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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. |
Niobium crystallizes in a body centred cubic structure. If density is 8.55 "g cm"^(-3), calculate atomic radius of niobium, given that its atomic mass is 93 u. |
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Answer» Solution :`a^3=(MxxZ)/(rhoxxN_0xx10^(-30))=(93 "g mol"^(-1)xx2)/("8.55 g cm"^(-3)xx6.02xx10^23 "mol"^(-1)xx10^(-30))=3.61xx10^7=36.1xx10^6` `therefore a=(36.1)^(1//3)xx10^2 "PM"=3.304xx10^2 "pm=330.4 pm"` `[x=(36.1)^(1//3), log x=1/3 "log"36.1 =1/3 XX"1.5575=0.519 or x =ANTILOG 0.519=3.304"]` For body-centred CUBIC, `r=sqrt3/4 a` =0.433 a =0.433 x 330.4 pm = 143.1 pm |
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| 2. |
Niobium crystallize in a body centredcubic structure. If density is8.55 " g cm"^(-3) , calculate atomic radius of niobium, given that its atomic mass is 93n. |
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Answer» Solution : ` a^(3) = ( M xx Z)/(p xx N_(0) xx 10^(-30))= ( 93 " g mol"^(-1) xxx2)/( 8.55 " g cm"^(-3) xx 6.02 xx 10^(23) "mol"^(-1) xx 10^(-30))= 3.61 xx 10^(7) = 36.1xx 10^(7) = 36.1 xx 10^(6)` ` a = (36.1)^(1//3) xx 10^(2) "pm" = 3.304 xx 10^(2)"pm" = 330.4 "pm"` ` [x = (36.1)^(1//3) , log x = 1/3 log 36.1 = 1/3 xx 1.5575 = 0.519or x = " antilog" 0.519 = 3.304`) For body - centred CUBIC , ` R= sqrt3/4 or = 0.433 a,= 0.433 xx 330.3 "pm" = 143.1 "pm" ` |
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| 3. |
Niobium crystallises in body-centred cubic structure. If the atomic radius is 143.1 pm, calculate the density of Niobium (Atomic mass=93u ) |
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Answer» `RHO=(ZxxM)/(a^3xxN_0)=(2xx93)/((330.5xx10^(-10))^3XX(6.02xx10^23))=8.6 "G cm"^(-3)` |
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| 4. |
Nicotinic acid (K_(a)=1.4xx10^(-5)) is represented by the formula HNiC. Calculate its percentage dissociation in a solution which contains 0.10 mole of nicotinic acid per 2.0 litre of solution. |
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Answer» Solution :Suppose degree of dissociation of NICOTINIC acid = `ALPHA` `{:(,HNiC,hArr,H^(+),+,NiC^(-)),("Initial amount",0.1 ",mole",,,,),("Amount",0.1-0.1 alpha,,0.1 alpha,,0.1 alpha),("at eqm.",=0.1 (1-alpha),,,,),("MOLAR conc.",=0.1(1-alpha)//2,,0.1 alpha//2,,0.1 alpha//2),("at eqm.",,,,,):}` `K_(a)=([H^(+)][NiC^(-)])/([HNiC])` or `1.4xx10^(-5)=((0.1 alpha)^(2)(0.1alpha//2))/(0.1(1-alpha)//2)` If `alpha lt lt 1`, then `1.4xx10^(-5)=((0.05 alpha)^(2))/(0.05)=0.05 alpha^(2)` or `alpha^(2)=2.8xx10^(-4)` or `alpha=1.67xx10^(-2)` % dissociation `= 1.67xx106(-2)xx100=1.67%` |
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| 5. |
Nicotinic acid (K_(a)=1.4xx10^(-5))is represented by the formula HNiC. Calculate its percentage dissociation in a solution which contains 0.10 mole of nicotinic acid per 2.0 litre of solution. |
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Answer» |
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| 6. |
Nickel atom can lose two electrons to form Ni^(2+) ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons |
| Answer» Solution :`._(28)Ni = [Ar]^(18) 3d^(8) 4S^(2)`. To form `Ni^(2+)` ION, it will lose ELECTRONS from 4s | |
| 7. |
Nickel atom can lose two electrons to form Ni^(2+) ion. The atomic number of nickel is 28. From w hich orb ital will nickel lose two electrons. |
