52671 + Interview Questions in Chemistry in Class 11

1.	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) = 6.02 xx 10^(23) mol^(-1)
Answer» Solution :`d=a/sqrt2 or a = sqrt2d = 1.414 xx 287 " pm " 406 " pm " , Z = 4 , M = 107 . 87 " G mol"^(-1)` ` p = ( Z xx M)/ (a^(3) xx N_(0))= ( 4 xx 63.5 " g mol" ^(-1))/ ((406 xx 10^(-10) "cm")^(3) xx (6.02 xx 10^(23) "mol"^(-1))) = 10.7o1 " g cm"^(-3)`

Discussion

2.	Calculate the density of N_2 gas at S.T.P ?
Answer» 1.250 g/L 0.628 g/L 2.450 g/L 1.42 g/L SOLUTION :As we KNOW at S.T.P. DENSITY of GAS (d) is `d (G.M.W)/(G.M.V ) = (28)/(22.4) = 1.25` g/L

Discussion

3.	Calculate the density of carbon dioxide at 97° C and 760 mm of Hg of pressure.
Answer» ANSWER :1.452 g/l

Discussion

4.	Calculate the DeltaH_("vaporization") [CH_(3)COOH(l)] in kJ//mol. Given data :
Answer» ANSWER :21

Discussion

5.	Calculate the density of carbon dioxide at 27^@C and 5 atmosphere pressure.
Answer» Solution :The molecular MASS of `CO_2 = 12 + (2 xx 16) = 44` amu In the present case, `P=5 " atm, " T = 27 + 273 = 300 K`, `M = 44, " and " R = 0.0821 " litre atm " k^(-1) mol^(-1)` `:. "" M=(dRT)/р` or`d=(MP)/(RT) = (44 xx 5)/(0.0821 xx 300) = 8.93 g L^(-1)` Hence, the density of `CO_2` in the given conditions is 8.93 grams per litre.

Discussion

6.	Calculate the Delta H for the isothermal reversible expansion of 1 mole of an ideal gas from initial pressure of 1.0 bar to a final pressure 0.1 bar at a constant temperature at 273K.
Answer» SOLUTION :Constant `T RARR DELTA H = 0`

Discussion

7.	Calculate the Delta H at 298 K for the reaction (1)/(2)N_(2)(g) + (3)/(2)H_(2)(g) rarr NH_(3)(g) given that Delta H for the formation of NH_(3) has a valve of -46.0 kJ mol^(-1) (R = 8.314 JK^(-1) mol^(-1)).
Answer» Solution :Using the eqn. `Delta H = Delta U + Delta nRT` `Delta H = -46.0 kJ MOL^(-1) = -46000 J mol^(-1), Delta N = 1 - 2 = -1, R = 8.31 JK^(-1) mol^(-1)`, `T = 298 K ""Delta H = ?` `Delta U = Delta H - Delta nRT` `= -46000 J mol^(-1) -[(-1) xx 8.314 JK^(-1) mol^(-1) xx 298 K]` `= 43523 J mol^(-1) = -43.52 kJ mol^(-1)` Change in the internal energy for this reaction `= -43.52 kJ mol^(-1)`.

Discussion

8.	Calculate the degree of ionization of 0.05 M acetic acid if its pK_(a) value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M HCl ?
Answer» Solution :`pK_(a) = 4.74, i.e., - log K_(a) = 4.74 or log K_(a) = - 4.74 = BAR(5).26 :. K_(a) = 1.82xx10^(-5)` `alpha=SQRT(K_(a)//C)=sqrt((1.82xx10^(-5))//(5xx10^(-2)))=1.908 xx 10^(-2)` In presence of HCl, DUE to high concentration of `H^(+)` ion, dissociation equilibrium will shift backward, i.e., dissociation of aceticacidwill decrease . (a) In presence of 0.01 MHCL, if x is the amount dissociated, then `{:(,CH_(3)CO OH ,hArr,CH_(3)CO O^(-) ,+,H^(+),,,),("Initial conc.",0.05 M,,,,,,,),("After disso.",0.05 - x ,,x,,0.01 + x,,,),(,~=0.05,,,,,~=0.01M,,),(,,,,,(0.01 M H^(+) "ions are obtained from 0.01 M HCl"),,,):}` `:. K_(a) = (x(0.01))/(0.05) or (x)/(0.05)=(K_(a))/(0.01)=(1.82xx10^(-5))/(10^(-2))=1.82xx10^(-3) . "But" (x)/(0.05)=alpha` Hence, `alpha=1.82xx10^(-3)`(`:' alpha = ("Amount dissociated")/("Amount taken")`) (b) In the presence of 0.1 M HCl, if y is the amount of acetic ACID dissociated, then at equilibrium, `[CH_(3)CO OH]=0.05-y~=0.05 M` `[CH_(3)CO O^(-)]=y, [H^(+)]=0.1M+y~=0.1M` `K_(a)=(y(0.1))/(0.05) or (y)/(0.05)=(K_(a))/(0.1)=(1.82xx106(-5))/(10^(-1))=1.82xx10^(-4), i.e., alpha = 1.82xx10^(-4)`

