94420 + Interview Questions in Chemistry in Class 12

1.	Equivalent conductivity of acetic acid at infinite dilution is 39.7 and for 0.1 M acetic acid the equivalent conductance is "5.2 mho.cm"^(2)."gm.equiv."^(-1). Calculate degree of dissociation, H^(+) ion concentration and dissociation constant of the acid.
Answer» Solution :`ALPHA=(lambda_(C ))/(lambda_(oo))=0.01333=1.33%` `{:(CH_(3)COOHhArrH^(+)+CHCOO^(-)),(C(1-alpha)""Calpha""Calpha):}` `therefore[H^(+)]=Calpha=0.1xx0.0133=0.00133M` `K=(alpha^(2)C)/(1-alpha)=(0.0133^(2)xx0.1)/((1-0.0133))=2.38xx10^(-5)M`

Discussion

2.	Equivalent conductivity of a weak acid HA at infinite dilution is 390 S cm^(2) eq^(-1).Conductivity of 1 xx 10^(-3) N HA solution is 4.9 xx 10^(-5) S cm^(-1). Calculate the extent of dissociation and dissociation constant of the acid.
Answer» Solution :Equivalent conductivity of `1 XX 10^(-3) N HA` solution `(Lambda_m) = k//"normality"` `Lambda_c = (4.9 xx 10^(-5) xx 1000)/(1 xx 10^(-3)) = 49 S cm^(2) eq^(-1)` Equivalent CONDUCTANCE at infinite dilution `(Lambda_0) = 390 S cm^(2) eq^(-1)` Extent of dissociation = `alpha = (Lambda_c)/(Lambda_0) = 49/390 = 0.126` Dissociation constant of ACID = `C alpha^(2) = 1 xx 10^(-3) (0.126)^(2) = 1.5 xx 10^(-5) mol L^(-1)`.

Discussion

3.	Equivalent conductivity of a weak acid HA at infinite dilution is390 S cm^(2) eq^(-1).Conductivity of 1xx10^(-3) N HA solution is 4.9xx10^(-5) S cm^(-1).Calculate the extent of dissociation and dissociation constant of the acid.
Answer» Solution :`Lambda_(c) = (4.9 xx 10^(-5) xx 1000)/(1xx10^(-3))=49 S cm^(2) eq^(-1)`. Equivalent CONDUCTANCE at infinite DILUTION `(Lambda_(0))=390 S cm^(2) eq^(-1)`. Extent of dissociation `= alpha = (Lambda_(c))/(Lambda_(0))=(49)/(390)=0.126`. Dissociation CONSTANT of ACID `= C alpha^(2) = 1xx10^(-3) (0.126)^(2)=1.5 x 10^(-5) "mol" L^(-1)`.

Discussion

4.	Equivalent conductivity at infinite dilution for sodium potassium oxalate, (COO^(-))_(2)Na^(+)K^(+), will be (given, molar conductivities of oxalate, K^(+) and Na^(+) ions at infinite diluton are 148.2,50.1,73.5" S "cm^(2)mol^(-1) respectively).
Answer» `271.8" S "cm^(2)eq^(-1)` `67.96" S "cm^(2)eq^(-1)` `543.6" S "cm^(2)eq^(-1)` `135.9" S "cm^(2)eq^(-1)`. Solution :`wedge_(m)^(@)` (Sod. Pot. Oxalate)`=lamda_(m)^(@)(Na^(+))+lamda_(m)^(@)(K^(+))+lamda^(2)`(oxalate) `=(50.1+73.5+148.2)" S "cm^(2)MOL^(-1)` `=271.8" S "cm^(2)mol^(-1)` `wedge_(eq)^(@)=(wedge_(m)^(@))/("TOTAL chrge on cations or ANIONS (n factor)")` `=(271.8)/(2)=135.9" S "cm^(2)eq^(-1)`.

Discussion

5.	Equivalent conductivity at infinite dilution for sodium potassium oxalate (COO^(-))_(2)Na^(+)K^(+) will be [given, molar conductivities of oxalate, K^+ and Na^(+) ions at infinite dilution are 148.2, 50.1, 73.5 S cm^(2)mol^(-1), respectively]
Answer» `271.8cm^(2)eq^(-1)` `67.95S" "cm^(2)eq^(-1)` `543.6S" "cm^(2)eq^(-1)` `135.9S" "cm^(2)eq^(-1)` Solution :`lamda_(M)^(OO)=lamda_(M)^(oo)("Oxalate")+lamda_(M)^(oo)(NA^(+))+lamda_(M)^(oo)(K^(+))` `lamda_(M)^(oo)=(148.2+50.1+73.5)S" "cm^(2)mol^(-1)` `lamda_(M)^(oo)=271.8S" "cm^(2)mol^(-1)` `thereforelamda_(Eq)^(oo)=(271.8)/(2)=135.9S" "cm^(2)eq^(-1)(lamda_(eq)^(oo)=(lamda_(M)^(oo))/("N. factor"))`