| Answer» SOLUTION :`_(28)Ni = ls^(2) , 2s^(2) , 2p^(6) , 3S^(2) , 3p^(6) , 3d^(8) , 4s^(2)` , NICKEL lose 2 electrons from 4s (outerm OST shell) to form `Ni^(2+)` ion. So, `28 Ni^(2+) = Is^(2) , 2s^(2) , 2p^(8) , 3s^(2) , 3p^(6) , 3d^(8) , 4s^(@)` | |
| 8. |
NH_(4)NO_(3)overset(Delta)rarrN_(2)O+2H_(2)O How many gram equivalents are present in one mole of Ammonium nitrate? |
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| 9. |
NH_(4)HS(s)hArrNh_(3)(g)+H_(2)S(g) K_(P_(1) NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) K_(P_(2) 2 mol NH_(4)HS(s) is taken & 50% of this is dissociated till at equilibrium in 1 litre container. Find (K_(P_(2)^(2)))/(K_(P_(1)^(6)) if 0.25 moles of N_(2) are found finally. |
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| 10. |
NH_4Cl_((s)) hArr NHCl_((g)) , this equilibrium is which physical reaction ? |
| Answer» Solution :This INDICATE the EQUILIBRIUM of sublimation of ammonium chloride in a CLOSE vessel. | |
| 11. |
NH_(4)CI +Na_(2)HPO_(4)+4Z rarr underset(("sparingly soluble salt"))(A)+ NaCI A overset(Delta)rarr A' +4Z A' overset(Delta)underset("High" T)rarr X+Y uarr +Z Compound (Z) is a neutral oxide and turns anhydrous CuSO_(4) blue. Gas (Y) is a basic gas which can also be obtained by treating ammonium sulphate with lime water. sequences mentioned above involve balanced reactions. The compound X is: |
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Answer» `NaHPO_(4)` `Y rarr NH_(3),A' rarr Na(NH)_(4)HPO_(4)` |
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| 12. |
NH_(4)CI +Na_(2)HPO_(4)+4Z rarr underset(("sparingly soluble salt"))(A)+ NaCI A overset(Delta)rarr A' +4Z A' overset(Delta)underset("High" T)rarr X+Y uarr +Z Compound (Z) is a neutral oxide and turns anhydrous CuSO_(4) blue. Gas (Y) is a basic gas which can also be obtained by treating ammonium sulphate with lime water. sequences mentioned above involve balanced reactions. Which of the following reactions would produce gas Y? |
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Answer» `Mg_(3)N_(2)+H_(2)O RARR` `Y rarr NH_(3),A' rarr Na(NH)_(4)HPO_(4)` |
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| 13. |
NH_(4)CI +Na_(2)HPO_(4)+4Z rarr underset(("sparingly soluble salt"))(A)+ NaCI A overset(Delta)rarr A' +4Z A' overset(Delta)underset("High" T)rarr X+Y uarr +Z Compound (Z) is a neutral oxide and turns anhydrous CuSO_(4) blue. Gas (Y) is a basic gas which can also be obtained by treating ammonium sulphate with lime water. sequences mentioned above involve balanced reactions. Formula of (A) is: |
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Answer» `Na_(2)(NH_(4))PO_(4).4H_(2)O` `Y rarr NH_(3),A' rarr Na(NH)_(4)HPO_(4)` |
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| 14. |
(NH_4)_2 CO_(3(s)) hArr 2NH_(3(g)) + CO_(2(g)) + H_2O_((g)) If the total pressure is P at equilibrium then what will be the volume of equilibrium constant, K_p in a closed vessel ? |
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Answer» <P>`P^4/64` `K_P=(P_(NH_3))^2 xx (P_(CO_2))xx(P_(H_2O))` `=((2P)/4)^2 xx (P/4) xx (P/4)=P^4/64` |
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| 15. |
NH_4 OH hArr NH_(4)^(+) + OH^(-) ,K_b = 2 xx 10^(-5) .If one litre of a solution contains 0.1 mole each of ammonia and ammonium sulphate, calculate the pH of buffer solution |
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Answer» Solution :`k_b` of `NH_4 OH= 2 XX 10^(-5)` HENCE `pK_a` of `NH_4 OH=- logk_b = 5- LOG 2` ` [NH_3] =0.1mol, [NH_(4)^(+)]=2[ (NH_4)_2 SO_4]=0.2mol` ` pOH= pK_b + log""([NH_(4)^(+) ])/([NH_3]) = 5- log 2 + log (0.2 )/( 0.1 ) = 5.0` ` pH =-pOH=14-5=9` |