Discussion

9.	Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74 . How is thedegree of dissociation affected when its solutionalso contains(a)0.01 M , (b) 0.1 M in HCl ?
Answer» SOLUTION :(i)Calculation of `K_a` on base of `pK_a` : `pK_a=-log (K_a)=-4.74 = bar5.26` So, `K_a=1.8197xx10^(-5)` `THEREFORE K_a= 10.0^(-4.74)` `=1.8197xx10^(-5)approx 1.82xx10^(-5)` (ii)Calculation of `K_a` on base concentration of `H^+` : `{:(,CH_3COOH_((aq))+H_2O_((l)) hArr, CH_3COO_((aq))^(-)+ , H_3O_((l))^(+)),("Inital M :" , 0.05 M, 0.0 M , 0.0 M),("At equilibrium M ", underset(approx 0.05 M) (0.05-x),CALPHA=xM, x M=Calpha):}` `therefore K_a=([CH_3COO^-][H^+])/([CH_3COOH])= 1.82xx10^(-5)` `therefore ((x)(x))/0.05 = 1.82xx10^(-5)` `therefore x^2=0.05 xx 1.82 xx 10^(-5) = 0.91 xx 10^(-6)` `therefore x=sqrt(0.91xx10^(-6))= 0.95396xx 10^(-3) M = [H^+]` (iii) Calculation of degree of ionization : At equilibrium `[H^+]= alpha C = 0.05 alpha =x` `therefore [H^+]=0.9539xx10^(-3)M =Calpha` `therefore alpha=([H^+])/C=(0.9539xx10^(-3))/0.05` `=1.9078xx10^(-2)` =0.019078 `approx` 0.0190 = Dissociation degree (iv) Degree of ionization in 0.01 M HCl : `H^+` is formed from HCl So, initial `[H^+]` = 0.01 M `{:(,CH_3COOH_((l)) hArr, H_((aq))^(+)+,CH_3COO_((aq))^(-)),("Initial M:",0.05,0.01 M,0.0),("Change M:",-Calpha,+Calpha,+Calpha),("At equilibrium M",(0.05-0.05alpha),0.01+Calpha,+Calpha),(,=0.05(1-alpha),(0.01+0.05alpha),0.05 alpha),(,approx 0.05,approx 0.01 M,0.05 alpha):}` `K_a=([H^+][CH_3COO^-])/([CH_3COOH])=1.82xx10^(-5)` `therefore ((0.05alpha)(0.01))/(0.05)=1.82xx10^(-5)` `therefore alpha=1.82xx10^(-3)`=0.00182 (v)Calculation of degree of ionization `alpha` of 0.01 M HCl : `((0.005alpha)(0.1))/(0.05)=1.82xx10^(-5) therefore alpha = 1.82xx10^(-4)`

Discussion

10.	Calculate the degree of ionization and pH of 0.05 M ammonia solution. The ionization constant of ammonia (K_(b)) is 1.77xx10^(-5). Also calculate the ionic constant of the conjugate acid of ammonia .
Answer» Solution :`{:(,NH_(3),+,H_(2)O,hArr,NH_(4)^(+),+,OH^(-)),("Initial conc.",0.05M,,,,,,),("Eqm. conc. ",0.05(1 -alpha),,,,0.05 alpha,,0.05 alpha):}` `K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])` `:. 1.77xx10^(-5)=((0.5 alpha)^(2))/(0.05(1-alpha))~=0.05 alpha^(2) or alpha^(2) = (1.77xx10^(-5))/(0.05) = 3.54xx10^(-4) or alpha = 1.88xx10^(-2)` Alternatively, applydirectly theformula : `alpha =sqrt(K_(b)//C)` (II) `[OH^(-)]=0.05 alpha = 0.05 xx 1.88xx10^(-2) = 9.4 xx 10^(-4) M` `pOH = log (9.04xx10^(-4)) =4-0.9562=3.04 :. pH = 14-3.04 = 10.96` (iii) For conjugate acid BASE pair, `K_(a) = (K_(w))/(K_(b))=(10^(-14))/(1.77xx10^(-5))=5.64xx10^(-10)`

