52671 + Interview Questions in Chemistry in Class 11

1.	In Lassaigne.s test, the reason behind the usage of sodium metal is
Answer» Sodium melts easily and reacts with elements easily Sodium SALTS are ionic and WATER soluble Sodium salts are slightly solule in water Sodium forms COVALENT compounds with elements of organic of organic compounds Solution :The COVALENTLY BONDED elements in the organic compound are converted into ionic, since ionic reactions are instantaneous.

Discussion

2.	In Lassaigne's test, the organic compound is fused with sodium metal as to
Answer» hydrolyse the compound forms a SODIUM derivative COVERT nitrogen, SULPHUR of halogens if present into soluble IONIC sodium compounds burn the compound. Solution :In S.F.E., the elements N, S halogens of the organic compound are converted to ionic STATE.

Discussion

3.	In Lassaigne's test for N,S and halogens, the organic compound is
Answer» FUSED with sodium dissolved with sodamide extracted with sodamide fused with calcium Answer :A

Discussion

4.	In Lassaigne's test for halogens, if colour of the precipitate is curdy white, the halogen present is ............ .
Answer» SOLUTION :CHLORINE

Discussion

5.	In Lassaigne's solution, pink/violet colouration is produced when sodium nitroprusside solution is added. It indicates the presence of:
Answer» sulphur nitrogen chlorine none of these Answer :A

Discussion

6.	In Lassaigne test for nitrogen, conc HNO_(3) is used to destroy..................and ................. .
Answer» ANSWER :NACN, `Na_(2)S`

Discussion

7.	In lanthanum (Z = 57), the 5th electron enters in a:
Answer» 6P orbital 5d orbital 6s orbital 4f orbital Answer :B

Discussion

8.	In lanthanides, the differentiating electron enters into
Answer» d-subshell f-subshell p-subshell s-subshell ANSWER :B

Discussion

9.	In Langmuir's model of adsorption of a gas on a solid surface,
Answer» the mass of GAS striking a given AREA of surface is proportional to the pressure of the gas. the mass of gas striking a given area of surface is independent of the pressure of the gas. The rate of DISSOCIATION of adsorbed molecules, from the surface does not depend on the surface covered the adsorption at a SINGLE site on the surface covered Solution :The mass of the gas absorbed depends UPON the pressure of the gas .

Discussion

10.	In laboratory, solvent can be separated from solute by the process:
Answer» decantation sedimentation distillation filtration Answer :C

Discussion

11.	In lab H_2O_2 is prepared by …..
Answer» COLD `H_2SO_4 + BaO_2` HCL+ `BaO_2` CONC. `H_2SO_4 + Na_2O_2` `H_2+O_2` SOLUTION :`H_2SO_4 + BaO_2 BaSO_4 + H_2O_2`

Discussion

12.	In Kjeldahl's method, the nitrogen present in the organic compound is converted into
Answer» GASEOUS AMMONIA AMMONIUM sulphate ammonium PHOSPHATE ammonium nitrate. Answer :B

Discussion

13.	In Kjeldahl's method of estimation of nitrogen, nitrogen is quantitatively converted to ammonium sulphate. It is then treated with standard solution of alkli. The nitrogen which is present is estimated as
Answer» `N_(2)` gas `NO_(2)` gas `NH_(3)` gas `(NH_(4))_(2)SO_(4)` ppt SOLUTION :when treated with STANDARD solution of alkali NITROGEN is liberated as `NH_(3)` which is absored in `H_(2)SO_(4)` to get percentage of nitrogen.

Discussion

14.	In Kjeldahl's method of estimation of nitrogen K_(2)SO_(4) acts as:
Answer» an OXIDISING agent catalytic agent hydrolysing agent boiling POINT elevator Solution :Addition of `K_(2)SO_(4)` INCREASES the boiling point of `H_(2)SO_(4)`

Discussion

15.	In Kjeldahl's method, nitrogen present in the organic compond is first converted into
Answer» `NH_(3)` `(NH_(4))_(2)SO_(4)` `N_(2)` `NO` Answer :B

Discussion

16.	In Kjeldahl's method for estimation of nitrogen presen in a soil sample, ammonia evolved from 0.75 g of a sample neutralized 10 mL of 1 M H_(2)SO_(4). The percentage of nitrogen in the soil is
Answer» 37.33 45.33 35.33 43.33 Solution :`%N = (1.4 xx "Volume" xx "Molarity" xx "Basicity of acid")/("Weight of the substance TAKEN")` `= (1.4 xx 10 xx 1 xx 2)/(0.75) = 37.33`

Discussion

17.	In Kjeldahl's method, estimation of nitrogen, from 3.88 milligram compound produce ammonia require 5.73 mL 0.011N HCl. Calculate % of N.
Answer» ANSWER :0.2274

Discussion

18.	In Kjeldahl's method, nitrogen is estimated as :
Answer» `N_(2)` `NH_(3)` `NO_(2)` `NH_(4)OH` SOLUTION :Nitrogen is ESTIMATED as AMMONIA.

