94420 + Interview Questions in Chemistry in Class 12

1.	At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar.If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be itsconcentration ?
Answer» Solution :APPLYING the relation : `pi = CRT` In the first case, ` = 36/1880 XX R xx 300 = 60 R `....(i) In the second case,`1.52 = C xx R xx 300 `(C is concentration)....(ii) Dividing (ii) by (i), we get `C = 0.061 M`

Discussion

2.	At 30^(@)C, the solubility of Ag_(2)CO_(3)(K_(sp) = 8 xx 10^(-12)) would be greatest in one liter of
Answer» `0.05 M Na_(2)CO_(3)` `0.05 M AgNO_(3)` Pure water `0.05 M NH_(3)` Solution :PRESENCE of common ion decreases the solubility of SALT.

Discussion

3.	At 300 K, 36 g of glucose, C_6H_12O_6present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of another glucose solution is 1.52 bar at the same temperature, calculate the concentration of the other solution.
Answer» Solution :Apply the relation `pi V = cRT`...(i) First case Given thatT = 300 K, P = 4.98 bar Mass of glucose = 36 g, MOLECULAR mass = 180 u `c = 36/180 = 0.2` SUBSTITUTING the values in equation (i), we get `4.98 XX 1 = 0.2 xx R xx 300 `...(ii) Let the concentration of glucose in the second solution be `c_1` moles per litre. Substituting the values in equation (i) again, we get `1.52 xx 1 = c_1 xx R xx 300`....(III) Dividing (ii) by (iii), we have `(4.98)XX1)/(1.52 xx 1) = (0.2 xx R xx 300)/(c_1 xx R xx 300) ` Simplifying the above equation, we get `c_1= (0.2 xx 1.52)/(4.98) = 0.061`moles/litre

Discussion

4.	At 300K the reaction A(g)+B(g)hArrC(s) is in equilibrium in a closed, vessel. At the beginning of the reaction, the partial pressures of A and B gases are 0.2 and 0.3 atrn respectively and total pressure of the equilibrium mixture is 0.3 atm. K_c, for the reaction is-
Answer» `6.06xx10^4L^2Mol^(-2)` `2.59xx10^3L^2Mol^(-2)` `3.03xx10^4L^2Mol^(-2)` `8.2xx10^(-2)L^2Mol^(-2)` ANSWER :C

Discussion

5.	At 300 K, 36 g of glucose, C_(6)H_(12)O_(6) present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of another glucose solution is 1.52 bar at the same temperature, calculate the concentration of the other solution.
Answer» SOLUTION :`pi=CRT"(C = MOLAR concentration)"` `(pi_(1))/(pi_(2))=(C_(1))/(C_(2)),""(4.98)/(1.52)=(36//180)/(C_(2))"or"C_(2)=(36)/(180)xx(1.52)/(4.98)="0.061 M"`

Discussion

6.	At 30^(@)C, K_(p) for the dissociation reaction :SO_(2)Cl_(2)(g) hArr SO_(2)(g)+Cl_(2)(g)is 2.9xx10^(-2) atm. If the total pressure is1 atm, the degree of dissociation of SO_(2)Cl_(2) is : (assume 1-alpha^(2)=1).
Answer» `87%` `13%` `17%` `29%` SOLUTION :`SO_(2)Cl_(2)hArr SO_(2)+Cl_(2)` If degree of DISSOCIATION is `alpha` `1-alpha "" alpha "" alpha` Total no. of MOLES `=1-alpha+alpha+alpha=1+alpha` `pSO_(2)=(alpha P)/(1+alpha)=(alpha)/(1+alpha)` `pCl_(2)=(alpha P)/(1+alpha)=(alpha)/(1+alpha)` `pSO_(2)Cl_(2)=(1-alpha)/(1+alpha)P=(1-alpha)/(1+alpha)` `K_(p)=((pSO_(2))xx(pCl_(2)))/(p(SO_(2)Cl_(2)))` `=(((alpha)/(1+alpha))((alpha)/(1+alpha)))/((1-alpha)/(1+alpha))` `K_(p)=(alpha^(2))/(1-alpha^(2))=alpha^(2)` `therefore alpha^(2)=K_(p)=2.9xx10^(-2)` `= sqrt(2.9xx10^(-12))=0.17=17%`