Discussion

11.	Calculate the degree of ionization and [H_(3)O^(+)] of 0.01 M CH_(3)CO OH solution. The equilibrium constantof acetic acid is 1.8 xx 10^(-5)
Answer» ANSWER :`4.24xx10^(-2),4.24xx10^(-4) "MOL"L^(-1)`

Discussion

12.	Calculate the degree of dissociation ofHI at 450^(2)Cif the equilibrium constant for the dissociation reaction is 0.263
Answer» Solution :SUPPOSE we start with 1 mole of HI and x is the degree of dissociation. Then at EQUILIBRIUM, `[HI] = (1-x) //V, [H_(2)]= x//2 V and [I_(2)] = x//2" moles per litre".` Put the values in the equation, `K_(c) = ([H_(2)][I_(2)])/([HI]^(2)` and CALCULATE x.

Discussion

13.	Calculate the degree of dissociation and concentration of H_(3)O^(+) ions in 0.01 M solution of formic acid. K_(a)=2.1xx10^(-4) at 298 K.
Answer» Solution :Formic acid is weakelectrolyte and ionizes inwaterto give `H_(3)O^(+)` ions ACCORDING to the equation : `HCO OH+H_(2)O HARR H_(3)O^(+)+HC O O^(-)` Let `alpha` be the degree of ionization. Then the concentration of the various species present at equilibrium wouldbe as under : `{:(,HCO OH + H_(2)O hArrH_(3)O^(+) +HCO O^(-)),("Initial conc. ",0.01"00"),("Conc. at eqm",0.01(1-alpha)~=0.01""0.01 alpha ""0.01alpha):}` [`:' alpha` is very small and can be neglected in comparison to 1] THUS, `K_(a)=(0.01 alpha xx 0.01 alpha)/(0.01)=0.01 xx alpha^(2) :. 0.01 alpha^(2)=2.1xx10^(-4)[K_(a)=2.1xx10^(-4), "given"]` or `alpha^(2)=(2.1xx10^(-4))/(0.01)=2.1xx10^(-2)` `:.` Degree of ionisation, `alpha = sqrt(2.1xx10^(-2))=0.14` Concentration of `H_(3)O^(+)` ions `=C alpha = 0.14 xx 0.01 = 1.4 xx 10^(-3) "mol" L^(-1)`

Discussion

14.	Calculate the degree of dissociation of 0.5 M NH_(3) at 25^(@)C in a solution of pH = 12.
Answer» Solution :`{:(,NH_(4)OH,hArr,NH_(4)^(+),+,OH^(-),,),("Initial conc.",C "MOL " L^(-1),,0,,0,,),("After disso.",C-Calpha ,,C alpha,,C alpha,,):}` pH = 12 means `[H^(+)]=10^(-12) or [OH^(-)]=10^(-2)` `:. [OH^(+)]=C alpha = 10^(-2) or alpha = (10^(-2))/(C) = (10^(-2))/(0.5) = 2 xx 10^(-2) or 2 % `.

Discussion

15.	Calculate the degree of dissociation and K_(P) for the following reaction. PCl_(5)(g)iffPCl_(3)(g)+ Cl_(2)(g){:(t=0,a,0,0),(t=t,a-x,x,x):} Since for a mole, x moles are dissociated
Answer» Solution :`THEREFORE"For"1 "MOLE", (x)/(a)"moles"=ALPHA` are DISSOCIATED `therefore x=a alpha`

Discussion

16.	Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1kV.
Answer» Solution :Energy acquired by the electron (as kinetic energy) after being aacelerated by a POTENTIAL DIFFERENCE of 1 kV. (i.e. 1000 volts) `= 1000 eV = 1000 XX 1.602 xx 10^(-19) J = 1.602 xx 10^(-16) J (1 eV = 1.602 xx 10^(-19)J)` i.e., Kinetic energy, `(1)/(2) mv^(2) = 1.602 xx 10^(-16) J or (1)/(2) xx 9.1 xx 10^(-31) v^(2)` or `v^(2) = 3.521 xx 10^(14) or v = 1.88 xx 10^(7) ms^(-1)` `:. lamda = (h)/(mv) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/((9.1 xx 10^(-31) kg) xx (1.88 xx 10^(7) ms^(-1))) = 3.87 xx 10^(-11) m`

