94420 + Interview Questions in Chemistry in Class 12

1.	At lower temperature, all gases show:
Answer» NEGATIVE deviation Positive deviation Positive and negative deviation None Answer :C

Discussion

2.	At low temperatures, the slow addition of molecular bromine to CH_(2)=CH-CH_(2)-=CH gives
Answer» `CH_2=CH-CH_2-CBr=CHBr`<BR>`BrCH_2-CHBr-CH_2-C-=CH` `CH_2=CH-CH_2-CH_2-CBr_3` `CH_3-CBr_2-CH_2-C-=CH` SOLUTION :`CH_2=CH-CH_2-C-=CH+Br_2 to CH_2=CH-CH_2-undersetunderset"Br"\|C=undersetunderset"Br"\|CH`

Discussion

3.	At low temperature phenol reacts with Br_2 in CS_2 to form
Answer» m-bromophenol o-and p-bromophenol p-bromophenol 2, 4, 6-tribromophenol Answer :B

Discussion

4.	At low pressure Vander Wall's equation for 3 moles of a real gas will have its simplified form
Answer» `( PV)/( RT - ( 3A)/( V)) = 3` `( PV)/( RT + Rb) = 3` `( PV)/( RT -3Pb) = 1` `( PV)/( RT - ( 9)/( V)) = 3` Answer :A

Discussion

5.	At low pressure, van der Waals' equation is reduced to[(P) + (a)/(V^2)]V = RT. The compressibility factor can be given as :
Answer» `1-(a)/(RTV)` `1-(RTV)/(a)` `1+ (a)/(RTV)` `1+ (RTV)/(a)` ANSWER :A

Discussion

6.	At low pressure, the vander Waals equation for 1 mole become:
Answer» <P>`PV_(m)=RT` `P(V_(m)-B)=RT` `(P+(a)/(V_(m)^(2)))V_(m)=RT` `P=(RT)/(V_(m))+(a)/(V_(m)^(2))` ANSWER :C

Discussion

7.	At low pressure, the fraction of the surfacecovered follows
Answer» <P>ZERO ORDER kinetics first order kinetics second order kinetics fractional order kinetics. SOLUTION :As `x/m = kP^(1//n)` At lowpressure, ` x/m prop P ` (graph is nearlya straightline ) Thus the reactionfollows first orderkinetics.

Discussion

8.	At low pressure the fraction of the surface covered follows
Answer» zero ORDER reaction second order reaction first order reaction FRACTIONAL order Solution :At low PRESSURE the extent of adsorption is DIRECTLY proportional to pressure which follows first order KINETICS.

Discussion

9.	At low pressure, the fraction of the surface covered follows.
Answer» zero-order kinetics FIRST order kinetics second order kinetics fractional order kinetics Solution :As `(X)/(m)=KP^(1//n)` At LOW pressure `(x)/(m) prop p` (graph is nearly a straight LINE) Thus, the reaction follows first order kinetics.

Discussion

10.	At low pressure and high temperature the Van der Waals equation is finally reduced (simplified) to
Answer» <P>`(p + (a)/(V_m^2))(V_m - B) = RT` `p(V_m - b) = RT` `(p + a/(V_m^2))V_m = RT` `pV_(m) = RT` ANSWER :D

Discussion

11.	At low concentrations, the statement that equimolal solutions under a given set of experimental conditions have equal osmotic pressures is true for :
Answer» All solutions Solutions of non-ELECTROLYTES which NEITHER dissociates nor associater Solutions of electrolytes only None Answer :B

Discussion

12.	At low concentrations, the statement that equimoala solutions under a given set of experimental conditions have equal osmotic pressure is true for
Answer» All SOLUTIONS Solutions of non-electrolytes only Solutions of electrolytes only NONE of these Solution :Equal OSMOTIC PRESSURE only applicable of non-electrolytes solution at concentration.

Discussion

13.	At low concentrations, the statement that equimolal solutions under a given set of experimental conditions have equal osmotic pressures is true for
Answer» all solutions solutions of non-ELECTROLYTES which neither dissociates nor associates solutions of electrolytes only none Solution :For ISOTONIC solutions `pi_(1)=pi_(2) , C_(1) = C_(2)` is valid only for solutes which neither DISSOCIATE (NONELECTROLYTES) nor associate.

Discussion

14.	At low concentrations, the statement that equimolal solutions under a given set of experimental conditions have equal osmotic pressure is true for
Answer» all solutions solutions of NON-ELECTROLYTES only solutions of electrolytes only NONE of these. Solution :Non electrolyets do not undergo dissociation.

Discussion

15.	At low concentration the statement that equimolal solutions under a given set of experiment conditions have equal osmoticpressure is true for
Answer» All SOLUTIONS Solutions of non-ELECTROLYTES only Solutions of electrolytes only Colloidal solutions Answer :B

Discussion

16.	At its melting point ice is lighter than water becouse
Answer» on melting of ICE the `H_(2)O`molecule shrink in size ice crystals have hollow HEXAGONAL arrangement of `H_(2)O`molecules `H_2O` molecules are more closely PACKED in SOLID state ice forms mostly HEAVY water on first melting Solution :Ice crystals have open cage like structures.

Discussion

17.	At infinite the equivalent conductances of CH_(3)COONa, HCl and CH_(3)COOH are 91, 426 and 391 mho cm^(2)eqv^(-1) respectively at 25^(@)C Q. The equivalent conductance of NaCl at infinite dilution will be:
Answer» 126 209 391 908 Answer :A

Discussion

18.	At infinite the equivalent conductances of CH_(3)COONa, HCl and CH_(3)COOH are 91, 426 and 391 mho cm^(2)eqv^(-1) respectively at 25^(@)C Q. The difference of equivalent conductances of H^(+) and Na^(+) is:
Answer» 300 35 335 340 Answer :A

Discussion

19.	At infinite dilution, the percentage ionisation for both strong and weak electrolytes is
Answer» `1%` `20%` `50%` `100%` SOLUTION :According to Ostwald's dilution law because degree of ionization is directly proportional to the dilution.

Discussion

20.	At infinite dilution, the percentage ionisation for both strong and weak electrolyte is…..
Answer» 0.01 0.2 0.5 1 Solution :According to Ostwaid.s DILUTION law, degree of ionisation is DIRECTLY PROPORTIONAL to the dilution.

Discussion

21.	At infinite dilution, the eq. conductances of CH_(3)COONa, HCl and CH_(3)COOH are 91, 426 and 391 mho cm^(2) respectively at 25^(@)C. The eq. conductance of NaCl at infinite dilution will be
Answer» 126 209 391 908 Answer :A

Discussion

22.	At infinite dilution the degree of dissociation for sucrose in aqueous solution is
Answer» 0 0.5 0.99 1

Discussion

23.	At infinite dilution molar conductivity ofBa^(2+)andCl^(-) ions are =127.32S cm^(2)//moland76.34S cm^(2)//mol respectively. What is Lambda_(m)^(oo)for BaCl_(2)at same dilution?
Answer» `280S CM^(2)//MOL^(-1)` `330.98S cm^(2)//mol^(-1)` `90.98S cm^(2)//mol^(-1)` `203.6S cm^(2)//mol^(-1)` ANSWER :A

Discussion

24.	At infiinite dilution, the aqueous solution of BaCl_(2). M olar conductivity of Ba^(2+) and Cl^(-) ions are =127.32 S cm^(2) /mol and 76.34S cm^(2)/mol respectively what is ^^_(m)^(@)for BaCl_(2) at same dilution?
Answer» `280Scm^(2)"MOL"^(-1)` `330.98S CM^(2)"mol"^(-1)` `90.98S cm^(2)"mol"^(-1)` `203.6S cm^(2) "mol"^(-1)` SOLUTION :`^^_(m)^(oo)=^^_(BA^(2+))^(oo)+2lambda_(Cl^(-))^(oo)` `=127.32+2xx76.34` `=280S cm^(2) "mol"^(-1)`