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| 16. |
NH_(4)^(+)and Br^(-) " ions" have ionic radii of 143 pm and 196pm respectively. The coordination number of NH_(4)^(+) "ion in" NH_(4)Br is |
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| 17. |
NH_4^+ and Br^- ions have ionic radii of 143 pm and 196 pm respectively. The coordination number of NH_4^+ ion in NH_4Br is |
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Answer» <BR> Solution :`r_(NH_4^+)//r_(Br^-)`=143/196=0.730 which lies in the range 0.414-0.732 . HENCE, coordination NUMBER =6. |
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| 18. |
NH_(3) polar but BF_(3) non-polar. |
| Answer» Solution :Ammonia has dipole MOMENT due to PYRAMIDAL STRUCTURE but in `BF_(3)` the dipole moment is zero due to SYMMETRICAL planar structure. | |
| 19. |
NH_(3) molecules remain associated through intermolecular hydrogen bonding but there is not such association among HCl molecules even though electronegativities of N and Cl are the same explain |
| Answer» Solution :Although the ELECTRONEGATIVITIES of nitrogen and chlorine are same, nitrogen can form hydrogen bond but Cl cannot. This is because N-atom is much smaller than Cl-atom. Due to large size of Cl-atom, the ELECTROSTATIC attraction between Cl-atom of one molecule and H-atom of another molecule becomes WEAK. hence, Cl does not form hydrogen bonds while `NH_(3)` molecules UNDERGOES association by intermolecular hydrogen bonds. | |
| 20. |
NH_(3) " is heated at 15 atm from "27^(@)C " to " 347^(@)C keeping the volume constnt . The new pressure becomes 50 atm at equilibrium of the reaction 2 NH_(3) hArr N_(2) + 3H_(2) Calculate % of mole of NH_(3) actually decomposed. |
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Answer» Solution :` {: ( ,2 NH_(3),HARR,N_(2),+,3H_(2),), ("Intial moles",a,,0,,0,), ("Moles at eqm.",a-2x,,x,,3x,"Total " = a+2 x ):} ` Pressure of a moles of `NH_(3) at 27^(@)C = 15 "atm".` Pressure of a moles of `NH_(3) "at" 347^(@)C ` = P atm (say) As volume remains constant ,` P_(1)/T_(1) = P_(2) /T_(2)` `15/300 = P/620 or P= 31 atm .` Now, at `327^(@)C` and constant volume , Pressure ` prop ` No. of moles ` :. 31 prop a` ` 50 prop a + 2 x ` ` :.(a + 2 x )/ a = 50/31 or x = 19/62 a ` ` % "of " NH_(3)" decomposed " = (2x)/a xx 100 = 2 xx (19 a)/62 xx 1/a xx 100 = 61* 3 % ` |
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| 21. |
NH_3 is not as Lewis and Bronsted-Lawry Base. Explain. |
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Answer» Solution :Bronsted BASE : `:NH_3 + H_2O hArr NH_4^(+)+OH^(-)` `NH_3` act as a bronsted base by accepting `H^+` (Proton) of `H_2O` `: NH_3 + H^(+) to NH_4^(+)` electron pair donor `NH_3` `NH_3` act as a Lewis base by GIVING electron pair present of N. |
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| 22. |
NH_(3) is more polar than NF_(3) because _________. |
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Answer» F is more electornegative than H |
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| 23. |
NH_(3) is formed in the following steps I. Ca+2CrarrCaC_(2) 50% yield II. CaC_(2)+N_(2)rarrCaCN_(2)+C 100% yield III. CaCN_(2)+3H_(2)Orarr2NH_(3)+CaCO_(3) 50% yield To obtain 2 moles NH_(3), calcium required is |
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Answer» Solution :`Ca+2CrarrCaC_(2)` `100rarr50` `CaC_(2)+N_(2)rarrCaCN_(2)+C` `50rarr50` `CaCN_(2)+3H_(2)Orarr2NH_(3)+CaCO_(3)` `50rarr100xx(50)/(100)=50` `therefore CararrNH_(3)` `100rarr50` `?rarr2 =4mol` |
