94420 + Interview Questions in Chemistry in Class 12

1.	In any reaction rate of reactant is double when concentration of reactant is increase in 4 time.If concentration of reaction is increase in 9 times than calciulate rate of reactant.
Answer» 4.5 TIMES 3 times 9 times 81 times Solution :Order of reaction =2 , So,`r_(1)=k[R]^(x)` and CONCENTRATION 4 times =`2^(2)` ,the rate =`2^(1)` Concentration 9 times =`3^(2)` ,then rate will BECOMES 3 times

Discussion

2.	In any group second period element exhibits anomalous properties. Correct statements about this is/are
Answer» generally maximum covalence of the first MEMBEROF each group cannot exceed four he first member of p-block element displaysgreater ability to form `ppi-ppi` multiple bonds to itself and to the other second PERIOD elements,compared to subsequent MEMBERS of the group anomalous behaviour is due to small size, largecharge/radius ratio and high electronegativity ofthe elements Lithium EXHIBITS DIAGONAL relationship withaluminium. Answer :A::C

Discussion

3.	In any natural process -
Answer» The entropy of the universe REMAINS constant The entropy of universe TENDS TOWARDS maximum. The entropy of universe tends towards minimum. Any of the above can happen Answer :B

Discussion

4.	In any chemical reaction, equlibrium is supposed to be estabilished when
Answer» Mutual reaction undergo in OPPOSITE reaction Concentration of reactants and resulting products are EQUAL Velocity of mutual reactions become equal The temperaturee of mutual opposite reactions become equal SOLUTION :When rate of FORWARD recton is requal to the rate of backeard reaction then equlibrium is supposed to be estblished.

Discussion

5.	In any group of the periodic table, atomic size and ionisation energy vary with the increase in atomic number. Which of the following trends is correct?
Answer» `{:("ATOMIC size""""IONISATION ENTHALPY"),("DECREASES""""Increases"):}` `{:("Atomic size""""Ionisation enthalpy"),("Decreases""""Decreases"):}` `{:("Atomic size""""Ionisation enthalpy"),("Increases""""Decreases"):}` `{:("Atomic size""""Ionisation enthalpy"),("Decreases""""Increases"):}` Answer :C

Discussion

6.	Inanti -configurationof n-butanethe twomethylgroupsare at an angle of :
Answer» `60^(@)` `45^(@)` `120^(@)` `180^(@)` ANSWER :D

Discussion

7.	In antiflourite structure, the negative ions:
Answer» OCCUPY TETRAHEDRAL voids Occupy OCTAHEDRAL voids Are ARRANGED in ccp Are arranged in hcp Answer :C

Discussion

8.	In anthrcene ,number of pi electrons is equal to x. the value of x is
Answer» 6 10 14 12 Answer :C

Discussion

9.	InAntarticaozonedepletion is duetotheformation of the followingcompound.
Answer» ACROLEIN PEROXYACETYL NITRATE `SO_(2)` and `SO_(3)` chlorinenitrate ANSWER :D

Discussion

10.	In Antarctica, ozone depletion is due to the formation of
Answer» acrolein peroxy ACETYL NITRATE `SO_2 and SO_3` CHLORINE nitrate. SOLUTION :chlorine nitrate.

Discussion

11.	In anomeric forms of fructose which carbon atom involved in ring formation
Answer» C-2 and C-5 C-3 and C-5 C-2 and C-4 C-1 and C-5 Answer :A

Discussion

12.	In aniline, the -NH_(2) group
Answer» ACTIVATES the benzene ring via both INDUCTIVE and resonance effects DEACTIVATES the benzene ring via both inductive and resonance effects activates the benzene ring via resonance effect and deactivates it via inductive-effect activates the benzene ring via inductive effect and deactivates it via resonance effect. Solution :Activates the benzene ring via resonance effect and deactivates it via inductive effect.