Discussion

17.	Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 keV.
Answer» Solution :`gamma =h/(mv)` Potendial difference of an ELECTRON=V=1keV. `THEREFORE "Potential energy "=1/2mv^(2)=eV` e=charge of an electron=`1.609xx10^(-19)c` `kv=1000V` `therefore "Potential energy"=1.609xx10^(-19)xx1000` `=1.609xx10^(-16)` `1/2mv^(2)=1.609xx10^(16)V``m=9.1xx10^(-31)kg` `gamma=h/(mv)` `v^(2)=(2xx1.609xx10^(-16))/(9.1xx10^(-31))` `v=sqrt(2xx1.609xx10^(-16))/(9.6xx10^(-31)` =`5.9xx10^(7)ms^(-1)` `gamma=h/(mv)=(6.626xx10^(-34))/(9.1xx10^(-31)xx5.93xx10^(7))` `=1.2xx10^(11)m` `gamma=1.2xx10^(-11)m`

Discussion

18.	Calculate the de Broglie wavelength of the electron in third orbit of hydrogen atom
Answer» Solution :`r_(n) = r_(0) xx n^(2)` `:. r_(3) = 3^(2)xx 0.529 Å = 4.761 Å` By Bohr's postulate of angular MOMENTUM `mv R = (n h)/(2pi)` `:. mv = (3H)/(2pi r)` `lamda= (h)/(mv) = (h)/(3h//2pir) = (2pi r)/(3) = (2(3.143) (4.761 Å))/(3) = 9.975 Å`

Discussion

19.	Calculate the De-Broglie wavelength of a particle whose momentum is 66.26 xx 10^(-28) kg ms^(-1).
Answer» SOLUTION :`lamda=h/p=(6.626xx10^(-34))/(6.626xx10^(-28))=1XX10^(-7)m`

Discussion

20.	Calculate the de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 kV.
Answer» Solution :Energy acquired = `1000 eV` `"" = 1000 xx 1.602 xx 10^(-19)` `= 1.602 xx 10^(-16) J` `1/2 mv^2 = 1.602 xx 10^(-16) J` `1/2 xx 9.1 xx 10^(-31) V^2 = 1.602 xx 10^(-16) J` `v^2 = (2 xx 1.602 xx 10^(-16))/(9.1 xx 10^(-31)) = 3.521 xx 10^(14)` `v = 1.876 xx 10^7 m s^(-1)` Now, `lambda = h/(mv)` `= (6.626 xx 10^(-34) kg m^2 s^(-1))/((9.1 xx 10^(-31) kg) xx (1.876 xx 10^(7) ms^(-1)))` `= 3.88 xx 10^(-11) m`

Discussion

21.	Calculate the de Broglie wavelength of an electron moving with 1% of the speed of light?
Answer» Solution :Accoding to DE de BROGLIE equation,`gamma=(H)/(mv)` MASSOF electron`=9.1xx10^(-31)kg`,Planck.s constant`=6.626xx10^(-34)kgm^(2)s^(-1)` Velocity of electron=1% speed of light`=3.0xx10^(8)xx0.01=3xx10^(6)ms^(-1)`Wavelength of electron`(gamma)=h/(mv)=((6.626xx10^(-34)kgm^(2)s^(-1)))/((9.1xx10^(-31)kg)xx(3xx10^(6)ms^(-1)))` `=2.43xx10^(-10)m`

Discussion

22.	Calculate the de Broglie wavelength associated with a helium atom in a helium gas sample at 27^@C and 1 atm pressure.
Answer» ANSWER :`7.3xx10^(-11)` METRE.

Discussion

23.	Calculate the concentrations of dichloroacetate and proton in a solution that is 0.01 M HCl and 0.01 M in CHCI_2 COOH. K_a for CHCI_2 COOH is 5 xx 10^(-2)
Answer» SOLUTION :`7.4 XX 10^(-3) M and1.74 xx 10^(-2) M`

Discussion

24.	Calculate the concentration of sugar (C_12H_22O_11) in mol L^-1 if its 20g are dissolved in water to make a final volume of 2L.
Answer» Solution :Molar mass`C_12H_22O_11=12+12+22+16+11=342g MOL^(-1)` VOLUME of solution =2L=2000 mL Molanty `W_B/M_Bxx1000/(VML)xx20/342xx1000/2000=0.0292molL^(-1)`

Discussion

25.	Calculate the concentration of silver ion, if water is saturated with AgCNS (K_(sp)=1 xx 10^(-12)) and AgBr (K_(sp)= 5 xx 10^(-13)).
Answer» SOLUTION :`8 XX 10^(-7) M`

Discussion

26.	Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL^(–1) and the mass per cent of nitric acid in it being 69%.
Answer» Solution :The given sample is 69%, i.e., 100 G of the solution contain only 69 g of nitric acid. Molar mass of nitric acid (`HNO_3) = 1 + 14 + (3 xx 16) = 63 g mol^(-1)` `therefore` Number of moles in 69 g nitric acid (100 g of the solution) `=69/63 =1.095` Volume of 100 g of `HNO_(3)` solution `=("mass")/("density") = 100/1.41` =70.92 ML = 0.07092 L HENCE, concentration of nitric acid solution = `1.095/(0.07092) = 15.44` moles/L =15.44 M

