94420 + Interview Questions in Chemistry in Class 12 Page 521 InterviewSolution

26001.	The pH of 0.001M HCl solution is
Answer» 3 2 1 10 Solution :`[H_3O^+]=0.01M` `pH=-log_10[H_3O^+]=-log_10[0.01]` `=-log_10[10^-2]` pH=2

Discussion

26002.	The pH of 0.001 M HCL solution is
Answer» 3 2 1 11 Solution :`[H3O^+]=0.001 MOL dm^-3` `pH=-log_10[H_3O^+]=-log_10[0.01]` `=-log_10[10^-2]` pH=2

Discussion

26003.	The pH of 0.001 M Ba(OH)_(2) solution will be :-
Answer» 2 8.4 11.3 13 Answer :3

Discussion

26004.	The pH of 0.0001 N solution of KOH will be
Answer» 4 6 10 12 Solution :`[OH^(-)] = 0.001 N, POH = 4, pH + pOH = 14` `pH = 14 - pOH = 14 - 4 = 10`.

Discussion

26005.	The pH is less than 7, of the solution of
Answer» `FeCl_(3)` NaCN NaOH NaCl Solution :`FeCl_(3) + 3H_(2)I hArr FE(OH)_(3) + 3HCl`. (Salts of strong acid and weak base have `pH lt 7`).

Discussion

26006.	The pH at which M(OH), will begin to precipitate from a solution containing 0.10 M M^2+ ions [K_sp of M(OH)_2 = 1 x 10^-9 M^3] 1 is
Answer» ANSWER :D

Discussion

26007.	The pH indicators are
Answer» SALTS of strong acid and strong bases Salts of WEAK ACIDS and weak bases Either weak acids or weak bases Either strong acids or strong bases Answer :C

Discussion

26008.	The pH at the equivalence point of a titration may differ from 7.0 because of
Answer» The self ionisation of WATER Hydrolysis of the salt formed The indicator used The CONCENTRATION of the standard SOLUTION Answer :B

Discussion

26009.	The pH at which an amino acid carries no net charge is called it's
Answer» ISOELECTRIC POINT inversion point neutralisation poin TRIPLE point Answer :A

Discussion

26010.	The petrol having octane number 80 has
Answer» 20% NORMAL HEPTANE +80% iso-OCTANE 80% normal heptane+ 20% isooctane 20% normal heptane +80% normal octane 80% normal heptane +20% normal octane Answer :A

Discussion

26011.	The periodicity of related to the electronic configuration.That is, all chemical and physical properties are manifestation of the electronic configurations of the elements. The atomic and ionic radii generally decrease in a period from left to right.As a consequence, the ionization enthalpies generally increase and electron gain enthalpies become more negative across a period.In other words, the ionization enthalpy of the extreme left element in a period is the least and the electron gain enthalpy of the element on the extreme right is the highest negative.This result into high chemical reactivity at the two extremes and the lowest in the centre.Similarly down the group, the increase in atomic and ionic radii result in gradual decrease in ionization enthalpies and a regular decrease (with exception in some third period elements) in electron gain enthalpies in the case of main group elements. These properties can be related with the : (i)reducing and oxidising behaviour of the elements (ii)metallic and non-metallic character of element (iii)acidic, basic, amphoteric, and neutral character of the oxides of the elements The correct order of the metallic character is :
Answer» BgtAlgtMggtK AlgtMggtBgtK MggtAlgtKgtB KgtMggtAlgtB Solution :As ionisation energy decreases electropositive character increases and thus the TENDENCY to FORM cation increases. THEREFORE, METALLIC character increases.

