52671 + Interview Questions in Chemistry in Class 11

1.	The heat of combustion of C_(2)H_(6) is -368.4 kcal. Calculate the heat of combustion of C_(2)H_(2), when the heat of combustion of H_(2) is 68.32 kcal mol^(-1). C_(2)H_(4)(g) + H_(2)(g) rarr C_(2)H_(6)(g) , Delta H = 37.1 kcal.
Answer» Solution :`C_(2)H_(4)(G) + H_(2)(g) RARR C_(2)H_(6)(g) , Delta H = -37.1 kcl` `Delta H_(c )C_(2)H_(6) = -368.4 kcal, Delta H_(c )C_(2)H_(4) = ? , Delta H_(c )H_(2(g)) = -68.32 kcal` `Delta H = SigmaDelta H_("c(reactants)") = -Sigma Delta H_("c(products)")` `Delta H = Delta H_(c )C_(2)H_(4) + Delta H_(c )H_(2(g)) - Delta H_(c )C_(2)H_(6(c ))` `-37.1 kcal = Delta H_(c )C_(2)H_(4) - 68.32 - (-368.4)` `Delta H_(c )C_(2)H_(4) = -337.18 K cal`.

Discussion

2.	The heatof combustion of benzene in a bomb calorimeter ( i.e., constant volume ) was found to be 3263.9 kJ mol^(-1) at 25^(@)C. Calculate the heatof combusion of benzene at constant pressure.
Answer» Solution :The reaction is `C_(6)H_(6) (l) +7(1)/(2)O_(2)(G) rarr 76CO_(2)(g) + 3H_(2)O(l)` In this reaction, `O_(2)` is the only gaseous reactant and `CO_(2)` is the only gaseous product. `:. Delta N_(g)= n_(p) - n_(r) = 6-7(1)/(2) = -1(1)/(2)= - (3)/(2)` Also, we are given`Delta U ( or q_(V))= - 3263.9 kJmol^(-1)` `T= 25^(@)C = 298 K` `R = 8.314 J K^(-1)mol^(-1) = (8.314)/(1000) KJ K^(-1) mol^(-1)` `Delta H ( or q_(p))= Delta U + Delta n_(g) RT = - 3263.9- 3.7 kJ mol^(-1) + (-(3)/(2)mol) ((8.314)/(1000) kJ K^(-1) mol^(-1)) ( 298K)` `= - 3263.9 - 3.7 kJ mol^(-1) = - 3267.6 kJ mol^(-1)`

Discussion

3.	The heat of combustion of benzene at 27^@C found by bomb calorimeter i.e., for the reaction C_6H_6_((l)) +7 1/2O_(2(g)) + 6CO_(2(g)) +3H_2O_((l)) is -780 K.Cal mol^(-1). The heat evolved on burning 39g of benzene in an open vessel will be
Answer» 390 K.Cal `780.9 K.Cal ` `390.45 K.Cal ` `780 K.Cal ` ANSWER :C

Discussion

4.	The heat of combustion of acetylene is 312kcal. If heat of formation of CO_(2)and H_(2)O are 94.38 and 68.38 kcal respectively. Calculate C-=C bond energy. Given that heat of atomisation of C and H are 150.0 and 51.5kcal respectively and C-H bond energy is 93.64kcal.
Answer» ANSWER :`160.86kcal;`

Discussion

5.	The heat of combustion of benzene at 27^@C and at constant volume is -3245 KJ. Then the heat of combustion of benzene at the same temperature and at constant pressure
Answer» `-3241.26 KJ` `-3252.48 KJ` `-3248.74 KJ` `3248.74 KJ` ANSWER :C

Discussion

6.	The heat of combustion a sugar is -790KJmol^(-1). A person requires 2370KJ of energy for his daily activities and he consumed 540 g of sugar to meet his activities. Calculate the molecular weight of the given sugar.
Answer» ANSWER :180

Discussion

7.	The heat of atomisation of PH_(3)(g) and P_(2)H_(4)(g) are 954 kJ"mol"^(-1) and 1485 kJ"mol"^(-1) respectively. The P - P bond energy in kJ"mol"^(-1) is
Answer» 213 426 318 1272 Solution :`PH(3)(g) to P(g) + 3H(g) Delta_(r)H^(@) = 954 kJ"mol"^(-1)` The bond energy is the average energy for 3 P-H bonds as : Average bond energy pf P-H bond = `(954)/(3) = 318 kJ"mol"^(-1)` `P_(2)H_(4)` has four P-H bonds and ONE P-P bond B.E (P-P) + 4 B.E(P-H) = `1485 kJ"mol"^(-1)` B.E(P-P) + `4 xx 318 = 1485 kJ"mol"^(-1)` B.E(p-p) = `1485 - 4 xx 318` `= 1485 - 1272` =`213 kJ "mol"^(-1)`.