Discussion

7.	At 298K, the conductivity of 0.2M KCI solution is 0.02485 ohm^(-1)cm^(-1). Calculate the molar conductivity of the solution.
Answer» Solution :`Delta_(m)=(kxx1000)/(M)` `=(0.02485xx1000)/(0.2)` `=124.25ohm^(-1)CM^(2)MOL^(-1)`

Discussion

8.	At 300K, the osmotic pressue of 300mL of a protein aqueous solution is 8.3 xx 10^(-5)bar. The molar mass of protein is 10^4 "gmol^(-1) . What is the weight (in g) of the protein present in this solution ? (R=0.083 L bar "mol"^(-1) K^(-1) )
Answer» 0.1 `1.0` `10` `0.01` ANSWER :D

Discussion

9.	At 298 K, the vapour pressure of water is 23.75 mm Hg. Calculate the vapour pressure at the same temperature over 5% aqueous solution of urea [CO(NH_(2))_(2)].
Answer» Solution :`5%` aqueous solution of urea means that Mass of solution = 100 g , Mass of SOLUTE, i.e, urea, `w_(2)=5G` `therefore"Mass of solvent, i.e., water "w_(1)=100-5=95 g,` Vapour pressure of pure water `(p^(@))` at 298 K = 23.75 mm Vapour pressure of urea solution =, i.e, `p_(s)=?,` Molar mass of water `(M_(1))="18 g mol"^(-1)` Molar mass of urea `CO(NH_(2))_(2)`, i.e., `M_(2)=12+16+14xx2+2="60 g mol"^(-1)` APPLYING Raoult's law, `(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2))/(M_(2))xx(M_(1))/(w_(1))` Substituting the values, we get `(23.75 - p_(s))/(23.75)=(5)/(60)xx(18)/(95)` `"or"23.75-p_(s)=(5)/(60)xx(18)/(95)xx23.75` `"or"23.75-p_(s)=0.375` `"or"p_(s)=23.375` Thus, the vapour pressure of `5%` urea solution = 23.375 mm

Discussion

10.	The standard reduction potentials at 298 K for the following half cell reactions are given belowZn^(2+) (aq) + 2e^(-) to Zn (s)-0.762Cr^(3+) (aq) + 3e^(-) to Cr (s)-0.7402H^(+) (aq) + 2e^(-) to H_(2) (g)-0.000Fe^(3+) (aq) + e^(-) to Fe^(3+) (aq)-0.770Which one is the strongest reducing agent?
Answer» `Zn_((s))` `H_(2(g))` `Cr_((s))` `Fe_((aq))^(2+)`

Discussion

11.	At 298 K, the solubility product of PbCl_(2) is 1.0 xx 10^(-6). What will be the solubility of PbCl_(2) in moles/litre
Answer» `6.3 xx 10^(-3)` `1.0 xx 10^(-3)` `3.0 xx 10^(-3)` `4.6 xx 10^(-14)` SOLUTION :`K_(SP) = 4S^(3)` `S = 3sqrt((K_(sp))/(4)) = 3sqrt((1.0 xx 10^(-6))/(4)) = 6.3 xx 10^(-3)`

Discussion

12.	At 300 K rate constant for Ato products at t=50 min is 0.02 s^(-1), then rate constant at t=75 min and 310 K will be (in s^(-1))
Answer» `(0.04)/(25)` `((0.02)/(25))` `0.04` `0.04xx25` ANSWER :C

Discussion

13.	At 298 K, the solubility product of MI is 10^(-6) (conc. Are expressed as mol kg^(-1)). The solubility of MI in 0.1 molal KI solution is :
Answer» `10^(-6)mol kg^(-1)` `10^(-5)mol kg^(-1)` `10^(-4)mol kg^(-1)` `10^(-3)mol kg^(-1)` ANSWER :B

Discussion

14.	At 300K for the reaction AB_3(g)
Answer» +2.19 KJ -2.52kJ +3.85kJ -3.26kJ Answer :B

Discussion

15.	At 298 K, the solubility of PbCl_(2) is 2 xx 10^(-2) mol/lit, then K_(sp)=
Answer» `1 XX 10^(-7)` `3.2 xx 10^(-7)` `1 xx 10^(-5)` `3.2 xx 10^(-5)` SOLUTION :`{:(PbCl_(2)hArr,Pb^(2+),+,2Cl^(-)),(,(S),,(2S)^(2)):}` `K_(sp) = 4S^(3) = 4 xx (2 xx 10^(-2))^(3) = 3.2 xx 10^(-5)`.