Discussion

26012.	The periodic system of the elements in our three-dimensional world is based on the four electron quantym numbers n = 1,2,3,….1 = 0,1..n -1,m = 0, +- 1, +-2…+-1 and s = +- 1//2. In Flatlandia a twodimensional world, the periodic system in thus based on three electron quantum numbers: n = 1,2,3...m_(1) = 0, +- 1,+-2...+-(n-1), and s = +-1//2 where m_(1) plays the combined role of 1 and m_(1) of the three dimensional world. The followingtakes relate to this two-dimensional world, where the chemical and physical experience obtained from our world is supposed to be still applicable. Which second period (row) element has the six ionization energies (IE in electron volts, eV) listed below? {:(IE_(1),IE_(2),IE_(3),IE_(4),IE_(5),IE_(6)),(11,24,48,64,392,490):}
Answer» `B` `C` `N` `O` ANSWER :B

Discussion

26013.	The periodic table consists of 18 groups. An isotope of Cu, on bombardment with protons undergoes a nuclear reaction yielding element, X as shown below. To which group element X belongs to in the periodic table? ""_(29)^(63)Cu + ""_(1)^(1)H rarr 6 ""_(0)^(1)n + ""_(2)^(4)alpha + 2 ""_(1)^(1)H + X
Answer» Solution :X is `""_(25)^(52)X`, ATOMIC no=26. Transition METAL `FE(""_(4)S^(2) ""_(3)d^(6))`

Discussion

26014.	The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic table ._(29)^(63)Cu + ._(1)^(1)H rarr 6 ._(0)^(1)n + alpha + 2 ._(1)^(1)H + X
Answer» SOLUTION :`._(29)^(63)Cu + ._(1)^(1)H rarr 6 ._(0)^(1)N + ._(2)^(1) alpha + 2 ._(1)^(1) H + X` `64 = 6 + 4 + 2 + A rArr A = 52` `29 + 1 = 30 = 0 + 2 + 2 + z rArr z = 26` element X should be iron in GROUP 8

Discussion

26015.	The periodic table consists of 18 groups. An isotope of copper, on bombardment with proton,undergoes a nuclear reaction yielding element as shown below. To which group element X belongs in the periodic table? ""_(29)^(63)Cu+""_(1)^(1)Hto6""_(0)^(1)n+alpha+2""_(1)^(1)H+X
Answer» Solution :`""_(29)C^(63)+""_(1)H^(1)to6(""_(0)n^(1))+1(""_(2)He^(4))+2(""_(1)H^(1))+""_(26)X^(52)`

Discussion

26016.	The periodic system of the elements in our three-dimensional world is based on the four electron quantym numbers n = 1,2,3,….1 = 0,1..n -1,m = 0, +- 1, +-2…+-1 and s = +- 1//2. In Flatlandia a twodimensional world, the periodic system in thus based on three electron quantum numbers: n = 1,2,3...m_(1) = 0, +- 1,+-2...+-(n-1), and s = +-1//2 where m_(1) plays the combined role of 1 and m_(1) of the three dimensional world. The followingtakes relate to this two-dimensional world, where the chemical and physical experience obtained from our world is supposed to be still applicable. Draw the first four periods of the Flatlandian periodic table of the elements. Number them according to their nuclear charge. Use the atomic numbers (Z) as symbols of the specific element. Write the electron configuration for each element.
Answer» ANSWER :In the two DIMENSIONAL world and not the electron QUANTUM numbers GIVEN, we OBTAIN the following Flatlandian periodic table: `(##RES_INO_CHM_XI_C01_E01_283_A01##)`

Discussion

26017.	The periodic table consists of 18 groups. An isotope ofcopper on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below . To which group, element X belongs in the periodic table? ltbregt ._(29)^(63) Cu + ._(1)^(1) H to 6 ._(0)^(1) n+a+2_(1)^(1)H+X
Answer» ANSWER :8

Discussion

26018.	The periodic table consists of 18 grooups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belogns in the periodic table? ._(29)^(63)Cu + ._(1)^(1)H rarr 6 ._(0)^(1)n + alpha + 2 ._(1)^(1)H + X
Answer» Solution :`{:(1,2,3,4,5,6,7,8,),(KR,CR,Sc,Ti,V,Cr,mn,FE,):}`