Discussion

27.	Calculate the concentration of nitric acid in moles per litre in a sample which has a density , 1.41 g mL^(-1) and the mass per cent of nitric acid in it being 69%.
Answer» Solution :Density `=1.41 gm//mL` Mass precent of `HNO_(3)= 69%` Here, `69 gm HNO_(3)` present in 100 gm solution MOLECULAR mass of `HNO_(3) = 1.0079+14.0067+3xx(16.00)` `= 63.0146` gm/mol MOLARITY `= ("WEIGHT of solute " xx 1000)/("Molecular massof solute" xx "Volume of solution")` `= (69xx1000xx1.41)/(63.0146xx100)=15.439 M`

Discussion

28.	Calculate the concentration of hypo (Na_(2)S_(2)O_(3)5 H_(2)O)solution in g dm^(-3) uf 10.0 ML of this solution decolourised 15 mL of (M)/(40) iodine solution
Answer» SOLUTION :The balanced equation for the REDOX reaction is `2S_(2)O_(3)^(2-)+I_(2) rarr2 I^(-)+S_(4)O_(6)^(2-)` From the balanced equation it is evident that 2 moles of `Na_(2)S_(2)O_(3)=1 "mole of" I_(2)` applying molarity equation we have `(M_(1)V_(1))/(n_(1))(Na_(2)S_(2)O_(3))=(M_(2)V_(2))/(n_(2))(I_(2)) or (M_(1)xx10)/(2)=(15xx1)/(1xx40) or M_(1)=(15xx1xx2)/(10xx40)=3/40` THUS the molarity of hte hypo solution `=3//40` M Mol MASS of `Na_(2)S_(2)O_(3)5 H_(2)O=248 g "mol"^(-1)` `therefore` Concentration of `Na_(2)S_(2)O_(3)5H_(2)O=(248xx3)/(40)=18.6 dx^(-3)`

Discussion

29.	Calculate the concentration of H_(3)O^(+) ions in a mixture of 0.02 M acetic acidand0.2 M sodiumacetate. Given that the ionization constant (K_(a)) foracetic acid is 1.8xx10^(-5).
Answer» Solution :`{:(,CH_(3)CO OH +H_(2)O,hArr,CH_(3)CO O^(-) +, H_(3)O^(+),),("Initial moles",0.02,,0,0,),("Moles at eqm.",0.02-x,,x,x,),(,CH_(3)CO ONa,rarr,0,0,),("Initial moles",0.2,,0,0,),("Moles at eqm.",0,,0.2,0.2,),(,,,,,):}` THUS, in the mixture solution, `[CH_(3)CO O^(-)]=0.2+x~=0.2 M (CH_(3)CO O^(-)" are obtained mainly from" CH_(3)CO ONa," therefore," x ltlt 0.2)` `[CH_(3)CO OH ] = 0.02 - x ~= 0.02 M` `K_(a)=([CH_(3)CO O^(-)][H_(3)O^(+)])/([CH_(3)CO OH]), i.e., 1.8xx10^(-5)=(0.2xx[H_(3)O^(+)])/(0.02)` or`[H_(3)O^(+)]=(1.8xx10^(-5)xx0.02)/(0.2) = 1.8xx10^(-6)M` Note. The GIVEN mixture is a buffer as discussed LATER inArt. 7.32 .

Discussion

30.	Calculate the concentration of H^(+) (aq) in 0.2 M solution of HCN. Given that the dissociation constant of HCN in water is 4.9 xx 10^(-10).
Answer» ANSWER :`9.9xx10^(-6)M`

Discussion

31.	Calculate the concentration of free silver ion [Ag^(+) ] in an aqueous solution that is prepared as 0.1 M AgNO_3 and 3.0 M NH_3 ,Ag^(+)(aq) +2 NH_3 (aq) hArr [Ag(NH_3) _2]^(+)(aq) K_(f) = 1.6 xx 10^(7). If answer isx xx 10^(-10)Mthen x= ______?
Answer» Solution :` Ag^(+) +2NH_3 hArr 2NH_3 hArr [Ag(NH_3]_2]^(+) ` ` ([Ag(NH_3)_2]^(+) )/([Ag^(+) ][NH_3]^(2) ) = K_f` ` (0.1)/( [Ag^(+)](3-0.2)^(2)) = 1.6 xx 10^(7) ` ` [Ag^(+) ]= (0.1)/(1.6 xx 10^(-7) xx (2.8) ^(2)) ~~8 xx 10 ^(-10) `

