52671 + Interview Questions in Chemistry in Class 11

1.	Calculate the standard enthalpy of formation of CH_(3) OH_((l)) from the following data : CH_(3) OH_((l)) + (3)/(2) O_(2(g)) to CO_(2(g)) + 2H_(2) O_((l)) , Delta_(r) H^( Theta ) = -726 "kJ mol"^(-1) C_("(graphite)") +O_(2(g)) to CO_(2(g)) , Delta_(r) H^( Theta ) = -393 "kJ mol"^(-1) H_(2(g)) + (1)/(2) O_((g)) to H_(2) O_((l)) , Delta_(f) H^( Theta ) = -286 "kJ mol"^(-1)
Answer» Solution :`CH_(3) OH_((l)) + (3)/(2) O_(2(G)) to CO_(2(g)) + 2H_(2) O_((l)) Delta_(r) H^( Theta ) = - 726 "kJ mol"^(-1)` `Delta_(r) H^( Theta ) = [2 Delta_(r) H^( Theta ) H_(2) O + Delta_(r) H CO_(2) ] - [ Delta_(r) H^( Theta ) CH_(3) OH]` `-726 = [2 xx (-286) + (-393) ] -[X]` `-726 = [-572-393]-x` `therefore x=-239 "kJ/mol"`

Discussion

2.	Calculate the standard enthalpy of formation of CH_3OH_((l)) from the following data : (i)CH_3OH_((l)) +3//2 O_(2(g)) to CO_(2(g)) + 2H_2O_((l)) , Delta_r H^(Ө)=-726 "kJ mol"^(-1) (ii)C_((s)) + O_(2(g)) to CO_(2(g)) , Delta_c H^(Ө)=-393 "kJ mol"^(-1) (iii) H_(2(g)) +1//2O_(2(g)) to H_2O_((l)) , Delta_f H^(Ө)=-286 "kJ mol"^(-1)
Answer» Solution :The equation we aim at : `C_((s)) +2H_(2(g)) + 1//2 O_(2(g)) to CH_3OH_((l)) , Delta_fH^(Ө)=pm` …(iv) Multiply equation (iii) by 2 and add to equation (ii) `C_((s)) +2H_(2(g)) to CO_(2(g)) +2H_2O_((l)) "" DELTAH`=-(393+522) =-965 KJ `"mol"^(-1)` Subtract equation (iv) from equation (i) `CH_3OH_((l)) +3//2 O_(2(g)) to CO_(2(g)) + 2H_2O_((l)) , DeltaH=-726 "kJ mol"^(-1)` Subtract : `C_((s)) + 2H_(2(g)) + 1//2 O_(2(g)) to CH_3OH_((l)) , Delta_fH^(Ө)=-239 "kJ mol"^(-1)`

Discussion

3.	Calculate the standard enthalpy change ( in kJ mol^(-1)) for the reaction , H_(2)(g) + O_(2)(g) rarrH_(2)O_(2)(g), given that bond enthalpy of H-H,O = O, O-H and O-O ( in kJ mol^(-1))are respectively 438, 498,464 and 138
Answer» `-1 30` `065` `+130` `- 334` SOLUTION :`H -H(g)+O=O(g) rarr H-O-O-H(g)` `Delta_(R)H= BE(H-H) + BE(O=O)-[2BE(O-H) +BE(O-H)]` `=438 +498 - [2 xx 464 +138 ]KJ mol^(-1)` `= 936- ( 928+ 138 )= - 130 kJ mol^(-1)`

Discussion

4.	Calculate the speed of a train ms^(-1) which covers 972 miles in 27 hours (1 mile = 1.60 km)
Answer» Solution :`("972 MILES")/("27 hrs")=(("972 miles")xx("CONVERSION factor"))/(("27 hrs")xx("conversion factor"))` `= (("972 miles")xx(1.6xx10^(3)m))/(("1 mile"))xx(1)/(("27 hrs"))xx(("1 HR"))/((3.6 xx10^(3)s))=16.0 "m s"^(-1)`.

Discussion

5.	Calculate the solubility product of silver bromide if the solubility of the salt in saturated solution is 5.7xx10^(-7) moles/litre.
Answer» ANSWER :`3.25xx10^(-13)`

Discussion

6.	Calculate the solubility of silver chloride in water at room temperature if the solubility product of AgCl is 1.6xx10^(-10).
Answer» ANSWER :`1.26xx10^(-5) "MOL" L^(-1)`

Discussion

7.	Calculate the solubility of silver chloride in water at room temperature if the K_(sp) " of " AgCl " is " 1.6 xx 10^(-10)
Answer» ANSWER :`1.26 XX 10^(-5)M`

