94420 + Interview Questions in Chemistry in Class 12

1.	H_2PO_4^(-) the conjugate base of……..
Answer» `PO_4^(3-)` `P_2O_5` `H_3PO_4` `HPO_4^(2-)` Solution :`H_3PO_4+H-OHleftrightarrowH_3O^(+)+H_2PO_4^(-)` ACID 1 base 1 ACI d 2 base 2 `therefore H_2Po_4^(-)` is the CONJUGATE base of `H_3PO_4`

Discussion

2.	H_(2)O(l)hArrH^(+)(aq)+OH^(-)(aq),DeltaH=+13.7 kcal/mol K_(eq) at 25^(@)C=10^(-14). Calculate K_(eq) at 60^(@)C?
Answer» Solution :`LOG(K_(2))/(K_(1))=(DeltaH)/(2.303R)[1/(T_(1))-1/(T_(2))]` `log(K_(2))/(10^(-14))=(13.7)/(2.303xx2xx10^(-3))[1/298-1/333]` `K_(2)=1.26xx10^(-12)`

Discussion

3.	{:([(H_(2)O)_(4)Co(O_(2))_(2)Co(H_(2)O)_(4)](SO_(4))_(2)overset[[X]"reagent"]to [(H_(2)O)_(4)Co(O_(2))_(2)Co(H_(2)O)_(4)](SO_(4))),(""(I)""""(II)"):} In both the complex Co hast_(2g)^(6)e_(g^(0)) configuration. Which option is incorrect ?
Answer» Complex (I) is Paramagnetic Complex (II) is Diamagnetic (X) is OXIDISING AGENT (X) is reducing agent Solution :`(I) O_(2)^(-)` is paramagnetic `(II) O_(2)^(2-)`is diamagnetic

Discussion

4.	[(H_(2)O)_(8)Fe_(2)(OH)_(2)].(SO_(4))_(2) Calculate the coordination number of central metal, E.A.N., Primary valencies. [If the coordination number is 4, EAN is 18, Primary valency = 2 than answer is 4182.]
Answer» Solution :Fe-Fe bondis PRESENT , C.N. of Fe=6 EAN=26-2+10+1=35,P.V. of 1 `OH^(-), 2SO_(4)^(2-)=3`

Discussion

5.	H_2Oboils at higher temperature than H_2S becauseit is capable of forming :
Answer» IONIC bonds Covalentbonds Hydrogen bonds Metallic bonds Answer :C

Discussion

6.	H_(2)O_(2)toH_(2)O+1/2O_(2)(1storder) Time to time the H_(2)O_(2) solutioni is titrated with standard acidified KMnO_(4). Thedata is Time (min)""0""15 KMnO_(4) ("consumed")""16ml""2ml What is half life in minutes?
Answer» SOLUTION :`C_(0)` ` C_(0)-X` `0` `K=2.303/15xxlog(16/2),K=2.303/15xx3xx0.3010=0.693/5` `t_(1/2)=(0.693/K),t_(1/2)=(0.693xx5)/0.693=5`

Discussion

7.	H_(2)O_(2)overset(1st"order")(to)H_(2)O+1/2O_(2). Pressure of O_(2) is recorded as: time (min)"" 10 min ""prop P_(O_(2)) (mm)""120""160 What is half life?
Answer» Solution :`H_(2)O_(2)O_(2)toH_(2)O+1/2O_(2)` `C_(0)"P_(0)=0` `(C_(0)-x)""P_(t)=120` `0""P_(alpha)=160` `KX2.303/10="LOG"160/40,t_(1/2)=0.693/k=5` `H_(2)O_(2)toH_(2)O+1/2O_(2)`

Discussion

8.	H_2O_2 when added to a solution containing KMnO_4 and H_2SO_4 acts :
Answer» As an oxidising AGENT As a REDUCING agent Both as an oxidising as well as a reducing agent None of these Answer :B

Discussion

9.	H_2O_2 restores the colour of old lead paintings , blackenedby the action of H_2S gas, by :
Answer» CONVERTING `PbO_2` TO Pb By OXIDISING PBS to `PbSO_4` Converting `PbCO_3` to Pb Oxidising `Pbso_3` to `PbSO_4` ANSWER :B

