52671 + Interview Questions in Chemistry in Class 11

1.	Gold number is a mesure of the
Answer» Protective action by a lyophilic colloid on a LYOPHOBIC colloid Protective action by a lyophobic colloid on a lyophilic colloid Number of mg of GOLD in a standard red gold SOL Stability of gold sol Answer :A

Discussion

2.	Gold number gives the indication of
Answer» gm-molecules of GOLD PER 1000 ml of colloidal solution percentage of gold in the suspension charge on the colloids protective POWER of a colloid Solution :Gold NUMBER is a measure of protective power of a colloid

Discussion

3.	Gold number is
Answer» The NUMBER of mg of LYOPHILIC colloid which should be ADDED to 10 ML of ferric hydroxide sol so as to prevent its coagulation by the ADDITION of 1 ml of 10% sodium chloride solution The number of mg of lyophilic colloid which should be added to 10 ml of standard gold sol so as to prevent its coagulation by the addition of 1 ml of 10% NaCl The mg of gold salt to be added to a lyophilic colloid to coagulate it The mg of gold salt to be added to a lyophilic colloid Answer :B

Discussion

4.	Gold has a close packed structure which can be reviewed as spheres occupying 0.74of the total volume. If the density of gold is 19.3 gg/cc, calculate the apparent radius of a gold atom in the solid (Au=197 amu)
Answer» Solution :As the packing FRACTION is 0.74 , packing can be hcp or FCC. Knowing that gold has fcc lattice, Z=4 `rho=(ZxxM)/(a^3xxN_0) "" THEREFORE 19.3=(4xx197)/(a^3xx6.023xx10^23) "or" a=4.07xx10^(-8)` CM For fcc, `r=a/(2sqrt2)=(4.07xx10^(-8))/(2xx1.414)=1.439xx10^(-8)` cm

Discussion

5.	Gold has a close packed structure which can be reviewed as sphere occupying 0.74 of the total volume. If the density of gold is 19.3 g/cc. Calculate the apparent radius of a gold atom in the solid(Au = 197 amu)/
Answer» SOLUTION :As the packing fraction is 0.74 , packing can be hcp or FCC. KNOWING that gold has fcc lattice, Z = 4 p = ` ( Z xx M)/ (a^(3) xx N_(0))therefore19.3 = ( 4 xx 197)/ (a^(3)xx 6.023 xx 10^(23))ora= 4.07 xx 10^(-8) ` CM For fcc, ` r = a / (2sqrt2)= ( 4.07 xx 10^(-8))/ (2xx 1.414)= 1.439 xx 10^(-8) ` cm

Discussion

6.	Gold crystallizes in a face - centred cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate the density of gold as well as its atomic radius assuming it to be spherical . Atomic mass of gold =197 amu.
Answer» SOLUTION :For FCC LATTICE , Z= 4 , Radius` r = a/ (2 sqrt2)= 0.3535 a `

Discussion

7.	Gold crystallizes in a face centred cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold = 197 amu
Answer» SOLUTION :For FCC LATTICE, Z=4 , RADIUS `r=a/(2SQRT2)`=0.3535 a

Discussion

8.	Gold can easily dissolve in
Answer» CONC.`HNO_3` conc.HCl KCN(AQ) + `O_(2)` NONE of these. Answer :C

Discussion

9.	Gold (atomic radius =0.144 nm) crystallises in a face centred unit cell. What is the length of the side of the cell ?
Answer» SOLUTION :For FCC, `a=2sqrt2r`=2 X 1.414 x 0.144 NM = 0.407 nm

Discussion

10.	Gold(atomic radius = 0.144 nm) crystallisesin a face centred unit cell.What is the length of theside of the cell.
Answer» SOLUTION :For FCC , ` a = 2 SQRT2 r = 2 XX 1.414 xx 0.144 NM= 0.407 nm ` ]

Discussion

11.	Gold (atomic mass 197 u, atomic radius = 0.144 m ) crystallizes in a face centred unit cell. Determine the density of gold
Answer» Solution :For fcc unit cell , `r=a/(2sqrt2)` or `a=2sqrt2 r=2xx1.414 XX 0.144`NM = 0.407 nm =`0.407xx10^(-7)` CM `rho=(ZxxM)/(a^3xxN_0)=(4xx197 "g MOL"^(-1))/((0.407xx10^(-7)cm)^3 xx (6.02xx10^23 mol^(-1)))=19.4 g cm^(-3)`

Discussion

12.	Go through the following flow sheet. Based on, it answer the questions given at the end of it. Yellow coloured solution (B) is of:
Answer» `Na_(2)CrO_(4)` formed by oxidation of `Cr^(3+)` by `H_(2)O_(2)` in ALKALINE medium `FeCI_(2)` formed by REDUCTION of `FeCI_(3)` by `H_(2)O_(2)` `FeCI_(3)` formed by oxidation of `FeCI_(2)` by `H_(2)O_(2)` `Cu(OH)_(2)` formed by precipitation of `CuCI_(2)` by `H_(2)O_(2)//NaOH` Solution :`{:((A)rarrCrCI_(3),,(B)rarrNa_(2)CrO_(4),),((C)rarrNa_(2)Cr_(2)O_(7),,(D)RARR(NH_(4))_(2)Cr_(2)O_(7)),((E)rarrCrO_(2)CI_(2),,(F)rarrPbCrO_(4),),((G)rarrN_(2),,(H)rarrCr_(2)O_(3),),((I)rarrLi_(3)N,,(J)rarrNH_(3)):}`

Discussion

13.	Go through the following flow sheet. Based on, it answer the questions given at the end of it. Select correct statement(s):
Answer» `CO_(2)` makes the solution of (B) ACIDIC and CONVERTS (B) into `FeCI_(2)(C)`. `CO_(2)` makes the solution of (B) basic and converts it into `Na_(2)Cr_(2)O_(7)` `CO_(2)` makes the solution of (B) acidic and converts it into `Na_(2)Cr_(2)O_(7)` NONE of the above is correct Solution :`{:((A)rarrCrCI_(3),,(B)rarrNa_(2)CrO_(4),),((C)rarrNa_(2)Cr_(2)O_(7),,(D)RARR(NH_(4))_(2)Cr_(2)O_(7)),((E)rarrCrO_(2)CI_(2),,(F)rarrPbCrO_(4),),((G)rarrN_(2),,(H)rarrCr_(2)O_(3),),((I)rarrLi_(3)N,,(J)rarrNH_(3)):}`

Discussion

14.	Go through the following flow sheet. Based on, it answer the questions given at the end of it. Formation of (F) using (G), KCI (solid) and concentrated H_(2)SO_(4) is called:
Answer» redox TEST of `Cr_(2)O_(7)^(2-)` chromy-chloride test of `CI^(-)` chromyl-chloride test of `Cr^(3+)` none of the above is correct Solution :`{:((A)rarrCrCI_(3),,(B)rarrNa_(2)CrO_(4),),((C)rarrNa_(2)Cr_(2)O_(7),,(D)rarr(NH_(4))_(2)Cr_(2)O_(7)),((E)rarrCrO_(2)CI_(2),,(F)rarrPbCrO_(4),),((G)rarrN_(2),,(H)rarrCr_(2)O_(3),),((I)rarrLi_(3)N,,(J)rarrNH_(3)):}`

Discussion

15.	Go through the following flow sheet. Based on, it answer the questions given at the end of it. Compounds D,E and F respectively are:
Answer» `{:(D,E,F),((NH_(4))_(2)CrO_(4),CrO_(2)CI_(2),PbCrO_(4)):}` `{:(D,E,F),((NH_(4))_(2)CrO_(7),CrO_(3),PbCrO_(4)):}` `{:(D,E,F),((NH_(4))_(2)Cr_(2)O_(7),CrO_(2)CI_(2),PbCrO_(4)):}` `{:(D,E,F),((NH_(4))_(2)CrO_(4),CrO_(3),PbCrO_(4)):}` Solution :`{:((A)rarrCrCI_(3),,(B)rarrNa_(2)CrO_(4),),((C)rarrNa_(2)Cr_(2)O_(7),,(D)rarr(NH_(4))_(2)Cr_(2)O_(7)),((E)rarrCrO_(2)CI_(2),,(F)rarrPbCrO_(4),),((G)rarrN_(2),,(H)rarrCr_(2)O_(3),),((I)rarrLi_(3)N,,(J)rarrNH_(3)):}`

Discussion

16.	Go through the following flow sheet. Based on, it answer the questions given at the end of it. Compound (A) is:
Answer» `CuCI_(2)` `FeCI_(3)` `FeCI_(2)` `CrCI_(3)` Solution :`{:((A)rarrCrCI_(3),,(B)rarrNa_(2)CrO_(4),),((C)rarrNa_(2)Cr_(2)O_(7),,(D)rarr(NH_(4))_(2)Cr_(2)O_(7)),((E)rarrCrO_(2)CI_(2),,(F)rarrPbCrO_(4),),((G)rarrN_(2),,(H)rarrCr_(2)O_(3),),((I)rarrLi_(3)N,,(J)rarrNH_(3)):}`

Discussion

17.	Glycrine can be purified by
Answer» VACUUM distillation simple distillation steam distillation fractional distillation Answer :A

Discussion

18.	Glycol on reacting with phthalic acid gives
Answer» Glycerol Glycollic ACID Glyptal Dacron Answer :C

Discussion

19.	Glycol is used in the manufacture of
Answer» Terylene Glyptal Antifreeze All Answer :D

Discussion

20.	Glycol is added to aviation petrol because
Answer» It prevents freezing of petrol It minimies the loss of petrol It increases the EFFICIENCY of fuel It prevents the engine from HEATING up Solution :Gylcol is used as an antifreeze in aeroplane AVIATION petrol under the NAME prestone.

Discussion

21.	Glycine is an alpha-amino acid which exists in the form of Z witter ion as overset(+)NH_(3) CH_(2)CO O^(-). Write the formula of its conjugate acid and conjugate base.
Answer» Solution :CONJUGATE acid `= overset(+)NH_(3)CH_(2)CO O^(-) + H^(+) = overset(+)NH_(3)CH_(2)CO OH` Conjugate base `= overset(+)NH_(3)CH_(2)CO O^(-) -H^(+)=NH_(2)CH_(2)CO O^(-)`.

Discussion

22.	Glycine exists as the zwitter ion, N^(+)H_(3), CH_(2),COO^(-). Its conjugate base is
Answer» `NH_2 CH_2COOH` `NH_2CH_2COO^(-)` ` NH_3CH_2COOH` `NH_3 CH_2COOH^(+) ` SOLUTION :` OVERSET( +) NH_3CH_2COO^(-)-H^(+)RARR NH_2CH_2COO^(-) `

Discussion

23.	Glycerol which decomposes at its boiling point can be purified by
Answer» STEAM distillation simple distillation distillation under REDUCED pressure fractional distillation Answer :C

Discussion

24.	Glycerol when trated with a mixture of conc. HNO_(3) and conc. H_(2)SO_(4) forms
Answer» Acrolein Glycerolmononitrate Nitroglycerine Glyceroldinitrate Solution :`{:(CH_(2)-OH""CH_(2)ONO_(2)),("\|""\|"),(CHOH+3HNO_(3) underset(-3H_(2)O)OVERSET(CONC. H_(2)SO_(4))rarr " "CHONO_(2)),("\|""\|"),(CH_(2)OH "" CH_(2)ONO_(2)),("Nitroglycerine"):}`

Discussion

25.	Glycerol on warming with a small amount of hydriodic acid gas converted into
Answer» `CH_(3)-CHI-CH_(3)` `CH_(2)=CH-CH_(2)I` `ICH_(2)-CHI-CH_(2)I` `CH_(3)-CH=CH_(2)` Answer :B

Discussion

26.	Glycerol on treatment with oxalic acid at 110^(@)C forms
Answer» Formic ACID Allyl ALCOHOL `CO_(2)` and `CO` Acrolein Answer :A

Discussion

27.	Glycerol on reaction with excess HI yields
Answer» ALKYL iodide n-Propyl iodide Propene Isopropyl iodide Answer :D

Discussion

28.	Glycerol on oxidation with periodic acid gives
Answer» `2HCHO + HCOOH` `2HCOOH + HCHO` `3HCOOH` `3HCHO` Solution :`{:(CH_(2)OH),("\|"),(CHOH+2HIO_(4)OVERSET(Delta)rarr2HCHO+HCOOH+2HIO_(3)+H_(2)O),("\|"),(CH_(2)OH):}`

Discussion

29.	Glycerol on heating with phosphorus pentaoxide or with KHSO_(4) or conc. H_(2)SO_(4) produces a compound with bad odour. The compound is
Answer» FORMIC ACID ETHYL alcohol acrolein isopropyl iodide Solution :

Discussion

30.	Glycerol on heating with a mixture of KMnO_(4) and H_(2)SO_(4) produces
Answer» Ethanedioicacid Methanoic acid Tartaronic acid Glyceric acid Solution :`{:(CH_(2)-OH""COOH),("\|" "\|"),(CH-CH underset(H_(2)SO_(4))overset(KMnO_(4))rarr" COOH"+CO_(2)+H_(2)O),("\|" "Oxalic acid"),(CH_(2)-OH "(Ethanedioic acid)"):}`

Discussion

31.	Glycerol, on heating with oxalic acid at 110^(@)C gives
Answer» Ethanol Methanoic acid Ether Acetone Answer :B

Discussion

32.	Glycerol is used
Answer» as a sweetening AGENT in BEVERAGES and confectionary in the manufacture of GOOD quality TOILET soaps and cosmetics in the manufacture of nitroglycerine used as an explosive All the above Answer :D

Discussion

33.	Glycerol is highyl viscous. It is due to the fact that
Answer» it is HIGHLY polar it contains THREE `-OH` groups it forms intramolecular H-bonds it forms EXTENSIVE INTERMOLECULAR H-bonds Solution :High viscosity of glycerol is ATTRIBUTED to extensive H-bonding.

Discussion

34.	Glycerol is a
Answer» Primary ALCOHOL Monohydric alcohol Secondary alcohol Trihydric alocohol Solution :`{:(.^(1^(@))CH_(2)-OH),("\|"),(.^(2^(@))CH-OH "(Trihydric alcohol)"),("\|"),(.^(1^(@))CH_(2)-OH),("Glycerol"):}`

Discussion

35.	Glycerol changes to acrolein in the presence of
Answer» Conc. `H_(2)SO_(4)` ANHY. `ZnCl_(2)` `KHSO_(4)` Anhy. `CaCl_(2)` Answer :A::B::C

Discussion

36.	Glycerol is separated in soap industries by :
Answer» Vacuum distillation STEAM distillation Sublimation CRYSTALLIZATION Solution :In vacuum, `P_("ext")=0,` Glycerol will start vapourising faster

Discussion

37.	Glycerol can be obtained from
Answer» FATS PROPYLENE Both None Solution :From fats and propylene both.

Discussion

38.	Glycerol can be purified by
Answer» STEAM distillation Simple distillation Vacuum distillation Fractional distillation Answer :C

Discussion

39.	Glycerine is purified by (a) vacuum distillation (b) simple distillation (c) steam distillation (d) sublimation (e) solvent extraction
Answer» ANSWER :A

Discussion

40.	Glucose underset((ii)P//KI)overset((i)HCN//H_(3)O^(o+))toP P is
Answer» n-heptonoic acid 2-methyl hexanoic acid n-heptane 2-methyl hexane Solution :

Discussion

41.	Glucose as well as fructose are oxidized by periodic acid. The number of moles of HCOOH formed from each mole of glucose and fructose are
Answer» 5 and 5 5 and 4 5 and 3 4 and 3 Answer :C

Discussion

42.	Global warming can be prevented by
Answer» Constructing more dams on rivers DEFORESTATION Growing more trees Carefully UTILIZING ground WATER Answer :C

Discussion

43.	Glassy bead is obtained by heating
Answer» `Na_2 B_4 O_7 10H_2 O` `H_3 BO_3` `B_2 H_6` `Ca_2 B_6 O_(11)` ANSWER :A

Discussion

44.	Glass reacts with IIF to produce
Answer» `SiF_4` `H_2SiF_6` `H_2SiO_3` `Na_3AlF_6` SOLUTION :`6HF + SiO_2 to H_2SiF_6 + 2H_2O`

Discussion

45.	Glass is considered as a supercooled liquid. Give reason.
Answer» SOLUTION :On being heated, glass has the PROPERTY to flow like LIQUIDS as shown by greater thickness of the WINDOW panes at the bottom. Hence, it is called SUPERCOOLED liquid.

Discussion

46.	Glass is considered as a supercooled liquid, Give reason.
Answer» Solution :On being HEATED, glass has the PROPERTY to flow like LIQUIDS as shown by graeter THICKNESS of the window panes at the bottom. Hence, it is called supercooled liquid.

Discussion

47.	Glass is an extremely viscous liquid. Why ?
Answer» Solution :GLASS is an extermely viscous liquid because it is so viscous that MEANS of its properties resemble solids. This observation PROVES that Glass is very thick liquid and glass is heavy at LOWER side.

Discussion

48.	Glass is a
Answer» micro-crystalline solid super-cooled LIQUID gel polymeric mixture Solution :Glass is a supercooled liquid with no SHARP melting POINT.

Discussion

49.	Glass is a ......
Answer» LIQUID solid SUPERCOOLED liquid. TRANSPARENT ORGANIC polymer. Solution :GLASS is a supercooled liquid.

Discussion

50.	Glacial acetic acid is obtained by
Answer» distilling VINEGAR crystallizing, separating and melting acetic ACID treating vinegar with dehyrating agent chemically separating acetic acid. Solution :Acetic acid freezes at `16.6^(@)C` while WATER freezes at `0^(@)C`. So glacial acetic acid is OBTAINED by crystallizing, separating and melting acetic acid

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Gold number is a mesure of the

Gold number gives the indication of

Gold number is

Gold has a close packed structure which can be reviewed as spheres occupying 0.74of the total volume. If the density of gold is 19.3 gg/cc, calculate the apparent radius of a gold atom in the solid (Au=197 amu)

Gold has a close packed structure which can be reviewed as sphere occupying 0.74 of the total volume. If the density of gold is 19.3 g/cc. Calculate the apparent radius of a gold atom in the solid(Au = 197 amu)/

Gold crystallizes in a face - centred cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate the density of gold as well as its atomic radius assuming it to be spherical . Atomic mass of gold =197 amu.

Gold crystallizes in a face centred cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold = 197 amu

Gold can easily dissolve in

Gold (atomic radius =0.144 nm) crystallises in a face centred unit cell. What is the length of the side of the cell ?

Gold(atomic radius = 0.144 nm) crystallisesin a face centred unit cell.What is the length of theside of the cell.

Gold (atomic mass 197 u, atomic radius = 0.144 m ) crystallizes in a face centred unit cell. Determine the density of gold

Go through the following flow sheet. Based on, it answer the questions given at the end of it. Yellow coloured solution (B) is of:

Go through the following flow sheet. Based on, it answer the questions given at the end of it. Select correct statement(s):

Go through the following flow sheet. Based on, it answer the questions given at the end of it. Formation of (F) using (G), KCI (solid) and concentrated H_(2)SO_(4) is called:

Go through the following flow sheet. Based on, it answer the questions given at the end of it. Compounds D,E and F respectively are:

Go through the following flow sheet. Based on, it answer the questions given at the end of it. Compound (A) is:

Glycrine can be purified by

Glycol on reacting with phthalic acid gives

Glycol is used in the manufacture of

Glycol is added to aviation petrol because

Glycine is an alpha-amino acid which exists in the form of Z witter ion as overset(+)NH_(3) CH_(2)CO O^(-). Write the formula of its conjugate acid and conjugate base.

Glycine exists as the zwitter ion, N^(+)H_(3), CH_(2),COO^(-). Its conjugate base is

Glycerol which decomposes at its boiling point can be purified by

Glycerol when trated with a mixture of conc. HNO_(3) and conc. H_(2)SO_(4) forms

Glycerol on warming with a small amount of hydriodic acid gas converted into

Glycerol on treatment with oxalic acid at 110^(@)C forms

Glycerol on reaction with excess HI yields

Glycerol on oxidation with periodic acid gives

Glycerol on heating with phosphorus pentaoxide or with KHSO_(4) or conc. H_(2)SO_(4) produces a compound with bad odour. The compound is

Glycerol on heating with a mixture of KMnO_(4) and H_(2)SO_(4) produces

Glycerol, on heating with oxalic acid at 110^(@)C gives

Glycerol is used

Glycerol is highyl viscous. It is due to the fact that

Glycerol is a

Glycerol changes to acrolein in the presence of

Glycerol is separated in soap industries by :

Glycerol can be obtained from

Glycerol can be purified by

Glycerine is purified by (a) vacuum distillation (b) simple distillation (c) steam distillation (d) sublimation (e) solvent extraction

Glucose underset((ii)P//KI)overset((i)HCN//H_(3)O^(o+))toP P is

Glucose as well as fructose are oxidized by periodic acid. The number of moles of HCOOH formed from each mole of glucose and fructose are

Global warming can be prevented by

Glassy bead is obtained by heating

Glass reacts with IIF to produce

Glass is considered as a supercooled liquid. Give reason.

Glass is considered as a supercooled liquid, Give reason.

Glass is an extremely viscous liquid. Why ?

Glass is a

Glass is a ......

Glacial acetic acid is obtained by