94420 + Interview Questions in Chemistry in Class 12 Page 579 InterviewSolution

28901.	The lowest degree of paramagnetism per mol of the compound in the following will be shown by :
Answer» `MnSO_4cdot4H_2O` `ZnSO_4cdot7H_2O` `FeSO_4cdot7H_2O` `NiSO_4cdot6H_2O` ANSWER :B

Discussion

28902.	The lowest bond energy exist in the following bonds for:
Answer» C-C N-N H-H O-O Answer :D

Discussion

28903.	The lowering of vapour pressure of the solvent takes place
Answer» only when the solute is non-volatile only when the solute is volatile only when the solute is a non-electrolyte in all the above THREE cases. Solution :Lowering of V.P. of SOLVENT TAKES place-the solute MAY be volatile, non-volatile or electrolyte or non-electroyte.

Discussion

28904.	The lowering of vapour pressure in a saturated aq. Solution of salt AB is found to be 0.108 torr. If vapour pressure of pure solvent at the same temperature is 300 torr, find the solubility product of salt AB.
Answer» `10^(-8)` `10^(-6)` `10^(-4)` `10^(-5)` ANSWER :C

Discussion

28905.	The lowering of vapour pressure of a solvent by the addition of a non-volatile solute to it , is directly proportional to :
Answer» The STRENGTH of the SOLUTION The nature of the solute in the solution The ATMOSPHERIC pressure All Answer :A

Discussion

28906.	The lowering of vapour pressure of 0.1M aqueous solutions of NaCl, CuSO_4, and K_2SO_4 are:
Answer» All equal In the RATIO of 1 : 1 : 1.5 In the ratio of 3 : 2 : 1 In the ratio of 1.5 : 1 : 2.5 Answer :B

Discussion

28907.	The lowering in vapour pressure when a solute is added
Answer» Will depend uponthe nature of the SOLUTE Will be the same irrespective of the solvent TAKEN Will be DIFFERENT when different solvent are taken Will depend on the form IDEAL solution Solution :Explanation : (a) `Delta P=oP_(A)^(@)x_(B)` (a) Hence, `Delta F` depends upon i which is dependent on nature of solute (Hence, choice (a) is correct). (c ) `Delta F` depends upon `P_(A)^(@)`. If `P_(A)^(@)` changes then `Delta F` changes. Hence when different solvents are taken `Delta F` value is different. Choice (c ) is also correct. (d) Is also correct. `Delta F` changes with mole fraction of solute. `Delta F` also changes with TEMPERATURE. (b) Is wrong because `Delta F` depends upon solvent.

Discussion

28908.	The low solubility of BaSO_(4) in water is due to
Answer» Low DISSOCIATION ENERGY Ionic bonds High VLAUE of LATTICE energy NONE of these Answer :C

Discussion

28909.	The low solubility of BaSO_4 in water is due to:
Answer» Ionic BOND in `BaSO_4` LOW lattice ENERGY of `BaSO_4` High lattice energy of `BaSO_4` Dissociation energy of `BaSO_4` ANSWER :C

Discussion

28910.	The low bond energy of F_(2) is explained by :
Answer» the attainment of noble gas configuration `(F^(-))` the low electron AFFINITY of F repulsion by electron PAIRS on F the SMALL size of F. Answer :C

Discussion

28911.	The longest lambda for the Lyman series is…. (Given R_H=109678 cm^(-1)):
Answer» 1215 1315 1415 1515 Answer :A

Discussion

28912.	The long form of periodic table is based on
Answer» SHAPE of the atom Mass of the atom Atomic NUMBER of the atom Electronegativity ANSWER :C

Discussion

28913.	The long form of periodic table has
Answer» Eight horizontal ROWS and seven VERTICAL columns Seven horizontal rows and EIGHTEEN vertical columns Seven horizontal rows and seven vertical columns Eight horizontal rows and eight vertical columns Answer :B

