52671 + Interview Questions in Chemistry in Class 11

1.	Calculate the enthalpy of formationof acetic acid if the enthalpy of combustionto CO_(2)(g) and H_(2)O(l) is - 867.0 kJ mol^(-1) and enthalpies of formation of CO_(2)(g) and H_(2)O(l)are respectively -393.5 and -285.9 kJ mol^(-1)
Answer» SOLUTION :AIM `: 2C(s) + 2H_(2) (g) + O_(2)(g)rarrCH_(3)COOH,DeltaH = ?`

Discussion

2.	Calculate the enthalpy of combustion of ethylene at 300K at constant pressure if its enthalpy of combustion at constant volume is -"1406 kJ mol"^(-1).
Answer» Solution :The complete ethylene combustion reaction can be written as, `C_(2)H_(4(g))+3O_(2(g))rarr 2CO_(2(g))+2H_(2)O_((l))` `DeltaH=DELTAE+RT Deltan_((g))`, where `Deltan_((g))=n_(p(g))-nP_(r(g))`. `therefore""Deltan_((g))=2-(3+1)=-2`. Enthalpy of combustion at constant VOLUME `=DeltaE=-"1406 kJ mol"^(-1)` `therefore` Overall enthalpy of combustion `=DeltaH_(c)=-1406+(-2xx8.134xx10^(-3)xx300)` `=-1406-4.9884` `DeltaH_(c )=-1410.9kJ mol^(-1)`.

Discussion

3.	Calculate the enthalpy of combustion of benezene from the following data:- (i)6C(s) + 3H_(2)(g) rarr C_(6)H_(6) (l), DeltaH =49.0 kJ mol^(-1) (ii) H_(2) (g) + (1)/(2) O_(2)(g) rarrH_(2)O(l), DeltaH = - 285.8 kJ mol^(-1) (iii) C(s) + O_(2) (g) rarr CO_(2)(g) , DeltaH = - 389. 3 kJ mol^(-1)
Answer» SOLUTION :Aim `: C_(6)H_(6) (L) + (15)/(2) O_(2)(g) rarr 6CO_(2)(g) +3H_(2)O(l) , DeltaH = ?` `6 xx `Eqn. (ii)`+ 3 xx `Eqn. (ii) `- `Eqn. (i) gives the REQUIRED result.

Discussion

4.	Calculate the enthalpy of combustion of acetic (1) when burnt in excess of O_(2) in a bomb calorimeter. Given that DeltaH_(f)^(@), H_(2)O_((l))=-285.84" KJ mol"^(-1) and Delta_(f)H^(@), CO_(2(g))=-"393.52 KJ mol"^(-1), Delta_(f)H^(@)CH_(3)COOH_((l))=-"463 Kj mol"^(-1).
Answer» SOLUTION :`Delta_(c )H^(@)=-"895.72 KJ. MOL"^(-1)`

Discussion

5.	Calculate the enthalpychangewhen 2.38 g of carbonmonoxide ( CO)vaporize at its normal boiling point.Given thatthe enthalpy of vaporisationof carbon monoxide is6.04 kJ mol^(-1) at its normal boiling point of 82.0 K.
Answer» SOLUTION :`Delta_(VAP) H ` for `CO = 6.04 kJ MOL^(-1)`, i.e., = 6.04 kJfor28 G `:. `Enthalpy change for vaporisation of2.38 g `= ( 6.04) /( 28) xx2.38 kJ = 0.5134 kJ= 513.4 J`

Discussion

6.	Calculate the enthalpy change when 6.80g of NH_(3) is passed over heated CuO. The standard heat enthalpies of NH_(3(g)),CuO_((s)) and H_(2)O_((l)) are -46.0,-155.0 and -285.0kJ mol^(-1) respectively and the change is NH_(3)+(3)/(2)CuO rarr (1)/(2) N_(2(g))+(3)/(2) H_(2)O_((l))+(3)/(2)Cu_((s)) .
Answer» ANSWER :`-59.6kJ;`