Discussion

26019.	The periodic system of the elements in our three-dimensional world is based on the four electron quantym numbers n = 1,2,3,….1 = 0,1..n -1,m = 0, +- 1, +-2…+-1 and s = +- 1//2. In Flatlandia a twodimensional world, the periodic system in thus based on three electron quantum numbers: n = 1,2,3...m_(1) = 0, +- 1,+-2...+-(n-1), and s = +-1//2 where m_(1) plays the combined role of 1 and m_(1) of the three dimensional world. The followingtakes relate to this two-dimensional world, where the chemical and physical experience obtained from our world is supposed to be still applicable. Which of the following elements has the largest third ionization energy?
Answer» `B` `C` `N` `MG` ANSWER :d

Discussion

26020.	The period number in the long form of the periodic table is equal to
Answer» magnetic quantum number of any ELEMENT of the period. atomic number of any element of the period. maximum principal quantum number of anyelement of the period. maximum AZIMUTHAL quantum number of any element of the period. Answer :C

Discussion

26021.	The period number and group number of "Tantalum" (Z=73) are respectively:
Answer» 5,7 6,13 6,5 none of these Solution :`{:("Period number","GROUP number","Element"),(5,7,TC),(6,13,TL),(6,5,TA):}`

Discussion

26022.	The perentageofC, H, Nin a disubtitutedaromaticcompound (A) is 71.11 , 6.67 and 10.37 respectively . Its moleculecontainsa singleatom of nitrogen . Compound(A) gives NH_(3)whenheatedwith NaOH to from a salt (B) form whichan acid (C) is obtained on acidifcations. Twoisomersare generatedwhen (C) undergoeselectrophilicaromatic substitution. (C) on treatementwithPCl_(5)generates (D) whichon reactionwith (B)generates(E) . Identify(A) to (E).
Answer» Solution :Firstly, we shouldcalculatedth empirical FORMULA of the compound. `{:("Element","Perecentage","At wt",5"by atom", "Simplest ratio"),(C,71.11,12,5.926,8),(H,6.67,1,6.67,9),(N,10.37,14,0.74,1),(O,11.85,16,0.74,1):}` Empirical formula of (A)is `C_(6)H_(0)NO` whichis similar molecular FORMUAL as THEMOLECULE containsonly oneN. Since(A) givesoff `NH_(3)`with `NaOH`is it clear thatit has` - CONH_(2)`group. So, thepossiblestrcuctureare . Since (C) givesonly twoisomericProductson electrophilicaromatic substitution (C)is .

Discussion

26023.	The perhalate ion with maximum oxidising power is
Answer» `ClO_4^-` `BrO_4^-` `IO_4^-` All have NEARLY same OXIDISING agent. Answer :B

Discussion

26024.	The percentage S-character of the hybrid orbitals in methane, ethene and ethyne are respectively
Answer» 25,33,50 25,50,75 50,75,100 10,20,40 Solution :25,33,50. The HYBRID orbitals used for bonding in methane `(sp^(3)),` ethene `(sp^(2))` and ETHYNE (sp) have 25, 33 and 50% s-character.

Discussion

26025.	The percentage of Zinc in the electron is there is Glauber's salt.
Answer» SOLUTION :Percentage of zinc in the electron is `5%` electron 95% `Mg+4.5% Zn+0.5 CU`

Discussion

26026.	The percentage packing efficiency of the two dimensional arrangement of sphere of plane ABCDEF shown below is :
Answer» `90.64%` `74.05%` `68.02%` `78.54%` SOLUTION :NA

Discussion

26027.	The percentage of these three isomeric anilines from the reaction show above are
Answer» 1-25%,2-50%, 3-25% 1-33%,2-33%,3-33% 1-50%,2-25%,3-25% 1-66%,2-33%,3-0% ANSWER :A

Discussion

26028.	The percentage of void space of a metallic element crystallising in a ABCABC .....type lattice pattern is
Answer» 0.24 0.26 0.34 0.74 Answer :B

Discussion

26029.	The percentage of sulphur in an organic compound whose 0.32g produces 0.233g of BaSO_(4) [At. Wt. Ba=137, S=32] is
Answer» `1.0` `10.0` `23.5` `32.1` SOLUTION :% of S`=(32)/(233)TIMES(0.33)/(0.32)times100=10%`

Discussion

26030.	The percentage of silver and chlorine in two samples of silver chloride prepared by heating silver foil in the current of chlorine and by the intercation of silver nitrate and hydrochloric acid were found to be identical . This illusrates the law of :
Answer» conservation of MASS constant proportion MULTIPLE proportion RECIPROCAL proportion Solution :multiple proportion

Discussion

26031.	The percentage of Se in peroxidase enzyme is 0.5% by weight (atomic weight = 78.4). Then minimum molecular weight of peroxidase anhydrous enzyme is
Answer» `1.568 xx 10^(4)` `1.568 xx 10^(3)` `15.68` `3.136 xx 10^(4)` Solution :0.5% by weight means 0.5 g SE is present in 100 g of PEROXIDASE ANHYDROUS ENZYME. As at LEAST one atom of Se must be present in the enzyme and atomic weight of Se = 78.4, therefore, 1 g atom of Se i.e. 78.4 g will be present in: `=100/0.5 xx 78.4= 1.568 xx 10^(4)` g in enzyme.

Discussion

26032.	The percentage of Se in peroxidase enzyme is 0.5% by mass (atomic mass of Se=78.4 amu). Then, the minimum molecular mass of enzyme which contains not more than one Se atom is:
Answer» `1.568xx10^(4)` amu `1.568xx10^(7)` amu `1.568xx10^(3)` amu `1.568xx10^(6)` amu Solution :`because 0.5g SE` is present in 100g ENZYME. `therefore 78.4g Se` will be present in `(100)/(0.5)xx78.4 g` enzyme =15680amu `=1.568xx10^(4)` amu.

Discussion

26033.	The percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (atomic weight = 78.4) . Then minimum molecular weight of peroxidase anhydrous enzyme is
Answer» `1.568xx10^(4)` `1.568xx10^(3)` `15.68` `3.136xx10^(4)` SOLUTION :`0.5%` by weight means 0.5 g Se is present in 100 g of peroxidase anhydrous enzyme. As at least ONE atom of Se MUST be present in the enzyme and atomic weight of Se = 78.4, therefore, 1 g atom of Se, i.e., 78.4 g will be present in enzyme `=(100)/(0.5)xx78.4g=1.568xx10^(4)g`

Discussion

26034.	The percentage of pyridine (C_5H_5N) that forms pyridinium ion (C_5H_5N^(+) H) in a 0.10M aqueous pyridine solution (K_b for C_5H_5N=1.7xx10^(-9) ) is
Answer» `0.0060%` `0.013%` `0.77%` `1.6%` ANSWER :B

Discussion

26035.	The percentage of pyridine (C_5H_5N) that forms pyridinium ion (C_5H_5NH) in a 0.10 M aqueous pyridine solution (K_b for C_5H_5N=1.7 times 10^-9) is………..
Answer» `0.006%` `0.013%` `0.77%` `1.6%` Solution :`C_5H_5N+H-OH leftrightarrowC_5H_5NH+OH^(-)` `(a^2C)/(1-a)=K_b` `a^2C approxK_b` `a=sqrt((K_b)/C)sqrt((1.7 times 10^-9)/(0.1))=sqrt1.7 times 10^-4` Percentage of DISSOCIATION `sqrt1.7 times 10^-4 times 100=1.3 times 10^-2=0.013%`

Discussion

26036.	The percentage of pyridine (C_(2)H_(5)N) that forms pyridinium ion (C_(2)H_(5)N^(+)H) in a 0.10 M aqueous pyridine solution (K_(b) for C_(2)H_(5)N = 1.7 xx 10^(-9)) is
Answer» `1.6%` `0.0060%` `0.013%` `0.77%` Solution :`C_(5)H_(5)N + H_(2)O = C_(5)H_(5)N^(+)H + OH^(-)` 0.1 `alpha = SQRT((K_(b))/(c)) = sqrt((1.7 xx 10^(-9))/(0.1)) = sqrt(1.7 xx 10^(-8)) = 1.3 xx 10^(-4)` `% prop = 1.3 xx 10^(-4) xx 100` `= 1.3 xx 10^(-2) = 0.013`.

Discussion

26037.	The percentage of P_(2)O_(5) in diammonium hydrogen phosphate [(NH_(4))_(2)HPO_(2))] is
Answer» 23.48 46.96 53.78 71 Answer :C

Discussion

26038.	The percentage of pi-character in the orbitals forming P-P bonds in P_(4) is
Answer» 25 33 50 75 Solution :In `P_(4)`, P is `sp^(3)` hybridized. SINCE p-orbitals form `pi-` bonds, therefore, `pi` CHARACTER of p-orbitals is 75%.

Discussion

26039.	The percentage of p-character in the orbital forming P–P bonds in P_(4) is :-
Answer» 25 75 33 100 Answer :B

Discussion

26040.	The percentage of p - character in the orbitals forming P-P bonds in P_(4) is :
Answer» <P>25 33 50 75 Solution :In `P_(4)`, each P is `sp^(3)` hybridised so that PERCENTAGE of p - character in these orbitals is `75%`.

Discussion

26041.	The percentage of ozone in ozonized oxygen is about:
Answer» `10%` `40%` `80%` `100%` ANSWER :A

Discussion

26042.	The percentage of nitrogen in urea (NH_(2)CONH_(2)), is:
Answer» 38.4 46.6 59.1 61.3 Answer :B

Discussion

26043.	The percentage of nitrogen in urea is about :
Answer» 38.4 46.6 59.1 61.3 Answer :B

Discussion

26044.	The percentage of nitrogen in urea is about
Answer» 28 46.7 35.8 8 Answer :D

Discussion

26045.	The percentage of nitrogen in air remains almost constant due to:
Answer» The FIXATION of NITROGEN The ACTIVITY ofsymbolic bacteria The effect of lightning and bacteria The nitrogen cyclic in nature Answer :D

Discussion

26046.	The percentage of nitrogen in a compound is determined by
Answer» Nessler's mathod Kjeldahl's method Carius method taking the DIFFERENCE between total PERCENTAGE and the SUM of percentage of all other ELEMENTS present. Solution :N//A

Discussion

26047.	thepercentageof nitrogenby massin ammoniumsulphate is closedis(atomicMasses H=1, N=14, O=16, S=32 )
Answer» 0.21 0.24 0.36 0.16 Solution : 5 Nitrogen `=(" WT of N") /( Wt "of" (NH_(4))_(2) SO_(4))` ` % N =(28)/(132 ) XX100` `=21.21%`

Discussion

26048.	The percentage of NaCl in sea water
Answer» `2.5 omega //w` `3.5 omega //v` `2.5 omega //v` `3.5 y omega //w` SOLUTION :`2.5 ((omega)/(v))^(y)`NACL in sea water, LAKES and rocks.

Discussion

26049.	The percentage of N_(2) in urea is about
Answer» 85 46 28 18 Solution :Molecular mass of urea `(NH_(2)CONH_(2)) = 60` Amount of `N_(2)` in ONE mole = 28 `RARR` % of `N_(2)` in urea `= (28)/(60) xx 100 = 46.6%`

Discussion

26050.	The percentage of lanthanides and iron, respectively , in which metal are
Answer» 50 , 50 75 , 25 90, 10 95, 5 ANSWER :D

Discussion

Explore topic-wise InterviewSolutions in .

The pH of 0.001M HCl solution is

The pH of 0.001 M HCL solution is

The pH of 0.001 M Ba(OH)_(2) solution will be :-

The pH of 0.0001 N solution of KOH will be

The pH is less than 7, of the solution of

The pH at which M(OH), will begin to precipitate from a solution containing 0.10 M M^2+ ions [K_sp of M(OH)_2 = 1 x 10^-9 M^3] 1 is

The pH indicators are

The pH at the equivalence point of a titration may differ from 7.0 because of

The pH at which an amino acid carries no net charge is called it's

The petrol having octane number 80 has

The periodic table consists of 18 groups. An isotope of Cu, on bombardment with protons undergoes a nuclear reaction yielding element, X as shown below. To which group element X belongs to in the periodic table? ""_(29)^(63)Cu + ""_(1)^(1)H rarr 6 ""_(0)^(1)n + ""_(2)^(4)alpha + 2 ""_(1)^(1)H + X

The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic table ._(29)^(63)Cu + ._(1)^(1)H rarr 6 ._(0)^(1)n + alpha + 2 ._(1)^(1)H + X

The periodic table consists of 18 groups. An isotope of copper, on bombardment with proton,undergoes a nuclear reaction yielding element as shown below. To which group element X belongs in the periodic table? ""_(29)^(63)Cu+""_(1)^(1)Hto6""_(0)^(1)n+alpha+2""_(1)^(1)H+X

The periodic table consists of 18 groups. An isotope ofcopper on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below . To which group, element X belongs in the periodic table? ltbregt ._(29)^(63) Cu + ._(1)^(1) H to 6 ._(0)^(1) n+a+2_(1)^(1)H+X

The periodic table consists of 18 grooups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belogns in the periodic table? ._(29)^(63)Cu + ._(1)^(1)H rarr 6 ._(0)^(1)n + alpha + 2 ._(1)^(1)H + X

The period number in the long form of the periodic table is equal to

The period number and group number of "Tantalum" (Z=73) are respectively:

The perhalate ion with maximum oxidising power is

The percentage S-character of the hybrid orbitals in methane, ethene and ethyne are respectively

The percentage of Zinc in the electron is there is Glauber's salt.

The percentage packing efficiency of the two dimensional arrangement of sphere of plane ABCDEF shown below is :

The percentage of these three isomeric anilines from the reaction show above are

The percentage of void space of a metallic element crystallising in a ABCABC .....type lattice pattern is

The percentage of sulphur in an organic compound whose 0.32g produces 0.233g of BaSO_(4) [At. Wt. Ba=137, S=32] is

The percentage of silver and chlorine in two samples of silver chloride prepared by heating silver foil in the current of chlorine and by the intercation of silver nitrate and hydrochloric acid were found to be identical . This illusrates the law of :

The percentage of Se in peroxidase enzyme is 0.5% by weight (atomic weight = 78.4). Then minimum molecular weight of peroxidase anhydrous enzyme is

The percentage of Se in peroxidase enzyme is 0.5% by mass (atomic mass of Se=78.4 amu). Then, the minimum molecular mass of enzyme which contains not more than one Se atom is:

The percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (atomic weight = 78.4) . Then minimum molecular weight of peroxidase anhydrous enzyme is

The percentage of pyridine (C_5H_5N) that forms pyridinium ion (C_5H_5N^(+) H) in a 0.10M aqueous pyridine solution (K_b for C_5H_5N=1.7xx10^(-9) ) is

The percentage of pyridine (C_5H_5N) that forms pyridinium ion (C_5H_5NH) in a 0.10 M aqueous pyridine solution (K_b for C_5H_5N=1.7 times 10^-9) is………..

The percentage of pyridine (C_(2)H_(5)N) that forms pyridinium ion (C_(2)H_(5)N^(+)H) in a 0.10 M aqueous pyridine solution (K_(b) for C_(2)H_(5)N = 1.7 xx 10^(-9)) is

The percentage of P_(2)O_(5) in diammonium hydrogen phosphate [(NH_(4))_(2)HPO_(2))] is

The percentage of pi-character in the orbitals forming P-P bonds in P_(4) is

The percentage of p-character in the orbital forming P–P bonds in P_(4) is :-

The percentage of p - character in the orbitals forming P-P bonds in P_(4) is :

The percentage of ozone in ozonized oxygen is about:

The percentage of nitrogen in urea (NH_(2)CONH_(2)), is:

The percentage of nitrogen in urea is about :

The percentage of nitrogen in urea is about

The percentage of nitrogen in air remains almost constant due to:

The percentage of nitrogen in a compound is determined by

thepercentageof nitrogenby massin ammoniumsulphate is closedis(atomicMasses H=1, N=14, O=16, S=32 )

The percentage of NaCl in sea water

The percentage of N_(2) in urea is about

The percentage of lanthanides and iron, respectively , in which metal are