Discussion

19.	Kjeldahl's method is used in the estimation of
Answer» `" PERCENTAGE of N "=(1.4xxVxxW)/N` `" percentage of N "=(1.4xxNxxW)/V` `" percentage of N "=(VxxNxxW)/1.4` `" percentage of N "=(1.4xxNxxV)/W` ANSWER :D

Discussion

20.	In Kjeldahl's method, 0.4422 g of an organic compound was digested with conc. H_(2)SO_(4). The ammonia evolved on distilling the resulting solution with excess of NaOH solution was absorbed in 50 mL of 1 N H_(2)SO_(4) solution. The residual acid required 65.5 mL of N//2 alkali solution for complete neutralisation. Find the percentage of nitrogen in the given organic compound.
Answer» Solution :Calculation of the volume of UNUSED acid. volume of `NaOH` solution =65.5 mL Normality of `NaOH` solution `=1//2 N` Normality of `H_(2)SO_(4)` solution `= 1 N` Volume of the acid can be calculated by applying normality equation, `ubrace(N_(1)V_(1))_("Acid")= ubrace(N_(2)V_(2))_("Base")` `(1N)xxV_(1)=(1//2N)xx(65.5 mL), V_(1)=((1//2N)xx(65.5 mL))/((1N))=32.75 mL` STEP II. Calculation of the volume of acid used volume of `H_(2)SO_(4)` added =50 mL Volume of `H_(2)SO_(4)` left unused =32.75 mL `:.` Volume of `H_(2)SO_(4)` used `= (50-32.75)=17.25 mL` Step III. Calculation of the percentage of nitogen Mass of COMPOUND =0.4422 G Volume of `H_(2)SO_(4)` used =17.25 mL Normality of `H_(2)SO_(4)` used = 1 N Percentage of `N=(1.4 xx" Volume of acid used"xx"Normality of acid used")/("Mass of the compound")` `=(1.4xx17.25xx1)/0.4422=54.61 %`

Discussion

21.	In Kjeldahl's estimation of nitrogen, the ammonia evolved from 0.5g of an organic compound neutralised 10ml of 1M H_(2)SO_(4). Calculate the percentage of nitrogen in the compound
Answer» Solution :N of compound is converted into `NH_(3)` and is estimated by TITRATION. 1M of 10mL`H_(2)SO_(4)` = 1M of 2mL `NH_(3)` , 1000 mL of 1M ammonia contains 14g nitrogen 20mL of 1 M ammonia contains `(14 xx 20)/(1000)g N_(2)` , Precentage of nitrogen = `(14 xx 20 xx 100)/(1000 xx 0.5) = 56.0`

Discussion

22.	In Kjeldahl method of 0.35g of organic compound the produce ammonia absorb in 100ml, 0.1 M, H_(2)SO_(4). Then 154mL, 0.1M, NaOH is used in titration of excess H_(2)SO_(4). Calculate the % of nitrogen
Answer» ANSWER :0.184

Discussion

23.	In Kjedhal's estimation of nitrogen, the ammonia obtained from 0.5of an organic subsatce was passed into 100cm^(3) of (M)/(10)H_(2)SO_(4). The excess of acid required 154cm^(3) of (M)/(10) NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.
Answer» Solution :Massof the organic compound = 0.5 , Volume of STANDARD `H_(2)SO_(4)` TAKEN = `100cm^(3)` of `(M)/(10)` solution Let volume fo unused `H_(2)SO_(4) = V_(1)` Volume of NaOH requried for excess acid = `154cm^(3)` of `(M)/(10)` solution Volume of `(M)/(10)H_(2)SO_(4)` solution unused `(V_(1)) = (154)/(2)=77CM^(3)` Volume of `(M)/(10)H_(2)SO_(4)` used by `NH_(3) = 23CM^(3)` , Millimoles of `H_(2)SO_(4)` used by `NH_(3) = (23)/(10) = 2.3` Millimoles of `NH_(3) = 2 xx "Millimoles of" H_(2)SO_(4)=2xx2.3 = 4.6` perecntage of NITROGEN = `(14 xx 4.6 xx 100)/(1000 xx 0.5) = 12.88`