Discussion

7.	Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°c to ice at -10.0°c. Delta_("fus") H= 6.-03 "kJ mol"^(-1) "at " 0^(@) C C_(p) [H_(2) O_((l)) ]= 75.3 "J mol"^(-1) K C_(p) [H_(2) O_((s))] =36.8 "J mol"^(-1) K
Answer» Solution :`Delta_("FUS") H= 6030 "J/mol"` `DeltaH=C_(p) [H_(2) O_((L)) ] xx Delta T + (Delta H)/( T_(f) ) + C_(p) [H_(2) O _((s)) ] xx Delta T` `= (75.3) (0-10) K + (-6.03) "J mol"^(-1) - 368 "J mol"^(-1)` `=-7151 "J/mol"= -7.151 "kJ/mol"`

Discussion

8.	Calculate the enthalpy change on freezing 1.0 mol of water at 10.0^(@)C to ice at - 10.0 ^(@)C, Delta_(fus) H= 6.03 kJ mol^(-1) mol^(-1) at 0^(@)C C_(p) [H_(2)O(l)]= 75.3J mol^(-1) K^(-1) C_(p)[ H_(2)O(s)] = 36.8 J mol^(-1) K^(-1)
Answer» Solution :Total `DeltaH = ` ( 1 mol water at `10^(@)C rarr ` 1 mol of waterat `0^(@)C) + (` 1 mol water at `0^(@)C rarr 1 `mol ice at `0^(@)C )+(` 1 mol ice at `0^(@) Crarr `1 mol ice at `- 10^(@)C) ``(DeltaT = T_(2) - T_(1))` `=C_(p) [H_(2)O(l) ]xx DeltaT + DeltaH _("freezing") +C_(p)[H_(2)O (s) ] xx DeltaT ` `= ( 75.3J K^(-1) mol^(-1) ) ( 0- 10) K( - 6.03 kJ mol^(-1)) + ( 36.8 J K^(-1) mol^(-1) ) ( - 10 K) ( DeltaH_("freezing" ) =-DeltaH_("fusion") )` `=-753 J mol^(-1) - 603 kJ mol^(-1) - 368 J mol^(-1)` `= - 0.753 kJ mol^(-1) - 6.03 kJ mol^(-1) -0.368kJ mol^(-1) kJ mol^(-1) =-7.151kJ mol^(-1)` Note `:` Directly also, as in each STEP , heat is evolved,each step will have a NEGATIVE SIGN with `DeltaH`.

Discussion

9.	Calculate the enthalpy change for the reaction H_(2)(g)+I_(2)(g) rarr 2HI(g) Given that the bond energies of H-H,I-I and H-I are433, 151 and 299 kJmol^(-1) respectively.
Answer» ANSWER :`-14 KJ`

Discussion

10.	Give the following information, calculate the lattice enthalpy of the sodium chloride lattice: Delta ("atomisation Na") = +107 kJ mol^(-1) "" DeltaH (1^(st) "ionisation Na") = + 496 kJ mol^(-1) Delta H ("bond dissociation " CI_(2)) = +242 kJ mol^(-1) "" Delta (1^(st) "electron affinity CI") =-349 kJ mol^(-1) DeltaH( NaCI) =-411 kJ mol^(-1)
Answer» SOLUTION :-367 kJ/mol

Discussion

11.	Calculate the enthalpy change for the reaction H_(2)(g) + Br_(2)rarr 2HBr (g) Give that the bond enthalpies ofH-H,Br-Br and H-Br are435, 192 and 364kJ mol^(-1) respectively.
Answer» Solution :Energy absorbed for dissociation of 1 mole of H-H bonds =`435 KJ` Energy absorbed for dissociation OF1 mole of Br -Br bonds `= 192kJ` TOTAL energy absorbed `=435+ 192 = 627kJ` Energy released in the formationof1 mole of H- Br bonds= 364kJ `:.`Energy released in the formationof 2 moles of H - Br bonds `= 2 xx 364 kJ = 728 kJ` Energy released ` gt` Energy absorbed Hence, net result is the release of energy Energy released `= 728 kJ - 627 kJ = 101 kJ ` i.e., for the given reaction , `Delta_(r) H = - 101kJ` Alternatively, the problem may be solved by applying Hess's lawor by applying the following RELATION directly`Delta_(r) H = Sigma` B.E. ( Reactants ) `- Sigma ` B.E. ( Products) `=[B.E. (H_(2)) + ` B.E. ` (Br_(2))] - 2 B.E. ( HBr) = 435 + 192 - 2 xx 364 = - 101 kJ`