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| 26. |
NH_(3) has higher proton affinity than PH_(3). Explain. Or NH_(3) is more basic than PH_(3). |
Answer» SOLUTION :Due the presence of a lone PAIR of electrons on N and P, both `NH_(3)andPH_(3)` act as Lewis bases and accept a proton to FORM and additional N-H and P-H bonds respectively  However, due to smaller size of N over P, N-H bond thus formed ismuch stronger than the P-H bond Therefore, `NH_(3)` has higher proton affinity than `PH_(3)`. In other words, `NH_(3)` is more basic than `PH_(3)`. |
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| 27. |
NH_(3) has a much higher boiling point than PH_(3)because |
| Answer» SOLUTION :DUE to INTERMOLECULAR HYDROGEN BONDING | |
| 28. |
NH_3 has exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15. Explain. |
| Answer» Solution :`NH_(3)` has higher boiling and melting point compared to all other hydrides fo grop 15 elements DUE to INTERMOLECULAR hydrogen BONDING. | |
| 29. |
NH_(3)gas is liquified more easily than N_(2) . Hence : |
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Answer» Vander Waal's constant a and b of `NH_3 gt `that of `N_2` |
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| 30. |
NH_(3)can be liquefied at ordinary temperature without the application of pressure. But O_(2)cannot, because |
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Answer» its critical TEMP. is very HIGH |
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| 31. |
NH_(3)+ 3CI_(2) hArr NCI_(3) + 3HCI, - Delta H_(1) N_(2) + 3H_(2) hArr 2NH_(3) , - Delta H_(2) H_(2) + CI_(2) hArrr 2HCI, Delta H_(3) Calculate the enthalpy of formation of NCI_(3) |
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Answer» `Delta H_(f) = - Delta H_(1) + (Delta H_(2) )/( 2) - (3)/(2) Delta H_(3)` |
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| 32. |
NH_(3) and HCI do not obey Henry's law. Why ? |
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Answer» Solution :(i)Henry's law is applicable at moderate TEMPERATUREPRESSURE only. Only the less soluble gases obeys Henry's law (III)The gases reacting with solvent do not obey Henry's law. For example, ammonia a HCI REACTS with water and hence does not obey this law. `NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-)` (iv)The gases obeying Henry's law should not associate or dissociate while dissolving in the solvent . |
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| 33. |
NH_(2)NH_(2) contains nitrogen but still it does not give positive Lassaigne's test ? To get positive test for nitrogen in NH_(2)NH, what modification do you suggest ? |
| Answer» Solution :If hydrazine `(NH_(2)NH_(2))` is fused with Na metal, it does form NACN since it does not contain carbon and hence will not give +ve test for NITROGEN. In ORDER to test the presence of N in such compounds, during FUSION with Na, some charcoal or preferably starch (which contains C but N, S, halogens, etc.) is added. Under these conditions, C of starch or charcoal combines with N of the compound to form NaCN which will now give a +ve test for nitrogen. | |
| 34. |
NH_(2)NH_(2) compound loses 10 mole e^(-) and form new compound x then calculate oxidation number of N_(2) in x compound. (Here oxidation number of H does not change.) |