Discussion

16.	At 300 K when a solute is added to a solvent its vapour pressure over the mercury reduceds from 50 mm to 45mm. The value of mole fraction of solute will be
Answer» `0.005` `0.010` `0.100` `0.900` Solution :`(P^(0)-P)/(P^(0))= X_(B)` `(5)/(50) =X_(B)`

Discussion

17.	At 298 K temperature the activation energy for the reaction X_(2)+Y_(2)toXY+20KJ is 15 KJ. What will be the activation energy for the reaction 2XYto X_(2)+Y_(2)?
Answer» `+35KJ` `-35KJ` `-5KJ` `-15KJ` SOLUTION :`Delta_(R)H=("Activation energy of FORWARD REACTION")-("Activation energy of reverse reaction")` `therefore` -20=15 -x `therefore` x=35 KJ

Discussion

18.	At 300 K, vapour pressure of pure benzene and pure toluene are 100mm and 30 mm of Hg respectively. Also benzene and toluene is prepared by mixing 3.0 moles of toluene in 2.0 moles of benzene at 300 K. Anser the following questions: If the given solution is distilled by lowering the external pressure at constant 300 K, what will be the mole fraction of benzene in the last drop of liquid condensed ?
Answer» `(3)/(5)` `(6)/(5)` `(1)/(3)` `(1)/(6)` ANSWER :D

Discussion

19.	At 298 K temperature, how much pressure of H_(2) is required to make hydrogen electrode potential of pure water be zero ?
Answer» `10^(-12)` atm `10^(-10)` atm `10^(-4)` atm `10^(-14)` atm Solution :`E=E^(@)-(0.0591)/(2)"log"(p_(H_(2)))/([H^(+)]^(2))` `=0-(0.0591)/(2)"log"(p_(H_(2)))/((10^(-7))^(2))` So, if `p_(H_(2))=10^(-14)` then POTENTIAL of hyddrogen ELECTRODE will be 0.

Discussion

20.	At 300 K, vapour pressure of pure benzene and pure toluene are 100mm and 30 mm of Hg respectively. Also benzene and toluene is prepared by mixing 3.0 moles of toluene in 2.0 moles of benzene at 300 K. Anser the following questions: If vapours which is in equilibrium with the solution is condensed, mole fraction of benzene in the first drop of liquid formed will be
Answer» `(9)/(29)` `(20)/(29)` `(11)/(29)` `(18)/(29)` ANSWER :B::C::D

Discussion

21.	At 298 K in a constant volume calorimeter 0.01 mole of TNT was detonated when 8180 cals of heat was released.Each mole of TNT gives 6 moles of gaseous products on detonation. What is DeltaH/mole of TNT exploded?
Answer» `-714` kcals/`"MOLE"^(-1)` `-814` kcals/`"mole"^(-1)` `-914` kcals `"mole"^(-1)` NONE of the above Answer :B

Discussion

22.	At 300 K, two solutions of glucose in water of concentration 0.01M and 0.01M are separated by semipermeable membrane with respect to water. On which solution, the pressuer need be applied to prevent osmosis ? Calculate magnitude of this applied pressure.
Answer» ANSWER :`0.01M, 0.2217 ATM`;

Discussion

23.	At 298 K, DeltaH_("combustion")^(@)("sucrose")=-5737KJ//mol, DeltaG_("combustion")^(@)("sucrose")=-6333KJ//mol. Estimate additional non-PV work that is obtained by raising temperature to 310 K. Assume Delta_(r)C_(P)=0 for this temperature change
Answer» SOLUTION :24 kJ/mol

