94420 + Interview Questions in Chemistry in Class 12

1.	One mole of nitrogen gas at 0.8 atm takes 38 second of diffuse through a pin hole whereas one mole of an unknown compound of Xenon with fluorine at 1.6 atm takes 56.26 second to diffuse through at same hole, then the molecular formula of the compound is (Atomic mass of Xenon : 131.3 u)
Answer» `XeF_2` `XeF_4` `XeF_5` `XeF_6` ANSWER :D

Discussion

2.	One mole of NaCl (s) on melting absorbed 30.5 KJ of heat and its entropy is increased by 28.8JK^(-1). What is the melting point of sodium chloride ?
Answer» SOLUTION :T = 1059K

Discussion

3.	One mole of NH_(3) is made to react with one mole of CH_(3) Cl, which of the following is/ are formed
Answer» `CH_(3)NH_(2)` `(CH_(3))_(2)NH` `(CH_(3))_(3)N` All the above Solution :Amines forms `4^(@)` AMM onium salts with R-X

Discussion

4.	One mole of NaCl (s) on melting absorbed 30.5 kJ of heat and its entropy is increased by 28.8 JK^(-1). The melting point of NaCl is
Answer» 1059 K 30.5 K 28.8 K 28800 K Solution :`NaCl(s) rarrNaCl(l)` Given that : `DeltaH=30.5 kJ mol^(-1)` `DeltaS=28.8 JK^(-1)=28.8xx10^(-3)kJ K^(-1)` By USING `DeltaS=(DeltaH)/(T)=(30.5)/(28.8xx10^(-3))=1059 K`.

Discussion

5.	One mole of N_(2)O_(4) is heated in a flask with a volume of 10dm^(3). At equilibrium 1.708 mole of NO_(2) and 0.146 mole of N_(2)O_(4) were found at 134^(@)C. The equilibrium constnt will be
Answer» 250 mol `dm^(-3)` 300 mol `dm^(-3)` 2 mol `dm^(-3)` 230 mol `dm^(-3)` SOLUTION :`N_(2)O_(4)hArr2NO_(2)` `{:("Moles at "eqb^(m),0.146,1.708),(Conc.,,),((mol//dm^(3)),[(0.146)/(10)],[(1.708)/(10)]):}` `thereforeK_(c)=([NO_(2)]^(2))/([N_(2)O_(4)])=([(1.708)/(10)])/([(0.146)/(10)])` `impliesK_(c)=(2.917264)/(100)xx(10)/(0.146)` `IMPLIES K_(c)=(2.917264)/(1.46)=2"mol dm"^(-3)`

Discussion

6.	One mole of N_2O_4 (g) at 300k is kept in a closed container under one atmospheric pressure. It is heated to 600 k when 20 % by mass of N_2O_4 (g) decomposes to NO_2(g) . The resultant pressure is :
Answer» 1.2 atm 2.4 atm 2.0 atm 1.0 atm Answer :B

Discussion

7.	One mole of N_(2)and 3 moles of H_(2)are mixed in a litre flask. If 50% N_(2)is converted into ammonia by the reaction.N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)then the total number of moles of gas at equilibrium is :
Answer» `1.5` `3.0` `4.5` `6.0` ANSWER :B

Discussion

8.	One mole of monoatomic gas taken through a cyclic process as shown in figure. Calculate sum_(ABCA) = (dq_(rev))/(T).
Answer» ANSWER :0

Discussion

9.	One mole of monoatomc ideal gas expands adiabatically at initial temp. T against a constant external pressure of 1 atm from one litre to two litre. Find out the final temp. (R = 0.0821 litre. "atm"K^(-1)mol^(-1)):
Answer» T `T/((2)^(5//3-1))` `T-2/(3xx0.0821)` `T+2/(3xx0.0821)` ANSWER :C

Discussion

10.	One mole of methanol when burnt in O_(2), gives out 723 kJ mol^(-1) heat. If one mole of O_(2) is used, what will be the amount of heat evolved
Answer» 723 kJ 924 kJ 482 kJ 241 kJ Solution :`CH_(3)OH(L)+(3)/(2)O_(2)(G)rarrCO_(2)(g)+2H_(2)O(l) DeltaH=-723 kJ` By burning with `(3)/(2)O_(2)(g)` heat evolved =-723 kJ `therefore` By burning with 1 mole `O_(2)(g)=(-723xx2)/(3)=-482 kJ`.