Discussion

12.	Calculate the enthalpy change for the reaction C_(2)H_(4)(g) + H_(2)(g) to C_(2)H_(6)(g) using the data given below : C_(2)H_(4)(g) + 3O_(2)(g) to 2CO_(2)(g) + 2H_(2)O(l)DeltaH = -1415 kJ C_(2)H_(6)(g) + 7/2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l)DeltaH = -1566 kJ H_(2)(g) + 1/2 O_(2)(g) to H_(2)O(l)DeltaH = -286 kJ
Answer» `-437` kJ `+35` kJ `-135` kJ none of these Solution :(i) `C_(2)H_(4)(g) + 3O_(2)(g) to 2CO_(2)(g) + 2H_(2)O(l)DeltaH = -1415 kJ` (ii) `C_(2)H_(6)(g) + 7/2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l)DeltaH = -1566 kJ` (iii) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) DeltaH = -286 kJ` Adding eq.(i) and (iii) `C_(2)H_(4)(g) + 7/2O_(2)(g) + H_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l)DeltaH = -1701 kJ` Subtracting eqn (ii) from eqn (IV) `C_(2)H_(4)(g) + H_(2)(g) to C_(2)H_(6)(g) DeltaH = -135 kJ`

Discussion

13.	Calculate the enthalpy change for the reaction Fe_2O_3 +3CO to 2Fe +3CO_2 from the following data. 2Fe +3/2 O_2 to Fe_2O_3 , DeltaH=-741 kJ C+1/2O_2 to CO , DeltaH=-137 kJ C+O_2 to CO_2 , DeltaH=-394.5 kJ
Answer» Solution :Given : `DeltaH_f(Fe_2O_3)=-741 "KJ mol"^(-1)` `DeltaH_f(CO)=-137 "kJ mol"^(-1)` `DeltaH_f(CO_2)=-394.5 "kJ mol"^(-1)` `Fe_2O_3 + 3CO to 2FE + 3CO_2 "" DeltaHr`=? `DeltaH_r=sum(DeltaH_f)_"products"-sum (DeltaH_f)_"reactants"` `DeltaH_r=[2DeltaH_f(Fe)+3DeltaH_f(CO_2)]-[DeltaH_f (Fe_2O_3) + 3DeltaH_f (CO)]` `DeltaH_r` =[0+3(-394.5)]-[-741+3(-137)] `DeltaH_r`=-1183.5 +1152 `DeltaH_r=-31.5 "kJ mol"^(-1)`

Discussion

14.	Calculate the enthalpy change for the reaction between CO_(2) and H_(2)O to produce one mole of glucose (C_(6)H_(12)O_(6)) . What wouldbe enthalpy change for the production of 18g of glucose ? The enthalpy of combustion of glucoseis 2840 kJmol^(-1).
Answer» SOLUTION :Given `: C_(6)H_(12)O_(6)(s) + 6O_(2)(g) rarr 6 CO_(2)(g) + 6H_(2)O(l), DeltaH = - 2849kJ mol^(-1)` Aim `:` Reverse Reaction for which`Delta_(r) H = + 2840 kJ mol^(-1)` This is the enthalpy change for PRODUCTIONOF 1 MOLE ( 180g) of glucose. HENCE, for 18 g glucose, `Delta H = + 284kJ`

Discussion

15.	Calculate the enthalpy change for the process "CC"I_(4(g)) + C_((g)) + 4CI_((g)) and calculate bond enthalpy of C - CI in "CC"_(4(g)) Delta_("vap")H^( Theta ) ("CC"I_(4) ) = 30.5 "kJ mol"^(-1) Delta_(f) H^( Theta ) ("CCI"_(4) ) = -135.5 "kJ mol"^(-1) Delta_(a) H^( Theta ) (C) = 715.0 "kJ mol"^(-1) where Delta_(a) H^( Theta ) is enthalpy of atomisationDelta_(a) H^( Theta ) (CI_(2) ) = 242 "kJ mol"^(-1)
Answer» SOLUTION :As per the given information. (i) `"CCI"_(4(l)) to "CCI"_(4(G)) , Delta H = 30.5 "kJ/mol"` (ii) `C_((s)) + 2CI_(2(g)) to "CCI"_(4(l)), DeltaH = -135.5 "kJ /mol"` (III) `C_((s)) to C_((g)) , Delta H = 715 "kJ/mol"` (iv) `CI_(2(g)) to 2CI_((g)) , Delta H = 242 "kJ/mol"` Now, for `"CCI"_(4(g)) to C_((g)) + 4CI_((g))` `DeltaH= [715 + 2(242)] - [30.5 +(-135.5)]` `=1304 "kJ/mol"` `therefore` Bond enthalpy of `C- CI=(1304)/(4)=326 "kJ/mol"`