Discussion

24.	In kinetic theory of gases, only translational motion of molecules is considered because:
Answer» there is no INTERMOLECULAR forces. the molecules are considred rigid spheres of negligible volume. different molecules may travel at different SPEEDS in normal conditions, rotational and VIBRATIONAL MOTION is not observed in GAS molecules. Answer :b

Discussion

25.	In IUPAC nomenchlature, the order followed for naming the compounds is:
Answer» PREFIX (ES) + ROOT WORD + PRIMARY suffix + secondary suffix Root word + prefix (es) + primary suffix + secondary suffix Primary suffix + Perfix (es) + root word + secondary suffix Primary suffix + root word + prefix (es) + secondary suffix Answer :A

Discussion

26.	In IUPAC nomenclature method………..is used instead of lowest sum rule
Answer» LOWEST SUBSTRACTION rule lower substituted position lowest position none Answer :B

Discussion

27.	In its compounds, tin exhibits the oxidation numbers
Answer» `+2` `+4` `+6` `+3` Solution :Tin can exhibits +2, +4 OXIDATION STATES `[SN^(+2),Sn^(+4)]`

Discussion

28.	Inisoalkanes , oneof thecarbonatomisattachedto ____ othercarbonatom (s) .
Answer» ONE two three four ANSWER :C

Discussion

29.	In ion exchange method of defluoridation of water, which one of the following is used
Answer» `CaOCl_(2)` defluoron -I defluoron -2 Both 2 and 3 ANSWER :D

Discussion

30.	In ion-dipole forces, the magnitude of the interaction energy (E)
Answer» `E = (Z^2 mu)/(r^2)` `E = (Zmu)/(r)` `E = (Zmu^2)/(r^2)` `E = (Z mu)/(r^2)` SOLUTION :ion-dipole FORCES = `(z mu)/(r^2)`

Discussion

31.	In India on the occasion of marriage the fireworks are used to give coloured flames. The salt of which one of the following metals is used to obtain green flame for this purpose ?
Answer» `NA` `BA` K `Ca` Answer :B

Discussion

32.	In India, ethyl alcohol is mainly manufactured by
Answer» CATALYTIC HYDROGENATION of CO FERMENTATION of MOLASSES hydration of ethylene destructive DISTILLATION of wood. Solution :Fermentation of molasses.

Discussion

33.	In ice each oxygen is surrounded by four oxygen atoms in ______ manner.
Answer» SQUARE planar Tetrahedral Trigonal planar Angular Answer :B

Discussion

34.	In I_(3)^(-) , Lewis base is
Answer» `I_(2)` `I_(2)^(-)` `I_(2)^(+)` `I^(-)` Solution :As`I^(-)` ION donates electron pair into EMPTY `sp^(3)d` ORBITAL, HENCE`I^(-)` ion acts as Lewis base.

Discussion

35.	In hyperconjugation, the atom involved is
Answer» `beta-H" ATOM"` `alpha-H" atom"` `gamma-H" atom"` All the above ANSWER :B

Discussion

36.	In hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen ?
Answer» `5 rarr 2` `4 rarr 1` `2 rarr 5` `3 rarr 2` Solution :The lines at the red end BELONG to Balmer SERIES. These are obtained for jumps to n= 2 from outer ORBITS (1st line from n = 3, 2nd line from n = 4, 3ED line from n = 5)

Discussion

37.	In hydrogen spectrum the different lines of Lyman series are present is
Answer» UV field IR field Visible field Far IR field Solution :ACCORDING to Hydrogen SPECTRUM series

Discussion

38.	In hydrogen atom the kinetic energy of electron is 3.4 eV. The distance of that electron from the nucleus
Answer» `2.11 A^@` `0.529 A^@` `1.587A^@` `21.16A^@` SOLUTION :3.4 eV is KE .` e^-` in `2^(nd)` ORBIT `THEREFORE r = 4(0.529) A^@ = 2.116 A^@`

Discussion

39.	In hydrogen spectrum L_(alpha) line arises due to transition of electron from the orbit n=3 to the orbit n=2. In the spectrum of singly ionized helium there is a line having the same wavelength as that of the L_(alpha) line. This is due to the transition of electron from the orbit:
Answer» `N = 3 to n=2` `n = 2 to n = 1` `n = 5 to n = 3` `n = 6 to n = 4` Solution :`""H_(alpha) ""He^+` `(1)^2 xx R(1/(2^2) - 1/(3^2)) = (2^2) xx R (1/(n_1^2 - 1/(n_2^2)))` On SOLVING ,`n_1 = 4 and n_2 = 6`

Discussion

40.	In hydrogen atom, energy of the first excited state is -3.4 eV. Then find out the K.E. of the same orbit of H-atom
Answer» `+3.4 EV` `+6.8 eV` `-13.6 eV` `+13.6 eV` SOLUTION :ENERGY in EXCITED STATE = K.E.