Discussion

10.	H_2O_2 reduces K_3Fe(CN)_6 in
Answer» NEUTRAL solution Acidic solution Alkaline solution Non-polar medium Answer :C

Discussion

11.	H_(2)O_(2) reduces chlorine to chloride. Write the coefficients of all substances in the equation.
Answer» Solution :The EQUATION is GIVEN as `H_(2)O_(2) + Cl_(2) RARR HCl + O_(2)` The balanced equation is `H_(2)O_(2) + Cl_(2) rarr 2HCL + O_(2)` The coefficient of `H_(2)O_(2), HCl` and `O_(2)` are = 1,1,2 and 1

Discussion

12.	H_2O_2 on treatment with chlorine gives:
Answer» `H_2` OXYGEN HYPOCHLOROUS ACID `ClO_2` ANSWER :B

Discussion

13.	H_(2)O_(2) on reaction with PbSgives
Answer» PbO `PbSO_(4)` `PbO_(2)` `PbHSO_(4)` SOLUTION :When HYDROGEN peroxide react with PbS then they form `PbSO_(4)`.

Discussion

14.	H_2O_2 is prepared in the laboratory when:
Answer» `MnO_2` is added to DILUTE cold `H_2SO_4` `BaO_2` is added to `CO_2` BUBBLING through cold WATER `PbO_2` is added to an acidified solution of `KMnO_4` `Na_2O_2` is added to boiling water Answer :B

Discussion

15.	H_(2)O_(2) is formed in the upper atmosphere through the following mechanism H_(2)O+(O) to 2" OH" to H_(2)O_(2) The overall enthalpy change and activation energy for the forward reaction are 72" kJ mol"^(-1) and 77" kJ mol"^(-1) respectively. The activation energy for the decomposition of H_(2)O_(2) to give back H_(2)O and (O) will be
Answer» `5" KJ mol"^(-1)` `-5" kJ mol"^(-1)` `149" kJ mol"^(-1)` `-149" kJ mol"^(-1)` SOLUTION :Activation ENERGY for bhackward reaction `=77-72=5 " kJ mol"^-1`.

Discussion

16.	H_2O_2 is concentrated by:
Answer» STEAM distillation Fractional distillation FREEZING if freezing mixture Distillation under REDUCED pressure Answer :D

Discussion

17.	H_2O_2 is
Answer» an oxidising AGENT a REDUCING agent both oxidising and reducing agnet neither oxidising nor reducing agent. Solution :`H_2O_2`can acts both oxidising and reducing agent. Oxidising NATURE of `H_2O_2`can be interpreted in account of the possession of LABILE oxygen atom. `H_2O_2rarrH_2O+O` Reducing behaviour is as `H_2O_2+O rarrH_2O+O_2`

Discussion

18.	H_(2)O_(2) is
Answer» ANTIBIOTIC anatisecptic analgesic antacid Answer :B

Discussion

19.	H_2O_2 IS :
Answer» Diamagnetic Paramagnetic Ferromagnetic None Answer :A

Discussion

20.	H_(2)O_(2)+H_(2)O+1/2O(2) (1st order). Time to time the H_(2)O_(2) solution is titrated with standard acidified KMnO_(4). The data is Time (min ) ""0""15 KMnO_(4)"consumed")""16ml""2ml What is half life in minutes?
Answer» SOLUTION :`H_(2)O_(2)toH_(2)O+1/2O_(2)&KMNO_(4)` `t_(1)=0""C_(0),t_(2)=15C_(0)-x,t_(3)=alpha""0` `Kxxt=2.303xxlog((C_(0))/(C_(t))),0.693/((t_(1/2)))xx15=2.303xxlog(16/2),(t_(1/2))=(0.693xx16)/(2.303xx3xx0.3010)=5`

Discussion

21.	H_(2)O_(2) decomposes in an aqueous solution to give H_(2)O and oxygen gas. The rate constant of disappearance is 2xx10^(-2) sec^(-1). Calculate the amount of heat liberated per second initially from 0.5 L "of" 2M H_(2)O_(2) solution (in kJ) H_(2)O_(2) (aq) to H_(2)O (l) + 1/2 O_(2) (g) DeltaH= 100 kJ//"mole"
Answer» ANSWER :2