Discussion

28914.	The lone pair present on N family hydrides more easily participates in bond formation in:
Answer» `AsH_3` `PH_3` `NH_3` `SbH_3` ANSWER :C

Discussion

28915.	Thelogarithmof rateconstant of areaction
Answer» decreaseslinearlywithincreasein inverseoftemperature INCREASES linearlywithincreasesin INVERSE oftemperature increases linearlywithincreasesin TEMPERATURE Decreaseslinearly withincreases in temperature Solution :STATEMENT

Discussion

28916.	The lisat product (B) formed in the following reaction is
Answer» `CH _3-CH_2-Br+ Mg ---rarr( ^dry ETHANOL) A --rarr(^CH_3COOH) B ` `CH_3- CH_2-OH` `CH_3- CH_3` `CH_3 -- CH_2 -- COOH` ANSWER :C

Discussion

28917.	The liquids A and B have vapour pressure 500 and 200 torr respectivelyat a certain temperature . In an ideal solution of the two , the mole fraction of A in vapour state, at which two liquids have equal partial pressure is
Answer» 0.5 0.2 0.286 0.714 Answer :A

Discussion

28918.	The liquid which has the highest rate of evaporation is
Answer» petrol nail-polish remover water alcohol Solution :As intermolecular forces are least in case of petrol. THUS, it has highest rate of EVAPORATION.

Discussion

28919.	The liquid is in equlibrium with its vapoursat its boiling point. On the average, the molecules in the two phases have equal
Answer» POTENTIAL energy total energy KINETIC energy intermoleclar forces Solution :At the boilding POINT the liquid and its vapiour have EQUAL kinetic energies.

Discussion

28920.	The liquid hydrazinium is denoted as
Answer» `N_2H_4` `N_2H_5^+` `N_2H_3^-` `N_2H_3^+` ANSWER :B

Discussion

28921.	The liquefaction behaviour of temporary gases like CO_(2) approaches that of permanent gases like N_(2),O_(2) etc, as we go to
Answer» below CRITICAL temperature above critical temperature above ABSOLUTE zero below absolute zero Answer :B

Discussion

28922.	The linkage present in two nucleotide is
Answer» AMIDE linkage peptide linkage phosphodiester linkage glycosidic linkage Answer :C

Discussion

28923.	The linkage between the two monosaccharide units in lactose is
Answer» `C_(1)`of `BETA`-D-glucose and `C_(4)` of `beta`-D-glactose `C_(1)` of `beta`-D-galactose and `C_(4)` of `beta-`D-glucose `C_(1)` of `alpha`-D-galactose and `C_(4)`of `beta`-D-glucose `C_(1)` of `beta`-D-glactose and `C_(4)` of `alpha`-D-glucose Solution :Lactose is obtained by the condensation of one molecule of `beta`-D-glactopyranose (`C_(1)-beta)` unit and one molecule of `beta`-D-glycopyranose `(C_(4)-beta)` units.

Discussion

28924.	The liner structure is not assumed by:
Answer» `SnCl_2` `CS_2` `NO_2^+` HCN Answer :A

Discussion

28925.	The line at 434nm in the Balmer series of the hydrogen spectrum corresponds to a transition of an electron from the n^(th) to second Bohr orbit. What is the value of n?
Answer» ANSWER :n=5

Discussion

28926.	The line spectrum observed when electron falls from the higher level into L level is known as :
Answer» BALMER series Paschen series Bracket series None of these Answer :A

Discussion

28927.	The limiting molar conductivity of an electrolyte is obtained by adding the limiting molar conductivities of cation and anion of the electrolyte. What is meant by limiting molar conductivity ?
Answer» SOLUTION :The MAXIMUM or limiting VALUE of molar conductivityof an electrolyte when its concentration approaches zero is CALLED the limiting molar CONDUCTIVITY `(^^_m^@)`of the electrolyte.