Discussion

16.	Calculate the enthalpy change for the process C Cl_(4)(g) rarrC(g) +4Cl(g) and calculate bond enthalpyofC-Cl in C Cl_(4)(g) Given :V_(vap) H^(@) (C C l_(4)) =30.5 kJ mol^(-1), Delta_(f)H^(@) (C C l_94))= -135.5 kJ mol^(-1) Delta_(a)H^(@) ( C ) = 715.0 kJ mol^(-1)whereDelta_(a)H^(@)is enthalpyof atomisation Delta_(a) H^(@)( Cl_(2)) = 242 kJ mol^(-1)
Answer» Solution :The given data imply as under `:(i)C Cl_(4) (l) rarr C C l_(4)(g) , DeltaH= 30.5 kJ mol^(-1)` (II)`C(s) + 2Cl_(2)(g) rarrC C l_(4)(l) , DeltaH =-135.5 kJ mol^(-1)` (iii) `C(s)rarr C(g) , DeltaH= 715 .0 kJ mol^(-1)` (IV) `Cl_(2)(g) rarr2Cl(g), DeltaH = 242 kJ mol^(-1)` AIM`:C C l_(4)(g)rarr C(g) + 4Cl(g) , DeltaH = ?` Eqn. (iii)`+2 xx` Eqn. (iv) - Eqn. (i). Eqn. (ii) gives the requiredequation with `DeltaH=715.0+ 2 ( 242)-30.5 - ( - 135.5) kJ mol^(-1) = 1304 kJ mol^(-1)` Bond enthalpy ofC-Cl in `C C l_(4)` ( average value)`= ( 1304)/( 4) = 326 kJ mol^(-1)`

Discussion

17.	Calculate the enthalpy change for the isomerization reaction as given: CH_(2)=CH-CH_(2)-CH=CH-CH=CH_(2) (A) (NaNH_(4))/(Delta)CH_(2)=CH-CH=CH-CH=CH-CH_(3) (B) "Use the following data": DeltaH_(f)^(@)(C_(2)H_(4))=-2275.5KJ//mol DeltaH_(f)^(@)(C_(2)H_(6))=-2839.2 KJ//mol Resonance energy of A=50Kj/mol Resonance energy of B=70KJ/mol
Answer» `-1692.6KJ//mol` `-1642.6KJ//mol` `-1622.6KJ/mol` `-20,000J//mol` ANSWER :d

Discussion

18.	Calculate the enthalpy change accompanying the transforming of C (graphite) to C(diamond) . Given that theenthalpies of combustion of graphite and diamond are 393.5 and 395.4 kJ mol^(-1) respectively.
Answer» Solution :We are given (i) C (GRAPHITE) `+ O_(2)(g) rarr CO_(2)(g) , Delta _(c ) H^(@)= - 393.5 KJ mol^(-1)` (ii) C (diamond) `+ O_(2)(g) rarr CO_(2)(g), Delta _(c ) H^(@) = - 395.4 kJ mol^(-1)` We aim at C( graphite ) `rarr ` C ( diamond) `, Delta _("trans") H^(@) = ?` Subtracting eqn. (ii) from eqn. (i), we get ORC(graphite)- C (diamond) `rarr 0 , Delta _(r) H^(@) = - 393.5 - ( - 395.4) = + 1.9 kJ`

Discussion

19.	Calculate the energy with 5th orbit of hydrogen atom.
Answer» Solution :`E=(-2.18xx10^(-18)Z^(2))/(a^(2))""z=1 n=5` `(-2.18xx10^(-18)XX1)/25=8.72xx10^(-20)J`