Discussion

22.	H_2O_2decomposes with first order kinetics in a 3 lit. container. If the pressure developed in 10 min. is 380 mm, the average rate at 27^@C is
Answer» `0.01 M. "MIN"^(-1)` `0.002 M. "min"^(-1)` `0.05 M. "min"^(-1)` `0.06 M."min"^(-1)` ANSWER :B

Discussion

23.	H_2O_2 converts potassium ferrocyanide to ferricyanide. The change observed in the oxidation state of iron is:
Answer» `FE^(2+)RARRFE^(3+)` `Fe^+rarrFe^(2+)` `Fe^(3+)rarrFe^(2+)` `Fe^(2+)rarrFe^+` ANSWER :A

Discussion

24.	H_2O_2 cannot oxidise
Answer» `O_(3)` PbS KI `Na_(2)SO_(3)` Solution :`H_(2)O_(2)` reduces `O_(3)` to `O_(2)`

Discussion

25.	H_(2)O_(2) can be produced by the Ammonium hydrogen sulphate. Reactions occuring in electrolytic cell. NH_(4)HSO_(4) rarr NH_(4)^(+) +H^(+) +SO_(4)^(2-) 2SO_(4)^(2-) rarr S_(2)O_(8)^(2-) +2e^(-) (Anode) 2H^(+) +2e^(-) rarr H_(2) (cathode) (NH_(4))_(2)S_(2)O_(8) +2H_(2)O rarr 2NH_(4)HSO_(4) +H_(2)O_(2) Hydrolysis of Ammonium persulphate Assume 100% yield of hydrolysis reaction. How many moles of electrons are to be passed in order to produce enough H_(2)O_(2) which when reacted with excess of KI then liberated iodine required 100 ml of centimolar Hypo solution.
Answer» `10^(-1)` `10^(-2)` `10^(-3)` `5 xx 10^(-4)` Answer :C

Discussion

26.	H_(2)O_(2) can be produced by the Ammonium hydrogen sulphate. Reactions occuring in electrolytic cell. NH_(4)HSO_(4) rarr NH_(4)^(+) +H^(+) +SO_(4)^(2-) 2SO_(4)^(2-) rarr S_(2)O_(8)^(2-) +2e^(-) (Anode) 2H^(+) +2e^(-) rarr H_(2) (cathode) (NH_(4))_(2)S_(2)O_(8) +2H_(2)O rarr 2NH_(4)HSO_(4) +H_(2)O_(2) Hydrolysis of Ammonium persulphate Assume 100% yield of hydrolysis reaction. What is the current efficiency when 100 Amp current is passed for 965 sec, in order to produce 17 gm of H_(2)O_(2).
Answer» `20%` `40%` `80%` `100%` ANSWER :D

Discussion

27.	H_2O_2 can be used
Answer» both an OXIDIZING and as a REDUCING AGENT ONLYAS an oxidizing agent only as a reducing agent neither as an oxidizing agent nor as a reducing agent Answer :A

Discussion

28.	H_(2)O_(2) cannot be synthesized by
Answer» ADDITION of ice COLD `H_(2)SO_(4) and BaO_(2)` Addition of ice cold `H_(2)SO_(4) and PbO_(2)` Aerial oxidation of 2 ethyl anthraquinol Electrolysis of `(NH_(4))_(2)SO_(4)` at a high current density Answer :B

Discussion

29.	H_2O_2 and heavy water were discovered respectively by
Answer» THENARD, Urey Urey, Rutherford Aston, Urey Aton, Chadwick Answer :A

Discussion

30.	H_2O_2 acts as a reducing agent in its:
Answer» REACTION with a FERROUS salt Reaction with iodides Reaction with LEAD sulphide Reaction with `KMnO_4` in ACID medium Answer :D

Discussion

31.	H_2O_2 acts as antiseptic due to its :
Answer» REDUCING PROPERTY Oxidising property Bleaching property Acidic property ANSWER :B