Discussion

24.	At 300 K, two solutions of glucose in water of concnetration 0.01 M and 0.001 M are separted by semipermeable membrane with respect to water. On which solution, the pressure should be applied to prevent osmosis? Calculate the magnitude of this applied presssure.
Answer» SOLUTION :`pi=CRT` For 0.01 M solution, `pi_(1)=0.01xx0.821xx300="0.2463 atm"` For 0.001 M solution, `pi_(2)=0.001xx0.821 xx300="0.02463 atm"` As movement of solvent molecules occurs from DILUTE to CONCENTRATION solution, pressure should be applied on CONCENTRATED solution, i.e., on 0.01 M solution to prevent osmosis. Magnitude of external pressure `=0.2463-0.0246="0.2217 atm"`

Discussion

25.	At 298 K , DeltaH_("combustio")^(@) (sucrose) = - 5737 KJ//mol & DeltaG_("combustio")^(@) ("sucrose") = - 6333 KJ//mol. Estimate additional non- PV workthat isobtainedby raisingtemperatureto 310 K, Assume Delta_(r) C_(p)= 0 for this temperature change
Answer» <P> Solution :`at" "298 K` `DeltaG_(1)^(@) = DeltaH^(@) = TDeltaS^(@)""...........(1)` `DeltaS^(@) = (DeltaH^(@) -DeltaG^(@))/(T_(1))` ` = - 2 KJ// mol- K` `at " " 310 K` `DeltaG_(2)^(@) = DeltaH^(@) - T_(2) DeltaS^(@)""..........(2)` `(because Delta_(r) C_(p) =0 "" thereforeDeltaH_(2) = DeltaH_(1) &DeltaS_(2) = DeltaS_(1))` `DeltaG_(2)- DeltaG_(2)^(@) = (T_(2) - T_(1)) DeltaS^(@)` `= + 12 xx 2 = + 24 KJ//mol`

Discussion

26.	At 300 K, the reactions which have following values of thermodynamic parameters occur spontaneously
Answer» `DeltaG^(@)=-400 kJ mol^(-1)` `DeltaH^(@)=200 kJ mol^(-1), DeltaS^(@)=-4 JK^(-1) mol^(-1)` `DeltaH^(@)=-200 kJ mol^(-1), DeltaS^(@)=4 JK^(-1) mol^(-1)` `DeltaH^(@)=200 J mol^(-1), DeltaS^(@)=40 JK^(-1)mol^(-1)` SOLUTION :When `DeltaH=+ve` and `DeltaS=-ve` than the REACTION is non-spontaneous.

Discussion

27.	At 300 K, the standard enthalpies of formation of C_(6)H_(5)COOH (s), CO_(2)(g) & H_(2)O (l) are , -408, -393 & -286" kJ mol"^(-1) respectively. Calculate the heat of combustion of benzoic acid at : (i) constant pressure (ii) constant volume.
Answer» SOLUTION :`C_(6)H_(5)COOH+15/2 O_(2) rarr 7CO_(2)+3H_(2)O(l)` `DeltaH=(7xx-393)+3(-286)-(408)=-3201` `DeltaU=DeltaH-Deltan_(g) RT` `Deltan_(g)=-1/2` `DeltaU=-3199.75 kJ mol^(-1)`

Discussion

28.	At 298 K a 0.1 M CH_(3)COOH solution is 1.34% ionized. The ionization constant K_(a) for acetic acid will be
Answer» `1.82 XX 10^(-5)` `18.2 xx 10^(-5)` `0.182 xx 10^(-5)` NONE of these Answer :A

Discussion

29.	At 300 K the equlibrium pressures of CO_(2),CO and O_(2) are 0.6, 0.4 and 0.2 atmosphere respectively. K_(p) for the reaction, 2CO_(2)hArr2CO+O_(2) is
Answer» <P>`0.089` `0.0533` `0.133` `0.177` SOLUTION :`K_(p)=([P_(CO)]^(2)[P_(O_(2))])/([P_(CO_(2))])=([0.4]^(2)xx[0.2])/([0.6]^(2))=0.0888.`

Discussion

30.	At 298 K, 100 ml solution containing 3.002 g of solute gave an osmotic pressure of 2.55 atmospheres. Find the molecular mass of the solute.
Answer» SOLUTION :`T = 298 K` `v=(100)/(1000)L` `M_B = ?` ` W_B = 3.002 g` `pi= 2.55` atm R = 0.0821 L atm। Kl mol `pi = (WB)/( M_B.V )RT.` `M_B =(W_B )/(M_B )`RT `=(3.002 xx 0.0821 xx 298 )/(2.55 xx (100) /(1000))` `= 288.2 `g/mol.