Discussion

20.	Calculatethe energy requiredto convert all theatoms of magnium to magnesium ions present in 24 mg ofmagnesium vapours : First and second ionization enthalpies of Mg are737.76 and1450.73 kJ mol^(1) respectively.
Answer» SOLUTION :According to THEDEFINITION of successiveionizationenthalpies. `Mg (g)+ Delta_(i) H_(1)to Mg^(+) (g) + e^(-) (g) ,Delta_(i) H_(1) 737.76 kJ mol^(-1)` `Mg^(+) (g)+ Delta_(i) H_(2) to Mg^(2+)(g) + e^(-) (g),Delta_(i) H_(2)= 1450 .73 kJ mol^(-1)` `:.` Totalamountof energyneededto CONVERT Mg (g)atom into `Mg^(2+) (g) ion = Delta_(i) H_(1) + Delta_(i) H_(2)` `=737.76 + 1450.73 kJ mol^(-1) =m 2188.49 kJ mol^(-1)` 24 mg of mg `=(24)/(1000) g= (24)/(1000xx 24) " mole " = 10^(-3) `mole `:.` Amountof energyneededto ionize`10^(-3)` mole ofMgvapours `=2188.49 xx 10^(-3) =2.188 kJ`

Discussion

21.	Calculatethe energyrequiredfor theprocess He_(g)^+ to He_(g)^(2+)+ e^(-) theionizationenergyfor the Hatom in thegroundstateis 2.18 xx 10^(-18)J atom^(-1)
Answer» Solution :Energyof norbitatom for`(H , He^(+) Li^(2+) ` is`E_(N))` `E_(n) = (2.18 XX 10^(18) Z^(2))/( n^(2)) J atom^(1)` for HZ-1and n=1 IP`2.18 xx 10^(18) J atom^(-1)` ionizationof`He^(+)` is`He^(+)to He^(2) + e^(-)` Hereionizationfor `He^(+)` `n_(1)1 ` ASND `n_(2)= PROP`z=-2 `Delta E = 2.18xx 10^(18) (2^(2))/(1^(2)) (1)/( 1^(2))- (1)/(prop^(2))` `=2.18 xx 10^(18) (4)` ionizationenertgy of `He^(+)` `=He^(+) to He^(2+)e^(-)`reactionenergy

Discussion

22.	Calculate the energy required for the process.He_(g)^(+)rarrHe_(g)^(2+)=e^(-)The ionization energy for the H atom in its ground state is -13.6eV atom^(-)
Answer» SOLUTION :The ionization energy for the H ATOM in Its ground STATE`=13.6eV "atom"^(1)` Ionization energy`=13.6z^(2)/(n^(2)eV)` z=atomic number n=principal quantum number or shell number For He,n=1,z=2 `I.E.=(-13.6xx2^(2))/(1^(2))eV` I.E. of`He^(+)` to `He^(2+)`=-54.4eV

Discussion

23.	Calculate the energy required for the process. He_(g)^(+) rarr He_((g))^(2+) + e^(-) The ionisation energy for the H atom in its ground state is -13.6 eV atom^(-1).
Answer» Solution :The ionization energy for the H ATOM in its GROUND state `= -13.6 eV "atom"^(-1)` Ionization energy = `(-13.6)/(n^2) z^2 eV` Z = atomic NUMBER n = PRINCIPLE QUANTUM number or shell number For `He n = 1, Z = 2` I.E = `(-13.6 xx 2^2)/(1^2) eV` I.E of the `He^(2+) = -54.4 eV`.

Discussion

24.	Calculate the energy required for the process He^(+) (g) rarr He^(2+) (g) + e^(-). The ionization energy for the H atom in the ground state is 2.18 xx 10^(-18) J "atom"^(-1)
Answer» Solution :For H-like particles, `E_(n) = - (2PI^(2) m Z^(2) e^(4))/(n^(2) h^(2))` For H-atom, I.E. `= E_(oo) - E_(1) = 0 - (-(2pi^(2) me^(4))/(1^(2) xx h^(2))) = (2pi^(2) me^(4))/(h^(2)) = 2.18 xx 10^9-18) J "atom"^(-1)` (Given) For the given PROCESS, Energy REQUIRED `= E_(oo) - E_(1) = 0 = (-(2pi^(2) m xx 2^(2) xx e^(4))/(1^(2) xx h^(2))) = 4 xx (2pi^(2) me^(4))/(h^(2)) = 4 xx 2.18 xx 10^(-18) J` `= 8.72 xx 10^(-18) J`