Discussion

8.	Calculate the solubility of CaF_(2) in water at 298 K which is 70% dissociated. K_(sp) of CaF_(2) is 1.7 xx 10^(-10). If answer is x xx 10^(-4) mol/ltr thenx = ____?
Answer» Solution :`CaF_2 hArr underset( 0.75)(CA^(+2) ) +underset( 1.4s)(2F^(-))` ` 1.7 XX 10^(-10)=(0.7S) (1.4S) ^(2) ` ` 1.7xx 10^(-10)=1.4 xx S^(3) rArr S= 5 xx 10 ^(_4) `

Discussion

9.	Calculate the solubility of AgCN in a buffer solution of pH 3*00. K_(sp) for AgCN is 2.2xx10^(-16)and K_(a) forHCN is 6.2xx10^(-12).
Answer» Solution :`AgCN rarr AG^(+) CN^(-)` pH = 3.0means `[H^(+)] = 10^(-3) M ` `CN^(-)` ions now combine with the `H^(+)` ion to form HCN but `[H^(+)]` remains almost CONSTANT because we have buffer solution . Now `HCN hArr H^(+) + CN^(-)` `K_(a) = ([H^(+)][CN^(-)])/([HCN]) or ([HCN])/([CN^(-)]) = ([H^(+)])/(K_(a)) = (10^(-3))/(6.2xx10^(-10)) = 1.6 xx 10^(6) ` ..(i) Suppose the solubility in the buffer solution is 'x' mol `L^(-1)`. Then `x=[Ag^(+)]=[CN^(-)]+ [HCN] ~= [HCN]` (as from eqn. (i) `[CN^(-)] = [ HCN]//(1.6xx10^(6))` is negligible in comparison to [HCN]) `:. [CN^(-)] = ([HCN])/(1.6xx10^(6))` `K_(sp) = [ Ag^(+)][CN^(-)]=x xx (x)/(1.6xx10^(6))=2.2xx10^(-16)` (Given) or `x^(2) = 3.52 xx 10^(-10)"" x=1.9xx10^(-5) "mol" L^(-1)`

Discussion

10.	Calculate the solubility of AgCl , Fe(OH)_3 Ag_2SO_4 and Hg_2Br_2 from their solubility products. K_(sp)(AgCl)=1.8xx10^(-10) ,K_(sp)[Fe(OH)_3]=1.0xx10^(-38), K_(sp)[Ag_2SO_4]=1.4xx10^(-5), K_(sp)=(Hg_2Br_2)=5.6xx10^(-23) .Also calculate the solubilities of salts in gL^(-1)and molarities of the ions.
Answer» SOLUTION :(i)`AgCl(s) hArr Ag^(+)(aq)+Cl^(-)(aq)` `K_(sp)=1.8xx10^(-10)` Let solubility be s moles per litre. Then the CONCENTRATION of various species at equilibrium are : `[Ag^+]=s, [Cl^-]=s` `K_(sp)=[Ag^+][Cl^-]=sxxs` `1.8xx10^(-10)=s^2` `therefore s=(1.8xx10^(-10))^(1//2) =1.34xx10^(-5) "mol L"^(-1)` Solubility in g `L^(-1)=1.34xx10^(-5)xx143.5` `=1.92xx10^(-3) gL^(-1)` `[Ag^+]=s=1.34xx10^(-5) "mol L"^(-1)` `[Cl^-]=s=1.34xx10^(-5) "mol L"^(-1)` (ii)For `Fe(OH)_3 : K_(sp)=1.0xx10^(-38)` `Fe(OH)_3 hArr Fe^(3+) + 3OH^-` If solubility is s , then `[Fe^(3+)]=s, [OH^-]=3s` `K_(sp)=[Fe^(3+)] [OH^-]^3` `1.0xx10^(-38)=sxx(3s)^3=27s^4` `27s^4=1.0xx10^(-38)` `s^4=(1.0xx10^(-38))/27` or `s=((1.0xx10^(-38))/27)^(1//4)` `=1.39xx10^(-10) "mol L"^(-1)` Solubility in g `L^(-1) =1.39xx10^(-10)xx106.5` `=1.48xx10^(-8) g L^(-1)` `[Fe^(3+)]=s=1.39xx10^(-10) "mol L"^(-1)` `[OH^-]=3s=3xx1.39xx10^(-10)` `=4.17xx10^(-10) "mol L"^(-1)` (iii)For `Ag_2SO_4 : K_(sp)=1.4xx10^(-5)` `Ag_2SO_4 hArr 2Ag^(+) + SO_4^(2-)` If solubility is s , then `[Ag^+]=2s, [SO_4^(2-)]=s` `K_(sp)=[Ag^+]^2 [SO_4^(2-)]` `1.4xx10^(-5) =(2s)^2 (s)` `1.4xx10^(-5)=4s^3` or `s^3=(1.4xx10^(-5))/4` `s=((1.4xx10^(-5))/4)^(1//3)` `=1.52xx10^(-2) "mol L"^(-1)` Solubility in `gL^(-1) =1.52xx10^(-2) xx312` `=4.74 g L^(-1)` `[Ag^+]=2s=2 xx 1.52xx10^(-2)` `=3.04xx10^(-2) "mol L"^(-1)` `[SO_4^(2-)]=s=1.52xx10^(-2) "mol L"^(-1)` (iv)For `Hg_2Br_2 , K_(sp) = 5.6xx10^(-23)` `Hg_2Br_2 hArr Hg_2^(2+)+ 2Br^-` If solubility is s, then `[Hg_2^(2-)]=s, Br^(-)=2s` `K_(sp)=(s)(2s)^2` `5.6xx10^(-23)=4s^3` or `4s^3 =5.6xx10^(-23)` `s=((5.6xx10^(-23))/4)^(1//3)` =`2.41xx10^(-8) "mol L"^(-1)` Solubility in g `L^(-1) = 2.41xx10^(-8) xx360.4` `=8.68xx10^(-6) gL^(-1)` `[Hg_2^(2+)]=2.41xx10^(-8)` `[Br^-]=2xx2.41xx10^(-8)` `=4.82xx10^(-8)` M