Discussion

13.	In aniline , the -NH_(2) group
Answer» activates the benzene ring via both inductive and RESONANCE effects deactivates the benzene ring via both inductive and resonance effects. activates the benzene ring via resonance effect and deactivates it via inductive effect, activates the benzene ring via inductive effect and deactivates it via resonance effect. Solution : Due to LONE pair on N ATOM, it activatesthe ring. On the other hand, electronegativity of N is more than C so it deactivates the benzene ring via inductive effect. Hence, (C) is the correct answer.

Discussion

14.	In angular structure of SO_2, the sigma bond between S-O is formed by
Answer» `SP^(3)-p` OVERLAPPING `sp^(2)-p` overlapping `sp^(3)-p` overlapping `sp^(2)-s` overlapping Answer :B

Discussion

15.	In analysis of third contains of mixture analysis, solid NH_(4)Cl is added prior to NH_(4)OH for the following
Answer» AVAILABILITY of `Cl^(-)` ions Availability of `NH_(4)^(+)` ions Complete DISSOCIATION of `NH_(4)OH` Controlled dissociation of `NH_(4)OH` ANSWER :D

Discussion

16.	In an sp-hybridized carbon atom,
Answer» two p orbitals remain unhybridized and ONE p orbital GETS HYBRIDIZED all the three p orbitals get hybridized one p orbital remains unhybridized and two p ORIBITALS get hybridized None of the above is true Solution :In CASE of sp hybridization, one s-orbital and one p-orbital hybridise.

Discussion

17.	In an S_(N^(1)) reaction on chiral centres, there is
Answer» INVERSION is more than RETENTION leading to PARTIAL racemisation 100% retention 100% inversion 100% racemisation. Solution :it is the correct ANSWER.

Discussion

18.	In an S_(N)1 reaction on chiral centres, there is :
Answer» 100 % retention 100 % inversion 100 % RACEMIZATION inversion more than retention leading to partial racemization. SOLUTION :`S_(N)1` reaction gives racemic MIXTURE with slight predominance of that isomer which corresponds to inversion because `S_(N)1` also depends UPON the degree of shielding of the front side of the reacting carbon.

Discussion

19.	In an SN1 reaction on centres there is
Answer» 100% RACEMISATION inversion more than ratention leading to partial recemisation 100% RETENTION 100% inversion Solution :In `S_(N)1` reaction EXTENT of inversion of configuration is SLIGHTLY higher than retention of configuration. This is due to the shielding effect of the leaving group on the front side of the C - atom undergoing NUCLEOPHILIC attack.

Discussion

20.	In an Ostwald viscometer, water takes 25 s to flow between the lower and upper marks, while the liquid X takes 38 s. Given d(H_(2)O) = 0.9983 kg//dm^(3), d(X) = 0.7894 kg//dm^(3), eta(H_(2)O) = 1.005centipoise. Calculate eta of liquid X
Answer» Solution :`eta_(1)/eta_(2) = (t_(1).d_(1))/(t_(2).d_(2))` Suppose 1 and 2 stand for X and water RESPECTIVELY. `eta_(1)/(1.005) = (38 XX 0.7894)/(25 xx 0.9983)` `eta_(1) = 1.208` centipose.

Discussion

21.	In an Ostwald -Walker experiment, dry air was first blown through a solution containing a certain amount of solute (M=278) in 150g of water, and then also through pure water. The loss in mass of water was found to be 0.0827 g while the mass of water absorbed in sulphuric acid was 3.317g. Calculate the amount of the solute.
Answer» ANSWER :`0.57g`