Discussion

25.	Calculate the energy requiredfor the processHe^(+) (g) to He^(2+) (g) + e^(-)The ionization energy for the H atom in the ground state is2.18 xx 10^(-18)J "atom"^(-1)
Answer» ANSWER :`8.72 XX 10^(-18)`

Discussion

26.	Calculate the energy released by the reaction 4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s) when a 55.8 g sample of iron reacts ccompletely with 1.00 moleof oxygen . The enthalpy ofiron reacts complately with 1.00 mole of Oxygen. Theenthalpy of formation (DeltaH_(f)^(@))"Of" Fe_(2)O_(3)(s),"is"-826 KJxxmol^(-1):
Answer» 206KJ 413KJ 826KJ 1650KJ Answer :B

Discussion

27.	Calculate the energy of one mole of photons of radiations whose frequency is 3 xx 10^(12) Hz
Answer» Solution :`E=h upsilon=6.625xx10^(-34)" JS "xx3xx10^(12)s^(-1)` `=1.9875xx10^(-21)"J for one photon "` `"ENERGY of one MOLE of photons is"` `="Energy of one photon"XX"AVOGADRO number "` `=1.9875xx10^(-21)Jxx6.023xx10^(23)` `=1.1970xx10^(3)" J mol"^(-1)`

Discussion

28.	Calculate the energy of one mole of photon of radiation whose frequency is 5xx10^(14)Hz(Given h=6.626xx10^(-34)Js).
Answer» SOLUTION :`E=h gamma` `E=6.626xx10^(-34)xx5xx10^(-14)=3.313xx10^(-19)J` `:.` ENERGY of 1 MOLE of photons `=3.313xx10^(-19)xx6.022xx10^(23)=199.51kJ"mol"(-1)`

Discussion

29.	Calculatethe energyrequiredfor theprocess He_(g)^e to He_(g)^(2+)+ e^(-) theionizationenergyfor the Hatom in thegroundstateis 2.18 xx 10^(18)J atom^(-1)
Answer» SOLUTION :

Discussion

30.	Calculate the energy of C-Cl bond from the following data : CH_(4)(g) + Cl_(2)(g) rarr CH_(3)(g) + HCl(g) +HCl(g), DeltaH=- 100 .3 kJ The bond energyof C-H, Cl- Cl and H-Cl bonds are 413, 243 and 431 kJ mol^(-1) respectively.
Answer» Solution :`H - underset( H ) underset(\|)OVERSET( H) overset(\|) (C) - H (g) + Cl- Cl(g) rarr H - underset(H) underset(\|) overset(H) overset(\|) (C) -Cl( (g) + H- Cl( g) , Delta H= - 100.3 KJ` `Delta H =[ 4 xx B.E. ( C-H) + B.E. ( Cl- Cl) -[ 3 xx B.E. ( C-H) + B.E.( C- Cl) + B.E. ( H -Cl)]` `=B.E. ( C-H)+ B.E. ( Cl- Cl) - B.E. ( C- Cl) - B.E. ( H- Cl) ` `:. -100.3 = 413 + 243 - B.E. ( C-Cl) - 431` or`B.E. ( C-Cl) = 413 + 243 - 431 + 100.3 = 325.3 kJ mol^(-1)`

Discussion

31.	Calculate the energy of a mole of photons of radiations whose frequency is 5 xx 10^(14)Hz ?
Answer» SOLUTION :Energy of ONE photon, `E = hv = (6.626 XX 10^(-34) Js) (5 xx 10^(14) s^(-1)) = 3.313 xx 10^(-19) J` `:.` Energy of one mole of PHOTONS `= (3.313 xx 10^(-19)J) xx (6.022 xx 10^(23) mol^(-1)) = 19951 J mol^(-1)` `= 19.951 kJ mol^(-1)`

