` (ii) n=5. l=3 If l=3. .m. values are -3, -2, -1, 0, +1, +2, +3 So, 7 orbitals such as `F_(l)^(2),f_(xz)^(2),f_(y,)^(2),f_(xyz),f_(l(x^(2)-y^(2))),f_(x(x^(2)-3Y^(2))^(2)),f_(y(3X^(2)-Z^(2))` (iii) n=7, l=0 If l=0, .m. values are 0. Only one value. So, I orbital such as 7s orbital. |
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| 37. |
For each of the following, give the sub level designation, the allowable m values and the number of orbitals (i) n = 4, l = 2. |
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Answer» Solution :`n = 4, l = 2` If `l = 2`, .m. values are `-2, -1, 0, +1, +2` So, 5 ORBITALS such as `d_(XY), d_(yz), d_(x^2-y^2) and d_(z^2)`. |
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| 38. |
For each of the following compounds, write a condensed formula and also their bond their formulae. (a) HOCH_(2)CH_(2)CH_(2)CH(CH_(3))CH(CH_(3))CH_(3)(b) {:(""OH),("|"),(N-=C-CH-C-=N):} |
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Answer» Solution :CONDENSED Formulae (a) `HO(CH_(2))_(3)CH(CH_(3))CH(CH_(3))_(2)`(B) `HOCH(CN)_(2)` Bond-line Formulae
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| 39. |
For each of the following compounds, write a condensed formula and also their bond-line formula. (a) HO CH_(2) CH_(2) CH_(2) CH (CH_(3)) CH (CH_(3)) CH_(3) (b) N -= C -overset(overset(OH)(|))(CH)- C -=N |
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Answer» Solution :CONDENSED formula: (a) `HO(CH_(2))_(3) CH(CH_(3))CH(CH_(3))_(2)` (b) `HOCH (CN)_(2)` Bond LINE STRUCTURE formula:
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| 40. |
For each of the following compounds, write acondensed formula and also their bond-line formula. (a) HOCH_(2)CH_(2)CH_(2)CH(CH_(3))CH(CH_(3))CH_(3) (b) N-=C-overset(OH)overset("|")(CH)-C-=N |
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Answer» SOLUTION :condensed formula (a) `HO(CH_(2))_(2)CH (CH_(3))CH(CH_(3))_(2)` (b) `HOCH(CN)_(2)` BOND LINE formula
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| 41. |
For dry cleaning, in the place of tetrachloroethene, liquefied carbon dioxide with suitable detergent is an alternative solvent. What type of harm to the environment will be prevented by stopping use of tetrachloroethene ? Will use of liquefied carbon dioxide with detergent be completely safe from the point of view of pollution ? Explain. |
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Answer» Solution :(i) Tetrachloroethene `(Cl_2C=C Cl_2)` is suspected to be CARCINOGENIC and contaminates the ground water This harmful effect will be prevented by USING liquefied `CO_2` along with suitable detergent. (ii) USE of liquefied `CO_2` along with detergent will not be completely safe because detergents also cause pollution as most of the detergents are non-biodegradable Also, liquefied `CO_2` will ULTIMATELY enter into the atmosphere and contribute to the greenhouse effect |
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| 42. |
For dry cleaning, in the place of tetrachloroethane, liquefied carbon dioxide with suitable detergent is an alternative solvent. What type of harm to the environment will be prevented by stopping use of tetrachloroethane ? Will use of liquified carbon dioxide with detergent be completely safe from the point of view of pollution? Explain. |
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Answer» Solution :Tetrachloroethane, `Cl_2CH - CHI_2` is suspected to be carcinogenic and it contaminates the ground WATER. It we use liquified `CO_2` ALONG with SUITABLE detergent. this harmful effect will be prevented. Use of liquifed `CO_2` along with detergent will not be completely safe because most of the detergents are non-biodegradable and they cause water pollution. Moreover, liquified `CO_2` will ultimately ENTER into the atmosphere and contributed to the green house effect. |
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| 43. |
For dissociation of NH_(3) giving N_(2) and H_(2) gases, the partial pressures at equilibrium are 100,80,80 torr respectively. If som N_(2) gas is removed and at new equilibrium partial pressure of H_(2) becomes 128 torr then the partial pressure of N_(2) remaining will be approx. |
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Answer» 9 torr |
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| 44. |
For diatomic molecules, the correct relation is/are |
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Answer» `C_(P) = 7//2R` |
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| 45. |
For detection of sulphur in an organic compound, sodium nitroprusside is added to the sodium extract. A violet colour is obtained which is due to the formation of |
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Answer» `Fe_(4)[FE(CN)_(6)]_(3)` |
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| 46. |
For decolourization of 1 mole of KMnO_(4) the moles fo H_(2)O_(2) required is |
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Answer» `1//2` `MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O` Oxid half reaction `H_(2)O_(2)rarrO_(2)+2H^(+)+2e^(-)` since 1 MOLE of `H_(2)O_(2)` requires `5e^(-)` therefore REDUCTION of 1 mole of `KMnO_(4)` will REQUIRE `5//2` molesof `H_(2)O_(2)` |
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| 47. |
For decolourisation of I mole of acidified KMnO_(4) the moles of H_(2)O_(2)required is ........... |
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Answer» `1/2` |
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| 48. |
For decolourization of 1 mole of KMnO_4 , the moles of H_2O_2 required is |
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Answer» `1//2` |
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| 49. |
For decolourisation of 1 mol of KMnO_4 the number of moles of H_2O_2 used is |
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Answer» `1/2` |
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