Discussion

11.	Calculate the solubility of Ag_(2)CrO_(4) in 0.1M AgNO_(3)K_(sp) of Ag_(3)CrO_(4)=1xx10^(-22)
Answer» Solution :`Ag_(2)CrO_(4)hArr2Ag^(+)+CrO_(2)^(2-)` `2xx0.1+x` `K_(SP)=[Ag^(+)]^(2)[CrO_(4)^(2-)]` `1XX10^(-12)=(2xx0.1)^(2)[CrO_(4)^(2-)]` `[CrO_(4)^(2-)]=(1xx10^(-12))/((2xx0.1)^(2))=(1xx10^(-12))/0.04=0.025xx10^(-10)`

Discussion

12.	Calculate the solubility of A2xx3 pure water. Assuming that neither king of ion reacts with water. The solubility product of A2X3Ksp=1.1xx10-23 Explain with equation.
Answer» Solution :`k_(sp)=[A^(3+)]^(2)[X^(-2)]^(3)=[2s]^(2).[3s]^(3)=108s^(5)` `S^(5)=(1.1xx10^(-23))/108=1xx10^(-25)` `S^(5)=1xx10.5"mol"DM^(-3)`

Discussion

13.	Calculate the solubility of A_(2)X_(3) in pure water, assuming that neither kind of ion reacts with water. The solubility product of A_(2)X_(3), K_(sp)=1.1xx10^(-23).
Answer» Solution :`{:(A_(2)X_(3),hArr,2A^(3+),+,3X^(2-),,,),(s,,s,,s,,,):}` `:. K_(sp)=(2S)^(2)(3s)^(3) = 108 s^(5)` `s=(K_(AP)/108)^(1//5)=(1.1xx10^(-23)//108)^(1//5)` `log s = (1)/(5)(log 1.1 + log 10^(-23)-log 108 ) = (1)/(5) (0.0414-23-2.0334)` or `log s = - 4.9984 = bar(5) .0016 :. s= 10^(-5) "mol" L^(-1)`

Discussion

14.	Calculate the solubility of A_2X_3 in pure water , assuming that neither kind of ion reacts with water. The solubility product of A_2X_3 , K_(sp)=1.1xx10^(-23).
Answer» Solution :Suppose , S mol `L^(-1)` is soluble in sparingly soluble salt of concentrated solution of `A_2X_3`. The following IONIC equilibrium of this concentrated solution. `{:(A_2X_(3(s)) to ,2A_((aq))^(3+)+, 3X_((aq))^(2-)),("Conc. according to stoichiometry:","2S M","3S M"):}` `therefore K_(SP)=[A^(3+)]^2 + [X^(2-)]^3 =(2S)^2 (3S)^3` `therefore K_(sp) =108 S^5 =1.1xx10^(-23)` `therefore 108 S^5 =111.0 xx 10^(-25)` `therefore S=(111xx10^(-25))^(1/5) =2.5645xx10^(-5)` M `approx 2.56xx 10^(-5)` M Here, `(111xx10^(-25))^(1/5) =1/5 LOG (111xx10^(-25))` `=1/5 log 111 + 1/5 log 10^(-25)` `=1/5 (2.0453)+1/5(-25)` =0.4090 + (-5) Antilog of `=2.5645xx10^(-5)`