Discussion

28928.	The limiting radius ratio for tetrahedral voids has the range:
Answer» `0.414` to `0.732 ` `0.225` to `0.414 ` `0.155` to `0.225 ` 0 to `0.155 ` ANSWER :C

Discussion

28929.	The limmiting molar conductivites ^^ ^(@) for NaCl_(2)Kbr and KCl are 126,152 and 150 S cm^(2) respectively the ^^ ^(@) for NaBr is
Answer» `128Scm^(2)"mol"^(-1)` `302Scm^(2)"mol"^(-1)` `278S CM^(2)"mol"^(-1)` `176Scm^(2)"mol"^(-1)` SOLUTION :`^^_(NaBr)^(@)=^^_(NaCl)^(@)+^^_(KBr)^(@)-^^_(KCL)^(@)` `=126+152-150=128 S cm^(2)"mol"^(-1)`

Discussion

28930.	The limiting radius ratio for tetrahedral shape is
Answer» `0" to "0.155` `0.155-0.225` `0.225-0.414` `0.414-0.732` ANSWER :C

Discussion

28931.	The limiting molar conductivities of KCl, NaCl and KNO_3 are 150, 126 and 109 S cm^2mol^(-1)respectively. What is the limiting molar conductivity of NaNO_3 ?
Answer» `385 S Cm^2 MOL^(-1)` `167 S Cm^2 mol^(-1)` `85 S Cm^2 mol^(-1)` `133 S Cm^2 mol^(-1)` ANSWER :C

Discussion

28932.	The limiting molar conductivity of an electrolyte is obtained by adding the limiting molar conductivities of cation and anion of the electrolyte. Name of the above law
Answer» SOLUTION :Kohlrausch.s LOW

Discussion

28933.	The limiting molar conductivities of HCl, CH_(3)COONa and NaCl are respectively 425, 90 and 250 mho cm^(2)mol^(-1) at 25^(@)C.t he molar conductivity of 0.1 M CH_(3)COOH solution is 7.8 mho cm^(2)mol^(-1) at the same temperature. The degree of dissociation of 0.1 M acetic acid solution at the same temperature is
Answer» `0.10` `0.02` `0.15` `0.03` SOLUTION :`wedge_(CH_(3)COOH)^(@)=wedge_(CH_(3)COONa)^(@)+wedge_(HCL)^(@)-wedge_(NaCl)^(@)` `=90+425-125=390" MHO "cm^(2)mol^(-1)` `alpha=(wedge_(m)^(@))/(wedge_(m)^(@))=(7.8)/(390)=.02`

Discussion

28934.	The limiting molar conductivities of HCl, CH_(3)COONa and NaCl are respectively 425,90 and 125 mho cm^(2)mol^(-1) at 25^(@)C. The molar conductivity of 0.1 M CH_(3)COOH solution is 7.8 mho cm^(2)mol^(-1) at the same temperature. The degree of dissociation of 0.1 M acetic acid solution at the same temperature is
Answer» 0.1 0.02 0.15 0.03 Solution :`WEDGE^(@)` for `CH_(3)COOH=lamda_(CH_(3)COO^(-))^(@)+lamda_(Na^(+))^(@)` `lamda_(H^(+))^(@)+lamda_(Cl^(-))^(@)-lamda_(Na^(+))^(@)-lamda_(Cl^(-))^(@)` `=lamda_(CH_(3)COO^(-))^(@)+lamda_(H^(+))^(@)` `=90+425-125=390mho" "cm^(2)mol^(-1)` DEGREE of DISSOCIATION`(alpha)=(wedge_(m)^(@))/(wedge_(m)^(@))=(7.8)/(390)=0.02`

Discussion

28935.	The limiting molar conductivities of HCl, CH_(3)COONa and NaCl are respective 425, 190 and 150 mho cm^(2)mol^(-1) at 25^(@)C. The molar conductivity of 0.1 M acetic acid is 9.2 mho cm^(2)mol^(-1). The degree of dissociation of 0.1 M acetic acid is
Answer» `0.10` 0.02 0.19 0.03 Solution :`wedge_(CH_(3)COOH)^(@)=wedge_(CH_(3)COONA)^(@)+wedge_(HCl)^(@)-wedge_(NaCl)^(@)` `""=190+425-150=465` `alpha=wedge_(m)^(C)/wedge_(m)^(@)=9.2/465=0.19`