Discussion

11.	One mole of methannol when burnt in oxygen gives out 723 kJ mol^(-1) heat. If one mole of oxygen is used, what will be the amount of heat evolved ?
Answer» 723 kJ 964 kJ 482 kJ 241 kJ Solution :`underset(2 mol)(2CH_(3)OH) + underset(3 mol)(3O_(2)) rarr 2CO_(2) + 4H_(2)O Delta H = - 723 kJ mol^(-1)` 1 mol of `O_(2) = (2)/(3)` mol of `CH_(3)OH` :. `(2)/(3)` mol of `CH_(3)OH` on COMBUSTION gives `= (723 xx 2)/(3)` = 482 kJ

Discussion

12.	One mole of magnesium in the vapour state absorbed 1200 kJ mol^(-1) of energy. If the first and second ionization energies of Mg are 750 and 1450 kJmol^(-1) respectively, the final composition of the mixture is
Answer» `31% Mg^(+)+69% Mg^(2+)` `69%Mg^(+)+31% Mg^(2+)` `86%Mg^(+)+14% Mg^(2+)` `14%Mg^(+)+86%Mg^(2+)` Solution :Number of moles of 1 g of Mg=1/24=0.0417 1 g of Mg (s) absorbs = `(1200)/(24)=50 kJ` `therefore` Energy required to CONVERT Mg (g) to `Mg^(+)`(g) `=0.0417xx750=31.275 kJ`. `therefore`Remaining energy = 50 - 31.275 = 18.725 kJ Number of moles of `Mg^(2+)` formed = `(18.725)/(1450)=0.013` Thus remaining `Mg^(+)` will be = `0.0417 - 0.013 = 0.0287` `therefore % Mg^(+)=(0.0287)/(0.0417)xx100=68.82%` `% Mg^(2+)=100-68.82 = 31.18%`.

Discussion

13.	Onemole of magnesium nitride on reaction with excess of water gives
Answer» TWO moles of ammonia one mole of NITRIC acid one mole of ammonia two moles of nitric acid Solution :`Mg_3N_2 + 3H_2O to 3MG(OH)_2 + 2NH_3` ` therefore ` 2 mol of `NH_3`

Discussion

14.	One mole of liquid A (vapour pressure =300 mm Hg ) and 2 moles of a liquid B (Vapour pressure =360 mm Hg) are mixed . The mole fraction of A in the vapour mixture if found to be 5//17 . Which of the following is correct ?
Answer» `(DeltaH)_("mix")gt0` `(DeltaS)_("mix")gt0` `(DELTAV)_("mix")=0` `(DeltaG)_("mix")gt0` ANSWER :B::C::D

Discussion

15.	One moleof ideal gas undergoes following cyclic process (i) Isochoric heating from (P_(1)V_(1)T_(1)) to double temperature.(II)Isobaric expansion from todouble voulme . (III) Linear expansion( on PV curve) to (P_(1),8V_(1)).Isobaric copmression to initial state.Calculate the magnitudeof work done in caloriesif initialtremperature of the gas in is300K?Given : R=2(Cal)/(mol- K)
Answer» Solution :Magniudeof work =`A=P_(1)V_(1)+(1)/(2)xxP_(1)xx 6V_(1)` `=4xx1xx2xx300` `=4xx1xx2xx300` `=2400 CAL`