Discussion

32.	H_2O_2 acts as an oxidising agent in:
Answer» NEUTRAL medium Acidic medium ALKALINE medium Acidic as WELL as in alkaline medium Answer :D

Discussion

33.	H_(2)O_2 solution used for hair bleaching is sold as a solution of approximately 5.0g H_(2)O_(2) per 100mL of the solution.The molecular mass of H_2O_2 is 34. The molarity of this solution is approximately
Answer» `1.5` `3.0` `4.0` `0.15` ANSWER :A

Discussion

34.	H_(2)O is reduce with Sn//HCl. Product formed
Answer» `RNH_(2)` `RNHO` `RNH_(3)^(+)` `R_(3)NH`. ANSWER :C

Discussion

35.	H_(2)O is hard if it contains
Answer» `NaHCO_(3)` `MgSO_(4)` KCl NaCl Solution :The HARDNESS of water is generally caused by the presence of bicarbonates. Chlorides and SULPHATES of calcium and magnesium .

Discussion

36.	H_(2)O has a net dipole moment while BeF_(2) has zero dipole moment because :
Answer» `F` is more electronegativity than oxygen `Be` is more electronegativity than oxygen `H_(2)O` molecule is LINEAR and `BeF_(2)` is BENT `BeF_(2)` molecule is linear and `H_(2)O` is bent Solution :

Discussion

37.	H_(2)O is dipolar whereas BeF_(2) is not. It is because
Answer» `H_(2)O` involves hydrogen BONDING whereas `BeF_(2)` is a DISCRETE MOLECULE. `H_(2)O` is linear but `BeF_(2)`is angular `H_(2)O` is angular and `BeF_(2)` is linear the electronegativity of F is GREATER than that of O. Answer :C

Discussion

38.	H_(2)O is a liquid while H_(2)S is a gas. This is due to :
Answer» DIFFERENCE in the STATE of hybridisations of O and S in their compounds high molecular mass of `H_(2)S` as compared to `H_(2)O`. strong HYDROGEN bonding in `H_(2)O` molecules as compared to `H_(2)S` molecule `H-O-H` bond angle in `H_(2)O` is `104.5^(@)` while `H-S-H` bond angle in `H_(2)S` is `92^(@)` Answer :C

Discussion

39.	H_(2)N-NH_(2)-Ph-overset(O)overset(\|\|)C-Cl" (excess)" overset(NaOH)toX,X is
Answer» `H-OVERSET(O)overset(\|\|)C-NH-Ph` `Ph-overset(O)overset(\|\|)C-NH-OH` `Ph-overset(O)overset(\|\|)C-NH_(2)` `Ph-overset(O)overset(\|\|)C-NH-NH-overset(O)overset(\|\|)C-Ph` ANSWER :D

Discussion

40.	H_2(g)+I_2(g)rarr2HI(g), triangleH=-12.40 kcals, Heat of formation of HI will be
Answer» 12.4 KCALS `-12.4 kcals` `-6.20 kacls` 6.20 kcals Answer :C

Discussion

41.	H_(2(g))+I_(2(g))hArr2HI_((g)) In this reaction when pressure increases, the reaction direction
Answer» Does not change Forward BACKWARD Decrease Solution :In that TYPE of reaction the state of equlibrium is not effect by change in VOLUME (hence pressure) of the reaction MIXTURE.

Discussion

42.	H_(2)(g) + I_(2) hArr 2HI(g): Delta H = 12.40 kcal. According to this reaction, heat of formation of HI will be
Answer» 12.4 KCAL `-12.4 kcal` `-6.20 kcal` 6.20 kcal Answer :D

Discussion

43.	H_(2)(g)+Cl_(2)(g)rarr2HCl(g).DeltaH=-44 kcal2Na(s)+2HCl(g)rarr2NaCl(s)+H_(2)(g), DeltaH=-152 kcalFor the reaction Na(s)+(1)/(2)Cl_(2)(g)rarrNaCl(s),DeltaH=
Answer» `-108 kcal` `-196 kcal` `-98 kcal` 54 kcal Solution :By (i) + (ii) , `Na+(1)/(2)Cl_(2)RARRNACL, DELTAH=-(196)/(2)kcal=-98kcal`.

Discussion

44.	H_(2(g))+Cl_(2(g))overset(hv)rarr 2HCl_((g)) The reaction proceeds with a uniform rate throughout. What do you conclude ?
Answer» SOLUTION :The REACTION is zero order reaction WHOSE RATE is independent on the concentration of REACTANTS.

Discussion

45.	H_2(g)+Cl_2(g)=2HCl(g),triangleH=-44.12 kcal The enthalpy of hydrogen cholride at 298 K is:
Answer» `-44.12` `-22.06` 44.12 22.06 Answer :B

Discussion

46.	H_(2)(g) and O_(2)(g)have a very strong tendency to react with each other. However, keepingthem in the same container at room temperature, do not produce water. Explain.
Answer» SOLUTION :Because the VALUE of activation ENERGY for the REACTION is very high

Discussion

47.	H_2(g)+Cl_2(g)→2HCl(g), triangleH298K=-22.06 kcals, For this reaction, △U is equal to:
Answer» `-22.06 + 2 xx 10^-3 xx298 xx 2 KCAL` `-22.06 -2xx 298 kcal` `-22.06 -2xx 298xx 4 kcal` `-22.06 kcl` Answer :D

Discussion

48.	H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l) , Delta H at 298 K = - 285.8 kJ The molar enthalpy of vapourization of water at 1 atm and 25^(@)C is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at is
Answer» `-241.8 KJ` 241.8 kJ 329.8 kJ `-329.8 kJ` ANSWER :A

Discussion

49.	{:[(H_(2)(g)+(1)/(2)O_(2)(g)=H_(2)O(g),DeltaH=-241.8kJ),(CO(g)+(1)/(2)O_(2)(g)=CO_(2)(g),DeltaH=-283 kJ):}] the heat evolved in the combustion of 112 litres of water gas (mixture of equal volumes of H_(2) and CO) is
Answer» `241.8` KJ `283` kJ `1312`kJ `1586` kJ Answer :C

Discussion

50.	H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l), DeltaH " at "298 K=-285.8 kJThe molar enthalpy of vaporization of water at 1 atm and 25^(@)C is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at 25^(@)C is
Answer» `-241.8` KJ 241.8 kJ 329.8 kJ `-329.8` kJ Solution :`H_(2)+(1)/(2)O_(2)rarrH_(2)O_((l)),DeltaH=-285.8 kJ` `H_(2)O_((l))rarrH_(2)O_((G)),DeltaH=44 kg` `THEREFORE H_(2)+(1)/(2)O_(2)rarrH_(2)O_((g)),DeltaH^(@)=-241.8 kJ`.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

H_2PO_4^(-) the conjugate base of……..

H_(2)O(l)hArrH^(+)(aq)+OH^(-)(aq),DeltaH=+13.7 kcal/mol K_(eq) at 25^(@)C=10^(-14). Calculate K_(eq) at 60^(@)C?

{:([(H_(2)O)_(4)Co(O_(2))_(2)Co(H_(2)O)_(4)](SO_(4))_(2)overset[[X]"reagent"]to [(H_(2)O)_(4)Co(O_(2))_(2)Co(H_(2)O)_(4)](SO_(4))),(""(I)""""(II)"):} In both the complex Co hast_(2g)^(6)e_(g^(0)) configuration. Which option is incorrect ?

[(H_(2)O)_(8)Fe_(2)(OH)_(2)].(SO_(4))_(2) Calculate the coordination number of central metal, E.A.N., Primary valencies. [If the coordination number is 4, EAN is 18, Primary valency = 2 than answer is 4182.]

H_2Oboils at higher temperature than H_2S becauseit is capable of forming :

H_(2)O_(2)toH_(2)O+1/2O_(2)(1storder) Time to time the H_(2)O_(2) solutioni is titrated with standard acidified KMnO_(4). Thedata is Time (min)""0""15 KMnO_(4) ("consumed")""16ml""2ml What is half life in minutes?

H_(2)O_(2)overset(1st"order")(to)H_(2)O+1/2O_(2). Pressure of O_(2) is recorded as: time (min)"" 10 min ""prop P_(O_(2)) (mm)""120""160 What is half life?

H_2O_2 when added to a solution containing KMnO_4 and H_2SO_4 acts :

H_2O_2 restores the colour of old lead paintings , blackenedby the action of H_2S gas, by :

H_2O_2 reduces K_3Fe(CN)_6 in

H_(2)O_(2) reduces chlorine to chloride. Write the coefficients of all substances in the equation.

H_2O_2 on treatment with chlorine gives:

H_(2)O_(2) on reaction with PbSgives

H_2O_2 is prepared in the laboratory when:

H_2O_2 is concentrated by:

H_2O_2 is

H_(2)O_(2) is

H_2O_2 IS :

H_(2)O_(2)+H_(2)O+1/2O(2) (1st order). Time to time the H_(2)O_(2) solution is titrated with standard acidified KMnO_(4). The data is Time (min ) ""0""15 KMnO_(4)"consumed")""16ml""2ml What is half life in minutes?

H_2O_2decomposes with first order kinetics in a 3 lit. container. If the pressure developed in 10 min. is 380 mm, the average rate at 27^@C is

H_2O_2 converts potassium ferrocyanide to ferricyanide. The change observed in the oxidation state of iron is:

H_2O_2 cannot oxidise

H_2O_2 can be used

H_(2)O_(2) cannot be synthesized by

H_2O_2 and heavy water were discovered respectively by

H_2O_2 acts as a reducing agent in its:

H_2O_2 acts as antiseptic due to its :

H_2O_2 acts as an oxidising agent in:

H_(2)O_2 solution used for hair bleaching is sold as a solution of approximately 5.0g H_(2)O_(2) per 100mL of the solution.The molecular mass of H_2O_2 is 34. The molarity of this solution is approximately

H_(2)O is reduce with Sn//HCl. Product formed

H_(2)O is hard if it contains

H_(2)O has a net dipole moment while BeF_(2) has zero dipole moment because :

H_(2)O is dipolar whereas BeF_(2) is not. It is because

H_(2)O is a liquid while H_(2)S is a gas. This is due to :

H_(2)N-NH_(2)-Ph-overset(O)overset(||)C-Cl" (excess)" overset(NaOH)toX,X is

H_2(g)+I_2(g)rarr2HI(g), triangleH=-12.40 kcals, Heat of formation of HI will be

H_(2(g))+I_(2(g))hArr2HI_((g)) In this reaction when pressure increases, the reaction direction

H_(2)(g) + I_(2) hArr 2HI(g): Delta H = 12.40 kcal. According to this reaction, heat of formation of HI will be

H_(2)(g)+Cl_(2)(g)rarr2HCl(g).DeltaH=-44 kcal2Na(s)+2HCl(g)rarr2NaCl(s)+H_(2)(g), DeltaH=-152 kcalFor the reaction Na(s)+(1)/(2)Cl_(2)(g)rarrNaCl(s),DeltaH=

H_(2(g))+Cl_(2(g))overset(hv)rarr 2HCl_((g)) The reaction proceeds with a uniform rate throughout. What do you conclude ?

H_2(g)+Cl_2(g)=2HCl(g),triangleH=-44.12 kcal The enthalpy of hydrogen cholride at 298 K is:

H_(2)(g) and O_(2)(g)have a very strong tendency to react with each other. However, keepingthem in the same container at room temperature, do not produce water. Explain.

H_2(g)+Cl_2(g)→2HCl(g), triangleH298K=-22.06 kcals, For this reaction, △U is equal to:

H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l) , Delta H at 298 K = - 285.8 kJ The molar enthalpy of vapourization of water at 1 atm and 25^(@)C is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at is

{:[(H_(2)(g)+(1)/(2)O_(2)(g)=H_(2)O(g),DeltaH=-241.8kJ),(CO(g)+(1)/(2)O_(2)(g)=CO_(2)(g),DeltaH=-283 kJ):}] the heat evolved in the combustion of 112 litres of water gas (mixture of equal volumes of H_(2) and CO) is

H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l), DeltaH " at "298 K=-285.8 kJThe molar enthalpy of vaporization of water at 1 atm and 25^(@)C is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at 25^(@)C is