Discussion

28936.	The limiting molar conductivities (Lambda^(0)) for NaCl, KBr and KCl are 126, 152 and 150 S.cm^(2) "mol"^(-1) respectively. Then A for NaBr is
Answer» `128 S cm^(2) MOL^(-1)` `302 S cm^(2) mol^(-1)` `278 S cm^(2) mol^(-1)` `176 S cm^(2) mol^(-1)` Answer :A

Discussion

28937.	The limiting molar conductivities ^^ for NaCl, KBr and KCl are 126,152 and 150 S cm^(2)mol^(-1) respectively. The ^^ for NaBr is
Answer» 278 S `cm^(2)mol^(-1)` 176 S `cm^(2)mol^(-1)` 128S `cm^(2)mol^(-1)` 302 S `cm^(2)mol^(-1)` Solution :`(126s" "cm^(2))wedge_(NaCl)^(o)=wedge_(Na^(+))^(o)+wedge_(Cl^(-))^(o)` . . .(i) `(152s" "cm^(2))wedge_(KBr)^(o)=wedge_(K^(+))^(o)+wedge_(Br^(-))^(o)` . . . (II) `(150s" "cm^(2))wedge_(KCL)^(o)=wedge_(K^(+))^(o)+wedge_(Cl^(-))^(o)` . . .(iii) by EQUATION (i)+(ii)-(iii) `because wedge_(NABR)^(o)=wedge_(Na^(+))^(o)+wedge_(Br^(-))^(o)` ltBrgt `=126+152-150=128s" "cm^(2)mol^(-1)`.

Discussion

28938.	The limiting molar conductances wedge^(infty) for NaCl, KBr and KCl are 126, 152 and 150 Scm^(2) mol^(–1) respectively. The wedge^(infty) for NaBr will be:
Answer» 302 S `CM^(2) MOL^(–1)` 176 S `cm^(2) mol^(–1)` 278 S `cm^(2) mol^(–1)` `128 S cm^(2) mol^(–1)` ANSWER :D

Discussion

28939.	The limiting molar conductance of sodium chloride, potassium chloride and potassium bromide are 126.45 , 149.86 and 151.92 ohm^(-1) cm^(2) mol^(-1) respectively . Calculate limiting molar ionic conductance of Na^(+) given that limiting molar ionic conductance of Br^(-) ion is 76.34 ohm^(-1) cm^(2) mol^(-1)
Answer» Solution :This can be calculated with the help of Kohlrausch's law because limiting MOLAR IONIC conductance are additive `Lambda_(m) ^(@) (KBr) = lambda_(m)^(@) (K^(+)) + lambda_(m)^(@) (Br^(-))` `Lambda_(m)^(@) (KBr) = 151.92 ohm^(-1) cm^(2) mol^(-1)`, `lambda_(m)^(@) (Br^(-)) = 76.34 ohm^(-1) cm^(2) mol^(-1)` `therefore lambda_(m)^(@) (K^(+)) = Lambda_(m)^(@) (KBr) - lambda_(m)^(@) (Br^(-))` = 151.92 - 76.34 =` 75.58 ohm^(-1) cm^(2) mol^(-1)` Now , `Lambda_(m)^(@) (KCl) = lambda_(m)^(@) (K^(+)) + lambda_(m)^(@) (Cl^(-))` `Lambda_(m)^(@) (KCl) = 149.86 ohm^(-1) cm^(2) mol^(-1)` , `lambda_(m)^(@) (K^(+)) = 75.58 ohm^(-1) cm^(2) mol^(-1)` `therefore lambda_(m)^(@) (Cl^(-)) = Lambda_(m)^(@) (KCl) - lambda_(m)^(@) (K^+)` `= 149.86 - 75.58` `= 74.28 ohm^(-1) cm^(2) mol^(-1)` Now `Lambda_(m)^(@) (NACL) = lambda_(m)^(@) (NA^(+)) + lambda_(m)^(@) (Cl^(-))` `Lambda_(m)^(@) (NaCl) = 126.45 ohm^(-1) cm^(2) mol^(-1)` Now `Lambda_(m)^(@) (NaCl) = lambda_(m)^(@) (Na^(+)) + lambda_(m)^(@) (Cl^(-))` `Lambda_(m)^(@) (NaCl) = 126.45 ohm^(-1) cm^(2) mol^(-1)` `lambda_(m)^(@) (Cl^(-)) = 74.28 ohm^(-1) cm^(2) mol^(-1)` `therefore lambda_(m)^(@) (Na^(+)) = Lambda_(m)^(@) (NaCl) - lambda_(m)^(@) (Cl^(-))` `=126.45 - 74.28` `=52.17 ohm^(-1) cm^(2) mol^(-1)`