Discussion

32.	Calculate the concentration in gram/litres of a 20 volume H_(2)O_(2) solution.
Answer» SOLUTION :60.7 g/l

Discussion

33.	Calculate the compressibility factor for CO_2 if one mole of it occupies 0.4 litre at 300 K and 40 atm. Comment on the result.
Answer» Solution :`0.65, " SINCE "Z lt 1, CO_2` is more compressible than IDEAL GAS

Discussion

34.	Calculate the compressibiity factor for CO_(2), if one mole of it occupies 0.4 litre at 300K and 40 atm. Comment on the result.
Answer» 0.40, `CO_(2)` is more COMPRESSIBLE than ideal GAS 0.65,`CO_(2)` is more compressible than ideal gas 0.55,`CO_(2)` is more compressible than ideal gas 0.62,`CO_(2)` is more compressible than ideal gas Answer :b

Discussion

35.	Calculate the circumference of Bohr's L shell.
Answer» SOLUTION :`"CIRCUMFERENCE "= 2pir = nlambda` `"WAVELENGTH "lambda=3.33nÅ=3.33xx2Å` `2pir=2lambda=13.32Å`

Discussion

36.	Calculate the charge of one electron?
Answer» SOLUTION :CHARGE of `6.022 xx 10^(23)` electrons=96,500 COUL. Charge of one electron `=(96500)/(6.022) xx 10^(-23)` `=1.602 xx 10^(-19)coul`.

Discussion

37.	Calculate the change in pH of water when 0.01 mole of NaOH are added in 10 litre water.
Answer» Solution :PH of water is 7 `M= ("no . of moles ")/(" sol . of solution (lit) ") =( 0.01)/(10 ) = 10 ^(_3) ` ` pOH=3, pH =11 , "change in pH"= 11- 7= 4`

Discussion

38.	Calculate the C-C bond energy from the following data : (i) 2 C ( graphite )+3H_(2)(g) rarr C_(2)H_(6)(g), DeltaH =- 84.67 kJ (ii) C ( graphite ) rarr C(g), Delta H = 716 .7 kJ (iii) H_(2)(g) rarr 2H(g), Delta H = 435.9 kJ Assume the C-H bond energy as 416 kJ
Answer» Solution :Let us calculate `DeltaH` for the reaction,`C_(2)H_(6)(g),i.e., H - underset(H) underset(\|) overset(H) overset(\|) (C)-underset(H) underset(\|) overset(H) overset(\|) (C) -H(g) RARR 2C(g) +6H(g)` ` 2 xx `EQN. (ii) `+3 xx ` Eqn. (iii) -Eqn. (i) GIVES `Delta _(r)H= 2825.77 kJ` Ifx is the bond energy of C-C bond,then `x+ 6 B.E. ( C-H)= 2825.77` `x +6 ( 416) = 2825.77` or `x=329.77kJ`

Discussion

39.	Calculate the bond order ofN_(2) , O_(2), O_(2)^(+) and O_(2)^(-)
Answer» Solution :(A) BOND order of `N_(2)` : In `N_(2) ` Z = 7 , so total electron = 14 Electron configuration of in MO for `N_(2) :(sigma 1s)^(2) (sigma^()1s)^(2) (sigma 2s)^(2) (PI 2p_(x))^(2) (pi 2p_(y))^(2) (sigma 2p_(z))^(2)` BO = `(1)/(2) (N_(b) - N_(a)) = (1)/(2) (10 - 4) = 3` (Triple bond ) (B) Bond order of `O_(2)` : In `O_(2), Z = 8` So, Total electron = 16 Electron configuration in MO for `O_(2) : (sigma 1s)^(2) (sigma^() 1s)^(2) (sigma 2s)^(2) (sigma^() 2s)^(2) (sigma 2p_(x))^(2) (pi 2p_(x))^(2) (pi 2p_(y))^(2)(pi^() 2p_(x))^(1) (pi^() 2p_(y))^(1)` BO = `(1)/(2) (N_(b) - N_(a)) = (1)/(2) (10 - 6) =` (Double bond ) (C) Bond order of `O_(2)^(+)`: ,br> Total electron in `O_(2) = 16 ` and total electron in Electron configuration in MO for `O_(2) : (sigma_(1s))^(2) (sigma_(1s)^() )^(2) (sigma_(2s))^(2) (sigma_(2s)^())^(2) (sigma_(2p_(z)))^(2) (pi_(2p_(x)))^(2) (pi_(2pi_(y)))^(2)(pi_(2pi_(x))^())^(1)(pi_(2p_(y))^())^(0)` BO = `(1)/(2) (N_(b) - N_(a)) = (1)/(2) (10- 5) = 2.5 ` (D) Bond order of `O_(2)^(-)` : Total electron in `O_(2)^(-) = 16 + 1 = 17` Electron configuration in MO for`O_(2)^(-) : (sigma_(1s))^(2) (sigma_(1s)^())^(2) (sigma_(2s))^(2) (sigma_(2s)^())^(2) (sigma_(2p_(z)))^(2) (pi _(2p_(x)))^(2) (pi_(2p_(y)))^(2) (pi_(2p_(x))^())^(2) (pi_(2p_(y))^(**))^(1) ` BO = `(1)/(2)(N_(b) - N_(a))` `= (1)/(2) (10 - 7) = (3)/(2) = 1.5 `

Discussion

40.	Calculate the bond order of Helium.
Answer» Solution :HELIUM contains four electrons. Its ELECTRONIC configuration is `H^(+)_(2)` Bond ORDER `=(1)/(2)(n_(b)-n_(a))` `n_(b)=` number of anti bonding electrons `(1)/(2)(2-2)=0` So, `He_(2)`, dons not EXIST.

Discussion

41.	Calculate the bond order of He_(2)^(+) molecular ion.
Answer» SOLUTION :`O_(2)^(+)`

Discussion

42.	Calculate the bond order and predict the magnetic nature of the following molecules.(i) H_(2)^(+)""(ii) Li_(2)^(-)""(iii) Li_(2)""(iv) B_(2)""(v) C_(2)^(+)
Answer» SOLUTION :

Discussion

43.	Calculate the bond order of H_(2)" and "B_(2)
Answer» SOLUTION :

Discussion

44.	Calculate the bond enthalpy of OH bond in water.
Answer» Solution :(i) In the case of polyatomic molecules with two or more same BOND types. the arithmetic mean of the bond energy value of the same TYPE of bonds is considered as average bond ENTHALPY. (ii) For e.g., in WATER, there are two OH bonds present and the energy needed to BREAK them are not same. (iii) `H_(2)O (g) rarr H(g) +OH""DeltaH_(1)= 502 " kJ mol"^(-1)` `OH (g) rarr H (g) + O(g) "" DeltaH = 427 " kJ mol"` The average bond enthalpy of OH bond in water `=(502+427)/(2)=464.5 " kj mol"^(-1)`

Discussion

45.	Calculate the bond energy of C-H bond,given that thee heat of formation ofCH_(4),heat of sublimation of carbon and heat of dissociation of H_(2) are-74.8, + 719.6 and 435.4 kJ mol^(-1) respectively.
Answer» SOLUTION :Here, we are GIVEN `C(s) +2H_(2)(g)rarr CH_(4) (g), Delta_(f)H^(@)= - 74.8 KJ`....(i) `C(s)rarr C(g), Delta_(r) H^(@) = +719.6kJ `....(ii) `H_(2)(g) rarr 2H(g), Delta_(r) H^(@) = + 435 .4 kJ`...(iii) We aim at `:` Eqn. (ii) `+ 2 xx`Eqn. (iii) - Eqn (i) gives `C(s) +2H_(2)(g) rarr C(g) + 4H(g)` `- C(s) -2H_(2)(g)rarr - CH_(4) (g) ` `0 =C(g)+4H(g)-CH_(4)(g), Delta_(r) H^(@)= 719.6+2 (435.4)- (- 74.8)` or ` CH_(4)(g) rarr C(g) + 4H(g), DELTA H = + 1665.2 kJ` This gives the enthalpy of dissociation of four moles of C-H bonds ( called enthalpy of atomization) . Hence, bond energy for C-H bond (average value ) i.e., `Delta_(C-H) H^(@) = (1665.2)/(4)=416.3 kJ mol^(_1)`