Discussion

32.	Calculate the energyin joulesrequired toconvert allthe atoms of sodiumtosodium ionspresentin 2.3mg or sodiumvapours : Ionizationenthalpy of sodiumis 495 kJ mol^(-1) (Atomic mass of Na =23)
Answer» SOLUTION :No of MOLESOF Na presentis 2.3 mg of Na = `(23)/(10) xx (1)/(1000) xx (1)/(23) =1 xx 10^(-4) ` mole `:.` REQUIRED energy=1 xx ` 10^(-4) xx 495 xx 10^(3)= 49.5 J`

Discussion

33.	Calculatethe energyof1 moleof photonwhosefrequencyis 5xx 10^(14) Hz
Answer» 19.951 *KJ 199.51 kJ 39.90 kj 399.0 kj Answer :C

Discussion

34.	Calculate the energy in joule corresponding to light of wavelength 45 nm (Planck's constant, h = 6.63 xx 10^(-34)Js, speed of light = 3 xx 10^(8) ms^(-1))
Answer» `6.67 XX 10^(15)` `6.67 xx 10^(11)` `4.42 xx 10^(-15)` `4.42 xx 10^(-18)` Solution :`E = hv = (HC)/(lamda)`. Given `lamda = 45 nm = 45 xx 10^(-9) m` `:. E = ((6.63 xx 10^(-34) Js) (3 xx 10^(8) ms^(-1)))/((45 xx 10^(-9) m))` `= 4.42 xx 10^(-18) J`

Discussion

35.	Calculate the energy in corresponding to light of wavelength 45 nm. (Planck’s constant h = 6.63 xx 10^(-34) JS, Speed of light C = 3 xx 10^(8)ms^(-1) )
Answer» `6.67xx10^(15)` `6.7xx10^(11)` `4.42xx10^(-15)` `4.42xx10^(-18)` SOLUTION :`E=(hc)/(LAMBDA)=(6.63xx10^(-34)xx3xx10^(8))/(45xx10^(-9))=4.4xx10^(-18)` J

Discussion

36.	Calculate the energy associated with the first orbit of He^+. What isthe radius of this orbit?
Answer» ANSWER :`-54.38 EV , 0.2645Å`

Discussion

37.	Calculate the energy associated with the first orbit of He^(+). What is the radius of this orbit ?
Answer» Solution :`E_(N) = - (2.18 XX 10^(-18) Z^(2))/(n^(2))J " atom"^(-1)` For `He^(+), n = 1, Z = 2 " " :. E_(1) = - ((2.18 xx 10^(--18) J) (2^(2)))/(12) = - 8.72 xx 10^(-18) J` Radius of H-like particles is given by `r_(n) = ((0.0529)n^(2))/(Z)nm` For `He^(+), n = 1, Z = 2 " " :. r_(1) = (0.0529 xx 1^(2))/(2^(2)) = 0.02645 nm`

Discussion

38.	Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom
Answer» Solution :`bar(v) = 109677 cm^(-1) ((1)/(2^(2)) - (1)/(3^(2))) = 109677 xx (5)/(36) = 15232.9 cm^(-1)` `Delta E = HV = h (c)/(LAMDA) = HC bar(v) = (6.626 xx 10^(-34) Js) (3.0 xx 10^(10) cm s^(-1)) (15239.9 cm^(-1)) = 3.028 xx 10^(-19) J` `v = c bar(v) = 3.0 xx 10^(10) cm s^(-1) xx 15232.9 cm^(-1) = 4.57 xx 10^(14) s^(-1)`

Discussion

39.	Calculate the E.N. of CI from the bond energy of CIF (61 K Cal//"mol"). Given that bond energies of F_(2) and CI_(2) are 38 and 58 Kcal//"mol" respectively. Given: Electrongativity of F = 4eV.
Answer» ANSWER :`[3.2]`

Discussion

40.	Calculate the empirical andmolecular formula of the compound containing 80% Carbon, 20%Hydrogen. If the molecular mass of the compound is 30 then determinethe molecular formula.
Answer» Solution :For `C rArr 80//12 = 6 . 6` For ` H rArr 20//1 = 20`now divide 6.6 and 20 by 6.6 to GET simple whole no. ratio of C and H which will come 1:3 so emperical FORMULAIS `CH_(3)`and its mass is 15 Nowto calculate N we have 30/15 = 2 somolecular formula is `CH_(3) xx 2 = C_(2) H_(6)`