Discussion

31.	At 298 K, 100 cm^(3) of a solution containing 3.002 g of an unidentified solute exhibits an osmotic pressure of 2.55 atmospheres. What is the molar mass of solute? (R = 0.0821 L atm. "mol"^(-1)K^(-1))
Answer» SOLUTION :`"288 G MOL"^(-1)`

Discussion

32.	At 300 K temperature, Beaker A containing 0.02 M solution of urea and Beaker B containing 0.002 M solution of sugar are separated by semipermiable membrane with repsect should be applied to prevent osmosis ? Molecular weight of urea = 60 g / mol and molecular weight of sugar = 342 g / mol. [R=0.082" L atm. mol"^(-1)k^(-1)]
Answer» 0.4428 ATM. on BEAKER B 0.4920 atm. on beaker A 0.4920 atm. on beaker B NONE of the above SOLUTION :`pi=CRT` `=0.02xx0.082xx300` = 0492 atm. on beaker A

Discussion

33.	At 298 K,atm among A. H_(2)+O_(2)to2H_(2)O B. H_(2)+Cl_(2)to2HCl c. N_(2)+O_(2)to2NO D. H_(2)SO_(4)+KOHtoK_(2)SO_(4) products, correct order of reaction rates is
Answer» `DgtAgtCgtB` `DltAltBltC` `DgtBgtAgtC` `DgtB=CgtA` ANSWER :C

Discussion

34.	At 291K, saturated solution of BaSO_(4) was found to have a specific conductivity of 3.648xx10^(-6)ohm^(-1)cm^(-1), that of water used being 1.25xx10^(-6)ohm^(-1)cm^(-1). Ioinc conductances of Ba^(2+) and SO_(4)^(2-) ions are 110 and 136.6 ohm^(-1)cm^(2)mol^(-1) respectively. calculate the solubility of BaSO_(4) at 291 K. (At masses: Ba=137,S=32,O=16)
Answer» SOLUTION :`kappa(BaSO_(4))=kappa("solution")-kappa("water")." SOLUBILITY"=(kappaxx1000)/(wedge_(m)^(@))=((3.648-1.250)10^(-6)xx1000)/((110+136.6))xx233" g "L^(-1)`.

Discussion

35.	At 291 K, the molar conductivities at infinite of NH_(4)Cl, NaOH and NaCl are 128.8,217.4 and 108.9 S cm^(2) respectively. If the molar conductivity of a centrinormal solution of NH_(4)OH is 9.33 S cm^(2), what is the percentage dissociation of NH_(4)OH at this dilution? Also calculate the dissociation constant of NH_(4)OH.
Answer» Solution :Here, we are given: `wedge^(@)` for `NA_(4)CL=129.8" S "cm^(2),wedge^(@)` for `NaOH=217.4" S "cm^(2)`, `wedge^(@)` for `NaCl=108.9" S "cm^(2)`, Kohlrausch's law, `wedge^(@)` for `NH_(4)OH=lamda_(NH_(4)^(+))^(@)+lamda_(OH^(-))^(@)=wedge^(@)(NH_(4)Cl)+wedge^(@)(NaOH)-wedge^(@)(NaCl)` `=129.8+217.4-108.9=238.3" S "cm^(2)` `wedge_(c)=9.33" S "cm^(2)` (given) `THEREFORE`Degree of DISSOCIATION `(alpha)=(wedge_(c))/(wedge^(@))=(9.33)/(238.3)=0.0392` or % age dissociation `=0.0392xx100=3.92%` Calculation of dissociation constant `{:(,NH_(3)OH,hArr,NH_(4)^(+),+,OH^(-)),("Initial CONC.",c,,,,),("Equilibrium conc.",c-calpha=c(1-alpha),,calpha,,calpha""K=(calphaxxcalpha)/(c(1-alpha))=(calpha^(2))/(1-alpha)):}` Putting c=0.01N=0.01M and `alpha=0.0392, ` we get `K=((0.01)(0.0392)^(2))/(1-0.0392)=(10^(-2)XX(3.92xx10^(-2))^(2))/(0.9608)=1.559xx10^(-5)`.