Discussion

46.	Calculate the bond enthalpyof HCl. Given that the bond enthalpies of H_(2) and Cl_(2) are 430 kJ mol^(-1) and242 kJ mol^(-1) respectively andDelta_(f)H^(@) for Hclis -91kJ mol^(-1)
Answer» SOLUTION :By using the relation. `Detla_(f)H^(@) = SigmaDelta_(f)H^(@)` ( Products) ` - Sigma Delta_(f)H^(@)`( Reactants) Here , we are given `H_(2) (G) rarr2H(g),Delta _(b) H^(@) = + 430 kJ mol^(-1)`....(i) `Cl_(2)(g) rarr 2CL(g) , Delta _(b) H^(@) = + 242kJ mol^(-1)`.....(ii) We aim at ` HCl(g) rarr H(g() + Cl(g) ,Delta_(b) H^(@) = ?`.....(iii) Evidently, for reaction (iii) `DeltaH = Sigma Delta_(f) H^(@) (`Poroduct) `- Sigma Delta_(f) H^(@) ` ( Reactants )` =[ Delta _(f) H^(@) (H)+ Delta_(f) H^(@) ( Cl) ] - [ Delta _(f) H^(@) (HCl)] ` ....(iv) From equation (i) and(ii) , `Delta H = Sigma( + 430 kJ ) = 215 kJ mol^(_1) , Delta _(f) H^(@) (Cl) = (1)/(2) (+242kJ ) = +121 kJmol^(-1)` Also ,we are given`Delta_(f) H^(@)(HCl) = -91 kJ mol^(-1)` Putting these value in eqn. (iv) , we GET`Delta _(H-Cl) H^(@) = [ + 215+121] - [-91] = 427kJ mol^(-1)` Secondmethod , By using Hess's law We are given `:(i) H_(2)(g) rarr 2H(g) , Delta_(r)H^(@)= + 430 kJ mol^(-1)` `(ii) Cl_(2)(g) rarr 2Cl(g), Delta_(r) H^(@) = + 242 kJ mol^(-1)` `(iii) (1)/(2) H_(2)(g) +(1)/(2) Cl_(2)(g) rarr HCl(g), Delta_(r)H^(@) = -91 kJ mol^(-1)` We aim at`HCl(g) rarr H(g) + Cl(g), Delta_(r) H^(@)=?` `(1)/(2) xx` eqn. (i)` + (1)/(2) xx `eqn. (ii) - eqn. (iii) GIVES the requiredresult. Third Method, By applying the relation `Delta_(r)H^(@) = Sigma` Bond Enthalpies of Reactants `- Sigma` Bond Enthalpies of Products For the formation of HCl, `(1)/(2) H_(2)(g)+(1)/(2) Cl_(2)(g) rarr HCl(g) , Delta_(r) H^(@) = Delta_(f) H^(@)` `:. Delta_(r) H^(@) = Sigma ` B.E. ( Reactants) `- Sigma ` B.E. ( Products ) `=(1)/(2) Delta_(H-H) H^(@) +(1)/(2) Delta _(Cl- Cl) H^(@) - Delta _(H-Cl) H^(@)` `- 91 = (1)/(2) xx 430 + (1)/(2) xx 242 - Delta_(H- Cl) H^(@)` `:. Delta _(H-Cl) H^(@) = 215+ 121 + 91 =427 kJ mol^(-1)`

Discussion

47.	Calculate the boiling point of a solution which is prepared by dissolving 68.4g of solute B in one kilogram of water . (Molar mass of solute B is 342 gmol^(-1) T_b=373.1 K and K_b (water ) = 0.52 Kkg mol ^(-1)
Answer» SOLUTION :` W_1 "" 1KG ` ` W_2 ""= 68.4g` `M_2 "" =342 g MOL^(-1) ` ` T_b "" =373.1K` ` K_b "" = 0.52K.kg mol ^(-1) ` ` T-T_b "" = K_b m ` `m ""= 0.2` ` T-T^(@) "" K_bxxm ` ` T-T^(@) =0.52 XX 0.2` ` T-T^(@) =0.104` ` T-373.1 =0.101` `T ="" 0.101+373.1` ` T""= 373.204`

Discussion

48.	Calculate the Avogadro's number from the following data of AB when AB has NaCl type structure : Density of AB =2.48 g cm^(-3), M=58 Distance between A^+ and B^- in AB =269 pm
Answer» ANSWER :`6.0xx10^23`

Discussion

49.	Calculate the average volume available to a molecule in a sample of oxygen gas at S.T.P. Also calculate the average distance between neighbouring molecules if a oxygen molecule is assumed to be spherical.
Answer» SOLUTION :`3.72 xx 10^(-20) CM^3, 4.14 xx 10^(-7) cm`

Discussion

50.	Calculate the average velocity of oxygen molecule at 27^@C.
Answer» Solution :The average velocity `bar(u)` is given by `bar(u) =sqrt((8RT)/(pi M))` In the present case, `T = 27^@C = 27 + 273 = 300 K, M = 32` and `R= 8.31 xx 10^7 " ERGS " K^(-1) mol^(-1)` `:. "" bar(u)=sqrt((8xx8.31xx10^7 xx 300)/2)=1.845xx10^5 " CM " s^(-1)` HENCE, the average velocity of oxygen molecule at `27^@C " is " 4.46 xx 10^4 " cm " s^(-1)`

Discussion

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