94420 + Interview Questions in Chemistry in Class 12

1.	H_(2)SO_(4) is used in
Answer» PETROLEUM REFINING manufactue of paints, PIGMENTS and dye stuff detergent INDUSTRY all of the above are the uses of `H_(2)SO_(4)` Solution :All of the above ae the uses of `H_(2)SO_(4)`

Discussion

2.	H_2SO_4 is added while preparing a standard solution of Mohr's salt to prevent :
Answer» Hydration Reduction Hydrolysis Complex formation Answer :C

Discussion

3.	H_(2)SO_(4) is a 1) Dehydratingagent 2) Sulphonating agent 3) Reducing agent 4) Highly viscousliquid Choose the correct set of choice from the options give below
Answer» 1, 2 & 3 2, 3 and 4 1, 3 and 4 1, 2 and 4 Solution :`H_(2)SO_(4)` never ACTS as reducing AGENT hence .d. is CORRECT.

Discussion

4.	H_(2)SO_(4) has very corrosive action on skin because
Answer» It REACTS with PROTEINS It acts as an oxidizing agent It acts as dehydrating agent It acts as dehydrating agent and ABSORPTION of WATER is highly EXOTHERMIC Answer :D

Discussion

5.	H_(2)SO_(4) has great affinity for water because
Answer» It hydrolysisthe acid It decomposes the acid Acid decomposeswater Acid FORMS hydrates with water Solution :It fumes STRONGLY strongly in moist and soluble in `H_(2)O` with evolution of heat due to the formation of hydrates. `H_(2)SO_(4).H_(2)O, H_(2)SO_(4). 2H_(2)O` ETC.,

Discussion

6.	H_2SO_4 has very corrosive action on skin because:
Answer» It reacts with proteins It ACTS as oxidising agent It acts as DEHYDRATING agent and absorption of water is highly exothermic None of the above Answer :C

Discussion

7.	H_(2)SO_(4) cannot be used to prepare HBr from NaBr as it
Answer» Reacts slowly with NaBr Oxidises HBR Reduces HBr DISPROPORTIONATE HBr Solution :HBr is not preparedby HEATING NaBr with conc. `H_(2)SO_(4)` because HBr is a strong REDUCING agent and reduce `H_(2)SO_(4)` to `SO_(2)` and get itself oxidised to bromine. `NaBr + H_(2)SO_(4) to NaHSO_(4) +HBr` `H_(2)SO_(4) +2HBr to SO_(2) +Br_(2) +2H_(2)O`

Discussion

8.	H_2SO_4 doesnot acts as dehydrating agent in its reaction with:
Answer» `BA(OH)_2` Zn KOH `H_2C_2O_4` ANSWER :D

Discussion

9.	H_2SO_4 and H_2SO_3 can be distinguished by the addition of:
Answer» LITMUS SOLUTION `FeCl_3` solution `NaHSO_4` solution Magnesium powder Answer :B

Discussion

10.	H_2SO_4 act as oxidising agent with
Answer» C `C_(6)H_(12)O_(6)` NaCl NaOH Answer :A

Discussion

11.	H_2SO_4 acid is known as "king of chemicals". Explain.
Answer» SOLUTION :Because there is HARDLY any INDUSTRY in which it is not USED in some FORM or the other.

Discussion

12.	H_(2)S wouldseparatethe following in pH lt 7
Answer» `ZN^(2+),Co^(2+)` `Cu^(2+),Cd^(2+)` `Cu^(2+),Cr^(3+)` `Cu^(2+),As^(3+)` Answer :C

Discussion

13.	H_2Swould separate the following at pH lt 7
Answer» `Cu^(2+), CD^(2+) ` `Pb^(2+), Cd^(2+) ` `Cu^(2+), CR^(3+)` `Cu^(2+), As^(3+)` ANSWER :C

Discussion

14.	H_(2)S will precipitate the sulphide of all the metals from the solution of chlorides of Cu,Zn and Cd if:
Answer» The solution is aqueous The solution is acidic The solution is DILUTE acidic Any of these solution is present Solution :DUE to COMMON ION EFFECT as `HCl to H^(+) +Cl^(-),H_(2)Sto2H^(+)+S^(2-)`