Discussion

36.	At 300 K temperature 2.5 gram unknown substance is dissolved in solvent and made the volume 4 liter of the solution. Its osmotic pressure is found to be 0.2 bar. Calculate the molar mass of unknown substance.
Answer» 19.95 gms/mole 77.94 gms/mole 199.5 gms/mole 779.4 gm/mole ANSWER :B

Discussion

37.	At 291 K, the molar conductivities at infinite dilution of NH_(4)Cl, NaOH and NaCl are 129.8, 217.4 and 108.9" S " cm^(2) mol^(-1) respectively. The molar conductivity of a centinormal solution of NH_(4)OH is 9.33" S "cm^(2)mol^(-1). The percentage dissociation of NH_(4)OH at this dilution and the dissociation constant of NH_(4)OH are :
Answer» `3.92%,1.599xx10^(-5)` `6.92%,3.599xx10^(-5)` `3.92%,4.599xx10^(-2)` `9.92%,1.599xx10^(-5)` Solution :(a) `Lambda_((NH_(4)OH))^(@)=Lambda_((NH_(4)CL))^(@)+Lambda_((NaOH))-Lambda_((HCL))^(@)` `=129.8+217.4-108.9=238.3" s "cm^(2)mol^(-1)` Degree of dissociation `(alpha)=(Lambda_(C ))/(Lambda^(@))=((9.33" S "cm^(2)mol^(-1)))/((238.3" S "cm^(2)mol^(-1)))` `=0.0392=0.0392xx100=3.92%` Dissociation constant `NH_(4)OH hArr NH_(4)^(+)+OH^(-)` `{:("Initial CONC".,C,0,0),("Equilibrium conc".,C-Calpha,Calpha,Calpha),(,C(1-alpha),Calpha,Calpha):}` `K_(a)=(CalphaxxCalpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)` `=((0.01)xx(0.0392))/((1.0.0392))=1.599xx10^(-5)`

Discussion

38.	At 300 K specific conductivity of ethanol is 4xx10^(-10)mhocm^(-1). The ionic conductances of H^(+),C_(2)H_(5)O^(-) at his temperature is 300 and 100 mhocm^(2)" equivalent"^(-1) respectively. Then the negative logarithm of ionic product of alcohol will be 18.
Answer» Solution :`lamda^(infty)=lamda_(H^(+))^(infty)+lamda_(C_(2)H_(5)O^(-))^(infty)=400` `thereforelamda=kxx(1000)/(C)""C=[H^(+)]=[.^(-)OC_(2)H_(5)]` `thereforeC=(4xx10^(-10)xx1000)/(400)=10^(-9)M` `thereforeK_("ALCOHOL")=[H^(+)][OC_(2)H_(5)]=(10^(-9))^(2)` `thereforepK_("alcohol")=-log(10^(-18))=18`

Discussion

39.	At 283 K, which of the following coil exist in solid state ?
Answer» `SO_2, I_2, KCL` NACL, KCl, CSCL `H_2O, I_2, NaCl` `H_2O, CaF_2, KCl` ANSWER :B

Discussion

40.	At 300 K, half life ofgaseous reactant initially at 58 K pa is 320 min. When the pressure is 29kpa, the half life is 160 min. The order of the reaction is
Answer» Solution :`t_(1/2)underset("MIN")(to)""320""160` `P_(0)to""58""29` `((t_(1/2))_(1))/((t_(1/2))_(2))=((P_(2))/(P_(1)))^((n-1)),(320/160)=(29/58)^(n-1),2^(1)=2^((1-n)),1=1-n,n=1-1,n=0`

Discussion

41.	At [300 K, delta G^0 (N_2O_5(g)) = -43 kJ mol^-1]. if at a particular instant of time partial pressure of N_2 ,O_2 and N_2 O_5 gases respectively are 2 atm, 1 atm and 2 atm then the value of deltaG for reaction [2N_2(g) +5O_2(g) implies 2N_2O_5(g)] Will be
Answer» `-94.4 KJ mol^-1` `-57.4 kJ mol^-1` `-77.6 kJ mol^-1` `-43 kJ mol^-1` ANSWER :1