Discussion

41.	Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47
Answer» Solution : Emperical formula=`C_(6(H_(6)O` VAPOUR DENSITY=47 `:.`Molecular mass= `2 xx` vapour density `=2xx47=94` Molecular formula= Empirical formula `xx n` `n=("molecular mass")/("Emperical frmular mass")=94/94=1` `:.` Molecular formula=`C_(6)H_(6)O`

Discussion

42.	Calculate the Effective nuclear charge of helium.
Answer» SOLUTION :`Z_("EFF")=Z-S` `Z_("eff")=2-0.30("for" 1se^(-)=0.30)` `Z_("eff")=1.70`

Discussion

43.	Calculate the effective nuclear charge of the last electron in an atom whose configuration is1s^(2)2s^(2)2p^(6)3s^(2)3p^(5)
Answer» Solution :Given:Z=17 `Z^()=Z-S` `=17-[(0.35xx"No"`.Of other electrons in `n^(TH)"SHELL")++(0.85xx"No"`.Of electrons in `(n-1)^(th)"shell")+(1.00xx"TOTAL number of electrons in the inner shells")]` `=17-[(0.35xx6)+(0.85xx8)+(1xx2)]` `=17-10.9=6.1` `Z^()=6.1`.

Discussion

44.	Calculate the effective molecular weight of air
Answer» Solution :Air is a homogenous MIXTURE. It has 79% NITROGEN and 21% OXYGEN by volume. If we take ONE mole of air, it WOULD contian 0.79 mole of nitrogen and 0.21 mole of oxygen. The weight of one mole of air `=0.79 xx 28+0.21 xx 32` =28.84 Effective molecular weight of air is 28.84

Discussion

45.	Calculate the distance of separation between the second and third orbits of hydrogen atom
Answer» SOLUTION :For H-atom, the radius of NTH ORBIT is given by : `r_(n) = 0.529 XX n^(2) Å` `:. r_(3) - r_(2) = 0.529 (3^(2) - 2^(2)) Å = 0.529 xx 5 = 2.645 Å`

Discussion

46.	Calculate the distance between (111) planes in a crystal of calcuim . Repeat the calculation for (222) planes . Which planes are closer? (a=0.556 nm)
Answer» SOLUTION :`d=a/(sqrt(h^2+k^2+l^2))` `d_111=0.556/(sqrt(1^2+1^2+1^2))=0.556/sqrt3`=0.321 nm `d_222=0.556/(sqrt(2^2+2^2+2^2))=0.556/sqrt12`=0.161 nm Thus (222) planes are CLOSER and their distance of separation is HALF of that of (111) planes

Discussion

47.	Calculate the dissociation constant of water at room temperature
Answer» Solution :DISSOCIATION of water is given as, `H_2 O hArr H^(+)_OH^(-)` Dissociation constant `K_a` is given as `K_a ([H^(+)][OH^(-)])/([H_2 O]) = (k_w )/( [H_2O]) =(1XX 10^(-14))/( 55.5 )` Dissociation constant of acid `= 1.8 xx10^(-16)mol L^(-1)`

Discussion

48.	Calculate the difference between DeltaE and DeltaH values for the combustion reaction of ethylene at 300K
Answer» ANSWER :4.99 KJ

Discussion

49.	Calculate the density of SO_2 " at " 27^@Cand 1.5 atm pressure.
Answer» SOLUTION :`3.9 G L^(_1)`

Discussion

50.	Calculate the density of silver which crystallizes in a face-centred cubic structure. The distance between the nearest silver atoms in this structure is 287 pm.(Molar mass of Ag =107.87" g mol"^(-1), N_A=6.02xx10^23 "mol"^(-1))
Answer» Solution :For FCC, `d=a/sqrt2` or a=`sqrt2`d=1414 x 287 pm = 406 pm , Z=4, M=`107. 87 "G mol"^(-1)` `RHO=(ZxxM)/(a^3xxN_0)=(4xx107.87 "g mol"^(-1))/((406xx10^(-10) CM)^3xx(6.02xx10^23mol^(-1)))=10.71 g cm^(-3)`

Discussion

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