Discussion

15.	Calculate the simultaneous solubility of AgCNS and AgBr ina solution of water. K_(sp) of AgCNS=1xx10^(-12)and AgBr=5xx10^(-13).
Answer» Solution :SUPPOSE the simultaneous solubility of AgCNS and AgBr in water are a and b respectively. `{:(AgCNS hArr Ag^(+) + CNS^(-) , AgBr hArr Ag^(+) + Br^(-)),("aa","bb"):}` Thus, `[Ag^(+)] = a+b , [ CNS^(-)] = a, [Br^(-) ] = b ` `K_(sp) ` of AgCNS `= [Ag^(+)] [ CNS^(-)]= a (a+b)` `:. a(a+b) = 5 xx 10^(-12) " " ...(i)` `K_(sp) "of" AgBr = [Ag^(+)][Br^(-)]=b (a+b)` `:. b(a+b) = 5 xx 10^(-13) " " ...(ii)` DIVIDING EQN.(i) by (ii), we get `(a)/(b) = (10^(-12))/(5xx10^(-13))=2 or a = 2b` PUTTING this valuein eqn. (i) `2b(2b+b) = 10^(-12) or 6B^(2) = 10^(-12) or b^(2) = 1/6 xx 10^(-12)` or, `b= 4.08xx10^(-7) "mol" L^(-1) ` `a=2b=2xx4.08xx10^(-7) "mol" L^(-1) = 8.16xx10^(-7) "mol " L^(-1)`.

Discussion

16.	Calculate the root mean square, average and most probable speeds of oxygen molecules at 27^(@)C.
Answer» Solution :Proceed as in SOLVED SAMPLE Problem 2(B), PAGE 5/41

Discussion

17.	Calculate the RMS velocity of hydrogen molecule at 0^@C.
Answer» Solution :The RMS velocity (U) is GIVEN by `u=sqrt((3RT)/M)` In the present case, `T=0^@C = 273 K, M = 2, R = 8.31 xx 10^7 "ERGS " K^(-1) MOL^(-1)` `:. ""u= sqrt((3xx8.31 xx 10^7 xx 273)/2) = 1.845 xx 10^5 " cm " s^(-1)` Hence, the RMS velocity of hydrogen molecules at `0^@C " is " 1.845 xx 10^5" cm " s^(-1)`.

Discussion

18.	Calculate the RMS velocity of chlorine molecules at 15^@C and 75 cm Hg pressure. (Given : density of Hg = 13.596 g cm , g = 980.6 " cm " s^(-2)
Answer» Solution :At S.T.P one mole of a GAS occupies 22400 `cm^3`. Let us FIRST convert this volume corresponding to the given conditions. `P_1 = 76 cm Hg`, `V_1 = 22400 cm^3`, `T_1 = 273 K` At given conditions `P_2 = 75 cm Hg`, `V_2` = ?, `T_2 = 15+273 = 288 K` ACCORDING to the gas EQUATION, `(P_1 V_1)/T_1 = (P_2 V_2)/T_2` Hence,`(76 xx 22400)(273) = (75 xx V_2)/(288)` `:. "" V_2 = 23945.8 cm^3` The RMS velocity (u) is given by `u=sqrt((3RT)/M) = sqrt((3PV)/M)`(for one mole of the gas) In the present case, `P = hd g = 75 xx 13.596 xx 980.6 = 9.999 xx 10^@ " dynes " cm^2` `V=23945.8 cm^3 " and " M=71` `:. "" u= sqrt((3xx9.999xx10^5xx23945.8)/71)` `=3.181xx10^4 " cm " s^(-1)` Therefore, the RMS velocity of chlorine molecules at `15^@C` and 75 cm pressure is `3.181xx10^4 " cm " s^(-1)`

Discussion

19.	Calculate the RMS velocity and most probable velocity of carbondioxide at 27^(@)C.
Answer» Solution :Absolute TEMPERATURE = T = 300 K Gram molecular WEIGHT`= M = 44 g mol^(-1)` RMS velocity (C ) ` = sqrt((3RT)/(M)) = sqrt((3 xx 8.31 xx 300 xx 10^(7))/(44))` `C = 4.12 xx 10^(4) cms^(-1) = 412 MS^(-1)` Most probable velocity `(C_(p)) = 0.81 xx C = 0.81 xx 412 = 335 ms^(-1)`

Discussion

20.	Calculate the rms speed of the molecules of ethane gas of colume 1.5 litre at 750 mm of Hg pressure
Answer» ANSWER :`1.225 XX 10^(4) cm//sec`