Discussion

42.	at 27^(@)C thereaction C_(6)H_(6)(l)+15/2O_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l) proceeds spontaneously becouse of magnitude of
Answer» `DELTAH=T .Delta S` `DeltaH GT T . DeltaS` `DeltaG lt T . DelatS` ` DeltaH gt 0 and T . DeltaSlt 0` Solution :in the REACTION , `DeltaS =- ve` Hencee , for the reaction to be SPONTANEOUS , `DeltaH` should be negative and higher than ` T . DeltaS i.e DeltaH gt T . DeltaS`

Discussion

43.	At 283 K, the osmotic pressure of 2% solution of X is 7.87 xx 10^4 Nm^(-2). Calculate molar mass of X.
Answer» ANSWER :598

Discussion

44.	At 300 K and 1 atm, 15mL of a gaseous hydrocarbon requires 375mL air containing 20% O_(2) by volume for complete combustion. After combination the gases occupy 330 mL. Assuming that the water formed isin liquid form and the volumes were measured at the Atomic mass of boron is 10.81. it has two isotopes namely ._(5)^(11)B and ._(5)^(x)B with their relative abundance of 80% and 20% respectively. the value of x is:
Answer» 10.05 10 10.01 10.02 Answer :B

Discussion

45.	At 27^(@)C, the ratio of rms velocities of ozone to oxygen is
Answer» `sqrt(3//5)` `sqrt(4//3)` `sqrt(2//3)` `0.25` Solution :`(u_(O_(3)))/(u_(O_(2)))=sqrt((M_(O_(2)))/(M_(O_(3))))=sqrt((32)/(48))=sqrt((2)/(3))`

Discussion

46.	At 27^(@)C , the ratio of r.m.s. velocities of ozone and oxygen is :
Answer» `SQRT( 3 //5 )` `sqrt( 4//3)` `sqrt( 2//3)` `0.25` Solution :`( "r.m.s. velocity" ( O_(3)))/("r.m.s. velocity"(O_(2))) = sqrt(( 32)/( 48)) = sqrt(( 2 )/( 3))`

Discussion

47.	At 300 K and 1 atm, 15mL of a gaseous hydrocarbon requires 375mL air containing 20% O_(2) by volume for complete combustion. After combination the gases occupy 330 mL. Assuming that the water formed isin liquid form and the volumes were measured at the same temperature and pressure. the formula of the hydrocarbon is:
Answer» `C_(3)H_(6)` `C_(3)H_(8)` `C_(4)H_(8)` `C_(4)H_(10)` Solution :Combustion of hydrocarbon `C_(x)H_(y)` is given as, `C_(x)H_(y)(g)+(x+(y)/(4))O_(2)(g)rarrxCO_(2)(g)+(y)/(2)H_(2)O(l)` 15ML hydrocarbon requires `15(x+(y)/(4))`mL oxygen. `therefore 15(x+(y)/(4))=75` or `x+(y)/(4)=5` `therefore` Hydrocarbon will be `C_(3)H_(8)`.

Discussion

48.	At 300 K , a gaseous reaction : A to B + C was found to follow first order kinetics . Starting with pure A , the total pressure at the end of 20 minutes was 100 mm of Hg . The total pressure after the completion of the reaction is 180 mm of Hg . The partial pressure of A (in mm of Hg) is
Answer» 100 90 180 80 Answer :d

Discussion

49.	At 27^(@)C temperature time required for 75 % completion of a first order reaction 20 seconds.What will be its rate constatn?
Answer» 0.693 `s^(-1) mol^(-1)` LT 0.0693 `s^(-1)` 0.693 `s^(-1)` 0.0693 `s^(-1) mol^(-1)` lt Solution :For =`(1)/(t)` in `(CO)/(ct)` at 75% completion :Ct =25% `K=(2.303)/(20)` log `(100)/(25)=0.0697 sec^(-1)`

Discussion

50.	At 27^@C, the osmotic pressure of a solution containing 4.0 g solute (molar mass = 246 )per litre at 27^@C is : R = 0.0821 atms. mol^-1k )
Answer» 0.1 atm 0.4 atm 0.2 atm 0.8 atm Answer :C

Discussion

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At 27^@C, the osmotic pressure of a solution containing 4.0 g solute (molar mass = 246 )per litre at 27^@C is : R = 0.0821 atms. mol^-1k )