Discussion

16.	One mole of ideal monatomic gas is carried through the reversible cyclic process as shown in figure. Calculate the max temperature attained by the gas during the cycle.
Answer» `(25)/(8)((P^(@)V^(@))/(R))` `-(25)/(8)((P^(@)V^(@))/(R))` `(35)/(8)((P^(@)V^(@))/(R))` `-(35)/(8)((P^(@)V^(@))/(R))` Solution :BC is a straight line & equation can be obtained by using `(y-y_(1)) = (Y_(2)-Y_(1))/(x_(2)-x_(1))(x-x_(1))` `(P-3P^(@))=(P^(@)-3P^(@))/(2V^(@)-V_(@))(V-V^(@)) implies (P-3P^(@)) =-(2P^(@))/(V^(@))(V-V^(@)) ""…..(1)` Replacing P by using `P=(RT)/(V)` in equation (1) `((RT)/(V)-3P^(@))=-(2P^(@))/(V^(@)) (V-V^(@)) implies T=(2P^(@)V(V-V^(@)))/(V^(@)R) + (3P^(@)V)/(R)` For T to be maximum `(dT)/(DV)=0` `(dT)/(dV) =-(2P^(@)(2V-V^(@)))/(V^(@)R)+ (3P^(@))/(R) =0 implies V=(5)/(4)V^(@)` `T_("max")=-(2P^(@)xx(5)/(4)V^(@)((5)/(4)V^(@)-V^(@)))/(V^(@)R)+(3P^(@)(5)/(4)V^(@))/(R)=(25)/(8)((P^(@)V^(@))/(R))`

Discussion

17.	One mole of idealmonatomicgas in takenin cycliceprocess ABCA as shownin the figure Calculate :(a) Theworkdone by thegas(b) The heatrejectedbythegas in the pathCA (c) The net heatabsorbedbe the bygas in the path BC (d)Maximum temperature obtainedby thegas duringthe cycle.
Answer» SOLUTION :N//A

Discussion

18.	One mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27^(@)C. If the work done by the gas in the process is 3 kJ, the final temperature will be equal to (C_(v)=20J//Kmol)
Answer» 100 K 450 K 150 K 400 K Answer :C

Discussion

19.	One mole of ice is coverted into water at 273 K. The entropies of H_(2)O(s) and H_(2)O(l) are 38.20 and 60.01 J mol^(-1) K^(-1) respectively. The enthaply change for the conversion is
Answer» `59.54 J mol^(-1)` `5954 J mol^(-1)` `595.4 J mol^(-1)` `320.6 J mol^(-1)` SOLUTION :`H_(2)O(s) hArr H_(2)O(L)` `DELTA S = S[H_(2)O(l)] - [H_(2)O(s)]` `60.01 - 38.20` `= 21.81 J mol^(-1)` `Delta H = T Delta S = 21.81 xx 273` `= 5954 J mol^(-1)`

Discussion

20.	One mole of ice is converted into water at 273 K. The entropies of H_2O(s) and H_2O(l) are 38.20 and 60.01 J mol^-1 K^-1 respectively. The enthalpy change for the conversion is:
Answer» `59.54 J mol^-1` `5954 J mol^-1` `595.4 J mol^-1` `320.6 J mol^-1` ANSWER :B

Discussion

21.	One mole of hydrocarbon (A) reacts with one mole of bromine giving a dibromocompound C_(5)H_(10)Br_(2). Substance (A) on treatment with cold, dilute alkaline KMnO_(4) solution forms a compound C_(5)H_(12)O_(2). On ozonolysis, (A) gives equimolar quantities of propanone and ethanal. Deduce the structural formula of (A)
Answer» SOLUTION :

Discussion

22.	One mole of hydrogen peroxide (H_2O_2)has a mass same as that of
Answer» 0.1 mol of sucrose ` (C_12H_22O_11)` 2.0 mol of ammonia 11.2 L of `SO_2`at N.T.P. 0.1 mol of `SO_3` Solution :Mass of 1 MOLE of `H_2O_2` = 34 G 2 mol of `NH_3 = 2 xx 17= 34 g `

Discussion

23.	one mole of hydrogen iodide is heated in a closed container of 2 litre. At equilibrium half mole of hydrogen iodide has dissociated. What is the value of the equilibrium constant ?
Answer» 1 5 0.25 0.75 Answer :C