Discussion

15.	H_(2)S will precipitate the sulphide of all the metals from thesolutionof chlorides of Cu,Zn and Cd if
Answer» The solution is aqueous The solution is acidic The solution is diute acidic Any of the above solution is present Solution :`H_(2)S` in neutralmedium provides sufficient `S^(2-)` toprecipitate `CU,CD,Zn, and MN `as SULPHIDE

Discussion

16.	H_(2)S may provide the colloidal sulphur by
Answer» Oxidation Reduction Neutralization Hydrolysis Solution :COLLOIDAL or `DELTA`- sulphur is prepared by passing `H_(2)S` through a solution of an oxidizing agent or WATER or by treating SODIUM thiosulphate with dil `HCL`.

Discussion

17.	H_(2)S when passed through dil HNO_(3) gives -
Answer» RHOMBIC sulphur monoclinic sulphur Colloidal sulphur Plastic sulphur Solution :`2HNO_(3)+H_(2)S to 2NO_(2)+2H_(2)O+S`

Discussion

18.	H_2S on passing through KMnO_4 solution gives:
Answer» `K_2SO_3` S `K_2MnO_4` `MnO_2` ANSWER :B

Discussion

19.	H_(2)S on incomplete combustion with oxygen forms mainly
Answer» `H_(2) and S` `H_(2) and SO_(3)` `H_(2)O` S SOLUTION :`2H_(2)S+O_(2) to 2H_(2)O+2S`

Discussion

20.	H_(2)S is passed through as acidified solution of Ag, Cu and Zn. Which forms precipitate
Answer» `Ag` Zn CU None of these Solution :`Cu^(2+)+H_(2)S to underset("black ppt.")(CUS)DARR`

Discussion

21.	H_(2)S is passed into one dm^(3) of a solution containing 0.1 mole of Zn^(2+) and 0.01 mole of Cu^(2+) till the sulphide ion concentration reaches 8.1xx10^(-19) moles .Which one of the following statements is true? [K_(sp) of ZnS and CuS are 3xx10^(-22) and 8xx10^(-36) respectively]
Answer» Only ZnS precipitates Both CuS and ZnS precipitate Only CuS precipitates No precipitation occurs Solution :Precipitation occures only when ionic product exceeds the VALUE of solubility product. 1`dm^(3)` of a solution containing 0.1 mole of `Zn^(2+)` 0.01 mole of `Cu^(2+)`and `8.1xx10^(-19)` mole of `S^(2+)` .Let Ionic product of ZnS=`[Zn^(2+)][S^(2-)]` `=0.1xx8.1xx10^(-19)=8.1xx10^(-20)` `K_(sp)` of ZnS`=3xx10^(-22)` `therefore` Ionic product`gtK_(sp)` Both ZnS and CuS have less `K_(sp)` values than their respective ionic products so ZnS and CuS both precipitates.

Discussion

22.	H_2S is passed through an acidified solution of copper sulphate and a black precipitate is fromen. This is due to :
Answer» Oxidation of `Cu^(2+)` REDUCTION of `Cu^(2+)` DOUBLE decomposition Reduction and oxidation Answer :C

Discussion

23.	H_2S is passed through acidified solution of CuSO_4 and black ppt is formed. This is due to:
Answer» Oxidation of `CU^(2+)` Reduction of `Cu^(2+)` DOUBLE decomposition Reduction and oxidation Answer :C

Discussion

24.	H_(2)S is more volatile than water because
Answer» S' is more electro negative than 'O' O' is more electro negative than 'S' `H_(2)O` SHOWS H-bonding `H_(2)O` bond angle is more than `H_(2)S` Solution :DUE to the formation of INTERMOLECULAR H-bonding by `H_(2)O` molecule.

Discussion

25.	H_(2)S is not a/an
Answer» Reducing agent ACIDIC OXIDISING agent None of these Solution :`H_(2)S` is not an oxidising agent. `PbO + H_2S PbS+ H_2O.` It acts as a reducing agent. Also in chalcogens the acidic NATURE of hydride increases from `H_2O" to "H_2Te.`

Discussion

26.	H_(2)S is less acidic than H_(2)Te. Why ?
Answer» SOLUTION :As the size of the element increases down the group, E-H BOND length increases and hence E-H bond dissociation energy DECREASES. In other words, H-S bond dissociation energy is HIGHER than that of H-Te bond dissociation energy and hence H-S bond BREAKS less easily than H-Te bond. Therefore, `H_(2)S` is a weaker acid than `H_(2)Te`.