Discussion

6.	Equivalent conductivity at infinite dilution for sodium- potassium oxalate ((COO^(-))_2 Na^(+)K^(+)) will be [ Given molar conductivities of oxalate, K^(+) and Na^(+) ions at infinite dilution are 148.2, 50.1, 73.5 S cm^(2) mol^(-2) respecitively '
Answer» `271.8 S cm^(2) eq^(-1)` `67.95 S cm^2 eq^(-1)` `543.6 S cm^(2) eq^(-1)` `135.9 cm^(2) eq^(-1)` SOLUTION :`lambda_(m)^(oo) = lambda_(m("oxalate"))^(oo) + lambda^(m(NA^+))^(oo) + lambda_(m(K^+))^(oo)` `lambda^(m)^(oo) =(148.2 + 50.1+73.5)S cm^(2) mol^(-1)` `lambda_(m)^(oo) =271. 8Scm^(2) mol^(-1)` `therefore lambda_(eq)^(oo) = (271.8)/(2) = 135.8 S cm^(2) eq^(-1)((lambda_(eq)^(oo) = (lambda_m)^(oo))/("N. factor"))`

Discussion

7.	Equivalent conductivity at infinite dilutionfor sodium-potassium oxalate ((COO^(-))_2 Na^(+) K^(+)) will be [given, molar conductivities of oxalate, K^(+) and Na^(+) ions at infinite dilution are 148.2, 50.1, 73.5S cm^(2) mol^(-2), respectively ] .
Answer» `271.8 S cm^2 EQ^(-1)` `67.95 S cm^2 eq^(-1)` `543. 6 S cm^2 eq^(-1) ` `135.9 S cm^2 eq^(-1)` ANSWER :A

Discussion

8.	Equivalent conductivity and infinite dilution of NaCl at C concentration is lamda_(C) and lamda_(oo) respectively. So, for lamda_(C) and lamda_(oo), which relation is true ? (where, constant B is solid).
Answer» `lamda_(C)=lamda_(OO)(B)SQRT(c)` `lamda_(C)=lamda_(oo)+(B)sqrt(c)` `lamda_(C)=lamda_(oo)+(B)C` `lamda_(C)=lamda_(oo)-(B)C` Solution :`lamda_(C)=lamda_(oo)(B)sqrt(c)`

Discussion

9.	Equivalentconductance of BaCI_(2), H_(2)SO_(4) and HCIare x_(1)x_(2) and x_(3) S cm^(2) "equiv"^(-1)atinfinitedilution if specific conductance of saturated BaSO_(4)solutionis of y S cm d^(-1) then k_("SP") of BaSO_(4) is
Answer» `(10^(3)y)/(2(x_(1)+x_(2)-2x_(3))` `(10^(6)y^(2))/(x_(1)+x_(2)-2x_(3))^(2)` `(10^(6)y^(2))/(4(x_(1)+x_(2)-2x_(3))^(2)` `(x_(1)+x_(2)-2x_(3))/(10^(6)y^(2))` Solution :`wedge_(BaSO_(4))^(@)=wedge_(BaCI_(2)^(@)+wedge_(H_(2)SO_(4)^(@)-2 wedge_(H_(2)SO_(4))^(@)-2wedge_(HCI)^(@)=(X_(1)+X_(2)-2x_(3))` `wedge_(BaSO_(4))^(@)=(1000xx"specific conductance")/("solubility" ("in SATURATED solution" ))` `X_(1)+x_(2)-2x_(3)=(1000Y)/("solubility")` `therefore`solubility `(BaSO_(4))=(1000y)/(x_(1)+x_(2)-2x_(3))N=(1000y)/(2(x_(1)+x_(2)-2x_(3))M` `K_(SP)(BaSO_(4))=[Ba^(2+)][SO_(4)^(2-)]M^(2)=(10^(6)y^(2))/(4(x_(1)+x_(2)-2_(3))^(2)`