Discussion

22.	In an organic compound phosphorus is estimated as
Answer» `Mg_(2)P_(2)O_(7)` `Mg_(3)(PO_(4))_(2)` `H_(3)PO_(4)` `P_(2)O_(5)` Solution :A pre-weighed amount of organic compound is heated with concentrated nitric acid `(HNO_(3))` in a tube. Phosphorus present in the compound GETS converted into phosphoric acid. Now the content of tube are heated with magnesia mixture i.e. `MgSO_(4)+NH_(4)Cl +NH_(4)OH , ` so formed which is collected , washed , DRIED and ignited to get magnesium pyrophosphate `Mg_(2)P_(2)O_(7)`, which is the weighed. `UNDERSET("acid")underset("Phosphoric")(H_(3)PO_(4)) overset("Magnesia mixture")to underset("phosphate (white ppt.)")underset("(Magnesium ammonium)")(MgNH_(4)PO_(4)+3HCl)` `2MgNH_(4)PO_(4) overset(DELTA)toMg_(2)P_(2)O_(7) DARR + 2NH_(3) +H_(2)O` Knowing the weight of `Mg_(2)P_(2)O_(7)` , percentage of phosphorus can be calculated.

Discussion

23.	In an ore, the only oxidizable material is Sn^(2+). This ore is titrated with a dichromate solution containing 2.5g of K_(2)Cr_(2)O_(7) in 0.5 "litre". A 0.40 g sample of the ore required 10.0 cm^(3) of titrant to reach equivalence point. Calculate the percentage of tin in ore.
Answer» Solution :`3SN^(2+) + Cr_2O_7^(2-) + 14H^(+) to 3Sn^(4+) + 2Cr^(3+) + 7H_2O`

Discussion

24.	In an orbital electrons are filled according to Aufbau principle, Pauli's exclusion principle and Hund's rule of maximum muliplicity . According to Aufbau principal the orbital are filled in order to their increasing energies . In the order of increase of energy of orbitals cna be calculated from ( n + l ) rule. Lower the value of ( n + l ) for an orbital the lowest energy hence orbital are filled in order of increasing ( n + l )value. If two orbitals have same ( n + l )value, the orbital with lower value of 'n' has lower energy hence it is filled first. According to Pauli's exclusion principle, an orbital can have maximum two electrons with opposite spin. According to Hunds rule pairing of electron in degenerate orbitals of the same sub shell does not take place until each orbital belonging to that sub shell has got one electron each i.e., singly occupied.Assertion : The energy of an electorn is largely determined by principal quantum number. Reason : The principle quantum number (n) is a measure of the most probable position finding the electron around the nucleus
Answer» If both (A) and ( R ) are CORRECT and ( R ) is a correct EXPLANATION of ( A ) If both ( A ) and ( R ) are correct and ( R ) is not the correct explanation of ( A ). If ( A ) is correct but ( R ) is incorrect If both ( A ) and ( R ) are incorrect. Answer :A

Discussion

25.	In an orbital electrons are filled according to Aufbau principle, Pauli's exclusion principle and Hund's rule of maximum muliplicity . According to Aufbau principal the orbital are filled in order to their increasing energies . In the order of increase of energy of orbitals cna be calculated from ( n + l ) rule. Lower the value of ( n + l ) for an orbital the lowest energy hence orbital are filled in order of increasing ( n + l )value. If two orbitals have same ( n + l )value, the orbital with lower value of 'n' has lower energy hence it is filled first. According to Pauli's exclusion principle, an orbital can have maximum two electrons with opposite spin. According to Hunds rule pairing of electron in degenerate orbitals of the same sub shell does not take place until each orbital belonging to that sub shell has got one electron each i.e., singly occupied. If n=4, how many element are possible?
Answer» 20 14 32 16 Answer :C

Discussion

26.	In an orbital electrons are filled according to Aufbau principle, Pauli's exclusion principle and Hund's rule of maximum muliplicity . According to Aufbau principal the orbital are filled in order to their increasing energies . In the order of increase of energy of orbitals cna be calculated from ( n + l ) rule. Lower the value of ( n + l ) for an orbital the lowest energy hence orbital are filled in order of increasing ( n + l )value. If two orbitals have same ( n + l )value, the orbital with lower value of 'n' has lower energy hence it is filled first. According to Pauli's exclusion principle, an orbital can have maximum two electrons with opposite spin. According to Hunds rule pairing of electron in degenerate orbitals of the same sub shell does not take place until each orbital belonging to that sub shell has got one electron each i.e., singly occupied. The electron identified by quantum number n & l (i) n = 4, l = 1(ii) n = 4, l = 0 ( iii) n = 3, l = 2(iv) n = 3, l = 1 can be placed in order of increasing energy from the lowest to the highest as
Answer» `(IV) LT (II) lt (III) lt (i)` `(ii) lt (iv) lt (i) lt (iii) ` `(i) lt (iii) lt (ii) lt (iv)` `(iii) lt (i) lt (iv) lt (ii) ` ANSWER :A

Discussion

27.	In an octahedral structure, the pair of d - orbitals involved in d^(2)sp^(3) hybridisation is :
Answer» 1. `d_(x^(2)-y^(2)), d_(XZ)` 2. `d_(z^(2)), d_(zx)` 3. `d_(xy), d_(yz)` 4. `d_(x^(2)-y^(2)), d_(z^(2))` Answer :D

Discussion

28.	In an octahedral crystal field, the correct set of low energy orbitals are
Answer» `d_(xy),d_(XZ),d_(Z2)` `d_(x^2-y^2), d_(z^2)` `d_(xy), d_(xz),d_(yz)` `d_(xy),d_(x^2-y^2)` ANSWER :C

Discussion

29.	In an octahedral crystal field, d - orbitals split into ….. Orbitals collectively called ….. And ….. Orbitals collectively called …..
Answer» Solution :`d_(XY)d_(YZ)d_(XZ),t_(2G) and d_(x^(2)-y^(2)), d_(z^(2)),e_(g)`

Discussion

30.	In an Ni-Cd battery
Answer» all the REACTANT and products in the overall reaction are in the solid state The voltage of the CELL changes repidly the electrolyte used is an `NaCl` solution all the above are true. Solution :The NICKEL-cadmium cell (nicad cell) is a com-mon storage BATTERY. It is a voltaic cell consisting of an anode of cadmimum and a cathode of hydrated nickel oxide (approximately `NiOOH`) on nicke, the electrolyte is potes-sium hydroxide. To see why a nicad battery produces a constant voltage, we can write the Nernst equation for its reaction and look at `Q`. No gaseous substance is involved.

Discussion

31.	In an isothermal process at 300 K, 1 mole of an ideal gas expands from a pressure 100 atm against an external pressure of 50 atm. Then total entropy change ("Cal K"^(-1)) in the process is -
Answer» `+0.39` `-0.39` `+1.59` `-1.59` ANSWER :A

Discussion

32.	In an isothermal process
Answer» Q=0 and `DeltaE=0` `q ne 0 and DeltaE = 0` `q=0 and DeltaE ne 0` `QNE0 and DeltaE ne 0` SOLUTION :For an isothermal process : `DeltaT=0` and `DeltaE=0` and `qne0` .

Discussion

33.	In an isothermal expansion of an ideal gas
Answer» Q = 0 `DELTA V = 0` `Delta U = 0` W = 0 Solution :For isothermal process, temperature remains constant. SINCE internal energy is a function of temperature, `Delta U` will be also ZERO.

Discussion

34.	In an isobaric process, the ratio of heat supplied to the system (dQ) and work done by the system (dW) for diatomic gas is
Answer» `1 : 1` `7 : 2` `7 : 5` `5 : 7` Solution :`-(dQ)/(DW)=(dQ)/(dQ-dE), dE=dW+dQ, -dW=dQ-dE` `=(nc_(p)DT)/(nc_(p)dT-nc_(v)dT)=(c_(p))/((c_(p)-c_(v)))` `=(7R)/(2R){"for diatomic of gas "c_(p)=(7R)/(2)}`.

Discussion

35.	In an isochoric process the increase in internal energy is
Answer» EQUAL to the heat absorbed Equal to the heat evolved Equal to the WORK done Equal to the sum of the heat absorbed and work done Solution :For isochoric process `DeltaV=0` so `q_(v)=DELTAE` i.e. heat GIVEN to a system under CONSTANT volume is used up in increasing `DeltaE`.

Discussion

36.	In an irreversible process, the value of triangleS_system+triangleS_surr is :
Answer» GREATER than 0 less than 0 equal 0 All Answer :A

Discussion

37.	In an irreversible process taking place at constant temperature and constant pressure and in which only p-V work is being done, the change in Gibbs free energy (dG) and the change in entropy (dS), satisfy the criteria
Answer» <P>`(DS)_(V,E)gt0,(dG)_(r,p)lt0` `(dS)_(V,E)=0,(dG)_(r,p)=0` `(dS)_(V,E)=0,(dG)_(r,p)gt0` `(dS)_(V,E)lt0,(dG)_(r,p)lt0` Answer :A

Discussion

38.	In an irreversible process taking place at constant T and p and in which only pressure volumee work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria.
Answer» `(dS)_(V,E) LT 0, (DG)_(T,p) lt0` `(dS)_(V,E) gt 0,(dG)_(T,p) lt 0` `(dS)_(V,E)=0,(dG)_(T,p)=0` `(dS)_(V,E)=0,(dG)_(T,p) gt 0` Solution :It is an IRREVERSIBLE and spontaneousprocess. `(dS)_((V,S))=+ve` `(dG)_(T,p)=-ve`

Discussion

39.	In an irreversible process taking place at constant T and P and in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria
Answer» `(DS)_(V,E)LT0,(dG)_(T,P)lt0` `(dS)_(V,E)GT0,(dG)_(T,P)lt0` `(dS)_(V,E)=0,(dG)_(T,P)=0` `(dS)_(V,E)=0,(dG)_(T,P)gt0` SOLUTION :`(dS)_(V,E)gt0, (dG)_(T, P) lt0`.

Discussion

40.	In an irreversible process taking place at constant T and P and in which only pressure volume work is being done , the change in Gibbs free energy (dG) and the change in entropy (dS) , satisfy the criteria :
Answer» `(dS)_(V , E) = 0 , (DG)_(T , P) = 0` `(dS)_(V , E) =0 , (dG)_(T.P) gt 0` `(dS)_(V.E) LT 0 , (dG)_(T.P) lt 0` `(dS)_(V.E) gt 0 , (dG)_(T.P) lt 0` Answer :D

Discussion

41.	In an ionic solid r_(c)=1.6Å and r_(a)=1.86Å.Find edge length of cubic unit cell is Å
Answer» Solution :`(r_(C))/(r_(a))=0.858` From above ratio it is CLEAR that cation OCCUPY cubic void in simple cubic unit cell, HENCE `SQRT(3)a=2(r_(c)+r_(a))` `a=4`

Discussion

42.	In an ionic crystal, both cations and an- ions are bound together by……………………
Answer» STRONG ELECTROSTATIC attractive FORCES Weak electrostatic attractive forces Vanderwaals forces of attraction Weak COHESIVE forces Solution :Strong electrostatic attractive forces

Discussion

43.	In an ionic compound the anion (N^(-)) form cubic close type of packing, while the cation (M^(+)) ions occupy one third of the tetrahedral voids. Deduce the empirical formula of the compound and the coordination number of (M^(+)) ions.
Answer» SOLUTION :Number of `N^(-)` ions `=8XX(1)/(8)=1` Number of `M^(+)` ions `=(1)/(3)xx2=(2)/(3)` EMPIRICAL formula `=M_(2//3)N_(1) or M_(2)N_(3)` Coordination number of `M^(+)` is 4.

Discussion

44.	In an ionic compound A+X', the radii of A^(+) and X ions are 1.0 pm and 2.0 pm respectively. The volume of the unit cell of the crystal AX will be
Answer» 27 `"PM"^(3)` 64 `"pm"^(3)` 125 `"pm"^(3)` 21 `"pm"^(3)` Answer :D

Discussion

45.	In an investigation of the kinetics of the reaction MnO_(4)^(-)+Cr^(3+) rarr CrO_(4)^(2-)+ Mn^(4+) at 25^(@) C False and Roller measured the times required to carry the reaction to various degrees of completion, first as a function of MnO_(4)^(-) concentration and then as a function of Cr^(3+) concentration. In each case after difinite intervals of time, 25 ml aliquots of the reaction mixture were removed and added rapidly to a solution containing 60 ml of 1.2M H_(2)SO_(4) and 25 ml of 0.5 M KBr to occur the following reaction 2MnO_(4)^(-)+10Br^(-)+16H^(+) rarr 2Mn^(2+)+5Br_(2)+8H_(2)O Free bromine instantly liberated was extracted by CS_(2)The Cr_(2)O_(7)^(2-) That had been formed by the reaction 2Cr_(4)^(2-)+2H^(+) rarr Cr_(2)O_(7)^(2-)+H_(2)O Was titrated with 0.01 N FeSO_(4) in the reaction Cr_(2)O_(7)^(2-)+6Fe^(2+)+16H^(+)+2SO_(4)^(2-) rarr 2Cr^(3+)+6Fe^(3+)+2HSO_(4)^(-)+7H_(2)O The number of millilitres of 0.01 N Cr_(2)O_(7)^(2-) Present in the mixture at the indicated reaction time is given in the table In a set of three experiments the result were as follows: {:(,"Experiment No",,"I",,"II",,"III"),(,[MnO_(4)^(-)](mol//l t),,1,,2,,1),(,[Cr^(3+)](mol//l t),,1,,1,,0.5),(,Volume of Cr_(2)O_(7)^(2-)(0.01N)present(ml),,"Time taken (min)",,,,),(,,,"I",,"II",,"III"),(,0.1,,22 min,,11 min,,45 min),(,0.2,,36 min,,18 min,,72 min),(,0.4,,60 min,,30 min,,121 min),(,0.6,,80 min,,40 min,,162 min):} What is the order of reaction with respect to [Cr^(3+)] ?
Answer» 1 2 `1.5` `0.5` ANSWER :A

Discussion

46.	In an investigation of the kinetics of the reaction MnO_(4)^(-)+Cr^(3+) rarr CrO_(4)^(2-)+ Mn^(4+) at 25^(@) C False and Roller measured the times required to carry the reaction to various degrees of completion, first as a function of MnO_(4)^(-) concentration and then as a function of Cr^(3+) concentration. In each case after difinite intervals of time, 25 ml aliquots of the reaction mixture were removed and added rapidly to a solution containing 60 ml of 1.2M H_(2)SO_(4) and 25 ml of 0.5 M KBr to occur the following reaction 2MnO_(4)^(-)+10Br^(-)+16H^(+) rarr 2Mn^(2+)+5Br_(2)+8H_(2)O Free bromine instantly liberated was extracted by CS_(2)The Cr_(2)O_(7)^(2-) That had been formed by the reaction 2Cr_(4)^(2-)+2H^(+) rarr Cr_(2)O_(7)^(2-)+H_(2)O Was titrated with 0.01 N FeSO_(4) in the reaction Cr_(2)O_(7)^(2-)+6Fe^(2+)+16H^(+)+2SO_(4)^(2-) rarr 2Cr^(3+)+6Fe^(3+)+2HSO_(4)^(-)+7H_(2)O The number of millilitres of 0.01 N Cr_(2)O_(7)^(2-) Present in the mixture at the indicated reaction time is given in the table In a set of three experiments the result were as follows: {:(,"Experiment No",,"I",,"II",,"III"),(,[MnO_(4)^(-)](mol//l t),,1,,2,,1),(,[Cr^(3+)](mol//l t),,1,,1,,0.5),(,Volume of Cr_(2)O_(7)^(2-)(0.01N)present(ml),,"Time taken (min)",,,,),(,,,"I",,"II",,"III"),(,0.1,,22 min,,11 min,,45 min),(,0.2,,36 min,,18 min,,72 min),(,0.4,,60 min,,30 min,,121 min),(,0.6,,80 min,,40 min,,162 min):} Overall order and type of reaction are respectively:
Answer» `1.5` , COMPLEX 2, elementary 1, complex `2.5` complex Answer :B

Discussion

47.	In an investigation of the kinetics of the reaction MnO_(4)^(-)+Cr^(3+) rarr CrO_(4)^(2-)+ Mn^(4+) at 25^(@) C False and Roller measured the times required to carry the reaction to various degrees of completion, first as a function of MnO_(4)^(-) concentration and then as a function of Cr^(3+) concentration. In each case after difinite intervals of time, 25 ml aliquots of the reaction mixture were removed and added rapidly to a solution containing 60 ml of 1.2M H_(2)SO_(4) and 25 ml of 0.5 M KBr to occur the following reaction 2MnO_(4)^(-)+10Br^(-)+16H^(+) rarr 2Mn^(2+)+5Br_(2)+8H_(2)O Free bromine instantly liberated was extracted by CS_(2)The Cr_(2)O_(7)^(2-) That had been formed by the reaction 2Cr_(4)^(2-)+2H^(+) rarr Cr_(2)O_(7)^(2-)+H_(2)O Was titrated with 0.01 N FeSO_(4) in the reaction Cr_(2)O_(7)^(2-)+6Fe^(2+)+16H^(+)+2SO_(4)^(2-) rarr 2Cr^(3+)+6Fe^(3+)+2HSO_(4)^(-)+7H_(2)O The number of millilitres of 0.01 N Cr_(2)O_(7)^(2-) Present in the mixture at the indicated reaction time is given in the table In a set of three experiments the result were as follows: {:(,"Experiment No",,"I",,"II",,"III"),(,[MnO_(4)^(-)](mol//l t),,1,,2,,1),(,[Cr^(3+)](mol//l t),,1,,1,,0.5),(,Volume of Cr_(2)O_(7)^(2-)(0.01N)present(ml),,"Time taken (min)",,,,),(,,,"I",,"II",,"III"),(,0.1,,22 min,,11 min,,45 min),(,0.2,,36 min,,18 min,,72 min),(,0.4,,60 min,,30 min,,121 min),(,0.6,,80 min,,40 min,,162 min):} What is the order of reaction with respect to [MnO_(4)^(-)] ?
Answer» 2 1 `0.5` `1.5` ANSWER :B

Discussion

48.	In an ideal mixture of liquids A and B the mole fraction of A is 0.25.If the ratio of P_A^@ to P_B^@ is 7//3.how many repeated distillations would be required as a "minimum" to obtain a small quantity to distillate which has a mole fraction A, better than 0.80 ?
Answer» 4 2 3 5 Solution :A=0.25B=0.75 `P_4^C/P_3^C=7/3implies P_B^0=3/7P_A^0` `P_T=0.25xxP_A^0+0.75xx3/7P_A^0=P_A^0(1/4+9/28)` `y_AP_T=P_A^0X_Aimplies y_A=(X0.25)/((1/4+9/28))=(((1//4)/(7+9))/28)=7/16 & y_B=9/16` `P_T^1=7/16xxP_A^0+9/16xx3/7 P_A^0=P_A^0(7/16+27/(16xx7))` `y_A^1P_T^1=P_A^0 X_A^1 implies y_A^1 =(7/16)/(((49+27)/(16xx7)))=(7xx16xx7)/(16(49+27))=49/76` `P^(11)=49/76xxP_A^0 +27/76xx3/7 P_A^0=P_A^0((49xx7+81)/(7xx76))` `y_A^1 P_T^11=P_A^0x_A^11implies y_A^11=P_A^0(P_A^0xx79//76)/((49xx7+81)/(7xx76))=(49xx7xx76)/(76(49xx7+81))` `y_A^1=0.8089 ge0.8` There will be 3 repeated distillation to GET `X_A=0.8`

Discussion

49.	In an industrial electrolytic cell it is desired to produce 36 kg of magnesium metal per hour. Calculate the current required. (Mg = 24)
Answer» SOLUTION :1 mole of electric charge (1 F) produces 1 EQ. of Mg or 1 mole of electric charge (1F) produces 12 g of Mg.`""(eq. wt. = (24)/(2) = 12)`

Discussion

50.	In an galvanic cell , the electrons flow
Answer» from cathode to anode from anode to cathode from anode to SOLUTION from solution to cathode Solution :The ELECTRONS liberated at anode during oxidation MOVE TOWARDS cathode where they are used up .

Discussion

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In an S_(N)1 reaction on chiral centres, there is :

In an SN1 reaction on centres there is

In an Ostwald viscometer, water takes 25 s to flow between the lower and upper marks, while the liquid X takes 38 s. Given d(H_(2)O) = 0.9983 kg//dm^(3), d(X) = 0.7894 kg//dm^(3), eta(H_(2)O) = 1.005centipoise. Calculate eta of liquid X

In an organic compound phosphorus is estimated as

In an ore, the only oxidizable material is Sn^(2+). This ore is titrated with a dichromate solution containing 2.5g of K_(2)Cr_(2)O_(7) in 0.5 "litre". A 0.40 g sample of the ore required 10.0 cm^(3) of titrant to reach equivalence point. Calculate the percentage of tin in ore.

In an octahedral structure, the pair of d - orbitals involved in d^(2)sp^(3) hybridisation is :

In an octahedral crystal field, the correct set of low energy orbitals are

In an octahedral crystal field, d - orbitals split into ….. Orbitals collectively called ….. And ….. Orbitals collectively called …..

In an Ni-Cd battery

In an isothermal process at 300 K, 1 mole of an ideal gas expands from a pressure 100 atm against an external pressure of 50 atm. Then total entropy change ("Cal K"^(-1)) in the process is -

In an isothermal process

In an isothermal expansion of an ideal gas

In an isobaric process, the ratio of heat supplied to the system (dQ) and work done by the system (dW) for diatomic gas is

In an isochoric process the increase in internal energy is

In an irreversible process, the value of triangleS_system+triangleS_surr is :

In an irreversible process taking place at constant temperature and constant pressure and in which only p-V work is being done, the change in Gibbs free energy (dG) and the change in entropy (dS), satisfy the criteria

In an irreversible process taking place at constant T and p and in which only pressure volumee work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria.

In an irreversible process taking place at constant T and P and in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria

In an irreversible process taking place at constant T and P and in which only pressure volume work is being done , the change in Gibbs free energy (dG) and the change in entropy (dS) , satisfy the criteria :

In an ionic solid r_(c)=1.6Å and r_(a)=1.86Å.Find edge length of cubic unit cell is Å

In an ionic crystal, both cations and an- ions are bound together by……………………

In an ionic compound the anion (N^(-)) form cubic close type of packing, while the cation (M^(+)) ions occupy one third of the tetrahedral voids. Deduce the empirical formula of the compound and the coordination number of (M^(+)) ions.

In an ionic compound A+X', the radii of A^(+) and X ions are 1.0 pm and 2.0 pm respecti­vely. The volume of the unit cell of the crystal AX will be

In an ideal mixture of liquids A and B the mole fraction of A is 0.25.If the ratio of P_A^@ to P_B^@ is 7//3.how many repeated distillations would be required as a "minimum" to obtain a small quantity to distillate which has a mole fraction A, better than 0.80 ?

In an industrial electrolytic cell it is desired to produce 36 kg of magnesium metal per hour. Calculate the current required. (Mg = 24)

In an galvanic cell , the electrons flow

In an ionic compound A+X', the radii of A^(+) and X ions are 1.0 pm and 2.0 pm respectively. The volume of the unit cell of the crystal AX will be