Discussion

25.	At infinite dilution, molar conductivities of Ba(OH)_(2), BaCl_(2) and NH_(4)Cl solution are 523.28, 280.0 and "129.8 ohm"^(-1)."cm"^(2)."mol"^(-1) respectively. Calculate the molar conductivity of NH_(4)Cl solution at infinite dilution.
Answer» Solution :`Lambda_(m)^(@)[BA(OH)_(2)]=lambda_(m)^(@)(B^(+2))+2lambda_(m)^(@)(OH^(-))` `=523.28" OHM"^(-1)."cm"^(2)."mol"^(-1)"...[1]"` `Lambda_(m)^(@)(BaCl_(2))=lambda_(m)^(@)(Ba^(+2))+2lambda_(m)^(@)(Cl^(-))` `="280.0 ohm"^(-1)."cm"^(2)."mol"^(-1)"...[2]"` and `Lambda_(m)^(@)(NH_(4)Cl)=lambda_(m)^(@)=lambda_(m)^(@)(NH_(4)^(+))+lambda_(m)^(@)(Cl^(-))` `="129.8 ohm"^(-1)."cm"^(2)."mol"^(-1)...[3]"` Subtracting equation [2] from [1] gives `2lambda_(m)^(@)(OH^(-))-2lambda_(m)^(@)(Cl^(-))="243.28 ohm"^(-1)."cm"^(2)."mol"^(-1)"...[4]"` Multiplying equation [3] by 2 and then adding the remit to equation [4], we get `2lambda_(m)^(@)(NH_(4)^(+))+2lambda_(m)^(@)(OH^(-))` `=2xx129.8+243.28=502.88" ohm"^(-1)."cm"^(2)."mol"^(-1)` `therefore""Lambda_(m)^(@)(NH_(4)OH)=(502.88)/(2)="251.44 ohm"^(-1).cm"^(2)."mol"^(-1)`

Discussion

26.	At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3sqrt(3) times that of a hydrocarbon having molecular formula C_(n)H_(2n-2). What is the value of ?
Answer» 1 4 3 8 Solution :`r_(1)PROP1/(sqrt(M_(1))`or `(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))` Given `r_(1)=3sqrt(3)r_(2)IMPLIES(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))=3sqrt(3)` or `3sqrt(3)=sqrt((M_(2))/2)` `M_(2)=27xx2=54` Now `12xxn+2n-2=54` `14n=56\>n=4`

Discussion

27.	At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3sqrt(3) times that of a hydrocarbon having molecular formula C_(n)H_(2n-n). What is the value of n?
Answer» Solution :`(r_(H_(2)))/(r_(HC))=SQRT((M_(HC))/(M_(H_(2))))IMPLIES 3sqrt(3) = sqrt((M_(HC))/(2))` or `M_(HC)=(3sqrt(3))^(2)xx2=54` `:. c_(n)H_(2n-2),12 xxn+(2n-2)=54 ` or `n=4`

Discussion

28.	At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3 sqrt( 3) times that of a hydrocarbon having molecular formula C_(n) H_(2n-2). What is the value of n ?
Answer» 1 4 3 8 Solution :`(r_(H_(2)))/(r_("Hydrocarbon")) = SQRT((M_("Hydrocarbon"))/(M_(H_(2))) ) = 3sqrt(3)` `:. sqrt((M_("Hydrocarbon"))/(2)) = 3 sqrt(3)` `(M_("Hydrocarbon"))/(2) = 27` `M_("Hydrocarbon") = 2 xx 27 = 54 ` `:. N = 4` and hydrocarbon is `C_(4) H_(6)`

Discussion

29.	At higher temperature, iodoform reaction is given by
Answer» `CH_(3)CO_(2)CH_(3)` `CH_(3)CO_(2)C_(2)H_(5)` `C_(6)H_(5)CO_(2)CH_(3)` `CH_(3)CO_(2)C_(6)H_(5)` Answer :D