Discussion

10.	Equivalent conductance _________ with dilution.
Answer» SOLUTION :INCREASES

Discussion

11.	Equivalent conductance of saturated BaSO_4 is 400 ohm^(-1) cm^(2) "equiv"^(-1) and specific conductance is8 xx 10^(-5) ohm^(-1), cm^(-1) . Hence K_(sp) of BaSO_4is
Answer» `4 XX 10^(-8) M^(2) ` `1 xx 10^(-8) M^(2)` `2 xx 10^(-4) M^(2)` `1 xx 10^(-4) M^(2)` SOLUTION :`^^_(EQ) = (K xx 1000)/(N^(0)) , N = (8 xx 10^(-5) xx 1000)/(400) = 2 xx 10^(-4) N` ` S = M = (N)/(2) = (2 xx 10^(-4))/(2) = 10^(-4) M , K_(sp) = S^(2) = (10^(-4))^(2) = 10^(-5)`

Discussion

12.	Equivalent conductance of 0.01 N Na_2SO_4 solution is 112.4ohm^(-1)cm^2eq^(-1).The equivalent conductance at infinite dilution is 129.9ohm^(-1)cm^2eq^(-1). What is the degree of dissociationin 0.01 N Na_2SO_4 solution
Answer»

Discussion

13.	Equivalent conductance for weak electrolyte on dilution ________.
Answer» SOLUTION :INCREASES RAPIDLY

Discussion

14.	Equivalent conductance for week electrolyte on dilution ________.
Answer» SOLUTION :INCREASES RAPIDLY

Discussion

15.	Equivalent conductance for strong electrolyte on dilution decreases rapidly.
Answer» SOLUTION :INCREASES SLOWLY.

Discussion

16.	Equivalent conductance for strong electrolyte on dilution decreases rapidly. Is it true or false?
Answer» SOLUTION :INCREASES SLOWLY.

Discussion

17.	Equinormal solution of two weak acids, HA(pK_(a) =3) and HB(pK_(a) =5) are each placed in contact with standard hydrogen electrode at 25^(@)C. When a cell is constructed by interconnecting them thorugh a salt bridge find the e.m.f. of the cell.
Answer» SOLUTION :The cell is, `Pt H_(2(1atm)) \|HA_(2)\|\|HA_(I) \|H_(2(1atm)) Pt` At `L.H.S.: E_(H//H^(+)) = E_(OP_(H//H^(+)))^(@) +(0.059)/(1) log_(10) [H^(+)]_(2)` `:' -log H^(+) =pH :. E_(H//H^(+)) = E_(OP_(H//H^(+)))^(@) -0.059(pH)_(2)` At `R.H.S. : E_(H^(+)//H) =E_(RP_(H^(+)//H))^(@) +(0.059)/(1)log[H^(+)]_(1)` `:. E_(H^(+)//H) =E_(RP_(H^(+)//H))^(@) -0.059(pH)_(1)` For Acid `HA_(1) HA_(1) hArr H^(+) +A_(1)^(-)` `[H^(+)] = C. alpha = sqrt(K_(a)/C)` `:. (pH)_(1) = (1)/(2) pK_(a_(1)) -(1)/(2) log_(10)C` SIMILARLY, `(pH)_(2) = (1)/(2) pK_(a_(1)) -(1)/(2) log_(10)C` `( :' C` are same) `E_(cell) = E_(OP_(H^(+)//H))^(@) +E_(RP_(H^(+)//H))^(@)` For II for I `= 0.059 [(1)/(2) pK_(a_(2)) -(1)/(2) pK_(a_(1))] =(0.059)/(2) [5-3]` `=+0.059`

Discussion

18.	Equivalent conductance for strong electrolyte on dilution _______.
Answer» SOLUTION :INCREASES SLOWLY

Discussion

19.	Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH.
Answer» `MgCl_(2)` `CaCl_(2)` `SrCl_(2)` `BaCl_(2)` Solution :`BaCl_(2)` is a SALT of strong BASE `Ba(OH)_(2)` and strong acid HCl and it will record HIGHEST pH.

Discussion

20.	Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value-
Answer» LICL `BaCl_2` `BeCl_2` `AlCl_3` Solution :In aqueous solution, metal cation `(M^(N+))` remains associated with water `[M(H_2O)_x]^(n+)`. The hydrated cation donates proton through the following reaction. `[M(H_2O)_x]^(n^+)X-1)OH]^(n-1)+H^+(aq)` A hydrated metal cation in which the cation possesses high charge density has high ability to donate proton. In the GIVEN salts, the charge densities of the cations follow the order `Ba^(2+)

Discussion

21.	Equimolar solutions of the following were prepared in water separately. Which of the solutions will have the highest pH-
Answer» `SnCl_2` `BaCl_2` `MgCl_2` `CaCl_2` ANSWER :B

Discussion

22.	Equimolar solutions of NaCl and BaCl_(2) are prepared . If the freezing point of NaCl is - 2^(@)C, the freezing point of BaCl_(2) is expected to be
Answer» `-2^(@)C` `-3^(@)C` `-1.5^(@)C` `-1.66^(@)C` Solution :`DeltaT_(f)=iK_(f)xxm` For NACL, `DeltaT_(f)=2, i=2` `2=2xxK_(f)xxm` For `BaCl_(2), DeltaT_(f)=?, i=3` `DeltaT_(f)=3xxK_(f)xxm` Since `K_(f)` and m are same for equimolar solutions, `(DeltaT_(f))/(2)=3/4` or `DeltaT_(f)=3` `:.` Freezing POINT of `BaCl_(2)` solution `=0-3=-3^(@)C`

Discussion

23.	Equimolar solutions of electrolytes with different number of ion in the same solvent have
Answer» Same BOILING POINT but DIFFERENT FREEZING point Same freezing point but different boiling point Same boiling and same freezing points Different boiling and different freezing point Answer :D

Discussion

24.	Equimolar solutions of electrolytes in the same solvent have
Answer» Same BOILING point but DIFFERENT freezing point Same freezing point but different boiling point Same boiling and same freezing POINTS Different boiling and different freezing point ANSWER :D

Discussion

25.	Equimolar solutions (A) of benzonic acid in bvenzene and (B) of benzoic acid in water are taken How are the Van't Hoff facdtors of the solutions related ?
Answer» Solution :In benzene solvet, benzoic acid exists as dimar while in water, it dissociates into IONS. `"Van't Hoff FACTOR (i)"=("Normal molar mass")/("Observed molar mass")` For solution A in benzene, `ilt1` and for solution B in water `igt1`. Thus, the Van't Hoff factor for solution A is less then the Van't Hoff factor for solution B.

Discussion

26.	Equimolar mixture of two gases A_(2) and B_(2) is taken in a rigid vessel at temperature 300 K. The gases reacts according to given equations : A_(2)(g)hArr2 A (g)K_(P1)=? B_(2)(g)hArr2 B (g)K_(P2)=? A_(2)(g)+B_(2)(g)hArr2 AB (g)K_(P3)=2 If the initial pressure in the container was 2 atm and final pressure developed at equilibrium is 2.75 atm. in which equilibrium partial pressure of gas AB was 0.5 atm, calculate the ratio of (K_(P2))/(K_(P1)) [Given : Degree of dissociation of B_(2) is greater than A_(2)].
Answer» 8 9 `1//8` NONE of these

Discussion

27.	Equimolar concentrations of H_(2)and I_(2) are heated to equilibrium in a 2 litre falsek. At equlibrium, the forward and the backward rate constants are found to be equal. What percentage of initial concentration of H_(2) has reacted at equilibrium
Answer» `33%` `66%` `50%` `40%` ANSWER :A

Discussion

28.	Equimolar mixture of alpha-D(+)-glucose has specific ([alpha]_(D)) is
Answer» `-92.4^@` `+112.5^@` `+52.5^@` `-19.2^@` ANSWER :C

Discussion

29.	Equimolar mixture of hydrogen and carbonmonoxide is called as………………………
Answer» ANSWER :WATER GAS

Discussion

30.	Equimolar aqueous solutions of NaCI and BACI_(2) are prepared. If the freezing point of NaCI is -2^(@)C, the freezing point of BaCI_(2) solution is expected to be:
Answer» `-2^(@)C` `-3^(@)C` `-1.5^(@)C` `-1.66^(@)C` Solution :`DeltaT_(f)("COLLIGATIVE property")` in linked with number of PARTICLES or IONS in solution `NaCItoNa_((aq))^(+)+CI_((aq))^(-)` ` BaCI_(2)toBa_((aq))^(2+)+2CI_((aq))^(-)` `"If "alph a"for " NaCI is -2^(@)C` `"Then " ALPHA "for" BaCI_(2) "is EXPECTED tobe" =-3^(@)C.`

Discussion

31.	Equimolal solutions will have the same elevation in boiling point, provided they do not show:
Answer» Electrolysis Association Dissociation Association or dissociation Answer :D

Discussion

32.	Equimolal solutions will have the same boiling point, provided they do not show
Answer» electrolysis ASSOCIATION dissociation association or dissociation Solution :The formula `Delta T =K_(B) xx `molality is valid when solute NEITHER DISSOCIATES nor ASSOCIATE. In case of dissociation or association. `Delta T=iK_(b) xx` `"molality" (iK_(b) xx "molality")`

Discussion

33.	Equimolal solutions of NaCl and BaCl_(2) are prepared in water. Freezing point of NaCl is foundto be -2^(@)C. What freezing point do you expect for BaCl_(2) solution?
Answer» Solution :`"i for NaCl = 2, i for BaCl"_(2)=3`. HENCE, `((DeltaT_(F))_("NACl"))/((DeltaT_(f))_("BaCl"_(2)))=(2)/(3)or(DeltaT_(f))_("BaCl"_(2))=(3)/(2)xx2=3^(@)" so that T"_(f)" for BACl"_(2)=-3^(@)C`.

Discussion

34.	Equimolal solutions of A and B show depression in freezing point in the ratio 2 : 1. A remains in its normal state in solutions. B will be in solution:
Answer» Normal Dissociated Associated Hydrolysed Answer :C

Discussion

35.	Equilibrium constants are given (in atm) for the following reactions at 0°C : SrCl_(2). 6H_(2)O(s)hArrSrCl_(2).2H_(2)O(s) + 4H_(2)O(g) K_(p) = 5 x× 10^(-12) Na_(2)HPO_(4).12H_(2)O(s)hArrNa_(2)HPO_(4). 7H_(2)O (s)+ 5H_(2)O(g)K_(p) = 2.43 x× 10^(-13) Na_(2)SO_(4).10H_(2)O(s)hArrNa_(2)SO_(4)(s) + 10 H_(2)O(g) K_(p) = 1.024 x× 10^(-27) The vapour pressure of water at 0° C is 4.56 torr. At what relative humidities will Na_(2)SO_(4) be deliquescent (i.e. absorb moisture) when exposed to the air at 0°C?
Answer» above 33.33 % below 33.33 % above 66.66% below 66.66% ANSWER :A

Discussion

36.	Equilibrium constants are given (in atm) for the following reactions at 0°C : SrCl_(2). 6H_(2)O(s)hArrSrCl_(2).2H_(2)O(s) + 4H_(2)O(g) K_(p) = 5 x× 10^(-12) Na_(2)HPO_(4).12H_(2)O(s)hArrNa_(2)HPO_(4). 7H_(2)O (s)+ 5H_(2)O(g)K_(p) = 2.43 x× 10^(-13) Na_(2)SO_(4).10H_(2)O(s)hArrNa_(2)SO_(4)(s) + 10 H_(2)O(g) K_(p) = 1.024 x× 10^(-27) The vapour pressure of water at 0° C is 4.56 torr. At what relative humidities will Na_(2)SO_(4). 10 H_(2)O be efflorescent (release moisture) when exposed to air at 0°C ?
Answer» above 33.33 % below 33.33 % above 66.66% below 66.66% ANSWER :B

Discussion

37.	Equilibrium constants are given (in atm) for the following reactions at 0°C : SrCl_(2). 6H_(2)O(s)hArrSrCl_(2).2H_(2)O(s) + 4H_(2)O(g) K_(p) = 5 x× 10^(-12) Na_(2)HPO_(4).12H_(2)O(s)hArrNa_(2)HPO_(4). 7H_(2)O (s)+ 5H_(2)O(g)K_(p) = 2.43 x× 10^(-13) Na_(2)SO_(4).10H_(2)O(s)hArrNa_(2)SO_(4)(s) + 10 H_(2)O(g) K_(p) = 1.024 x× 10^(-27) The vapour pressure of water at 0° C is 4.56 torr. Which is the most effective drying agent at 0°C
Answer» `SrCl_(2).2H_(2)O` `Na_(2)HPO_(4). 7H_(2)O` `Na_(2)SO_(4)` all equally Answer :A

Discussion

38.	The equilibrium constant for the reaction is 10 at 300K. What will be the value of triangleG^@ ?
Answer» Standard free energy CHANGE`traingleG^@` TEMPERATURE T Heat enthalpy None Answer :A

Discussion

39.	Equilibrium constant of a reaction is related to
Answer» Standard FREE energy CHANGE `DELTAG^(@)` Free energy change `DeltaG` Temperature T None Solution :`DeltaG^(@)=-2.303 RT log k`.

Discussion

40.	Equilibrium constant K_(p) for the reaction: CaCO_(3)(s)
Answer» Solution :`CaCO_(3)(s) `K_(p) = P_(CO_(2)) = 0.82` ATM, `n_(CO_(2)) =(PV)/(RT) = (0.82 xx 20)/(0.082 xx 1000) = 0.2` mole Mole of `CaCO_(3)` dissociated `=n_(CO_(2)) =0.2` AMOUNT dissociated `=0.2 xx 100 = 20g`

Discussion

41.	Equilibrium constant is related to E^(@) but not to E_(cell). Explain why?
Answer» Solution :When equilibrium is reached in the cell REACTION, BECOMES equal to ZERO. However `E_(cell)^(@)` is a CONSTANT quantitiy. Hence, applying Nernst equation to the cell reaction, e.g., to the reaction: `Zn+Cu^(2+)hArrZn^(2+)+Cu`, `E_(cell)=E_(cell)^(@)-(RT)/(nF)"LN"([Zn^(2+)])/((Cu^(2+)))=E_(cell)^(@)-(RT)/(nF)"ln "K_(c)`, At equilibrium `E_(cell)=0`. hence, `E_(cell)^(@)=(RT)/(nF)"ln "K_(c)`.

Discussion

42.	Equilibrium constant for the following reactions at 1200 K are given : 2H_(2)O_((g))iff2H_(2(g))+O_(2(g)),K_(1)=6.4xx10^(-8) 2CO_(2(g))iff2CO_((g))+O_(2(g)),K_(2)=1.6xx10^(-6) The equilibrium constant for the reaction H_(2(g))+CO_(2(g))iffCO_((g))+H_(2)O_((g)) at 1200 K will be
Answer» `0.05` 20 `0.2` `5.0` Solution :Given : `2H_(2)O_((G))iff2H_(2(g))+O_(2(g))` `K_(1)=6.4xx10^(-8)""...(i)` `2CO_(2(g))iff2CO_((g))+O_(2(g))` `K_(2)=1.6xx10^(-6)""...(ii)` Required equation is, `H_(2(g))+CO_(2(g))iffCO_((g))+H_(2)O_((g)),K=?` By reversing equation (i) and by multiplying it with 1/2, we get `H_(2(g))+(1)/(2)O_(2(g))iffH_(2)O_((g)),K._(1)=sqrt((1)/(6.4xx10^(-8)))""...(iii)` And by multiplying equation (ii) with 1/2, we get `H_(2(g))+CO_(2(g))iffCO_((g))+H_(2)O_((g)),K.=K._(1)xxK._(2)` `K.=sqrt((1.6xx10^(-6))/(6.4xx10^(-8)))=sqrt(25)=5`

Discussion

43.	Equilibrium constant for the following reactions have been determined at 823K. CaO(s)+H_(2)(g)hArrCO(s)+H_(2)O(g) K_(1)=60 CaO(s)+CO(g)hArrCO(S)+CO_(2)(g) K_(2)=400 Using this information, calculate, equilibrium constant (at the same temperature) for :- CO_(2)(g)+H_(2)(g)hArrCO(s)+H_(2)O(g)K_(3)=? CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g)K_(4)=?
Answer» `K_(3)=0.15,K_(4)=6.66` `K_(3)=1.5,K_(4)=66.6` `K_(3)=15,K_(4)=666` None Solution :`K_(3)=(K_(1))/(K_(2))` `K_(4)=(K_(2))/(K_(1))`

Discussion

44.	Equilibrium constant for the given reaction is K=10^(20) at temperature 300 K A(s)+2B(aq.)hArr2C(s)+D(aq.) K=10^20 The equilibrium conc. of B starting with mixture of 1 mole of A and1//3 mole/litre of B at 300 K is
Answer» `~4XX10^(-11)` `~2xx10^(-10)` `~2xx10^(-11)` `~10^(-11)` Solution :`{:(,A(s)+,2B(aq)""hArr,2C(s)+,D(aq)),("INITIAL",1,1/3,0,0),(At_(eq),1-x,underset(~~a)(1/3-x),2x,x):}` `x~~1//3` `10^(20)=(1/3)/[B]^2 " " implies 10^20=(1/3)/a^2" " implies a^2=1/(3xx10^20)=10^(-20)/3` `a=10^(-10)/sqrt3~~4xx10^(-11)M`

Discussion

45.	Equilibrium constant for reaction NH_(4)OH(aq)+H^(+)(aq)hArr NH_(4)^(+)(aq)+H_(2)O(l) 1.8xx19^(9). Hence equilibrium constant for ionization NH_(3)+H_(2)OhArr NH_(4)^(+)(aq)+OH^(-)(aq) is x xx 10^(-6). The value of 'x' is
Answer» ANSWER :18

Discussion

46.	Equilibrium concentrations of A and B involved in following equilibrium are 0.01 M and 0.02 M respectively: [A(g)
Answer» `6.67` X `10^-2` M `2.67` x `10^-2` M `1.33` x `10^-2` M `1.33` x `10^-1` M Answer :4

Discussion

47.	Equilibrium concentration of HI,I_(2) and H_(2) is 0.7, 0.1 and 0.1 M respectively. The equlibrium constant for the reaction I_(2)+H_(2)hArr2HIis
Answer» 36 49 `0.49` `0.36` SOLUTION :`K_(C)=([HI]^(2))/([H_(2)][I_(2)])=([0.7]^(2))/([0.1][0.1])=49`

Discussion

48.	Equilibrium concentration of HI, I_(2) and H_(2) is 0.7, 0.1 and 0.1 moles/litre. Calculate the equilibrium constant for the reaction : I_(2(g))+H_(2(g))hArr2HI_((g))-
Answer» 0.36 36 49 0.49 Answer :C

Discussion

49.	Equilibrium concentration of HI, I_2 and H_2 are 0.7, 0.1 and 0.1 M respectively.The equilibrium constant forthe reaction, I_2 + H_2 ⇌ 2HI is :
Answer» 0.36 36 49 0.49 Answer :C

Discussion

50.	Equations have been applied to the above problems for hydrogen atom. Can these equations be applied to calculate r,E, v, etcfor He^(+) and Li^(2+) ions?
Answer» Solution :As the Bohr theory is applicable to a ONE-electron system, the said equations can be APPLIED to `He^(+) and LI^(2+)` ions as these species have only one electron each. The value of Z will be taken as 2 and 3 for `He^(+) and Li^(2+)` respectively.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Equivalent conductivity of acetic acid at infinite dilution is 39.7 and for 0.1 M acetic acid the equivalent conductance is "5.2 mho.cm"^(2)."gm.equiv."^(-1). Calculate degree of dissociation, H^(+) ion concentration and dissociation constant of the acid.

Equivalent conductivity of a weak acid HA at infinite dilution is 390 S cm^(2) eq^(-1).Conductivity of 1 xx 10^(-3) N HA solution is 4.9 xx 10^(-5) S cm^(-1). Calculate the extent of dissociation and dissociation constant of the acid.

Equivalent conductivity of a weak acid HA at infinite dilution is390 S cm^(2) eq^(-1).Conductivity of 1xx10^(-3) N HA solution is 4.9xx10^(-5) S cm^(-1).Calculate the extent of dissociation and dissociation constant of the acid.

Equivalent conductivity at infinite dilution for sodium potassium oxalate, (COO^(-))_(2)Na^(+)K^(+), will be (given, molar conductivities of oxalate, K^(+) and Na^(+) ions at infinite diluton are 148.2,50.1,73.5" S "cm^(2)mol^(-1) respectively).

Equivalent conductivity at infinite dilution for sodium potassium oxalate (COO^(-))_(2)Na^(+)K^(+) will be [given, molar conductivities of oxalate, K^+ and Na^(+) ions at infinite dilution are 148.2, 50.1, 73.5 S cm^(2)mol^(-1), respectively]

Equivalent conductivity at infinite dilution for sodium- potassium oxalate ((COO^(-))_2 Na^(+)K^(+)) will be [ Given molar conductivities of oxalate, K^(+) and Na^(+) ions at infinite dilution are 148.2, 50.1, 73.5 S cm^(2) mol^(-2) respecitively '

Equivalent conductivity at infinite dilutionfor sodium-potassium oxalate ((COO^(-))_2 Na^(+) K^(+)) will be [given, molar conductivities of oxalate, K^(+) and Na^(+) ions at infinite dilution are 148.2, 50.1, 73.5S cm^(2) mol^(-2), respectively ] .

Equivalent conductivity and infinite dilution of NaCl at C concentration is lamda_(C) and lamda_(oo) respectively. So, for lamda_(C) and lamda_(oo), which relation is true ? (where, constant B is solid).

Equivalentconductance of BaCI_(2), H_(2)SO_(4) and HCIare x_(1)x_(2) and x_(3) S cm^(2) "equiv"^(-1)atinfinitedilution if specific conductance of saturated BaSO_(4)solutionis of y S cm d^(-1) then k_("SP") of BaSO_(4) is

Equivalent conductance _________ with dilution.

Equivalent conductance of saturated BaSO_4 is 400 ohm^(-1) cm^(2) "equiv"^(-1) and specific conductance is8 xx 10^(-5) ohm^(-1), cm^(-1) . Hence K_(sp) of BaSO_4is

Equivalent conductance of 0.01 N Na_2SO_4 solution is 112.4ohm^(-1)cm^2eq^(-1).The equivalent conductance at infinite dilution is 129.9ohm^(-1)cm^2eq^(-1). What is the degree of dissociationin 0.01 N Na_2SO_4 solution

Equivalent conductance for weak electrolyte on dilution ________.

Equivalent conductance for week electrolyte on dilution ________.

Equivalent conductance for strong electrolyte on dilution decreases rapidly.

Equivalent conductance for strong electrolyte on dilution decreases rapidly. Is it true or false?

Equinormal solution of two weak acids, HA(pK_(a) =3) and HB(pK_(a) =5) are each placed in contact with standard hydrogen electrode at 25^(@)C. When a cell is constructed by interconnecting them thorugh a salt bridge find the e.m.f. of the cell.

Equivalent conductance for strong electrolyte on dilution _______.

Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH.

Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value-

Equimolar solutions of the following were prepared in water separately. Which of the solutions will have the highest pH-

Equimolar solutions of NaCl and BaCl_(2) are prepared . If the freezing point of NaCl is - 2^(@)C, the freezing point of BaCl_(2) is expected to be

Equimolar solutions of electrolytes with different number of ion in the same solvent have

Equimolar solutions of electrolytes in the same solvent have

Equimolar solutions (A) of benzonic acid in bvenzene and (B) of benzoic acid in water are taken How are the Van't Hoff facdtors of the solutions related ?

Equimolar concentrations of H_(2)and I_(2) are heated to equilibrium in a 2 litre falsek. At equlibrium, the forward and the backward rate constants are found to be equal. What percentage of initial concentration of H_(2) has reacted at equilibrium

Equimolar mixture of alpha-D(+)-glucose has specific ([alpha]_(D)) is

Equimolar mixture of hydrogen and carbonmonoxide is called as………………………

Equimolar aqueous solutions of NaCI and BACI_(2) are prepared. If the freezing point of NaCI is -2^(@)C, the freezing point of BaCI_(2) solution is expected to be:

Equimolal solutions will have the same elevation in boiling point, provided they do not show:

Equimolal solutions will have the same boiling point, provided they do not show

Equimolal solutions of NaCl and BaCl_(2) are prepared in water. Freezing point of NaCl is foundto be -2^(@)C. What freezing point do you expect for BaCl_(2) solution?

Equimolal solutions of A and B show depression in freezing point in the ratio 2 : 1. A remains in its normal state in solutions. B will be in solution:

The equilibrium constant for the reaction is 10 at 300K. What will be the value of triangleG^@ ?

Equilibrium constant of a reaction is related to

Equilibrium constant K_(p) for the reaction: CaCO_(3)(s)

Equilibrium constant is related to E^(@) but not to E_(cell). Explain why?

Equilibrium constant for the following reactions at 1200 K are given : 2H_(2)O_((g))iff2H_(2(g))+O_(2(g)),K_(1)=6.4xx10^(-8) 2CO_(2(g))iff2CO_((g))+O_(2(g)),K_(2)=1.6xx10^(-6) The equilibrium constant for the reaction H_(2(g))+CO_(2(g))iffCO_((g))+H_(2)O_((g)) at 1200 K will be

Equilibrium constant for the given reaction is K=10^(20) at temperature 300 K A(s)+2B(aq.)hArr2C(s)+D(aq.) K=10^20 The equilibrium conc. of B starting with mixture of 1 mole of A and1//3 mole/litre of B at 300 K is

Equilibrium constant for reaction NH_(4)OH(aq)+H^(+)(aq)hArr NH_(4)^(+)(aq)+H_(2)O(l) 1.8xx19^(9). Hence equilibrium constant for ionization NH_(3)+H_(2)OhArr NH_(4)^(+)(aq)+OH^(-)(aq) is x xx 10^(-6). The value of 'x' is

Equilibrium concentrations of A and B involved in following equilibrium are 0.01 M and 0.02 M respectively: [A(g)

Equilibrium concentration of HI,I_(2) and H_(2) is 0.7, 0.1 and 0.1 M respectively. The equlibrium constant for the reaction I_(2)+H_(2)hArr2HIis

Equilibrium concentration of HI, I_(2) and H_(2) is 0.7, 0.1 and 0.1 moles/litre. Calculate the equilibrium constant for the reaction : I_(2(g))+H_(2(g))hArr2HI_((g))-

Equilibrium concentration of HI, I_2 and H_2 are 0.7, 0.1 and 0.1 M respectively.The equilibrium constant forthe reaction, I_2 + H_2 ⇌ 2HI is :

Equations have been applied to the above problems for hydrogen atom. Can these equations be applied to calculate r,E, v, etcfor He^(+) and Li^(2+) ions?