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Answer» `-3` `thereforeN_(2)H_(4)-10baretox` `therefore2x+4=0` `therefore2x=-4` Here, 10 electrons release during reaction. So, such electrons ADDED in this reaction. `2x=-4+10` `thereforex=+3` |
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| 35. |
Newton mechanicsis applicable towhichobjects andnot applicable to whichobjects ? |
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Answer» SOLUTION :NEWTON mechanicsis applicableto macroscopicparticlese.grollingballfallingstonemotionof planesin ORBIT ETC . BUTIT is not applicableto subatomicparticleslikeprotonelectron etc. Whichhave very lessmassSo,uncertaintyismere. |
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| 36. |
Newly shaped glass articles when cooled suddenly become brittle, these are cooled slowly, this process is known as |
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Answer» tempering |
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| 37. |
Neutrons was dicovered by |
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Answer» RUTHERFORD |
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| 38. |
Neutrons can be found in all atomic nuclei except in one case. Which is this atomic nucleus and what does it consist of ? |
| Answer» SOLUTION :The nucleus of HYDROGEN ATOM does not contain any NEUTRONS. It consists of only one proton. | |
| 39. |
Neutralisation constant of HCOOH with a strong bse is 10^(8). What is the pH of 0.01 M HCOOK solution at 25^(0)C? |
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Answer» `K_("neu") = (1)/(K_(h)) = (K_(a))/(K_(w))` `K_(a) = 10^(8) xx 10^(-14) = 10^(-6) , pH = 7+3-1 = 9` |
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| 40. |
Neutralisation constant of HCOOH with a strong base is 10^(8). What is the pH of 0.01 M HCOOK solution? At 25^(0)C |
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Answer» |
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| 41. |
Neutral solutions have pH = 7 at 298 K. A sample of pure water is found to have pH lt7. Does it mean thatit is acidic ? Explain. |
| Answer» Solution :`pH lt 7` for pure `H_(2)O` shows that water is at a temperature HIGHER than 298 K. It is neutral at all temperatures. At higher temperature, `H_(2)O` dissociates more to give larger CONCENTRATIONS of `H^(+)` ions and `OH^(-)` ions. HENCE, `pH lt 7`. HOWEVER, `[H^(+)]=[OH^(-)]` at all temperatures. | |
| 42. |
Nerve toxins are |
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Answer» `NaClO_(3)` |
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| 43. |
Neon is naturally available as ""^(20)Ne and ""^(22)Ne with average atomic mass 20.2. Calculate the relative abundance of heavier isotope. |
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Answer» Solution :LET the percentage ABUNDANCE of `""^(22)NE` is x. The percentage abundance of `""^(20)Ne` is 100-x. `20.2 =((100-x)20+22x)/(100)` Percent abundance of `""^(20)Ne=10%` |
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| 44. |
Neon gas is generally used in the sign boards. If it emits with wavelength of 616 nm, calculate (i) the frequency of emission (ii) energy of quantum and (iii) distance travelled by this radiation in 30 s. |
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Answer» Solution :`lambda_("Neon") = 16nm=616xx10^(-9)m` i) `v_("Neon")=(c)/(lambda)=(3XX10^(8)m.s^(-1))/(616xx10^(-9)m)=0.00487xx10^(17)s^(-1)` `=4.87xx10^(14)s^(-1)" (or Hertz)"` II) `E=hv=6.625xx10^(-34)J.s xx4.87xx10^(14)s^(-1)` `=32.26xx10^(-20)J` iii) `"distance travelled "= "Velocity "xx "time "=` `3xx10^(8) m.s^(-1)xx30s = 90 xx10^(8) m = 90 xx 10^(5) KM`. |