Discussion

8.	The heat of atomisation of PH_(3(g)) is 228 kCal mol^(-1) and that of P_(2)H_(2(g)) is 355 kCal mol^(-1). The energy of the P-P bond is (kn kCal)
Answer» 102 51 26 204 Solution :`PH_(3) rarr P + 3H, Delta H = 228` `3(P- H) = 288 rArr (P - H) = 76` `P_(2) H_(4) rarr 2P + 4H, Delta H = 355` `(P-P) + 4(P-H) = 355` `(P-P) = 355 -4 XX 76 = 51`

Discussion

9.	The heat of atomisation of PH _(3 (g))is118 K. Cal mol ^(-1)and that of P _(2) H _(4 (g)) is 355 K.Cal mol ^(-1)The energy of the P-P bond is (in K.Cal),
Answer» 102 51 26 204 Answer :B

Discussion

10.	The heat liberated when 1.89 g of benzoic acid is burnt in abomb calorimeterat25^(@)Cincreases the temperature of18.94kgof water by 0.632^(@)C. If the specific heat of water at25^(@)C is 0.998cal// g// deg, the value of heat of combustion of benzoic acid is
Answer» `881.1kcal` `771.4 KCAL` `981.1kcal` `871.2kcal` Solution :Heat absorbed by water `= m xx s xx DELTAT` `= 18940 g xx 0.998 cal^(-1) g deg^(-1) xx 0.632 ^(@) C` `=11946.14 cal` `:. ` Heat produced on combustion of1.89g of benzoic ACID`= 11946.14cal` Molar mass of benzoic acid `( C_(6)H_(5) COOH) = 122g` `:. `Molar heat of combustion `= ( 11946.14)/( 1.89) xx 12 cal` `=771126.5 cal = 771.12 kcal`.

Discussion

11.	The heat evolved in the combustion of methane is given by the equation : CH_(4)(s) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l), DeltaH = - 890 .3 kJ mol^(-1) (a ) How manygrams of methane would be required to produce 445.15 kJ of heat of combustion ? (b ) How many grams of carbon dioxidewould beformed when 445.15 kJ of heat is evolved ? (c ) What volume of oxygen at STP would be used in the combustion process (a ) or (b) ?
Answer» SOLUTION :(a) From the GIVEN equation , 890.3kJ of heatis produced from 1 mole of`CH_(4), i.e., 12+4 = 16 g ` of `CH_(4)` (b ) `:.445.15kJ` of heat is produced from 8g of `CH_(4)` (b) From the given equation, when 890.3 kJ of heat is evolved , `CO_(2)` FORMED `=1`mole `=44 g ` `:. `When 445.15 kJ of heatis evolved , `CO_(2)` formed `= 22g` (c ) From the equation , `O_(2)` used in the production of 890.3 kJ of heat `=2` moles`= 2 xx 22.4 ` litres at STP `= 44.8` litres at STP Hene, `O_(2)` used in the production of 445.1 5 kJ of heat = 2 22.4 litres at STP.

Discussion

12.	The heat of formation of crystalline sodium chloride is -410 kJ mol^(-1). The heat of sublimation of sodium metal is 180.8 kJ mol^(-1). The heat of dissociation of chlorine gas into atoms is 242.7 kJ mol^(-1). The ionisation energy of Na and electron affinity of Cl are 493.7 kJ and -368.2 kJ respectively. Calculate the lattice energy of NaCl.
Answer» Solution :Based on the Hess LAW, Born - HABER cycle is constructed. `-Q=+S+1/2D+I-E+U` Substituting the RESPECTIVE VALUES, we get `410=108.8+1/2xx242.7+493.7-368.2+U` Lattice energy of sodium chloride, `U=-765.65KJmol^(-1)`.

Discussion

13.	The heat evolved in the combustion of glucose is shown in the equation : C_(6)H_(12)O_(6)(s) + 6O_(2)(g) rarr 6CO_(2)(g) + 6H_(2)O (g) , Delta_(c ) H = -2840 kJ mol^(-1) What is the energy requirement for production of 0.36g of glucose by the reverse reaction ?
Answer» Solution :The given equation is `:C_(6)H_(12)O_(6)+ 6O_(2)(G) rarr6CO_(2)(g) +6H_(2)O(g), Delta_(r)H= -2849 k J MOL^(-1)` WRITTEN the reverse reaction, we have `6CO_(2)(g)+6H_(2)O(g) rarr C_(6)H_(12)O_(6)(s)+ 6O_(2)(g), Delta_(r)H= + 2840 kJ mol^(_1)` Thus, for production of 1 mole of `C_(6)H_(12)O_(6)( = 72 + 12 + 96= 180 g )`, heat required( absorbed) `= 2840kJ`ltbgt `:. `For production of0.36 g of glucose , heat absorbed`=(2860)/(180) XX 0.36 = 5.68 kJ`

Discussion

14.	The heat evolved in the combustion of 112 L ofwater gas( CO + H_(2) in the ration of1:1 by volume) is ( Giventhat heats of combustion of H_(2) andCO are -241.8and -283 kJ mol^(-1) respectively.
Answer» `- 1414 . 0 kJ` `-1209.0 kJ` `- 1312.0 kJ` `- 524.8 kJ` SOLUTION : `H_(2)+(1)/(2) O_(2) rarr H_(2)O, DeltaH =- 241.8kJ` `CO +(1)/(2) O_(2) rarr CO_(2), DeltaH = - 283kJ` Since, `H_(2)` and CO have same volume in the MIXTURE, THEREFORE, each `= 56L ` `DeltaH` for combustion of `56L ` of `H_(2)` `=-( 241.8) /( 22.40 xx56kJ = - 604.5 kJ` `DeltaH`for combustionof `56 L ` of CO `= - ( 283)/( 22.4) XX 56 kJ = - 707 .5 kJ` `:. `Total `DeltaH = - 604 + ( - 707.5) = -1312kJ`

Discussion

15.	The heat change for the following reaction C_((s)) + 2S_((s)) rarr CS_(2(l)) is known as
Answer» HEAT of formation of `CS_2` Heat of fusion of `CS_2` Heat of VAPOURISATION of `CS_2` Heat of TRANSITION of CARBON Answer :A

Discussion

16.	The heat capacities of the diatomic molecules attain a limiting value at higher temperature At room temperature the translational and rotational degrees of freedom contribute to the heat capacities of gases while the vibrational degree of freedom becomes active only at higher temperature .
Answer» SOLUTION :EXPLANTION is CORRECT REASON for STATEMENT

Discussion

17.	The heat associated with combustion of liquid benzene, at constant volume is -3268 kilo Joule/mole""^(-1) calculate the change in enthalpy, when this reaction occurs at 300 K, temperature (R=8.314 Joule)
Answer» SOLUTION :`DELTAH = - 3271.74` kJ/mol

Discussion

18.	The heat absorbed at constant volume is equal to the system's change in ___
Answer» SOLUTION :INTERNAL ENERGY

Discussion

19.	The Henry's law constant for the solubility of Nitrogen gas in water at350 K is8xx 10^(4)atm. The mole fraction of nitrogen in air is0.5.The numberof moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressureis
Answer» `4 xx 10 ^(-4)` `4 xx 10 ^(x)` `2xx 10 ^(-2)` `2.5 xx 10 ^(-4)` Solution :`K _(H) = 8 xx 10 ^(4)` `(X_(N_(2))) _("in air") = 0.5` TOTAL pressure = 4 atm Partial pressure of nitrogen = Mole fraction `xx` Total pressure `0.5 xx 4 =2` `(P _(N _(2))) = K _(H) xx` Mole fraction of `N_(2)` in solution `2 = 8 xx 10 ^(4) xx ("Number of moles of nitrogen")/("Total number of moles")` `(10 + "No. of moles of" N _(2))/( "No. of moles of" N _(2)) = (8xx 10 ^(4))/(2)` `(10)/("No. of moles of" N_(2)) +1 = 4 xx 10 ^(4)` `(10)/("No. of moles of" N_(2)) = 4000 -1` `THEREFORE ` No. of moles of `N _(2) = (10)/(39999) =2.5 xx 10 ^(-4)`

Discussion

20.	The hardness of water sample containing 0.002 of water is expressed as:
Answer» 20 PPM 200 ppm 2000 ppm 120 ppm Solution :Amount of `MgSO_(4)=0.002xx120xx1000` `=120` MG Now 120 mg `MgSO_(4)-=100 mg CaCO_(3)` `therefore` 240 mg `MgSO_(4)-=200 mg CaCO_(3)` 1 L of water CONTAINS =200 mg of `CaCO_(3)` or `10^(6)` mg `H_(2)O` contains =200 mg of `CaCO_(3)` `therefore ` DEGREE of hardness =200 ppm.

Discussion

21.	The hardness of water can be determined by volumetrically using the reagent
Answer» SODIUM thio sulphate tassium permanganate hydrogen PEROXIDE EDTA Answer :d

Discussion

22.	The hardness of water can be determined by volumetrically using the reagent ………………… .
Answer» soldium thio SULPHATE POTASSIUM PERMANGANATE hydrogen peroxide EDTA Solution :EDTA

Discussion

23.	The hardest substance among the following is
Answer» BeC graphite titanium SIC Solution :Out of the GIVEN FIVE CHOICES SiC is the HARDEST.

Discussion

24.	The halogen that shows same oxidation state in all its compounds with other element is
Answer» `I_2` `F_2` `Cl_2` `Br_2` Solution :`BaO_(2)+H_(2)SO_(4) rarr OVERSET(-2)(BaSO_4)+H_(2)overset(-1)(O_2)`

Discussion

25.	The halide that is highly soluble in H_(2)O among the following is
Answer» `BeF_(2)` `CaF_(2)` `MgF_(2)` `BaF_(2)` Solution :The solubility of halides of IIA group elements decreases from Be to BA DUE to decreases in hydration energy because of increasing size of metal cation. SO, `BeF_(2)` only soluble in water

Discussion

26.	The halide of which alkaline earth metal is covalent and soluble in organic solvents ?
Answer» Be Mg Ca Sr Solution :Because of SMALL size , `Be^(2+)` has the HIGHEST polarising power and hence `BeCl_(2)` has the highest covalent character and is THUS soluble in ORGANIC SOLVENTS .

Discussion

27.	The half reaction with their oxidation potential are Pb(s)rarrPb^(2+)(aq)+2 e^(-)E_("xi")^(@)=+0.13 V, Ag (s) rarr (aq)^(+)(aq)+e^(-),E_("xi")^(@)=-0.80 V write the cell reaction and calcualte its EMF
Answer» Solution :Rewrite the two equatin in the reduction form by reversing the sings of oxidatin potention thus `Pb^(2+)(AQ)+2 e^(-) rarr Pb(s) , E^(@)=-0.13V` `Ag^(+)(aq)+e^(-)rarr Ag(s),E^(@)=+0.80V` To obtain the equation for the cell reaction MULTIPLY Eq (ii) with 2 and SUBTRACT Eq (i) from it we have `Pb(s)+2 Ag^(+)(aq) rarr Pb^(2+)(aq)+2AG(s),E_(cell)^(@)=-0.80 -(0.13)=+0.93 V`

Discussion

28.	The half lives of two radioactive nuclides A and B are 1 and 2 min. respectively . Equal weights of A and B are taken separatelyand allowed to disintegrate for 4 min. What will be the ratio of weigths of A and B disintegrated ?
Answer» `1:1` `5:4` `1:2` `1:3` SOLUTION :For A T=4 min `t_(1//2)=1` min `N_(0)=x`(say) `n=4//1=4` `N=(N_(0))/(2^(n))=(x)/(2^(4))=(x)/(16)` A disintegrated `=x-(x)/(16)=(15x)/(16)` For B T=4 min `t_(1//2)=` 2 min `N_(0)=x` (say) `n=4//2=2` `N=(N_(0))/(2^(n))=(x)/(2^(2))` `=(x)/(4)` B disintegrated `x-(x)/(4)=(3x)/(4)` Ratio of A : B disintegrated `(15x)/(16):(3x)/(4)` or 5:4

Discussion

29.	The half life time of ._1^3H is 12.33 years, what is the decay constant ?
Answer» 12.33 YEARS 0.5620 `"years"^(-1)` 0.05620 `"years"^(-1)` 0.05620 `"years"^(-1)` SOLUTION :`t_(1//2)`=12.33 years `THEREFORE` DECAY constant `lambda=0.693/t_(1//2)=0.693/12.33` years `=0.05620 "years"^(-1)`

Discussion

30.	The half life period of tritium is _________
Answer» SOLUTION :12.33 YEARS

Discussion

31.	The half life period of francium is _______
Answer» SOLUTION :21 MINUTES

Discussion

32.	The half life of Pb^(212) is 10.6 hour. It undergoes decay to itsdaughter(unstable)element Bi^(212) of half-,ofe 60.5 mintue. Calcualte the time at whichdaughter elementwil have maximum activity.
Answer» ANSWER :`227.1` MINUTE

Discussion

33.	The half life of a radio isotope is four hours. If the initial mass of the isotope was 200 g the mass remaining after 24 hours undecayed is
Answer» 1.042 G 4.167 g 3.125 g 2.084g Solution :`t_(1//2)=4` hr No. of half lives in 24 hours, `n=(24)/(4)=6` `N_(0)=200` g `N=(N_(0))/(2^(n))=(200)/(2^(6)) g` `=(200)/(64)=3.125 g`

Discussion

34.	The half-life of .^(32)P si 14.3 day. Calculate the specific activity of a phosphours containingseciment having1.0 part per million .^(32)P (Atomic weight of p = 31)
Answer» ANSWER :`0.295 CI` PER `G`

Discussion

35.	The half life can serve as the molecular clock to determine the age of different products i.e., mineral, rocksand matter of vegetable origin such as wood, charcoal etc. because the half-life of a particlar nucleide is constant. C-14 being radio-isotope , it is also used to determine the age of different products. which is known as radio-carbon dating. C-14 is a neutron rich nucleide that decays by beta-emission with a half-life of 5730 years as under : ""_(6)^(14)C to ""_(7)^(14)N + ""_(-1)^(0)e Just after World War II . WIllcard F. Libby proposed a way to use this rection to determine the age of the carbon containing substances. The C-14 dating technique for which Libby received nobel is based upon the following assumption . (i) C-14 is produced in atmosphere at a more or less constant rate which is also equal to its rate of decay. As a result, there is a constant concentration of C-14 in the atmosphere and all living things. (ii) After death, organisms no longer pick C-14 (iii) By comparing activity of a sample with acitivity of living tissue we can estimate the age of organisms . One of the Libby's assumptions is questionable, the amount of C-14 has not been constant with time. It has varied as much as +- 5% . What will be the weight ratio of living tissue to dead tissue for the age of 22920 years , if it is dated by radio-carbon technique ?
Answer» `16:1` `8:1` `10:3` `5:7` ANSWER :A

Discussion

36.	The half life can serve as the molecular clock to determine the age of different products i.e., mineral, rocksand matter of vegetable origin such as wood, charcoal etc. because the half-life of a particlar nucleide is constant. C-14 being radio-isotope , it is also used to determine the age of different products. which is known as radio-carbon dating. C-14 is a neutron rich nucleide that decays by beta-emission with a half-life of 5730 years as under : ""_(6)^(14)C to ""_(7)^(14)N + ""_(-1)^(0)e Just after World War II . WIllcard F. Libby proposed a way to use this rection to determine the age of the carbon containing substances. The C-14 dating technique for which Libby received nobel is based upon the following assumption . (i) C-14 is produced in atmosphere at a more or less constant rate which is also equal to its rate of decay. As a result, there is a constant concentration of C-14 in the atmosphere and all living things. (ii) After death, organisms no longer pick C-14 (iii) By comparing activity of a sample with acitivity of living tissue we can estimate the age of organisms . One of the Libby's assumptions is questionable, the amount of C-14 has not been constant with time. It has varied as much as +- 5% . Charcoal sample emits 62.3% of disintegrations per gram of carbon per minute for living tissue. Then, what will be the age of charcoal ?
Answer» 4910 years 3910 years 6000 years 5050 years Answer :B

Discussion

37.	The half life can serve as the molecular clock to determine the age of different products i.e., mineral, rocksand matter of vegetable origin such as wood, charcoal etc. because the half-life of a particlar nucleide is constant. C-14 being radio-isotope , it is also used to determine the age of different products. which is known as radio-carbon dating. C-14 is a neutron rich nucleide that decays by beta-emission with a half-life of 5730 years as under : ""_(6)^(14)C to ""_(7)^(14)N + ""_(-1)^(0)e Just after World War II . WIllcard F. Libby proposed a way to use this rection to determine the age of the carbon containing substances. The C-14 dating technique for which Libby received nobel is based upon the following assumption . (i) C-14 is produced in atmosphere at a more or less constant rate which is also equal to its rate of decay. As a result, there is a constant concentration of C-14 in the atmosphere and all living things. (ii) After death, organisms no longer pick C-14 (iii) By comparing activity of a sample with acitivity of living tissue we can estimate the age of organisms . One of the Libby's assumptions is questionable, the amount of C-14 has not been constant with time. It has varied as much as +- 5% . It activity of C-14 in living matter is 15.3 dpm, then what will be the age of the object which has activity of 4 d.p.m. ?
Answer» 11100 YEARS 22200 years `5.1 XX 10^(5)` years `2.1 xx 10^(6)` years Answer :A

Discussion

38.	The half life can serve as the molecular clock to determine the age of different products i.e., mineral, rocksand matter of vegetable origin such as wood, charcoal etc. because the half-life of a particlar nucleide is constant. C-14 being radio-isotope , it is also used to determine the age of different products. which is known as radio-carbon dating. C-14 is a neutron rich nucleide that decays by beta-emission with a half-life of 5730 years as under : ""_(6)^(14)C to ""_(7)^(14)N + ""_(-1)^(0)e Just after World War II . WIllcard F. Libby proposed a way to use this rection to determine the age of the carbon containing substances. The C-14 dating technique for which Libby received nobel is based upon the following assumption . (i) C-14 is produced in atmosphere at a more or less constant rate which is also equal to its rate of decay. As a result, there is a constant concentration of C-14 in the atmosphere and all living things. (ii) After death, organisms no longer pick C-14 (iii) By comparing activity of a sample with acitivity of living tissue we can estimate the age of organisms . One of the Libby's assumptions is questionable, the amount of C-14 has not been constant with time. It has varied as much as +- 5% . The activity of the C-14 in given tissue is 15.3 dpm per gram of carbon. The limit for the reliable determination of C-14 is 0.1 dpm per gram of carbon. The maximum age of the sample, that can be dated accurately , is
Answer» 41600 years 42700 years 21600 years 20500 years Answer :A

Discussion

39.	The hair dyes available in the market generally contain two bottles, one containing the dye and the other hydrogen peroxide. Before applying the dye, the two solutions are mixed. The hydrogen peroxide.
Answer» is added to dilute the solution of the dye oxidises the dry to give the desired COLOUR reduces the dye to give the desired colour acidifies the solution of the dye Solution :The hydrogen PEROXIDE oxidises the dye to give the desired colour

Discussion

40.	The H-OO bondangle and O-H bond lengths are 101.9^(@)and 98.8p, resectvly in solid phase instead of 94.8^(@) and 95 pm in gaseous phase instedad tat the structure of H_(2)O_(2)this indicates that the struture of H_(2)O_(2) in solid and gasous phases are different .this is due to
Answer» INTERMOLECULAR HYDROGEN bonding Interamolecular hydrogen bonding VAN der Waal 's bonding All ERE true Answer :A

Discussion

41.	The H-O-H angle in water molecules is about
Answer» `90^(@)` `104.5^(@)` `109^(@)`28' `180^(@)` ANSWER :B

Discussion

42.	The H^(+)ion concentration of a solution is4xx10^(-5)M. Then the OH^(-)ion concentration of the same solution is
Answer» `4xx10^(-5)M` ` 2.5xx10^(-9) M` ` 1.0xx 10^(-7) M` ` 2.5 xx 10^(-10)M` Solution :` [H^(+) ] = 4xx 10 ^(-5)M, {OH^(-) }=(K_W)/( [H^(+)]) ` ` = ( 1xx10 ^(-14))/( 4xx10 ^(-5)) =0.25 xx1 0 ^(-9) ` ` THEREFORE [OH^(-) ] = 2.5 xx 10 ^(10)M`

Discussion

43.	The [H^(+)] in a solution containing 0.1 M HCOOH and 0.1 M HOCN [Ka for HCOOH and HOCN are 1.8 xx 10^(-4) and 3.3 xx 10^(-4)] respectively will be
Answer» `7.13 xx 10^(-2) M` `7.13 xx 10^(-3) M` `7.13 xx 10^(-5) M` `7.13 xx 10^(-6) M` Solution :`UNDERSET((0.1-x))(HCOOH)hArr underset((x+y))(H^(o+))+underset(x)(HCOOO^(o+))` `underset((0.1-y))(HOCN)hArr underset((x+y))(H^(o+))+underset(y)(OCN^(o+))` `K_(HCOOH)= ([H^(+)][HCOO^(-)])/([HCOOH]) = 1.8 xx 10^(-4)` `K_(HOCN) = ([H^(+)][OCN^(-)])/([HOCN]) = 3.3 xx 10^(-4)`

Discussion

44.	The H-H bond dissociation enthalpy of H_2 is _________, is the highest for a single bond between two atoms of any elements.
Answer» ANSWER :435.88`"KJ MOL"^(-1)`

Discussion

45.	The H-B-H bridged angle in diborane is
Answer» `121.5^@` `97^@` `119^@` `133^@` ANSWER :B

Discussion

46.	The H - C -H bond angle in ethene is…………
Answer» ANSWER :`120^(@)`

Discussion

47.	The group which exhibits -M effect is
Answer» `-CHO` `-C EQUIV N` `-NO_(2)` All Solution :All GROUPS are -M

Discussion

48.	The group responsible for the removal of cations in ion exchange resin is
Answer» `-NH_(2)OH` `-COOH` `-OH` `-SH` SOLUTION :Cation EXCHANGE of resin `=R-COOH` anion exchange resin `=R-NH_(3)-OH`

Discussion

49.	The group present in waxes are
Answer» ACID group ester group alcohol group ether group Solution :REFER to COMPREHENSIVE REVIEW

Discussion

50.	The group of elements in which the differentiating electron enters the anti-penultimate shell of atoms are called
Answer» p-block elements d-block elements s-block elements f-block elements Answer :d

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The heat of combustion of C_(2)H_(6) is -368.4 kcal. Calculate the heat of combustion of C_(2)H_(2), when the heat of combustion of H_(2) is 68.32 kcal mol^(-1). C_(2)H_(4)(g) + H_(2)(g) rarr C_(2)H_(6)(g) , Delta H = 37.1 kcal.

The heatof combustion of benzene in a bomb calorimeter ( i.e., constant volume ) was found to be 3263.9 kJ mol^(-1) at 25^(@)C. Calculate the heatof combusion of benzene at constant pressure.

The heat of combustion of benzene at 27^@C found by bomb calorimeter i.e., for the reaction C_6H_6_((l)) +7 1/2O_(2(g)) + 6CO_(2(g)) +3H_2O_((l)) is -780 K.Cal mol^(-1). The heat evolved on burning 39g of benzene in an open vessel will be

The heat of combustion of acetylene is 312kcal. If heat of formation of CO_(2)and H_(2)O are 94.38 and 68.38 kcal respectively. Calculate C-=C bond energy. Given that heat of atomisation of C and H are 150.0 and 51.5kcal respectively and C-H bond energy is 93.64kcal.

The heat of combustion of benzene at 27^@C and at constant volume is -3245 KJ. Then the heat of combustion of benzene at the same temperature and at constant pressure

The heat of combustion a sugar is -790KJmol^(-1). A person requires 2370KJ of energy for his daily activities and he consumed 540 g of sugar to meet his activities. Calculate the molecular weight of the given sugar.

The heat of atomisation of PH_(3)(g) and P_(2)H_(4)(g) are 954 kJ"mol"^(-1) and 1485 kJ"mol"^(-1) respectively. The P - P bond energy in kJ"mol"^(-1) is

The heat of atomisation of PH_(3(g)) is 228 kCal mol^(-1) and that of P_(2)H_(2(g)) is 355 kCal mol^(-1). The energy of the P-P bond is (kn kCal)

The heat of atomisation of PH _(3 (g))is118 K. Cal mol ^(-1)and that of P _(2) H _(4 (g)) is 355 K.Cal mol ^(-1)The energy of the P-P bond is (in K.Cal),

The heat liberated when 1.89 g of benzoic acid is burnt in abomb calorimeterat25^(@)Cincreases the temperature of18.94kgof water by 0.632^(@)C. If the specific heat of water at25^(@)C is 0.998cal// g// deg, the value of heat of combustion of benzoic acid is

The heat evolved in the combustion of glucose is shown in the equation : C_(6)H_(12)O_(6)(s) + 6O_(2)(g) rarr 6CO_(2)(g) + 6H_(2)O (g) , Delta_(c ) H = -2840 kJ mol^(-1) What is the energy requirement for production of 0.36g of glucose by the reverse reaction ?

The heat evolved in the combustion of 112 L ofwater gas( CO + H_(2) in the ration of1:1 by volume) is ( Giventhat heats of combustion of H_(2) andCO are -241.8and -283 kJ mol^(-1) respectively.

The heat change for the following reaction C_((s)) + 2S_((s)) rarr CS_(2(l)) is known as

The heat capacities of the diatomic molecules attain a limiting value at higher temperature At room temperature the translational and rotational degrees of freedom contribute to the heat capacities of gases while the vibrational degree of freedom becomes active only at higher temperature .

The heat associated with combustion of liquid benzene, at constant volume is -3268 kilo Joule/mole""^(-1) calculate the change in enthalpy, when this reaction occurs at 300 K, temperature (R=8.314 Joule)

The heat absorbed at constant volume is equal to the system's change in ___

The Henry's law constant for the solubility of Nitrogen gas in water at350 K is8xx 10^(4)atm. The mole fraction of nitrogen in air is0.5.The numberof moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressureis

The hardness of water sample containing 0.002 of water is expressed as:

The hardness of water can be determined by volumetrically using the reagent

The hardness of water can be determined by volumetrically using the reagent ………………… .

The hardest substance among the following is

The halogen that shows same oxidation state in all its compounds with other element is

The halide that is highly soluble in H_(2)O among the following is

The halide of which alkaline earth metal is covalent and soluble in organic solvents ?

The half reaction with their oxidation potential are Pb(s)rarrPb^(2+)(aq)+2 e^(-)E_("xi")^(@)=+0.13 V, Ag (s) rarr (aq)^(+)(aq)+e^(-),E_("xi")^(@)=-0.80 V write the cell reaction and calcualte its EMF

The half lives of two radioactive nuclides A and B are 1 and 2 min. respectively . Equal weights of A and B are taken separatelyand allowed to disintegrate for 4 min. What will be the ratio of weigths of A and B disintegrated ?

The half life time of ._1^3H is 12.33 years, what is the decay constant ?

The half life period of tritium is _________

The half life period of francium is _______

The half life of Pb^(212) is 10.6 hour. It undergoes decay to itsdaughter(unstable)element Bi^(212) of half-,ofe 60.5 mintue. Calcualte the time at whichdaughter elementwil have maximum activity.

The half life of a radio isotope is four hours. If the initial mass of the isotope was 200 g the mass remaining after 24 hours undecayed is

The half-life of .^(32)P si 14.3 day. Calculate the specific activity of a phosphours containingseciment having1.0 part per million .^(32)P (Atomic weight of p = 31)

The hair dyes available in the market generally contain two bottles, one containing the dye and the other hydrogen peroxide. Before applying the dye, the two solutions are mixed. The hydrogen peroxide.

The H-OO bondangle and O-H bond lengths are 101.9^(@)and 98.8p, resectvly in solid phase instead of 94.8^(@) and 95 pm in gaseous phase instedad tat the structure of H_(2)O_(2)this indicates that the struture of H_(2)O_(2) in solid and gasous phases are different .this is due to

The H-O-H angle in water molecules is about

The H^(+)ion concentration of a solution is4xx10^(-5)M. Then the OH^(-)ion concentration of the same solution is

The [H^(+)] in a solution containing 0.1 M HCOOH and 0.1 M HOCN [Ka for HCOOH and HOCN are 1.8 xx 10^(-4) and 3.3 xx 10^(-4)] respectively will be

The H-H bond dissociation enthalpy of H_2 is _________, is the highest for a single bond between two atoms of any elements.

The H-B-H bridged angle in diborane is

The H - C -H bond angle in ethene is…………

The group which exhibits -M effect is

The group responsible for the removal of cations in ion exchange resin is

The group present in waxes are

The group of elements in which the differentiating electron enters the anti-penultimate shell of atoms are called