Discussion

21.	Calculate the RMS, average and most probable velocity of SO_2 at 27^@C.
Answer» Solution :Absolute temperature = T = 300 K Gram MOLECULAR weight = M = 64 g `mol^(-1)` Gas CONSTANT `= R = 8.314 xx 10^7 ERG k^(-1)mol^(-1)` RMS velocity (C) = ` sqrt((3RT)/(M)) = sqrt(( 3 xx 8.314 xx 300 xx 10^7 )/( 64 )) = 3.42 xx 10^4cm s^(-1)` `=342 ms^(-1)` Most probable velocity `(C_p) = 0.8166 xx C ` `= 0.8166 xx 342 = 279.27ms^(-1)` Average vclocity `= vecC - 0.9213 XXC -0.9213 xx 342` `= 315 ms^(-1)`

Discussion

22.	Calculate the resonance energy of N_(2)O from the following data Delta H_(f) " of " N_(2)O= 82 kJ mol^(-1). Bond energies of N -= N, N = N, O = O and N = O bond are 946, 418 and 607 kJ mol^(-1) respectively
Answer» `-88 kJ mol^(-1)` `-170 kJ mol^(-1)` `-82 kJ mol^(-1)` `-258 kJ mol^(-1)` Solution :`Delta H_(f) = Sigma` B.E of reaction - `Sigma B. E` of PRODUCTS R. E = `Delta H_(f)` of OBSERVED - `Delta H` of calculated

Discussion

23.	Calculate the resonance energy of C_(6)H_(6) using Kerkule formula for C_(6)H_(6) from the following data. (i) DeltaH_(f)^(@) fo r C_(6)H_(6)=-358.5kJ mol^(-1) (ii)Heat of atomisation of C=716.8kJ mol^(-1) (iii) Bond energy of C-H,C-C,C=C and H-H are 490,340,620,436.9kJ mol^(-1) respectively.
Answer» ANSWER :`-150.0kJ MOL^(-1);`

Discussion

24.	Calculate the relative rates of diffusion of""^235UF_6 and ""^238UF_6in gaseous form (F= 19).
Answer» Solution :ACCORDING to Graham.s law of diffusion, `r_1/r_1 = sqrt(d_2/d_1) = sqrt(M_2/M_1)` The molecular weight of `""235UF_6 = 235 + (6 XX 19) = 349 (M_1)` The molecular weight of `""238UF_6 = 238 + (6 xx 19) = 352 (M_2)` `:. "" r_1/r_2 = sqrt(352/349)= 1.0043/1` HENCE, the rates of diffusion of`""^235UF_6 and ""^238UF_6 " are in the ratio " 1.0043 : 1`.

Discussion

25.	Calculate the relative molecular mass of the following. (i) Ethanol (C_(2)H_(5)OH)(ii) Potassium permanganate (KMnO_(4))(iii) Potassium dichromate (K_(2)Cr_(2)O_(7))(iv) Sucrose (C_(12)H_(22)O_(11))
Answer» Solution :(i) ETHANOL `(C_(2)H_(5)OH)` Molar mass = `(2 xx 12) + (6 xx 1) + (1 xx 16)` = 24 + 6+ 16 = 46 g `mol^(-1)` . (II) Potassium permanganate `(KMnO_(4))` Molar mass = 39 + 55 + `(4 xx 16)` = 39 - 55 + 64 = 158 g `mol^(-1)` . (iii) Potassium dichromate `(K_(2)Cr_(2)O_(7))` Molar mass = `(39 xx 2) + (2 xx 52) + (7 xx 16)` = 78+ 104+ 112 = 294 g `mol^(-1)` . (iv) Sucrose `(C_(12)H_(22)O_(11))` Molar mass = `(12 xx 12) + (22 xx l) + (H xx 16)` = 144 + 22 + 176 = 342 g `mol^(-1)`

Discussion

26.	Calculate the real mass of one methane molecule
Answer» ANSWER :`2.66xx10^(-26)KG`

Discussion

27.	Calculate the real mass of one carbon atom
Answer» SOLUTION :`1.99 XX 10^(-24)G`

Discussion

28.	Calculate the ratio of the time periods (T_(1)//T_(2)) in second orbit of hydrogen atom to third orbit of He^(+) ion
Answer» Solution :TIME PERIOD `(T_(n)) = alpha (n^(2))/(Z^(2))` `:. (T_(1))/(T_(2)) = (n_(1)^(3))/(Z_(1)^(2)) xx (Z_(2)^(2))/(n_(2)^(3)) = (2^(3))/(1^(2)) xx (2^(2))/(3^(3)) = (32)/(27)`

Discussion

29.	Calculate the ratio of solubility of AgCl " at "25 ^(@) Cin 3.0 M NH_3 (K_(sp)of AgCl in NH_3 " is "3.1 xx 10^(-3) ) and " in "H_2O (K_(sp)of AgCl in H_2O" is"1.8 xx 10^(-10)) .If answer is 1.15 xx 10 ^(x)then x=________?
Answer» Solution :` AgCl + 2NH_3 hArr [Af(NH_3)_2 ]^(+)+CL^(-), K_1` ` ((X)/(3-2x) )^(2) K_1 RARR (x) /( 3-2x) =sqrt( K_1) ~~(x)/(3) ` `AgCl hArr Ag^(+)+Cl^(-),K_2` ` (x)/(y)=sqrt(( K_1)/(K_2)) xx 3 =sqrt((3.1xx 10^(-3))/(1.8 xx 10^(-10))) xx 3 =1.2 xx 10^(4) `

Discussion

30.	Calculatethe ratioof radiusof 2nand 3rdorbitof hydrogenatom(Note r_(n)propn^(2))
Answer» SOLUTION :`4: 9` and 0.4444

Discussion

31.	Calculate the ratio of pH of a solution containing 1 mole of CH_(3)CO Ona + 1mole of HCl per litre to that of a solution containing 1 mole of CH_(3)CO Ona + 1mole of CH_(3)CO OHper litre.
Answer» SOLUTION :Case I. Calculation of pH of solution sontaining 1 mole of `CH_(3)CO ONA +1` mole of HCl per litre `{:(,CH_(3)CO ON a ,+,HCl,rarr,CH_(3)CO OH,+,NaCl ),("Initial moles",1 "mole",,1 "mole",,0,,0),("Moles after reaction ",0,,0,,1,,1):}` i.e., `[CH_(3)CO OH ] = 1 "mol"L^(-1)` `{:(,CH_(3)CO OH ,hArr,CH_(3)CO O^(-),+,H^(+),,),("Initial conc.",C "mol" L^(-1),,,,,,),("After dissociation",C - C alpha,,C alpha,,C alpha ,,):}` `:. [H^(+)]=C alpha . "But" alpha = sqrt((K_(a))/(C))` `:. [H^(+)]=C sqrt((K_(a))/(C))=sqrt(K_(a)C)=sqrt(K_(a))=K_(a)^(1//2)`(`:' C = 1 "mol" L^(-1)`) `:. - log [H^(+)]= - (1)/(2) log K_(a), i.e., (pH)_(1) = - (1)/(2)log K_(a)"" ...(i)` Case II. Calculation of pH of solution containing 1 mole of `CH_(3)CO ONa + 1 ` mole of `CH_(3)CO OH` per litre Applying Henderson equation, `(pH)_(2)=pK_(a) + log .(["Salt"])/(["ACID"]) = pK_(a)= - log K_(a)`...(ii) [[Salt]=[Acid]=1 mol `L^(-1)`] From equations (i) and (ii), `(pH)_(1)//(pH)_(2)=1//2`.

Discussion

32.	Calculate the ratio of number of pure and hybrid orbitals used for bonding in an acetylene molecule
Answer» Solution :ACETYLENE has TWO C atoms and two H atoms. Carbon undergoes sp hybridisation. Number of hybrid orbitals `=2xx2=4` Number of PURE orbitals =4 (from C) +2 (from H)=6 Ratio of number of pure and hybrid orbitals `=4:6=2:3`

Discussion

33.	Calculate the ratio of kinetic energies of 3g of hydrogen and 4g of oxygen at an given temperature.
Answer» ANSWER :`12:1`

Discussion

34.	Calculatethe ratioof diameterof atom and nucleus .
Answer» SOLUTION :DIAMETER of NUCLEUS: `10^(12` to `10^(13) cm` Diamterof atom `:10^(8)cm ` In this waywithrespectto nuclessdiameterofatom is100000 more thannucleus .

Discussion

35.	Calculate the ratio of average velocity of helium atoms at 27^@C and methane molecules at 127^@C.
Answer» Solution :`( vecC_(1))/(vecC_(2)) = SQRT((T_1M_2)/(T_2 M_1))` SUBSTITUTING the values, ` ( vecC_1)/( vecC_2)= sqrt((300 xx 16 )/( 400 xx 4 )) = sqrt(3) = 1.73` The ratio of average velocity of He and `CH_4 `=1.73:1

Discussion

36.	Calculate the ratio between the wavelength of an electron and a proton if the proton is moving with half the velocity of electron (mass of proton = 1.67 xx 10^(-27)kg and mass of electron = 9.11 xx 10^(-31) kg)
Answer» <P> Solution :`lamda_(E) = (h)/(m_(e) v_(e)), lamda_(p) = (h)/(m_(p) v_(p)), (lamda_(e))/(lamda_(p)) = (m_(p) v_(p))/(m_(e) v_(e)) = (1)/(2) XX (1.67 xx 10^(-27) KG)/(9.11 xx 10^(-31) kg) = 916.6`

Discussion

37.	Calculate the radius of Bohr’s third orbit for hydrogen atom.
Answer» ANSWER :4.761 NM

Discussion

38.	Calculate the radius of Bohr's fifth orbit for hydrogen atom. Also calculate the radius of third orbit of He^(+) ion
Answer» SOLUTION :For H-atom, `r_(N) = 0.529 xx n^(2) Å`. For 5th ORBIT, n = 5 `:. r_(5) = 0.029 xx 5^(2) = 13.225 Å= 1.3225nm` For hydrogen like particles `r_(n) = (n^(2))/(Z) xx a_(0)` (`a_(0)`= Bohr radius `= 0.529 Å`) For `He^(+), Z = 2`. For 3rd orbit, n = 3 `:. r_(3) = (3^(2))/(2) xx 0.529 Å = 2.3805 Å = 0.23805nm`

Discussion

39.	Calculate the quantity of slaked lime Ca(OH)_(2) required to soften 6000 litres of well water containing 16.2 g of calcium bicarbonate per 100 litres.
Answer» Solution :Step I. Mass of calcium BICARBONATE in well water 100 L of well water contain calcium bicarbonate = 16.2 g 60000 L of well water contain calcium bicarbonate `= (16.2)/(100) xx 60000 = 9720 g` Step II. Quantity of slaked lime of `CA(OH)_(2)` required `underset(40+2+24+96+(=162g))(Ca(HCO_(3))_(2))+underset(40+34(=74g))(Ca(OH)_(2))rarr2CaCO_(3)+2H_(2)O` 162 g of `Ca(HCO_(3))_(2)` require slaked lime = 74 g 9720 g of `Ca(HCO_(3))_(2)` require slaked lime `= (74xx9720)/(162)g=4440g`.

Discussion

40.	Calculate the radial distance between two peaks in the radial probability of 2s orbital.
Answer» Solution :The DISTANCE of the first PEAK is `0.53Å`. The distance of the 2nd peak is `2.6Å` The distance between TWO PEAKS `= 2.6 - 0.53` `=2.07Å`.

Discussion

41.	Calculate the quantities of reagents required to soften 10^(3) L of water containing Ca(HCO_3)_2 Mg(HCO_3)_2 and CaSO_4 as 20.0 g, 15.0 g and 5.0 g per litre respectively by lime soda process.
Answer» SOLUTION :`20.0 g of Ca(HCO_(3))_(2)-=(20xx100)/(162)=12.34 g of CaCO_(3)` `15.0 g of Mg(HCO_(3))_(2)-=(15xx100)/(146)=10.27g of CaCO_(3)` `5.0 of CaCO_(4)-=(5xx100)/(136)=3.67g of CaCO_(3)` Lime requirement: `=(74)/(100)` [ Temporary Ca- Hardness`+2xx` Temporary Mg-Hardness `+` Permanent Mg-hardness] `=(74)/(100)[12.34+2xx10.27]` `=(24.33g)/(L)` of `CaCO_(3)EQ` `=24.33xx(10^(3)g)/(10^(3)L)` `=(24.33gk)/(10^(3)L) of CaCO^(3) eq` Soda requirement: `=(106)/(100)` [permanent Ca-hardnes`s^(+)` Permanent Mg-hardness] `=(106)/(100)[3.67]` `k=(3.89g)/(L) of CaCO_(3)` equivalent `=3.89xx(10^(3)g)/(10^(3)L)` of `CaCO_(3)eq` `=(3.89kg)/(10^(3)L) of CaCO_(3) eq`

Discussion

42.	Calculate the quantities fo reagents required to soften 100 L of water containing the following impurities per litre in it. (a). CaCO_3=20, (b). MgCl_2=8.0g, (c). MgSO_4=7g, (d). MgCO_3=4.5g (e). CaSO_4=2.5g (f). NaCl=6.0g Purity of lime =90% purity of soda=99.5%
Answer» Solution :(a). `20 g CaCO_(3)=20g CaCO_(3)` (b). `8 g MgCl_(2)=(8xx100)/(95)=8.42 g of CaCO_(3)` (c). `7g MgSO_(4)=(7xx100)/(120)=5.83 g of CaCO_(3)` (d). `4.5 of MgCO_(3)` (Present as bicarbonate)`-=(4.5xx100)/(84)` `-=5.35g of CaCO_(3)` (e). `2.5g CaSO_(4)=(2.5xx100)/(136)=1.83g of CaCO_(3)` (f). NaCl does not contribute to hardness. Lime requirement: `=(74)/(100)[20+2xx5.35+8.42+5.83]` `=33.26 g of CaCO_(3)(eq)/(L) of H_(2)O` `=33.26xx100g of CaCO_(3)(eq)/(100L) of H_(2)O` `=3.326 kg` Weight of lime of `90%` PURITY`=(3.326xx100)/(90)=3.69kg` Soda requirement: `=(106)/(100)[8.42+5.83+1.83]` `=17.04g of CaCO_(3)(eq)/(L) of H_(2)O` `=17.04xx100g of CaCO_(3)(eq)/(100L) of H_(2)O` `=1.704 kg` Weight of soda of `95.5%` purity`=(1.704xx100)/(95.5)=1784kg`