Discussion

24.	One mole of Helium gas undergoes a reversible cyclic process ABCDA as shown in the figure.Assuming gas to be ideal, answer the following questions : Polytropic process for ideal gas is given as PV^(th)=constant. For polytropic process for an ideal gas.the expression for work obtained is : w=(P_1V_1[(V_2/V_1)^x-1])/((y)) Report your answer as (x+y).
Answer» 0 1 2 3 Solution :`W=-intPdV` `=-INT K/V^n dV=-K/(1-n)[V^(-n+1)]_(V_1)^(V_2)=K/(n-1)[V_2^(1-n)-V_1^(1-n)]` `=(P_1V_1^n)/(n-1)[V_2^(1-n)-V_1^(1-n)]=(P_1V_1^(n+1-n))/(n-1)[(V2/V_1)^(1-n)-1]`

Discussion

25.	One mole of Helium gas undergoes a reversible cyclic process ABCDA as shown in the figure.Assuming gas to be ideal, answer the following questions : What is the value of 'q' for the overall cyclic process:
Answer» `-100 Rln2` `+100Rln2` `+200Rln2` `-200Rln2` Solution :`Q=q_(AB)+q_(BC)+q_(CD)+q_(DA)` `=-1Rxx30ln2+1xx(5R)/2xx(400-300)+1Rxx400ln2+1xx(5R)/2xx(300-400)` (`because q_(AB)=-W_(AB)=-1Rxx300ln2` since process is reversible isothermal for which `DeltaU=0`) (`because q_(BC)=DeltaH_(BC)=1xx(5R)/2xx(400-300)` since process is reversible ISOBARIC ) (`because q_(CD)=-W_(CD)=1Rxx300ln2` since process is reversible isothermal for which `DeltaU=0` ) (`because q_(DA)=DeltaH_(AB)=1xx(5R)/2xx(300-400)` since process is reversible isobaric ) So, q=100 Rln2

Discussion

26.	One mole of Helium gas undergoes a reversible cyclic process ABCDA as shown in the figure.Assuming gas to be ideal, answer the following questions : What is the net work involved in the process A to C :
Answer» `-100 R(1-ln8)` `300Rln2` `-100 R(1+ln8)` 200 J Solution :NA

Discussion

27.	One mole of Helium gas undergoes a reversible cyclic process ABCDA as shown in the figure.Assuming gas to be ideal, answer the following questions : What is the value of DeltaH for the overall cyclic process:
Answer» `-100 Rln2` `+100Rln2` `+200Rln2` Zero Solution :SINCE, `DeltaH` is a STATE function, and the final state attained by the GAS is same as its INITIAL state, so value of `DeltaH=0`

Discussion

28.	One mole of fluorine is reacted with two moles of hot and connectrated KOH . The products formed are KF, H_(2)O and O_(2). The molar ratio of KF, H_(2)O and O_(2) respectively is
Answer» `1 : 1 : 2` `2 : 1 : 0.5` `1: 2 : 1` `2 : 1 : 2` Solution :`2Fe_(2)+4KOH to 4KF+O_(2)+2H_(2)O ` for 1 mole of `F_(2)` the molar RATIO. `{:(F_(2),KOH,KF,O_(2),H_(2)O),(1,2,2,(1)/(2),1):}`

Discussion

29.	One mole of glucose on respiration produces :
Answer» 36 MOLE of ATP 34 mole of ATP 40 mole of ATP 38 mole of ATP ANSWER :A

Discussion

30.	One mole of ethyl alcohol was treated with one mole of acetic acid at 25^(@)C. Two-third of the alcohol change into easter at equlibrium. The equlibrium constant for the reaction will be
Answer» 1 2 3 4 Solution :`C_(2)H_(2)OH+COOHtoCH_(3)COOC_(2)H_(5)+H_(2)O` `{:("Initial CONC.","1 mol","1 mol","0 MOLE","0 mole"),("At equlibrium", 1-2/3=1/3 "mol",1/3"mol",2/3"mol",2/3"mol"):}` `K_(C)=((2)/(3)xx(2)/(3))/(1/2xx1/3)=4`