Discussion

41.	In hydrogenatomfor whichchangehigherenergyrequired ?
Answer» N= 3TO n= 5 n=1 to n=2 n=2 ton=3 n=2 to n=1 Answer :B

Discussion

42.	In hydrogen atom, energy of the first excited state is -3.4 eV. Calculate the kinetic energy of the same orbit of hydrogen atom
Answer» Solution :Total energy of the electron in n =2 (1st excited state) `= -3.4 eV` `E_("total") = K.E. + P.E. = (1)/(2) m v^(2) + ((-Ze^(2))/(R))` But force of altraction on the NUCLEUS = CENTRIPETAL force i.e., `(Ze^(2))/(r^(2)) = (mv^(2))/(r) or mv^(2) = (Ze^(2))/(r)` `:. E_("total") = (1)/(2) (Ze^(2))/(r) - (Ze^(2))/(r) = - (Ze^(2))/(2r) = -3.4 eV` (GIVEN) `:. K.E. = (1)/(2) mv^(2) = (1)/(2) (Ze^(2))/(r) = 3.4 eV`

Discussion

43.	In hydrogen atom, an electron jumps from 3d orbit to the 2nd orbit. Calculate the wavelength of the radiatoin emitted. (h = 6.63 xx 10^(-34) J sec)
Answer» Solution :`E_(N) = - (21.8 xx 10^(-19))/(n^(2)) J "atom"^(-1)` `Delta E = E_(3) - E_(2) = 21.8 xx 10^(-19) ((1)/(2^(2)) - (1)/(3^(2))) = 3.03 xx 10^(-19) J` `Delta E = hv = H (c)/(lamda)` `lamda = (HC)/(Delta E) = ((6.63 xx 10^(-34) Js) (3 xx 10^(8) MS^(-1)))/((3.03 xx 10^(-19) J)) = 6.564 xx 10^(-7) m = 6564 xx 10^(-10) m =Å`

Discussion

44.	In hydrogenatomelectronjumpsfrom 3^(rh) to2^(nd)energylevelthe energyreleasedis
Answer» `3.03xx 10^(12)Jatom^(-1)` `1.03xx 10^(19) Jatom^(-1)` `6.06 xx 10^(19)J atom^(-1)` `3.03xx 10^(19)J atom^(-1)` Answer :C

Discussion

45.	In hydrogen an electron transition takes place from n=2 level n=3 level. The wavelength of the line in the hydrogen spectrum would be
Answer» `5.485xx10^(7)` `6.56xx10^(-7)m` `4.57xx10^(14)m` `0.529xx10^(-10)m` Solution :`1/lambda=109677[1/n_(1)^(2)-1/n_(2)^(2)]=6.56xx10^(-7)`m

Discussion

46.	In hydrofluorosilicic acid the covalency of Si is
Answer» 2 4 6 8 Answer :C

Discussion

47.	What is the covalency of silicon in H_(2)SiF_(6)?
Answer» 2 4 6 8 Answer :C

Discussion

48.	In hybridisation, mixing of ________ takes place.
Answer» orbitals of different PRINCIPAL QUANTUM number orbitals of same principal quantum number orbitals of different azimuthal quantum number both (B) and (C) Answer :D

Discussion

49.	In hydrazine (N_(2)H_(4)) the hybridization of nitrogen is
Answer» `SP` `sp^(2)` `sp^(3)` `dsp^(2)` ANSWER :C

Discussion

50.	In humans ,NaClis asourceof ______ forgastricjuices .
Answer» Polynucleic acid chlorine GAS HYDROCHLORICACID PROTEIN ANSWER :C

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

In Lassaigne.s test, the reason behind the usage of sodium metal is

In Lassaigne's test, the organic compound is fused with sodium metal as to

In Lassaigne's test for N,S and halogens, the organic compound is

In Lassaigne's test for halogens, if colour of the precipitate is curdy white, the halogen present is ............ .

In Lassaigne's solution, pink/violet colouration is produced when sodium nitroprusside solution is added. It indicates the presence of:

In Lassaigne test for nitrogen, conc HNO_(3) is used to destroy..................and ................. .

In lanthanum (Z = 57), the 5th electron enters in a:

In lanthanides, the differentiating electron enters into

In Langmuir's model of adsorption of a gas on a solid surface,

In laboratory, solvent can be separated from solute by the process:

In lab H_2O_2 is prepared by …..

In Kjeldahl's method, the nitrogen present in the organic compound is converted into

In Kjeldahl's method of estimation of nitrogen, nitrogen is quantitatively converted to ammonium sulphate. It is then treated with standard solution of alkli. The nitrogen which is present is estimated as

In Kjeldahl's method of estimation of nitrogen K_(2)SO_(4) acts as:

In Kjeldahl's method, nitrogen present in the organic compond is first converted into

In Kjeldahl's method for estimation of nitrogen presen in a soil sample, ammonia evolved from 0.75 g of a sample neutralized 10 mL of 1 M H_(2)SO_(4). The percentage of nitrogen in the soil is

In Kjeldahl's method, estimation of nitrogen, from 3.88 milligram compound produce ammonia require 5.73 mL 0.011N HCl. Calculate % of N.

In Kjeldahl's method, nitrogen is estimated as :

Kjeldahl's method is used in the estimation of

In Kjeldahl's estimation of nitrogen, the ammonia evolved from 0.5g of an organic compound neutralised 10ml of 1M H_(2)SO_(4). Calculate the percentage of nitrogen in the compound

In Kjeldahl method of 0.35g of organic compound the produce ammonia absorb in 100ml, 0.1 M, H_(2)SO_(4). Then 154mL, 0.1M, NaOH is used in titration of excess H_(2)SO_(4). Calculate the % of nitrogen

In Kjedhal's estimation of nitrogen, the ammonia obtained from 0.5of an organic subsatce was passed into 100cm^(3) of (M)/(10)H_(2)SO_(4). The excess of acid required 154cm^(3) of (M)/(10) NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.

In kinetic theory of gases, only translational motion of molecules is considered because:

In IUPAC nomenchlature, the order followed for naming the compounds is:

In IUPAC nomenclature method………..is used instead of lowest sum rule

In its compounds, tin exhibits the oxidation numbers

Inisoalkanes , oneof thecarbonatomisattachedto ____ othercarbonatom (s) .

In ion exchange method of defluoridation of water, which one of the following is used

In ion-dipole forces, the magnitude of the interaction energy (E)

In India on the occasion of marriage the fireworks are used to give coloured flames. The salt of which one of the following metals is used to obtain green flame for this purpose ?

In India, ethyl alcohol is mainly manufactured by

In ice each oxygen is surrounded by four oxygen atoms in ______ manner.

In I_(3)^(-) , Lewis base is

In hyperconjugation, the atom involved is

In hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen ?

In hydrogen spectrum the different lines of Lyman series are present is

In hydrogen atom the kinetic energy of electron is 3.4 eV. The distance of that electron from the nucleus

In hydrogen spectrum L_(alpha) line arises due to transition of electron from the orbit n=3 to the orbit n=2. In the spectrum of singly ionized helium there is a line having the same wavelength as that of the L_(alpha) line. This is due to the transition of electron from the orbit:

In hydrogen atom, energy of the first excited state is -3.4 eV. Then find out the K.E. of the same orbit of H-atom

In hydrogenatomfor whichchangehigherenergyrequired ?

In hydrogen atom, energy of the first excited state is -3.4 eV. Calculate the kinetic energy of the same orbit of hydrogen atom

In hydrogen atom, an electron jumps from 3d orbit to the 2nd orbit. Calculate the wavelength of the radiatoin emitted. (h = 6.63 xx 10^(-34) J sec)

In hydrogenatomelectronjumpsfrom 3^(rh) to2^(nd)energylevelthe energyreleasedis

In hydrogen an electron transition takes place from n=2 level n=3 level. The wavelength of the line in the hydrogen spectrum would be

In hydrofluorosilicic acid the covalency of Si is

What is the covalency of silicon in H_(2)SiF_(6)?

In hybridisation, mixing of ________ takes place.

In hydrazine (N_(2)H_(4)) the hybridization of nitrogen is

In humans ,NaClis asourceof ______ forgastricjuices .