Discussion

43.	Calculate the proton concentration and hydroxyl ion concentration of centimolar sulphuric acid?
Answer» Solution :`2 xx 10^(-2) MH^(+ ), 5 xx 10 ^(-13)M OH^(-)`

Discussion

44.	Calculate the proton affinity of NH_(3) (g) from the following data (in kJ/mol): {:(DeltaH^(@)"dissociation":H_(2)(g),=436),(DeltaH^(@)"formation":NH_(3)(g),=-46),("Lattic energyof"NH_(4)Cl(s),=-683),("Ionisation enorgy of"H,=130),("Electron affinity of"Cl,=380),(DeltaH^(@)"dissociation" :Cl_(2)(g),=240),(DeltaH^(@)"formation of"NH_(4)Cl(s),=-314):}
Answer» ANSWER :327

Discussion

45.	Calculate the proporation of O _(2) N_(2) dissolved in water at 298K. When air containing 20% O _(2)and 80% N_(2) by volume is in equilibrium with water at 1 atm pressure. Henry's law constants for two gases are K_(H) (O_(2)) =4.6 xx 10^(4)atm and K_(H) (N_(2)) = 8.5 xx 10 ^(4)atm.
Answer» Solution :`C_(1) V_(1) = C_(2) V_(2)` `6M (V_(1)) = 0.25 M xx 500 ML` `V _(1) = (0.25 xx 500)/(6)` ` V _(1) = 20.83mL`

Discussion

46.	Calculate the product of uncertainty in position and velocity for an electron of mass 9.1 xx 10^(-31)kg according to Heisenberg uncertainty principle
Answer» <P> Solution :`Delta X. Delta p = (h)/(4PI) or Delta x. (m Delta v) = (h)/(4pi) or Delta x. Delta x = (h)/(4pi m)`

Discussion

47.	Calculate the pressure of 4.0 mol of a gas occupying 5 dm^(3) at 3.32 bar pressure. (R = 0083 bar dm^(3) K^(-1) mol^(-1))
Answer» SOLUTION :50 K or `-233^(@)C`

Discussion

48.	Calculate the pressure of 11g of CO_2 at 40^@C in a 3 litre flask.
Answer» ANSWER :2.14 ATM

Discussion

49.	Calculate the pressure exerted by one mole of methane in a 450 mL container at 25^@C using van der Waals' equation. What pressure will be predicted by ideal gas equation ? (Given : a = 2.253 " atm " L^2mol^(-2), b = 0.0428 L mol^(-2), R =0.0821 L " atm " K^(-1) mol^(_1))
Answer» <P> Solution :Using van der WAALS. equation : For ONE mole of a gas, van der Waals. equation can be written as `(p + a/V_2) (v - b) = RT " or " P= (RT)/(V-b) - a/V_2` Substituting `a = 2.253 " atm " L_2mol^(-2), b = 0.0428 L mol^(-1)`, `V= 450/1000 = 0.450 L, T = 25 + 273 = 298 K` and `R = 0.0821 L " atm " K^(-1) mol^(-1)`, we have `P = (0.0821 xx 298)/(0.450-0.0428)- (2.253)/((0.450)^2)" or " P = 0.0428` atm Using ideal gas equation : `PV = nRT " or " P =(nRT)/V` Substituting, `n = 1, R = 0.0821 L " atm " K^(-1) mol^(-1)`, `T = 298 K " and " V = 0.450 L`, we have `P= (1 xx 0.0821 xx 298)/(0.450) = 54.4` atm Hence, the PRESSURE of methane under given conditions is 48.96 atm. Ideal gas equation predicts the pressure 54.4 atm under similar conditions.

Discussion

50.	Calculate the pressure in a mixture of 8g of O_(2)(g)and 4 g of H_(2)(g) confined in a vessel of1 dm^(3) "at" 27^(@)C (R=0*083 dm^(3) "bar" K^(-1) "mol"^(-1))
Answer» <P> Solution :No. of moles `=("mass")/("moleculear mass")` `NO_(2)=(8)/(32)=025 "" NH_(2)=(4)/(2)` `n_("TOTAL") =025+2=225, T=27+273=300K` `PV= nRT :. P(nRT)/(v)=(225xx0083xx300)/(1)` `:. P=025` bar

Discussion

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