Discussion

30.	At high temperature carbon reacts with water to prodcue a mixture of carbon monoxide, CO and hydrogen, H_(2). C+H_(2)O overset("red heat")to CO+H_(2) CO is separted from H_(2) and then used to separtee nickel from cobalt by forming a volatile compound, nickel tetracarbonyl, Ni(CO__(3)) Ni+4CO to Ni(CO)_(4) How many moles of Ni(CO)_(4) could be obatined from the CO produced by the reaction of 75g of carbon? Assume 100% reaction and 100% recovery in both steps.
Answer» ANSWER :B

Discussion

31.	At higher altitudes, the boiling point of water lowers because :
Answer» ATMOSPHERICPRESSURE is low Temperature is low Atmospheric PRESSURE is high None of these Answer :A

Discussion

32.	At the higher altitudes the boiling point of water lowers because
Answer» ATMOSPHERIC pressure is low Temperature is low Atmospheric pressure is HIGH NONE of these Solution :The boiling OCCURS at lower temperature if atmospheric pressure is lower than 76 cm Hg.

Discussion

33.	At high temperature, S_(4)N_(4) decomposes into sulphur vapour and nitrogen. If 1 mL of S_(4)N_(4) is decomposed and 2.5 mL of mixture are obtained, the forinula of sulphur vapour is
Answer» `S_(2)` `S_(4)` `S_(6)` `S_(8)` ANSWER :D

Discussion

34.	At high temperature nitrogen can combine directly with:
Answer» Fe Zn Mg Cr Answer :C

Discussion

35.	At high temperature nitrogen combines with CaC_2 to give:
Answer» CALCIUM cyanide Calcium cyanamide Calcium carbonate Calcium nitride Answer :B

Discussion

36.	At high pressure, the following reaction is of zero order. 2" NH"_(3)(g)overset(1130"K")underset("Platinum catalyst")to" N"_(2)(g)+3" H"_(2)(g) Which of the following options are correct for this reaction ?
Answer» Rate of reaction = Rate CONSTANT Rate of the reaction depends on concentration of ammonia Rate of decomposition of ammonia will REMAIN constant until ammonia disappears completely Further increase in pressure will change the rate of reaction. Solution :For zero order reaction, Rate `=k[A]^(0)=k.` Hence, (a) is correct. Rate of zero order reaction is independent of concentration of reactants. Hence, (b) is WRONG. For the same reason, ( c ) is correct. Rateof decomposition of a gas depends upon pressure. Hence, (d) is correct.

Discussion

37.	At high pressure the following reaction is zero order. 2NH_(3(g))(1130K)/("Platinum catalyst")N_(2(g))+3H_(2(g)) Which of the following options are correct for this reaction?
Answer» Rate of reaction =Rate constant Rate of the reaction depends on concentration of ammonia Rate of DECOMPOSITION of ammonia will remain constant until ammonia disappears completely Further increase in pressure will change the rate of reaction Solution :For zero order REATION,Rate =`k[A]^(0)=k` Hence (A) is correct.Rate of zero order reaction is INDEPENDENT of concentration of REACTANTS. Hence ,(B) is wrong . For the same reaction (C ) is correct Rate of decomposition of a gas depends UPON pressure.Hence (D) is correct.

Discussion

38.	At high concentration of soap in water, soap behaves as __________ .
Answer» MOLECULAR COLLOID ASSOCIATED colloid macromolecular colloid lyophilic colloid Solution :associated colloid

Discussion

39.	At high pressure, the compressibility factor for one mole of van der waals gas will be
Answer» 1 + `FRAC{B}{RT}` 1 - `frac{Pb}{RT}` 1 + `frac{Pb}{RT}` 1 + `frac{a}{VRT}` Answer :B

Discussion

40.	At high concentration of soap in wate, soap behave as……….
Answer» MOLECULAR COLLOID associated colloid MACROMOLECULAR colloid LYOPHILIC colloid Answer :B

Discussion

41.	At high concentration of soap in water, soap bchaves as
Answer» MOLECULAR colloid associated colloid macromolecular colloid lyophilic colloid SOLUTION :At a higher concentration, the soap molecules aggregates to form a micelles. At a LOW concentration it BEHAVES as an electrolyte.

Discussion

42.	At high concentration of soap in water, soap behaves as
Answer» MOLECULAR COLLOID ASSOCIATED colloid macromolecular colloid lyophilic colloid Solution :associated colloid

Discussion

43.	At high altitudes the boiling point of water gets lowered because
Answer» 1. TEMPERATURE is low 2. atmospheric pressure is low 3. pressure is HIGH 4. none of these Answer :B

Discussion

44.	At high altitudes the boiling point of water decreases because:
Answer» ATMOSPHERIC PRESSURE is low Temperature is low Atmospheric pressure is high None Answer :A

Discussion

45.	At high altitude ,the boiling point of water decreases because
Answer» the ATMOSPHERIC PRESSURE is high the TEMPARATURE is low the atmospheric pressure is low the temparature is high Answer :C

Discussion

46.	At first glance the conversion of bromobenzene to benzenitrile looks simple just cary out a nucleophilic substitution using cyanide ion as the nucleophile. Then we remember that bromobenzene does not undergo either an S_(N)1 or an S_(N)2 reaction (Section 6.14A). The conversion can be accomplishes. however. thogh it involves severals steps. Outline possible steps.
Answer» SOLUTION :

Discussion

47.	At equimolar concentrations of Fe^(2+) and Fe^(3+), what must [Ag^(+)] be so that voltage of the galvanic cell made from Ag^(+)//Ag and Fe^(3+)//Fe^(2+) electrodes equals zero ? The reaction is Fe^(2+)+Ag^(+) hArr Fe^(3+)+Ag. Determine the equilibrium constant at 25^(@)C for the reaction. ("Given: "E_(Ag^(+)//Ag)^(@)=0.799" volt and "E_(Fe^(3+)//Fe^(2+))^(@)=0.771" volt")
Answer» Solution :`E_(cell)^(@)=E_(FE^(2+)//Fe^(3+))^(@)+E_(Ag^(+)//Ag)^(@)` `=-0.771+0.799=0.028` VOLT At equilibrium, `E_(cell)=0` `0=E_(cell)^(@)-0.0591/1 "log" ([Fe^(3+)])/([Fe^(2+)][Ag^(+)])` `=E_(cell)^(@)-0.0591"log"1/([Ag^(+)])` `[Ag^(+)]=0.34` `log K=(NE^(@))/(0.0591),""K=3.0`

Discussion

48.	At given temperature and pressure adsorption of which gas of the following will take place the most ?
Answer» DI HYDROGEN Di OXYGEN AMONIA Di nitrogen Answer :C

Discussion

49.	At given temperature and pressure nitrogenn gas is more soluble I water than helium gas. Which one of them has higher K_(H) value.
Answer» SOLUTION :HELIUM GAS.

Discussion

50.	At equillibrium if K_p =1 then :
Answer» `DeltaG^o = 0` `DeltaG^o GT 1` `DeltaG^o LT 1` None Answer :A

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

At lower temperature, all gases show:

At low temperatures, the slow addition of molecular bromine to CH_(2)=CH-CH_(2)-=CH gives

At low temperature phenol reacts with Br_2 in CS_2 to form

At low pressure Vander Wall's equation for 3 moles of a real gas will have its simplified form

At low pressure, van der Waals' equation is reduced to[(P) + (a)/(V^2)]V = RT. The compressibility factor can be given as :

At low pressure, the vander Waals equation for 1 mole become:

At low pressure, the fraction of the surfacecovered follows

At low pressure the fraction of the surface covered follows

At low pressure, the fraction of the surface covered follows.

At low pressure and high temperature the Van der Waals equation is finally reduced (simplified) to

At low concentrations, the statement that equimolal solutions under a given set of experimental conditions have equal osmotic pressures is true for :

At low concentrations, the statement that equimoala solutions under a given set of experimental conditions have equal osmotic pressure is true for

At low concentrations, the statement that equimolal solutions under a given set of experimental conditions have equal osmotic pressures is true for

At low concentrations, the statement that equimolal solutions under a given set of experimental conditions have equal osmotic pressure is true for

At low concentration the statement that equimolal solutions under a given set of experiment conditions have equal osmoticpressure is true for

At its melting point ice is lighter than water becouse

At infinite the equivalent conductances of CH_(3)COONa, HCl and CH_(3)COOH are 91, 426 and 391 mho cm^(2)eqv^(-1) respectively at 25^(@)C Q. The equivalent conductance of NaCl at infinite dilution will be:

At infinite the equivalent conductances of CH_(3)COONa, HCl and CH_(3)COOH are 91, 426 and 391 mho cm^(2)eqv^(-1) respectively at 25^(@)C Q. The difference of equivalent conductances of H^(+) and Na^(+) is:

At infinite dilution, the percentage ionisation for both strong and weak electrolytes is

At infinite dilution, the percentage ionisation for both strong and weak electrolyte is…..

At infinite dilution, the eq. conductances of CH_(3)COONa, HCl and CH_(3)COOH are 91, 426 and 391 mho cm^(2) respectively at 25^(@)C. The eq. conductance of NaCl at infinite dilution will be

At infinite dilution the degree of dissociation for sucrose in aqueous solution is

At infinite dilution molar conductivity ofBa^(2+)andCl^(-) ions are =127.32S cm^(2)//moland76.34S cm^(2)//mol respectively. What is Lambda_(m)^(oo)for BaCl_(2)at same dilution?

At infiinite dilution, the aqueous solution of BaCl_(2). M olar conductivity of Ba^(2+) and Cl^(-) ions are =127.32 S cm^(2) /mol and 76.34S cm^(2)/mol respectively what is ^^_(m)^(@)for BaCl_(2) at same dilution?

At infinite dilution, molar conductivities of Ba(OH)_(2), BaCl_(2) and NH_(4)Cl solution are 523.28, 280.0 and "129.8 ohm"^(-1)."cm"^(2)."mol"^(-1) respectively. Calculate the molar conductivity of NH_(4)Cl solution at infinite dilution.

At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3sqrt(3) times that of a hydrocarbon having molecular formula C_(n)H_(2n-2). What is the value of ?

At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3sqrt(3) times that of a hydrocarbon having molecular formula C_(n)H_(2n-n). What is the value of n?

At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3 sqrt( 3) times that of a hydrocarbon having molecular formula C_(n) H_(2n-2). What is the value of n ?

At higher temperature, iodoform reaction is given by

At higher altitudes, the boiling point of water lowers because :

At the higher altitudes the boiling point of water lowers because

At high temperature, S_(4)N_(4) decomposes into sulphur vapour and nitrogen. If 1 mL of S_(4)N_(4) is decomposed and 2.5 mL of mixture are obtained, the forinula of sulphur vapour is

At high temperature nitrogen can combine directly with:

At high temperature nitrogen combines with CaC_2 to give:

At high pressure, the following reaction is of zero order. 2" NH"_(3)(g)overset(1130"K")underset("Platinum catalyst")to" N"_(2)(g)+3" H"_(2)(g) Which of the following options are correct for this reaction ?

At high pressure the following reaction is zero order. 2NH_(3(g))(1130K)/("Platinum catalyst")N_(2(g))+3H_(2(g)) Which of the following options are correct for this reaction?

At high concentration of soap in water, soap behaves as __________ .

At high pressure, the compressibility factor for one mole of van der waals gas will be

At high concentration of soap in wate, soap behave as……….

At high concentration of soap in water, soap bchaves as

At high concentration of soap in water, soap behaves as

At high altitudes the boiling point of water gets lowered because

At high altitudes the boiling point of water decreases because:

At high altitude ,the boiling point of water decreases because

At given temperature and pressure adsorption of which gas of the following will take place the most ?

At given temperature and pressure nitrogenn gas is more soluble I water than helium gas. Which one of them has higher K_(H) value.

At equillibrium if K_p =1 then :