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| 45. |
Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of the emission(b) distance travelled by this radiation in 30s (c) energy of quantum (d) number of quanta present if it produces 2J of energy. |
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Answer» Solution :`lamda = 616 nm = 616 XX 10^(-9) m` (a) Frequency, `v = (c)/(lamda) = (3.0 xx 10^(8) ms^(-1))/(616 xx 10^(-9) m) = 4.87 xx 10^(14) s^(-1)` (b) Velocity of the radiation `= 3.0 xx 10^(8) ms^(-1)` `:.` Distance travelled in `30s = 30 xx 3 xx 10^(8) m = 9.0 xx 10^(9) m` (c) `E = HV = h (c)/(lamda) = ((6.626 xx 10^(-34) Js) xx 3.0 xx 10^(8) ms^(-1))/(616 xx 10^(-9) m) = 32.27 xx 10^(-20)J` (d) No. of quanta in 2J of energy `= (2)/(32.27 xx 10^(-28)) = 6.2 xx 10^(18)` |
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| 46. |
Neongas isgenerally used in the signboards.ifit emitsstronglyat 616nmcalculate(a ) thefrequencyof emission( b)distancetraveledbythis radiationin 30s (c )energyof quantumand ( d)numberof quantapresentif itproduces2 J ofenergy . |
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Answer» SOLUTION :(a ) Frequencyof emission `V=(C )/(lambda)= (3.0 xx 10^ (8) MS^(-1))/(616 xx 10^(9) m)` ( b) Distancetraveledby thisradiationin 30s (d ) d=velocityof light`xx ` time `=3.0 xx 10^(8)ms^(-1)xx 30 s` `=9.0 xx 10^(9)m` =`900` carormeter= `(xx 10^(8) km)` (c )Quantum energy(E ) : `E= (hc )/( lambda )= (6.626 xx 10^(34)J s3.0 xx 10^(8) ms^(-1))/(616 xx 10^(9) m)` (d ) Numberof quanta in 2 J of energy (N ) : N=`("TOTAL energy ")/(" 1 quantum energy ") = (2J )/( 3.23 xx 10^(19))` `=6.192 xx 10^(18)` Numberof quantum |
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| 47. |
Neomycin, amino glycoside antibiotic cream contains 300 mg of neomycin sulphate the active ingredient in 30 g of ointment base. The mass percentage of neomycin is …………….. |
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Answer» Solution :`1%` The mass PERCENTAGE of nemycin `=("Mass of neomycine sulphare")/("Mass of solution in G") XX 100` `= (0.3g)/(30g) xx 100 =1%` |
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| 48. |
neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly. Justify. |
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Answer» SOLUTION :(i) `CH_3 - underset(CH_3)underset(|)overset(CH_3)overset(|)C - CH_2Br` neo-pentyl bromide UNDERGOES nucleophilic substitution reactions very slowly due to stéric hindrance of many alkyl groups. When bromide is attached to neo-pentyl carbon atom, the heterolytic cleavage of C-Br takes place very slowly and substitution is also a very SLOW REACTION. (ii) Due to bulky neo-pentyl group, it becomes difficult for a nucleophile to attack from the back side of carbon having C-Br BOND. (iii) Splitting of C-Br gives a primary carbocation which is very less stable. So neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly. |
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| 49. |
Neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly-justify. |
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Answer» SOLUTION :(i) Neopentyl bromide being a primary a primary hailde REACTS slowly through `S_(N)1`. (ii) BULKY neopentyl group difficult for a nuelcophilic attack from backside at C of C-Br BOND. (iii) Cleavage of C-Br bond gives `1^(@)` carbocation which is less stable. |
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