Discussion

31.	One mole of ethyl alcohol was treated with one mole of acetic acid at 25^@C. 2/3of the acid changes into ester at equilibrium Calculate the equilibrium constant for the reaction :
Answer» 1 2 3 4 Answer :D

Discussion

32.	One mole of ethyl acetate on treatment with an excess of LiAIH_4 in dry ether and subsequent acidification produces
Answer» 1 mol ACETIC acid + 1 mol ETHYL alcohol 1 mol ethyl alcohol + 1 mol methyl alcohol 2 MOLES of ethyl alcohol 1 mol of 2 - BUTANOL. Answer :D

Discussion

33.	One mole of each PdCl_(2).4NH_(3), PtCl_(2).2NH_(3),PtCl_(4).2HCl and CoCl_(3).4NH_(3) gave 2, 0, 0 and 1 mole of AgCl, respectively when excess of AgNO_(3) solution is added. Assign the secondary valences to the metals in these complexes.
Answer» SOLUTION :4, 4, 6 and 6

Discussion

34.	One mole of each of the following alkenes is catalytically hydrogenated. The quantity of heat evolved will be the lowest in the case of
Answer» 1-butene trans-2-butene cis-2-butene 1,3-butadiene Answer :D

Discussion

35.	One mole of CrCl_(3).6H_(2)O compound reacts with excess AgNO_(3) solution to yield two moles of AgCl (s). The structural formula of the compound is
Answer» `[Cr(H_(2)O)_(5)Cl]Cl_(2).H_(2)O` `[Cr(H_(2)O)_(3)Cl_(3)].3H_(2)O` `[Cr(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O` `[Cr(H_(2)O)_(6)]Cl_(3)` Answer :A

Discussion

36.	One mole of Co(NH_(3))_(5)Cl_(3) gives 3 moles of ions on dissolution in water. One mole of this reacts with two moles of AgNO_(3) to give two moles of AgCl. The complex is :
Answer» `[Co(NH_(3))_(4)Cl_(2)]Cl.NH_(3)` `[Co(NH_(3))_(4)Cl]Cl_(2).NH_(3)` `[Co(NH_(3))_(5)Cl]Cl_(2)` `[Co(NH_(3))_(3)Cl_(3)].2NH_(3)` ANSWER :C

Discussion

37.	One mole of complex compound Co(NH_(3))_(5)Cl_(3) gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles ofAgNO_(3) solutionto yield two moles of AgCl(s). The structure of the complex is :
Answer» `[Co(NH_(3))_(3)Cl_(3)].2NH_(3)` `[Co(NH_(3))_(4)Cl_(2)]Cl.NH_(3)` `[Co(NH_(3))_(4)Cl]Cl_(2). NH_(3)` `[Co(NH_(3))_(5)Cl]Cl_(2)` ANSWER :D

Discussion

38.	One mole of CO_(2) gas at 27^(@) C is compressed suddenly so that its pressure becomes 8/27 of its original pressure. Final temperature of gas would be
Answer» `300 XX (2/3)^(0.33)` `300 xx (2/3)^(-1.33)` `300 xx (2/3)^(1/1.33)` `300 xx (3/2)^(1/1.33)` ANSWER :C

Discussion

39.	One mole of CO_2corresponds to :
Answer» 22.4 L at 1 ATM and `25^@C` 44G 1g of carbon dioxide. ` 6.02 xx 10^23`C-atoms and `6.02 xx 10^23`O atoms SOLUTION :44g

Discussion

40.	One mole of Cl_(2(g)) which may be assumed to obey the ideal gas law, initially at 300 K and 1.01325 xx 10^(7) Pa, is expanded against a constant exteranl pressure of 1.01325 xx 10^(5) Pa to a final pressure of 1.01325 xx 10^(6) Pa . As a result of the expansion , the gas cooled to a temperature of 239 K (which is the normal boiling point of Cl_(2)) and 0.100 mol of Cl_(2) condensed. The enthalpy of vaporization of Cl_(2(l)) is 20.42 KJ "mol"^(-1) at the normal volume is C_(v)=28.66 JK^(-1) "mol"^(-1) and the density of Cl_(2(l)) is 1.56 g cm^(-1)(at 239 K) . Assume that the molar heat capacity at constant pressure for Cl_(2(g)) is C_(p)=C_(v)+R. (1 atm = 101325 xx 10^(5) Pa , R=8.314510 Jk^(-1) "mol"^(-1) = 0.0820584 L atm K^(-1)"mol"^(-1)) (i) Either draw a complete molecular orbital energy diagram or write the complete electronic configuration of Cl_(2) Predict the bond order of Cl_(2) and thus whether this molecules will be diamagnetic , ferromagnetic ,or paramagnetic. (ii) For the changes decribed above, calculate the change in the internal energy (DeltaE) and the change in the entropy (DeltaS_("sys")) of the system.
Answer» SOLUTION :N//A

Discussion

41.	One mole of CO_2contains
Answer» ` 6.02 XX 10^23`atoms of C `6.02 xx 10^23`atoms of O `18.1 xx 10^23` molecules of `CO_2` 3 g atoms of `CO_2` Solution :1 mole of `CO_2= 6.02 xx 10^23` molecules of `CO_2` ` = 6.02 xx 10^23` atoms of C

Discussion

42.	One mole of calcium phosphide on reaction with excess water gives:
Answer» ONE mole of PHOSPHINE Two mole of phosphine acid Two moles of phosphine One mole of phosphorus pentoxide Answer :C

Discussion

43.	One mole of calcium phosphide on reactionwith excess of water gives
Answer» ONE mole of phosphine Two MOLES of phosphoric ACID Two moles of phosphine One mole of PHOSPHORUS pentoxide Solution :Two moles of phosphine

Discussion

44.	Onemole of calcium phosphideon reaction with excess of water gives
Answer» ONE mole ofphosphine two MOLES of phsophine two moles of phosphine one mole of phosphorus pentoxide Solution :`Ca_(3)P_(2)+6H_(2)Orarr3Ca(OH)_(2)+2PH_(3)`

Discussion

45.	One mole of As_(2)O_(5) on treatment with furning hydrochloric acid gives how much mole of Cl_(2)?
Answer» ANSWER :2

Discussion

46.	One mole of any substance contains 6.022xx10^(23) atoms/molecules. Number of molecules of H_(2)SO_(4) present in 100 mL of 0.02 M H_(2)SO_(4) solutions is …………….. .
Answer» `12.044xx10^(20)" molecules"` `6.022xx10^(23)" molecules"` `1xx10^(23)" molecules"` `12.044xx10^(23)" molecules."` SOLUTION :`"1000 mL of 1 M "H_(2)SO_(4)" contain = 1 mole of "H_(2)SO_(4)` `"100 mL of 0.02 M "H_(2)SO_(4)" contain "H_(2)SO_(4)=(1)/(1000)xx100xx0.02" mole"=0.002" mole"=2xx10^(-3)" mole"` `=2xx10^(-3)xx(6.022xx10^(23))" molecules"=12.044xx10^(20)" molecules."`

Discussion

47.	One mole of anhydrous salt AB dissolves in water and liberates 15 J mol^(-1) of heat .The value of Delta H_(hydration)^° of AB is -20.05 J mol^(-1) .Hence the enthalpy of dissolution of hydrated salt AB.3H_2O(s) is
Answer» `-5.5J` `MOL^(-1)` `5.5J` `mol^(-1)` `35.5J` `mol^(-1)` `-35.5J` `mol^(-1)` ANSWER :B

Discussion

48.	One mole of aniline warmed with the mixture of NaNO_(2)+HCl. If we assume 100% yield, volume of N_(2) gas liberated at S.T.P is
Answer» `11.2 L` `22.4 L` `33.6 L` `44.8 L` ANSWER :B

Discussion

49.	One mole of an unsaturated hydrocarbon on reductive hydrolysis gives one mole each of formaldehyde, acetaldehyde and methyl glyoxal. The hydrocarbon is
Answer» `CH_(2)=CH-CH=CH-CH_(3)` `CH_(3)-CH=CH-CH=CH-CH_(3)` `CH_(2)=C(CH_(3))-CH=CHCH_(3)` `CH_(2)=CH-C(CH_(3))=CHCH_(3)` Solution :

Discussion

50.	One mole of an organic compound requires 0.5 mole of oxygen to produce an acid.The compound may be:
Answer» Alcohol Ether Ketone Aldehyde Answer :D

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

One mole of nitrogen gas at 0.8 atm takes 38 second of diffuse through a pin hole whereas one mole of an unknown compound of Xenon with fluorine at 1.6 atm takes 56.26 second to diffuse through at same hole, then the molecular formula of the compound is (Atomic mass of Xenon : 131.3 u)

One mole of NaCl (s) on melting absorbed 30.5 KJ of heat and its entropy is increased by 28.8JK^(-1). What is the melting point of sodium chloride ?

One mole of NH_(3) is made to react with one mole of CH_(3) Cl, which of the following is/ are formed

One mole of NaCl (s) on melting absorbed 30.5 kJ of heat and its entropy is increased by 28.8 JK^(-1). The melting point of NaCl is

One mole of N_(2)O_(4) is heated in a flask with a volume of 10dm^(3). At equilibrium 1.708 mole of NO_(2) and 0.146 mole of N_(2)O_(4) were found at 134^(@)C. The equilibrium constnt will be

One mole of N_2O_4 (g) at 300k is kept in a closed container under one atmospheric pressure. It is heated to 600 k when 20 % by mass of N_2O_4 (g) decomposes to NO_2(g) . The resultant pressure is :

One mole of N_(2)and 3 moles of H_(2)are mixed in a litre flask. If 50% N_(2)is converted into ammonia by the reaction.N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)then the total number of moles of gas at equilibrium is :

One mole of monoatomic gas taken through a cyclic process as shown in figure. Calculate sum_(ABCA) = (dq_(rev))/(T).

One mole of monoatomc ideal gas expands adiabatically at initial temp. T against a constant external pressure of 1 atm from one litre to two litre. Find out the final temp. (R = 0.0821 litre. "atm"K^(-1)mol^(-1)):

One mole of methanol when burnt in O_(2), gives out 723 kJ mol^(-1) heat. If one mole of O_(2) is used, what will be the amount of heat evolved

One mole of methannol when burnt in oxygen gives out 723 kJ mol^(-1) heat. If one mole of oxygen is used, what will be the amount of heat evolved ?

One mole of magnesium in the vapour state absorbed 1200 kJ mol^(-1) of energy. If the first and second ionization energies of Mg are 750 and 1450 kJmol^(-1) respectively, the final composition of the mixture is

Onemole of magnesium nitride on reaction with excess of water gives

One mole of liquid A (vapour pressure =300 mm Hg ) and 2 moles of a liquid B (Vapour pressure =360 mm Hg) are mixed . The mole fraction of A in the vapour mixture if found to be 5//17 . Which of the following is correct ?

One mole of ideal monatomic gas is carried through the reversible cyclic process as shown in figure. Calculate the max temperature attained by the gas during the cycle.

One mole of idealmonatomicgas in takenin cycliceprocess ABCA as shownin the figure Calculate :(a) Theworkdone by thegas(b) The heatrejectedbythegas in the pathCA (c) The net heatabsorbedbe the bygas in the path BC (d)Maximum temperature obtainedby thegas duringthe cycle.

One mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27^(@)C. If the work done by the gas in the process is 3 kJ, the final temperature will be equal to (C_(v)=20J//Kmol)

One mole of ice is coverted into water at 273 K. The entropies of H_(2)O(s) and H_(2)O(l) are 38.20 and 60.01 J mol^(-1) K^(-1) respectively. The enthaply change for the conversion is

One mole of ice is converted into water at 273 K. The entropies of H_2O(s) and H_2O(l) are 38.20 and 60.01 J mol^-1 K^-1 respectively. The enthalpy change for the conversion is:

One mole of hydrogen peroxide (H_2O_2)has a mass same as that of

one mole of hydrogen iodide is heated in a closed container of 2 litre. At equilibrium half mole of hydrogen iodide has dissociated. What is the value of the equilibrium constant ?

One mole of Helium gas undergoes a reversible cyclic process ABCDA as shown in the figure.Assuming gas to be ideal, answer the following questions : What is the value of 'q' for the overall cyclic process:

One mole of Helium gas undergoes a reversible cyclic process ABCDA as shown in the figure.Assuming gas to be ideal, answer the following questions : What is the net work involved in the process A to C :

One mole of Helium gas undergoes a reversible cyclic process ABCDA as shown in the figure.Assuming gas to be ideal, answer the following questions : What is the value of DeltaH for the overall cyclic process:

One mole of fluorine is reacted with two moles of hot and connectrated KOH . The products formed are KF, H_(2)O and O_(2). The molar ratio of KF, H_(2)O and O_(2) respectively is

One mole of glucose on respiration produces :

One mole of ethyl alcohol was treated with one mole of acetic acid at 25^(@)C. Two-third of the alcohol change into easter at equlibrium. The equlibrium constant for the reaction will be

One mole of ethyl alcohol was treated with one mole of acetic acid at 25^@C. 2/3of the acid changes into ester at equilibrium Calculate the equilibrium constant for the reaction :

One mole of ethyl acetate on treatment with an excess of LiAIH_4 in dry ether and subsequent acidification produces

One mole of each PdCl_(2).4NH_(3), PtCl_(2).2NH_(3),PtCl_(4).2HCl and CoCl_(3).4NH_(3) gave 2, 0, 0 and 1 mole of AgCl, respectively when excess of AgNO_(3) solution is added. Assign the secondary valences to the metals in these complexes.

One mole of each of the following alkenes is catalytically hydrogenated. The quantity of heat evolved will be the lowest in the case of

One mole of CrCl_(3).6H_(2)O compound reacts with excess AgNO_(3) solution to yield two moles of AgCl (s). The structural formula of the compound is

One mole of Co(NH_(3))_(5)Cl_(3) gives 3 moles of ions on dissolution in water. One mole of this reacts with two moles of AgNO_(3) to give two moles of AgCl. The complex is :

One mole of complex compound Co(NH_(3))_(5)Cl_(3) gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles ofAgNO_(3) solutionto yield two moles of AgCl(s). The structure of the complex is :

One mole of CO_(2) gas at 27^(@) C is compressed suddenly so that its pressure becomes 8/27 of its original pressure. Final temperature of gas would be

One mole of CO_2corresponds to :

One mole of CO_2contains

One mole of calcium phosphide on reaction with excess water gives:

One mole of calcium phosphide on reactionwith excess of water gives

Onemole of calcium phosphideon reaction with excess of water gives

One mole of As_(2)O_(5) on treatment with furning hydrochloric acid gives how much mole of Cl_(2)?

One mole of any substance contains 6.022xx10^(23) atoms/molecules. Number of molecules of H_(2)SO_(4) present in 100 mL of 0.02 M H_(2)SO_(4) solutions is …………….. .

One mole of anhydrous salt AB dissolves in water and liberates 15 J mol^(-1) of heat .The value of Delta H_(hydration)^° of AB is -20.05 J mol^(-1) .Hence the enthalpy of dissolution of hydrated salt AB.3H_2O(s) is

One mole of aniline warmed with the mixture of NaNO_(2)+HCl. If we assume 100% yield, volume of N_(2) gas liberated at S.T.P is

One mole of an unsaturated hydrocarbon on reductive hydrolysis gives one mole each of formaldehyde, acetaldehyde and methyl glyoxal. The hydrocarbon is

One mole of an organic compound requires 0.5 mole of oxygen to produce an acid.The compound may be: