52671 + Interview Questions in Chemistry in Class 11

1.	In each of the following questions, a statementof Assertion is given followed by a corresponding statementof Reason just below it. Of the statements, mark the correctanswer as Assertion . If standard free energy change of a reaction is zero , this implies that equilibrium constant of the reaction is unity . Reason . For a reaction in equilibrium , equilibrium constant is always unity .
Answer» If both assertion and reason are true, and reason is the true explanation of the assertion . If both assertion and reason are true but reason is the true explanation of the assertion . If assertion is true, but reason is false. If both assertion and reason are false . SOLUTION :`DELTAG^(@) = -2 *303 RT LOG K ` When ` DeltaG^(@) =0, log K=0 :. K=1` Hence, Assertion is correct . Correct R. For a reaction in EQUILIBRIUM , the value of equilibrum constant depends upon the relative amounts of the reactants and products .

Discussion

2.	In each of the following questions, a statementof Assertion is given followed by a corresponding statementof Reason just below it. Of the statements, mark the correctanswer as Assertion . The equilibrium constant of a reaction increases if temperature is increased . Reason . The forward reaction becomes faster with increase of temperature .
Answer» If both assertion and reason are TRUE, and reason is the true explanation of the assertion . If both assertion and reason are true but reason is the true explanation of the assertion . If assertion is true, but reason is FALSE. If both assertion and reason are false . SOLUTION :Correct A and R . Efffectof temperature on equilibrium constatn K depends upon whetehr the REACTION is exothermic or endother MIC .

Discussion

3.	In each of the following questions, a statementof Assertion is given followed by a corresponding statementof Reason just below it. Of the statements, mark the correctanswer as Assertion. The equilibrium constant is fixed and characterstic for any given chemical reactionat a specified temperature . Reason . The composition of the final equilibrium mixture at a particular temperature depends upon the starting amount of reactants .
Answer» If both assertion and reason are true, and reason is the true explanation of the assertion . If both assertion and reason are true but reason is the true explanation of the assertion . If assertion is true, but reason is false. If both assertion and reason are false . SOLUTION :If the STARTING AMOUNTS are changed , the composition of the equilibrium mixture, CHANGES is such a way that Kremains constant at constantat CONSTANTTEMPERATURE .

Discussion

4.	In each of the following questions, a statementof Assertion is given followed by a corresponding statementof Reason just below it. Of the statements, mark the correctanswer as Assertion.In heterogenous equilibrium , the active massesof oure solidsor oure liquids are taken as constant . Reason. Pure solids and pure liquids have fixed densities and definite molecular masses.
Answer» If both ASSERTION and reason are true, and reason is the true EXPLANATION of the assertion . If both assertion and reason are true but reason is the true explanation of the assertion . If assertion is true, but reason is false. If both assertion and reason are false . Solution :R is the CORRECT explanation of A

Discussion

5.	In each of the following questions, a statementof Assertion is given followed by a corresponding statementof Reason just below it. Of the statements, mark the correctanswer as Assertion .For the reaction ,N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) , " units of " K_(c) = L^(2) mol^(-2) Reason . For the reaction ,N_(2) (g) + 3 H_(2) (g) hArr2 NH_(3) (g), Equilibrium constant , K_(c) - ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3)).
Answer» If both assertion and reason are TRUE, and reason is the true explanation of the assertion . If both assertion and reason are true but reason is the true explanation of the assertion . If assertion is true, but reason is false. If both assertion and reason are false . Solution :` K_(c) = [NH_(3)]^(2)/([N_(2)][H_(2)]^(3))=( " MOL" L^(-1))^(2)/(("mol"L^(-1))("mol"L^(-1))^(3))` ` = (mol L^(-1))^(-2)=L^(2) mol^(-2)` Hence , R is the correct explanation of A.

Discussion

6.	In each of the following questions, a statementof Assertion is given followed by a corresponding statementof Reason just below it. Of the statements, mark the correctanswer as Assertion. A reversible reaction cannot be carried out in an open vessel. Reason. When equilibrium is reached , no more vapour are formed .
Answer» If both assertion and reason are true, and reason is the true explanation of the assertion . If both assertion and reason are true but reason is the true explanation of the assertion . If assertion is true, but reason is false. If both assertion and reason are false . Solution :Correct A. A REVERSIBLE reaction CONNOT be carried out in an open vessel if at least one of the reactants or products is gaseous . Correct R. Only a gaseous reactantor PRODUCT can escape in an open vessel.

Discussion

7.	In each of the following questions, a statementof Assertion is given followed by a corresponding statementof Reason just below it. Of the statements, mark the correctanswer as Assertion . The vapour pressure of a pure liquid has a fixed value at a particular temperature . Reason . When equilibrium is reached , no more vapour are formed .
Answer» If both ASSERTION and REASON are true, and reason is the true explanation of the assertion . If both assertion and reason are true but reason is the true explanation of the assertion . If assertion is true, but reason is FALSE. If both assertion and reason are false . Solution :Correct R . When equilibrium is REACHED , rate of evaporation becomes equal to rate ofcondensation . Vapours are formed but same amount condenses in the same time.

Discussion

8.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion. Thehighestoxidationstateof Os is + 8. Reason . Osmiumis a 5d- block element .
Answer» Solution :Correctexplanation .Theelectronicconfigurationof Os is `[Xe] 6s^(2)4f^(14)5d^(6)6s^(2) `. Sinceit has8 electrons in thevalenceshell`(5d 6s^(2))` thereforeit has thehighestoxidation STATEOF + 8 .

Discussion

9.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion.F is more electronegativethan C1 . Reason . F hashigherelectronaffinitythan C1.
Answer» SOLUTION :CORRECTREASON. FHAS greatertendencythan C1to attractthe sharedpair ofelectronsof a COVALENT BOND.

Discussion

10.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion . Noblegases havepositiveelectron gainenthalpy. Reason. Noblegases havestableclosedshellelectronicconfiguration .
Answer» SOLUTION :NA

Discussion

11.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion. Firstionizationenergy fornitrogen islowerthan thatof oxygen . Reason . Acrossa periodeffectivenuclearchargedecreases.
Answer» SOLUTION :CORRECTASSERTION . Firstionizationenergyofnitrogenis HIGHERTHAN THATOF OXYGEN. Correctreason.Nitrogen hasstableexactlyhalf - filledelectronicconfigurationbut oxygendoes nothave.

Discussion

12.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion. The firstionizationenergyof Be isgreaterthan thatof B. Reason. 2p- Orbitalsis lowerinenergythan 2s- orbital.
Answer» SOLUTION :CORRECT REASON . 2P- ORBITAL has higherenergythan2s- orbital

Discussion

13.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion. F atomhas lessnegativeelectrongain enthalpythan C1atom. Reason . Additional electrons arerepelledmore effectivelyby 3p-electronsin C1thanby 2p- electronsin F atom
Answer» Solution :Correct reason.Additionalelectron isrepelledmoreeffectivelyby 2p- ELECTRONS in F than3p-electrons in C1.

Discussion

14.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion. The firstionizationenthalpy ofaluminiumis lowerthan thatof magnesium . Reason. Ionicradius ofof aluminium issmallerthan thatof magnesium .
Answer» SOLUTION :CORRECTEXPLANATION . InA1 ,a lessstronglyheld3p- electronis to be removedwhilein Mga morestrongly held 3s-electronis to be REMOVED.

Discussion

15.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion .The ionizationof s- electronrequiresmore energy thanionizationof p- electronof the sameshell Reason. s- electrons arecloser to thenucleusthan p- electronsand henceare morestronglyattractedby thenuclues .
Answer» SOLUTION :NA

Discussion

16.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer asAssertion.The elementwithelectronicconfiguration[Xe]^(54)4f^(1)5d^(1)6s^(2)isd- blockelement Reason . the lastelectronentersthe d-orbital.
Answer» Solution :Correctassertion . Elementwith ELECTRONICCONFIGURATION`[Xe] ^(54)4f^(1) 5d^(1) 6s^(2) ` is a f-blockelement . Correct reason. The lastelectronentersthe f- blockorbital

Discussion

17.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as AssertionNa^(+) and A1^(3+) are isoelectronicbut themagnitude of theionicradiusof A1^(3+) is lessthan thatofNa^(+) Reason. The magnitudeof effectivenuclearcharge ofthe outershellelectronsin A1^(3+) is greaterthan in Na^(+)
Answer» SOLUTION :NA

Discussion

18.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion. Whenthe atoms of firsttransitionseriesionize the 4s- orbitalelectronsare ionized beforethe 3d-orbitalelectrons . Reason . Theenergyof 3d- orbital electronis lower than thatof 4s-orbitalelectronic .
Answer» SOLUTION :NA

Discussion

19.	In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer asAssertion .Heliumand beryllimhave similarouterelectronicconfigurationof the typens^(2) Reason . both arechemciallyinert .
Answer» Statement -1 isTrue Statement-2 isTrue, Statement-2 iscorrectexplanationfor Statement -5 If bothassertionand reasonare truebutreasonis NOTTHE trueexplanation of theassertion If assertion is truebut REASONIS false If bothassertionand reasonare false Solution :Correctreason.AlthoughHe and Behavesimilar outerelectronicconfigurationof the TYPE `ns^(2)` ONLYHE `(1s^(2))`has inertgas configurationand henceis chemicallyinertbut Be `(1s^(2)2S^(2))`does nothaveinertgas configurationand henceis notchemicallyinert.

Discussion

20.	In each of the following pairs which is more stable:
Answer» Answer :[(a) 2; (B) 2; (C) 1; (d) 2; (E) 1]

Discussion

21.	In each of the following pairs which is more stable.:
Answer» ANSWER :[(a) 1; (B) 1;(C) 1 ;(d) 1; (E) 1]

Discussion

22.	In each of the following pairs of salts, which one is more stable and why ? (i) Ferrous and ferric salts (ii) Cuprous and cupric salts.
Answer» Solution :(i) Electronic configuation Fe atom (Z = 26) is `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6) 4s^(2)` `:.` Electronic configuration of `Fe^(2+)` ion `= 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6)` Electronic configuration of `Fe^(3+) " ion " = 1s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5)` As half-filled `3d^(5)` configuration is more stable, THEREFORE, ferric salts are more stable then ferrous salts. (ii) Electronic configuration of Cu atom `(Z = 29) = 1s^(2) 2^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(10) 4s^(1)` `:.` Electronic configuration of `Cu^(+) " ion " 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(10)` Electronic configuration of `Cu^(2+) " ion " = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(9)` Although `Cu^(+)` has fully-filled d-orbital, YET cuprous salts are LESS stable. This is because the nuclear charge is not sufficient enough to hold 18 electrons of `Cu^(+)` ion present in the outermost shell.

Discussion

23.	In each of the following pairs of compounds, which will give iodoform test ? (a) Sec-butyl alcohol and tert-butyl alcohol (b) Ethyl alcohol and isopropyl alcohol (c) Formaldehyde and acetaldehyde (d) Methylpropyl ketone and diethyl ketone.
Answer» SOLUTION :(i) sec-butyl ALCOHOL (ii) ETHYL alcohol (iii) acetaldehyde (IV) methylpropyl ketone

Discussion

24.	In each of the following pairs of compounds, which one is more covalent and why ? (i) AgCl, AgI (ii) BeCl_(2), MgCl_2 (iii) SnCl_(2), SnCl_(4) (iv) CuO,CuS
Answer» Solution :Applying Fajan's rules , the RESULT can be obtained in each CASE as follows : (i) `AgI` is more covalent than `AgCl`. This is because `I^(-)` ion is larger in size than`Cl^(-)` ion and hence `I^(-)` is more polarized than`Cl^(-)`ion. (ii) `BeCl_(2)` is more covalent than `MgCl_(2)`. This is because `Be^(2+)` ion is smallerin size than `MG^(2+)` ion andhence `Be^(2+)` ion has greater polarizing power . (iii) `SnCl_(4)` is more covalent than `SnCl_(2)` . This is because `Sn^(4+)` ion has greater charge and smaller size than `Sn^(2+)` ion and hence `Sn^(4+)` has greater polarizing power. (iv) `CuS` is more covalent than `CuO`. This is because`S^(2-)` ion has larger size than `O^(2-)` ion and hence `S^(2-)` is more polarized than `O^(2-)` ion .

Discussion

25.	In each of the following pairs of compounds, Identify the one with more covalent nature and why ? LiBe and LiI
Answer» Solution :According to Fajan's rule, LI I in more covalent than LiBr. Since `I^(-)` is larger in size than `Br^(-)`, the larger sized anion PULLS the CATION EFFECTIVELY towards the nucleus. This results in more polarisation effect . GREATER the polarisation effect greater will be the covalent character .

Discussion

26.	In each of the following pairs of compounds, Identify the one with more covalent nature and why ? LiCl and NaCl
Answer» Solution :ACCORDING to Fajan's rule, LICL is more covalnet than NaCl Always the cation of the smaller size causes a great extent of polarisation. Greater the POLARIZATION effect, greater will be the covalent CHARACTER . Since` Li^(+)` is smaller in size than `Na^(+)`, Li possess higher polarization effect which in turn increases the covalnet character.

Discussion

27.	In each of the following pairs of compounds, Identify the one with more covalent nature and why ? FeCl_(2)" and " FeCl_(3)
Answer» Solution :According to Fajan's RULE, `FeCl_(3)` is morecovalent than `FeCl_(2)` . If the oxidation stateof the cation is HIGHER, the polarization of anion will be more. Thus more will be covalent NATURE in the bonding of the molecule.

Discussion

28.	In each of the compounds : NaCl, ZnS andCaF_(2) , Write (i) ions occupying thevoids (ii)types of voidsoccupied (iii) fraction of voidsoccupied.
Answer» SOLUTION :(a) NaCl = `Na^(+)`ions in all the OCTAHEDRAL voids. (B)` ZNS = Zn^(2+) ` ions in altenate tetrahedral vodis. ( c)`CaF_(2) = F^(-)`ions in all the tetrahedral voids.

Discussion

29.	In each of the compouds : NaCl , ZnS and CaF_2 , write (i) ions occupying the voids (ii)types of voids occupied (iii) fraction of voids occupied.
Answer» Solution :(a)`NaCl=Na^+`ions in all the OCTAHEDRAL VOIDS (b)`ZnS=Zn^(2+)` ions in alternate tetrahedral voids (C ) `CaF_2=F^-` ions INALL the tetrahedral voids.

Discussion

30.	In -E effect pi electron from multiple bond migrate to the …….
Answer» SOLUTION :ADJECENT CARBON

Discussion

31.	In +E effect pi electron from multiple bond migrate to the …….
Answer» ANSWER :CARBON

Discussion

32.	In Dunna's method, the nitrogen present in the organic compound is converted into.............. .
Answer» ANSWER :`N_(2)` GAS

Discussion

33.	In Duma's method, the gas collected in nitrometer is :
Answer» `N_(2)` `NO` `NH_(3)` `H_(2)` Solution :Nitrogen GAS is COLLECTED in SCHIFF's nitrometer.

Discussion

34.	In Dumas method of estimation of nitrogen 3.88gm compound at 293K temperature and 746mm pressure 1.31 mL N_(2) gas. Calculate percentage of nitrogen. (Aqueor tension 6mm)
Answer» ANSWER :0.2197

Discussion

35.	In Dumas' method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K = 15mm)
Answer» SOLUTION :Laboratory temperature `=T_(1) = 300K` Pressure of DRY `N_(2)` GAS = `(("Given"),("atmospheric"),("pressure"))- (("Aquery"),("Tension"))` `:. p_(1) = (715-15) = 700mm` 22400 mL of `N_(2)` at STP weight = 28g Volume of `N_(2)` at STP `= (p_(1)V_(1))/(T_(1)) XX (273)/(760)` `V= (700 xx 50)/(300) xx (273)/(760) mL` = 41.9 mL Mass of `N_(2)` at STP `= (28 xx 41.9)/(22400)` %N `= (28 xx 41.9)/(22400) xx (100)/(m)` `=(28 xx 41.9 xx 100)/(22400 xx 0.3)` = 17.16% OR %N `= (0.0499 xx p_(1)V_(1))/(m T_(1)) = (0.0499 xx 700 xx 50)/(0.3 xx 300)` = 17.46%

Discussion

36.	In Dumas method for estimation of nitrogen, 0.30g of an organic compound gave 50 mL of nitrogen collected at 300 K and mm pressure. Calculate the percentage composition of nitrogen in the compound (Aqueous tension at 300 K is 15 mm).
Answer» Solution :Here, mass of the SUBSTANCE taken = 0.30g Volume of NITROGEN collected = 50 ML, Atmospheric pressure = 715mm Hg Room temperature = 300K Aqueous tension at 300 K = 15mm `:.` Actual pressure of the gas (dry gas) = `715 - 15 = 700mm Hg` Step 1. To convert the volume at experimental conditions to volume at STP. `{:("Experimental values",,"At STP"),(P_(1)=700 mm,,P_(2) = 760 mm),(V_(1) = 50 mL,,V_(2) = ?),(T_(1) = 300 K,,T_(2) = 273 K):}` Substituting these values in the gas equation. `(P_(2)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`, we get `(700 mm xx 50 mL)/(300 K) = (760 mm xx V_(2) mL)/(273 K)` or `V_(2) = (273 xx 700 xx 50)/(300 xx 760) = 41.9 mL`. Step 2.To convert the volume at STP into mass. According to the definition of GMV, 22400 mL of nitrogen at STP weigh = 28 g `:. 41.9 mL` of nitrogen at STP will weigh `= (28 xx 41.9)/(22400) g` Step 3. To calculate the percentage of nitrogen. Percentage of nitrogen `= ("Mass of "N_(2) " at STP")/("Mass of the substance taken") xx 100 = (28 xx 41.9 xx 100)/(22400 xx 0.3) = 17.46`

Discussion

37.	In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm)
Answer» 0.228 0.378 0.428 0.578 Solution :`V_0 = (732.3 XX 50 xx 273)/(288 xx 760) = 45.67 " ML "` `" % of " N_2 = (28xx 45.67xx100)/(22,400 xx 0.25)=22.8 `

Discussion

38.	In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is
Answer» `16.76` `15.76` `17.36` `18.20` Solution :Let us calculate volume of `N_(2)` EVOLVED at N.T.P. Experimental conditions N.T.P. conditions `{:(V_(1)=40 mL,V_(2)=?),(P_(1)=(725-25)=700mm,P_(2)=760 mm),(T_(1)=300 K,T_(2)=273 K):}` Byapplying gas equation : `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` `V_(2)=(P_(1)V_(1)T_(2))/(P_(2)T_(1))` `V_(2)=((700mm)XX(40 mL)xx(273 K))/((760mm)xx(300 K))` `=33.52 mL` `%` of `N=28/22400xx("Volume of "N_(2)" at N.T.P.")/("Mass of compound")XX100` `=28/22400xx((33.52 mL))/((0.25 g))xx100` `=16.76 %`

Discussion

39.	In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mLof nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25mm, the percentage of nitrogen in the compound is
Answer» 16.76 15.76 17.36 `18.20` Solution :`V_(1) = 40 mL, V_(2)` at NTP = ? `P_(1) = 725 - 25 = 700 mm, P_(2) = 760 mm` `T_(1) = 300 K, T_(2) = 273 K` `:. V_(2) = (P_(1)V_(1)T_(2))/(P_(2)T_(1)) = (700 xx 40 xx 273)/(760 xx 300) = 33.52 mL` `%N = (28)/(22400) xx (33.52)/(0.25) xx 100 = 16.76`

Discussion

40.	In Duma's method 0.52g of an organic compound on combustion gave 68.6 mL N_(2) at 27^(@)C and 756mm pressure. What is the percentage of nitrogen in the compound?
Answer» 0.1222 0.1493 0.1584 0.1623 Solution :`V_(1)=68.6mL,P_(1)=756mm,T_(1)=300K` `V_(2)=?,P_(2)=760MM,T_(2)=273K` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` At NTP, vol. of `N_(2),V_(2)=(P_(1)V_(1))/(T_(1))*(T_(2))/(P_(2))=(756xx68.6)/(300)xx(273)/(760)` =62.09 mL Percentage of nitrogen in organic compound `=(28)/(22400)xx(V_(2))/(w)xx100=(28)/(22400)xx(62.09)/(0.52)xx100=14.93%`

Discussion

41.	In Dumas method , 0.3 g of an organic compound gave 45 ml of nitrogen at STP. The percentage of nitrogen is
Answer» 16.9 18.7 23.2 29.6 Answer :B

Discussion

42.	In Duma's method, 0.3 g of an organic compound gave 50 cm^(3) of nitrogen collected at 300 K and 715 mm pressure. Calculate the percentage of nitrogen in the compound. Aqueous tension of water at 300 K is 15 mm.
Answer» Solution :Step I. Volume of `N_(2)` at N.T.P. `{:("EXPERIMENTAL Conditions","N.T.P. Conditions"),(V_(1)=50 cm^(3),V_(2)=?),(P_(1)=715-15=700 mm,P_(2)=760 mm),(T_(1)=300 K,T_(2)=273 K):}` `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` or `V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((700mm)xx(50 cm^(3))xx(273 K))/((300 K)xx(760 mm))=41.9 cm^(3)` Step II. Percentage of NITROGEN `=28/22400xx("Volume of "N_(2)" at N.T.P.")/("MASS of COMPOUND")xx100=28/22400xx41.9/0.3xx100=17.46 %`

Discussion

43.	In Duma's method 0.206 g of an organic compound gave 18.8 cm^(3) of moist N_(2) at 17^(@)C and 760 mm Hg pressure. If aqueous at 17^(@)C is 14.5 mm Hg, calculate the percentage of nitrogen in the given organic compound.
Answer» Solution :Mass of ORGANIC compound =0.206 G Calculation of the volume of `N_(2)` at N.T.P. `{:("Experimental Conditions","N.T.P. conditions"),(V_(1)=18.8 cm^(3),V_(2)=?),(P_(1)=(760-14.5)mm Hg,P_(2)=760 mm Hg),(=745.5 mm Hg,),(T_(1)=(273+17)K=290 K,T_(2)=273 K):}` By applying gas EQUATION, `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` `V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((745.5mmHg)xx(18.8 cm^(3))xx(273 K))/((290 K)xx(760 mm Hg))=17.37 cm^(3)` Step II. Calculation of percentage of nitrogen `22400 cm^(3)` of `N_(2)` at N.T.P. has a mass =28 g `17.37 cm^(3)` of `N_(2)` at N.T.P. has a mass `=((28g)xx(17.37 cm^(3)))/((22400 cm^(3)))=0.0217 g` Percentage of nitrogen `=("Mass of "N_(2)" produced "xx100)/("Mass of organic compound")=((0.0217g)xx100)/((0.206 g))=10.54`

Discussion

44.	In the Duma's method of estimation of nitrogen,the nitrogen,the nitrogen in the organic compound is finally converted into
Answer» ANSWER :NITROGEN

Discussion

45.	In dry air, lithium and sodium react to give
Answer» `Li_(2)O,Li_(3)N,Na_(2)O` `Li_(2)O,Na_(2)O` `Li_(2)O,Li_(3)N,NH_(3),Na_(2)O` `Li_(2)O,Li_(3)N,Na_(2)O,Na_(3)N` Solution :In AIR, alkali metals REACT with oxygen to form OXIDE

Discussion

46.	In DNA and RNA, nitrogen atom is present in the ring system. Can Kjetdahl. Method be used for the estimation of nitrogen present in these?
Answer» Solution :In DNA and RNA, NITROGEN is present in heterocyclic rings, since Kjeldahl method cannot be used to estimate nitrogen present in rings, 920 nad nitro GROUPS because nitrogen present in these system cannot be completely CONVERTED into `(NH_(4))_(2)SO_(4)` during DIGESTION. Therefore, Kjeldahl method cannot be used to estimate nitrogen present in DNA and RNA

Discussion

47.	In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these ? Give reasons.
Answer» Solution :In DNA and RNA, nitrogen is present in heterocyclic rings. Since Kjeldahl method cannot be USED to ESTIMATE nitrogen present in rings, azo and NTRO groups because nitrogen present in these systems/groups cannot be completely converted into `(NH_(4))_(2)SO_(4)` during digestion. THEREFORE, Kjeldahl method cannot be used to estimate nitrogen present in DNA and RNA.

Discussion

48.	In DNA and RNA, nitrogen atom is predent in the ring system. Can kjeldahl's method be used for the estimation of nitrogen present in these ? Give reasons.
Answer» SOLUTION :In both DNA and RNA, the nitrogen atom is present in the heterocyclic RING system. Kjeldahl's method cannot be USED for such compounds. Nitrogen can be estimated by DUMA's method.

Discussion

49.	In dihorane total number of bonding electrons are
Answer» 12 18 24 20 Answer :1

Discussion

50.	In differential extraction method in separating funnel, two layers are there. What is the difference in mix before shaking the solution and at the end time?
Answer» Solution :At initial the compound is in aquous layer. At the END TIME the compound is more in upper solvent layer and LESS in aquous layer.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion. Thehighestoxidationstateof Os is + 8. Reason . Osmiumis a 5d- block element .

In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion.F is more electronegativethan C1 . Reason . F hashigherelectronaffinitythan C1.

In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion . Noblegases havepositiveelectron gainenthalpy. Reason. Noblegases havestableclosedshellelectronicconfiguration .

In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion. Firstionizationenergy fornitrogen islowerthan thatof oxygen . Reason . Acrossa periodeffectivenuclearchargedecreases.

In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer as Assertion. The firstionizationenergyof Be isgreaterthan thatof B. Reason. 2p- Orbitalsis lowerinenergythan 2s- orbital.

In each of thefollowingquestionsa Statementof Assertion(A ) isgivenfollowedby acorrespondingstatementof Reason(R ) Justbelow it . Of thestatementmarkthe correctanswer asAssertion .Heliumand beryllimhave similarouterelectronicconfigurationof the typens^(2) Reason . both arechemciallyinert .

In each of the following pairs which is more stable:

In each of the following pairs which is more stable.:

In each of the following pairs of salts, which one is more stable and why ? (i) Ferrous and ferric salts (ii) Cuprous and cupric salts.

In each of the following pairs of compounds, which will give iodoform test ?﻿ (a) Sec-butyl alcohol and tert-butyl alcohol (b) Ethyl alcohol and isopropyl alcohol﻿ (c) Formaldehyde and acetaldehyde﻿ (d) Methylpropyl ketone and diethyl ketone.﻿

In each of the following pairs of compounds, which one is more covalent and why ? (i) AgCl, AgI (ii) BeCl_(2), MgCl_2 (iii) SnCl_(2), SnCl_(4) (iv) CuO,CuS

In each of the following pairs of compounds, Identify the one with more covalent nature and why ? LiBe and LiI

In each of the following pairs of compounds, Identify the one with more covalent nature and why ? LiCl and NaCl

In each of the following pairs of compounds, Identify the one with more covalent nature and why ? FeCl_(2)" and " FeCl_(3)

In each of the compounds : NaCl, ZnS andCaF_(2) , Write (i) ions occupying thevoids (ii)types of voidsoccupied (iii) fraction of voidsoccupied.

In each of the compouds : NaCl , ZnS and CaF_2 , write (i) ions occupying the voids (ii)types of voids occupied (iii) fraction of voids occupied.

In -E effect pi electron from multiple bond migrate to the …….

In +E effect pi electron from multiple bond migrate to the …….

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

In Duma's method, the gas collected in nitrometer is :

In Dumas method of estimation of nitrogen 3.88gm compound at 293K temperature and 746mm pressure 1.31 mL N_(2) gas. Calculate percentage of nitrogen. (Aqueor tension 6mm)

In Dumas' method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K = 15mm)

In Dumas method for estimation of nitrogen, 0.30g of an organic compound gave 50 mL of nitrogen collected at 300 K and mm pressure. Calculate the percentage composition of nitrogen in the compound (Aqueous tension at 300 K is 15 mm).

In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm)

In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is

In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mLof nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25mm, the percentage of nitrogen in the compound is

In Duma's method 0.52g of an organic compound on combustion gave 68.6 mL N_(2) at 27^(@)C and 756mm pressure. What is the percentage of nitrogen in the compound?

In Dumas method , 0.3 g of an organic compound gave 45 ml of nitrogen at STP. The percentage of nitrogen is

In Duma's method, 0.3 g of an organic compound gave 50 cm^(3) of nitrogen collected at 300 K and 715 mm pressure. Calculate the percentage of nitrogen in the compound. Aqueous tension of water at 300 K is 15 mm.

In Duma's method 0.206 g of an organic compound gave 18.8 cm^(3) of moist N_(2) at 17^(@)C and 760 mm Hg pressure. If aqueous at 17^(@)C is 14.5 mm Hg, calculate the percentage of nitrogen in the given organic compound.

In the Duma's method of estimation of nitrogen,the nitrogen,the nitrogen in the organic compound is finally converted into

In dry air, lithium and sodium react to give

In DNA and RNA, nitrogen atom is present in the ring system. Can Kjetdahl. Method be used for the estimation of nitrogen present in these?

In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these ? Give reasons.

In DNA and RNA, nitrogen atom is predent in the ring system. Can kjeldahl's method be used for the estimation of nitrogen present in these ? Give reasons.

In dihorane total number of bonding electrons are

In differential extraction method in separating funnel, two layers are there. What is the difference in mix before shaking the solution and at the end time?

In each of the following pairs of compounds, which will give iodoform test ? (a) Sec-butyl alcohol and tert-butyl alcohol (b) Ethyl alcohol and isopropyl alcohol (c) Formaldehyde and acetaldehyde (d) Methylpropyl ketone and diethyl ketone.