Discussion

27.	H_2S is less acidic than H_(2)Te. Why ?
Answer» Solution :Due to the decrease in BOND (E-H) DISSOCIATION enthalpy down the GROUP, acidic CHARACTER increases.

Discussion

28.	H_(2)S is less acidic than H_(2)Te. Give reason.
Answer» SOLUTION :Due to the DECREASE in bond DISSEMINATION enthalpy down the group, ACIDIC character increases.

Discussion

29.	H_2S is far more volatile than water because:
Answer» Sulphur ATOM is more electronegative than OXYGEN atom Oxygen atom is more electronegative than sulphur atom `H_2O` has BOND angle of nearly `105^@` Hydrogen is loosely BONDED with sulphur Answer :B

Discussion

30.	H_(2)S in the presence of HCIprecipitates group IIradicalsbut not of the group IV because
Answer» HCI ACTIVATE `H_(2)S` HCIincreses cone of `CI^(THETA)` dueto commonion effect HCIdecreses cone of `S^(2-)` dueto commonion effect HCI LOWERS the solubility of `H_(2)S`in soin Answer :c

Discussion

31.	H_2S is a:
Answer» WEAK DIBASIC acid Weak MONOBASIC acid Strong dibasic acid Strong monobasic acid Answer :A

Discussion

32.	H_(2)S gas when passed through solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because
Answer» presence of HCl decreases the sulphide ION concentration sulphides of group IV CATIONS are unstable in HCl solubility PRODUCT of group II sulphides is more than that of group IV sulphides presence of HCl INCREASE the sulphide ion concentration Answer :A

Discussion

33.	H_(2)S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group because :
Answer» Sulphides of GROUP IV cations are unstable in HCI Presence of HCI decreases sulphide ION CONCENTRATION SOLUBILITY product of group II sulphides is more than that of group IV sulphides. Presence of HCI increases sulphide ion concentration. Answer :B

Discussion

34.	H_(2)S gas on passingthrough an alkline solution , forms a white precipitate .The solution contains ions of
Answer» Pb Zn Cu Ni Solution :ZNS is WHITE

Discussion

35.	H_2S gas is passed through a solution of X to obtain a colloidal solution of As_2S_3. Name X.
Answer» SOLUTION :`As_2O_3`

Discussion

36.	H_(2)S gas is passed into aqueous solution of Zn(CH_(3)COO)_(2) and ZnCl_(2) in test tubes I and II separately. Then ZnS is precipitated:
Answer» In I In II In both In NONE of these ANSWER :A

Discussion

37.	H_2S exhibits:
Answer» OXIDISING properties Reducing properties Basic properties None Answer :B

Discussion

38.	H_2S does not produce metallic sulphide with
Answer» `CdCl_2` `ZnCl_2` `CoCl_2` `CuCl_2` SOLUTION :`CoCl_2`

Discussion

39.	H_2S does not produce metallic sulphide with:
Answer» `CdCl_2` `ZnCl_2` `COCl_2` `CuCl_2` ANSWER :C

Discussion

40.	H_(2)S contains 5.88% hydrogen, H_(2)O contains 11.11% hydrogen while SO_(2) contains 50% sulphur. These figures illustrate the law of :
Answer» CONSERVATION of mass constant PROPORTIONS MULTIPLE proportions RECIPROCAL proportions Answer :D

Discussion

41.	H_(2)S combines with O_(2)F_(2) to give:
Answer» `SF_(6), OF_(2),HF` `HF,O_(2),SF_(6)` `OF_(2),SO_(2),HF` `HF,SO_(2),SF_(2)` Answer :B

Discussion

42.	H_(2)S can be dreid by how many of the following reagents? Anhydrous CaCl_(2) conc. H_(2)SO_(4), P_(2)O_(3), KOH solutons, Na_(2)CO_(3) soluton.
Answer» SOLUTION :FACT

Discussion

43.	H_2S cannot be dried by passing over conc. H_2SO_4 because:
Answer» The ACID OXIDISES it The acid COMBINES with `H_2S` to form a salt Both form complex It dissolves in the acid Answer :A

Discussion

44.	H_2S cannot be dried by passing over conc. H_2SO_4 because
Answer» The acid oxidises `H_2S` into S The acid COMBINES with `H_2S` to FORM a ppt Both form complex It dissovles in the acid SOLUTION :`H_2S` exhibits reducing PROPERTY and acidic property.

Discussion

45.	H_(2)S and SO_(2) can be distingnisted by
Answer» Limus paper `MnO_(4)^(Θ)` `Pb(CH_(3)COO)_(2)` `HCI` SOLUTION :`H_(2)S` reacts with lead ACETATE to give BLACK PbS

Discussion

46.	H_(2)S acts only as a reducing agent while SO_(2) acts as an oxidising as well as a reducing agent. Why?
Answer» Solution :(i) `H_(2)S`. Oxidation number (O.N.) of S in `H_(2)s` is -2. Maximum O.N. of S is +6 and minimum is -2. Since the O.N. of S in `H_(2)S` is minimum (-2), therefore, it can only increase by losing electrons. Hence, `H_(2)S` acts only as a reducing AGENT. (ii) `SO_(2)`. O.N. of S in `SO_(2)` = +4 Maximum O.N. of S is +6 and minimum is -2. Since O.N. of S in `SO_(2)` is +4, therefore, it can either increase by losing electrons or DECREASE by accepting electrons. Hence, `SO_(2)` can act both as a reducing agent as well as an oxidising agent.

Discussion

47.	H_(2)S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H_(2)S in water at STP is 0.195 m, calculate Henry's law constant.
Answer» Solution :It is given that the solubility of `H_(2)S` in water at STP is 0.195 m, i.e., 0.195 mol of `H_(2)S` is dissolved in 1000 G of water. Moles of water `= (1000)/(18)=55.56` mol `therefore` Mole fraction of `G_(2)S, X_(H_(2)O)` `= ("Moles of "H_(2)S)/("Moles of "H_(2)S+"Moles of water")` `= (0.195)/(0.195+55.56)=0.0035` At STP, pressure (p) = 0.987 BAR ACCORDING to Henry.s law : `p=K_(H).x` `K_(H)=(p)/(x)=(0.987)/(0.0035)=282` bar.

Discussion

48.	H_2S , a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H_2S in water at STP is 0.195 m, calculate Henry's law constant
Answer» <P> Solution :Solubility of `H_2S`gas = 0.195 m i.e., 0.195 MOLE in 1 kg of the solvent (water) 1 kg of the solvent (water) = 1000 g = 1000/18g = 55.55 moles ` therefore ` Mole fraction of `H_2S`gas in the solution (X) = `(0.195)/(0.195 + 55.55) = 0.0035` Pressure at STP = 0.987 bar Applying Henry.s law, `p_(H_2S) = K_H xx x_(H_2S) " or" K_H = (p_(H_2S))/(x_(H_2S)) = (0.987 bar)/(0.0035) = 282 bar `

Discussion

49.	H_(2)S, a toxis gas with rotten egg like small, is used for qualitative analysis. If the solubility of H_(2)S in water at STP is 0.195 m, calculate Henry's law constant.
Answer» Solution :Solubility of `H_(2)S` gas = 0.195 m = 0.195 mole in 1 kg of the solvent (water ) `"1 kg of the solvent (water) = 1000 g "=(1000g)/("18 g mol"^(-1))="55.55 moles"` `THEREFORE"Mole fraction of "H_(2)S" gas in the solution (x)"=(0.195)/(0.195+55.55)=(0.195)/(55.745)=0.0035` Pressure at STP = 0.987 bar `"Applying Henry's LAW, "p_(H_(2)S)=K_(H)xxx_(H_(2)S) or K_(H)=(p_(H_(2)S))/(x_(H_(2)S))=("0.987 bar")/("0.0035")="282 bar."`

Discussion

50.	H_2S + 4F_2 to 2HF + B The shape of molecule of compound B is
Answer» OCTA hedral TETRA hedral TRIGONAL planar linear ANSWER :A

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

H_(2)SO_(4) is used in

H_2SO_4 is added while preparing a standard solution of Mohr's salt to prevent :

H_(2)SO_(4) is a 1) Dehydratingagent 2) Sulphonating agent 3) Reducing agent 4) Highly viscousliquid Choose the correct set of choice from the options give below

H_(2)SO_(4) has very corrosive action on skin because

H_(2)SO_(4) has great affinity for water because

H_2SO_4 has very corrosive action on skin because:

H_(2)SO_(4) cannot be used to prepare HBr from NaBr as it

H_2SO_4 doesnot acts as dehydrating agent in its reaction with:

H_2SO_4 and H_2SO_3 can be distinguished by the addition of:

H_2SO_4 act as oxidising agent with

H_2SO_4 acid is known as "king of chemicals". Explain.

H_(2)S wouldseparatethe following in pH lt 7

H_2Swould separate the following at pH lt 7

H_(2)S will precipitate the sulphide of all the metals from the solution of chlorides of Cu,Zn and Cd if:

H_(2)S will precipitate the sulphide of all the metals from thesolutionof chlorides of Cu,Zn and Cd if

H_(2)S may provide the colloidal sulphur by

H_(2)S when passed through dil HNO_(3) gives -

H_2S on passing through KMnO_4 solution gives:

H_(2)S on incomplete combustion with oxygen forms mainly

H_(2)S is passed through as acidified solution of Ag, Cu and Zn. Which forms precipitate

H_(2)S is passed into one dm^(3) of a solution containing 0.1 mole of Zn^(2+) and 0.01 mole of Cu^(2+) till the sulphide ion concentration reaches 8.1xx10^(-19) moles .Which one of the following statements is true? [K_(sp) of ZnS and CuS are 3xx10^(-22) and 8xx10^(-36) respectively]

H_2S is passed through an acidified solution of copper sulphate and a black precipitate is fromen. This is due to :

H_2S is passed through acidified solution of CuSO_4 and black ppt is formed. This is due to:

H_(2)S is more volatile than water because

H_(2)S is not a/an

H_(2)S is less acidic than H_(2)Te. Why ?

H_2S is less acidic than H_(2)Te. Why ?

H_(2)S is less acidic than H_(2)Te. Give reason.

H_2S is far more volatile than water because:

H_(2)S in the presence of HCIprecipitates group IIradicalsbut not of the group IV because

H_2S is a:

H_(2)S gas when passed through solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because

H_(2)S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group because :

H_(2)S gas on passingthrough an alkline solution , forms a white precipitate .The solution contains ions of

H_2S gas is passed through a solution of X to obtain a colloidal solution of As_2S_3. Name X.

H_(2)S gas is passed into aqueous solution of Zn(CH_(3)COO)_(2) and ZnCl_(2) in test tubes I and II separately. Then ZnS is precipitated:

H_2S exhibits:

H_2S does not produce metallic sulphide with

H_2S does not produce metallic sulphide with:

H_(2)S contains 5.88% hydrogen, H_(2)O contains 11.11% hydrogen while SO_(2) contains 50% sulphur. These figures illustrate the law of :

H_(2)S combines with O_(2)F_(2) to give:

H_(2)S can be dreid by how many of the following reagents? Anhydrous CaCl_(2) conc. H_(2)SO_(4), P_(2)O_(3), KOH solutons, Na_(2)CO_(3) soluton.

H_2S cannot be dried by passing over conc. H_2SO_4 because:

H_2S cannot be dried by passing over conc. H_2SO_4 because

H_(2)S and SO_(2) can be distingnisted by

H_(2)S acts only as a reducing agent while SO_(2) acts as an oxidising as well as a reducing agent. Why?

H_(2)S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H_(2)S in water at STP is 0.195 m, calculate Henry's law constant.

H_2S , a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H_2S in water at STP is 0.195 m, calculate Henry's law constant

H_(2)S, a toxis gas with rotten egg like small, is used for qualitative analysis. If the solubility of H_(2)S in water at STP is 0.195 m, calculate Henry's law constant.

H_2S + 4F_2 to 2HF + B The shape of molecule of compound B is