Discussion

28940.	The limit of a particular detection system is 0.002 dps for 1g sample. Find the maximum t_((1)/(2)) that this system could detect in a 1-g sample of a nuclide of mass number 200.
Answer» SOLUTION :We know, N= no. of ATOMS= moles `xx` Av. Const.`=(1)/(200) xx 6.022 xx 10^(23)` `-(d(N))/(dt)= lamda (N)` `0.002 = (0.6932)/(t_((1)/(2)))xx (1)/(200) xx 6.022 xx 10^(23)` `therefore t_((1)/(2))=1.0436 xx 10^(24)` SECONDS `=3.334 xx 10^(16)` years.

Discussion

28941.	The limiting line in Paschen series corresponds to
Answer» `n_(1)=2,n_(2)=3` `n_(1)=3,n_(2)=4` `n_(1)=3,n_(2)=10` `n_(1)=3,n_(2)=OO` Answer :D

Discussion

28942.	The limiting density of hydrogen bromide is 3.6108 at 0^(@)C. The exact atomic weight ofbromine is (At. wt. of H = 1.008)
Answer» 80.92 79.92 89.29 79.29 Solution :This can be calculated by limiting density method which is BASED on the fact that at LIMIT of zero pressure the ideal gas equation hecomes exact for all gases i.e., PV=nRT Molecular weight of HBr on this BASIS is `M=RT(p/P)_(Pto0)` `M=0.0821xx273.15xx3.6108[0^(@)C=273.15K]=80.93` `therefore` 80.93 = atomic weight of hydrogen+ atomic weight of bromine =1.008 + atomic weight of bromine or atomic weight ofbromine= 80.93 - 1.008 = 79.92

Discussion

28943.	The lightning bolts in atmosphere cause the formation of:
Answer» `NO` `O_2` `CO_2` `H_2O_2` ANSWER :A

Discussion

28944.	The lightest, non-inflammable gas is:
Answer» `H_2` He `N_2` Ar Answer :B

Discussion

28945.	The lightest particle is:
Answer» Electron Proton Neutron `BETA`-particle Answer :A

Discussion

28946.	The lightest metal in the periodic table is :
Answer» H Mg Ca Li Answer :D

Discussion

28947.	The lightest metal is :
Answer» Li Mg Ca Na Answer :A

Discussion

28948.	The lightest metal I
Answer» Li Mg Ca Na ANSWER :A

Discussion

28949.	The light yellow compound produced when actone reacts with iodine and alkali, is
Answer» `CH_(3)COCH_(2)I` `CH_(3)I` `CHI_(3)` None of these Answer :C

Discussion

28950.	The lightest gas is:
Answer» Nitrogen Helium Oxygen Hydrogen Answer :D

Discussion

Explore topic-wise InterviewSolutions in .

The lowest degree of paramagnetism per mol of the compound in the following will be shown by :

The lowest bond energy exist in the following bonds for:

The lowering of vapour pressure of the solvent takes place

The lowering of vapour pressure in a saturated aq. Solution of salt AB is found to be 0.108 torr. If vapour pressure of pure solvent at the same temperature is 300 torr, find the solubility product of salt AB.

The lowering of vapour pressure of a solvent by the addition of a non-volatile solute to it , is directly proportional to :

The lowering of vapour pressure of 0.1M aqueous solutions of NaCl, CuSO_4, and K_2SO_4 are:

The lowering in vapour pressure when a solute is added

The low solubility of BaSO_(4) in water is due to

The low solubility of BaSO_4 in water is due to:

The low bond energy of F_(2) is explained by :

The longest lambda for the Lyman series is…. (Given R_H=109678 cm^(-1)):

The long form of periodic table is based on

The long form of periodic table has

The lone pair present on N family hydrides more easily participates in bond formation in:

Thelogarithmof rateconstant of areaction

The lisat product (B) formed in the following reaction is

The liquids A and B have vapour pressure 500 and 200 torr respectivelyat a certain temperature . In an ideal solution of the two , the mole fraction of A in vapour state, at which two liquids have equal partial pressure is

The liquid which has the highest rate of evaporation is

The liquid is in equlibrium with its vapoursat its boiling point. On the average, the molecules in the two phases have equal

The liquid hydrazinium is denoted as

The liquefaction behaviour of temporary gases like CO_(2) approaches that of permanent gases like N_(2),O_(2) etc, as we go to

The linkage present in two nucleotide is

The linkage between the two monosaccharide units in lactose is

The liner structure is not assumed by:

The line at 434nm in the Balmer series of the hydrogen spectrum corresponds to a transition of an electron from the n^(th) to second Bohr orbit. What is the value of n?

The line spectrum observed when electron falls from the higher level into L level is known as :

The limiting molar conductivity of an electrolyte is obtained by adding the limiting molar conductivities of cation and anion of the electrolyte. What is meant by limiting molar conductivity ?

The limiting radius ratio for tetrahedral voids has the range:

The limmiting molar conductivites ^^ ^(@) for NaCl_(2)Kbr and KCl are 126,152 and 150 S cm^(2) respectively the ^^ ^(@) for NaBr is

The limiting radius ratio for tetrahedral shape is

The limiting molar conductivities of KCl, NaCl and KNO_3 are 150, 126 and 109 S cm^2mol^(-1)respectively. What is the limiting molar conductivity of NaNO_3 ?

The limiting molar conductivity of an electrolyte is obtained by adding the limiting molar conductivities of cation and anion of the electrolyte. Name of the above law

The limiting molar conductivities of HCl, CH_(3)COONa and NaCl are respective 425, 190 and 150 mho cm^(2)mol^(-1) at 25^(@)C. The molar conductivity of 0.1 M acetic acid is 9.2 mho cm^(2)mol^(-1). The degree of dissociation of 0.1 M acetic acid is

The limiting molar conductivities (Lambda^(0)) for NaCl, KBr and KCl are 126, 152 and 150 S.cm^(2) "mol"^(-1) respectively. Then A for NaBr is

The limiting molar conductivities ^^ for NaCl, KBr and KCl are 126,152 and 150 S cm^(2)mol^(-1) respectively. The ^^ for NaBr is

The limiting molar conductances wedge^(infty) for NaCl, KBr and KCl are 126, 152 and 150 Scm^(2) mol^(–1) respectively. The wedge^(infty) for NaBr will be:

The limit of a particular detection system is 0.002 dps for 1g sample. Find the maximum t_((1)/(2)) that this system could detect in a 1-g sample of a nuclide of mass number 200.

The limiting line in Paschen series corresponds to

The limiting density of hydrogen bromide is 3.6108 at 0^(@)C. The exact atomic weight ofbromine is (At. wt. of H = 1.008)

The lightning bolts in atmosphere cause the formation of:

The lightest, non-inflammable gas is:

The lightest particle is:

The lightest metal in the periodic table is :

The lightest metal is :

The lightest metal I

The light yellow compound produced when actone reacts with iodine and alkali, is

The lightest gas is: