94420 + Interview Questions in Chemistry in Class 12 Page 1733 InterviewSolution

86601.	A volume of 50.7 m_(3) can be expressed in litres as
Answer» 50.7 LITRES 507 litres 5070 litres 50700 litres ANSWER :D

Discussion

86602.	A voltaic cell is set up at 25^(@)C with the helf cell, Al^(3+)(0.001M)and Ni^(2+)(0.50M) Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential. (Given E_(Ni^(2+)//NI)^(@)=-0.25V,E_(Al^(3+)//Al)^(@)=-1.66V)
Answer» Solution :`NI^(2+)+2e^(-)toNi,E_(Ni^(2+)//Ni)=0.25-(0.0591)/(2)"LOG"(1)/(0.50)=-0.259V` `Al^(3)+3e^(-)toAl,E_(Al^(3+)//Al)=-1.66-(0.0591)/(3)"log"(1)/(10^(-3))=-1.719V` For EMF to be +ve, oxidation should OCCUR on Al-electrode. Cell reaction: `2Al+3Ni^(2+)to2Al^(3+)+3Ni` `E_(cell)=-0.259+1.719V=1.46V`

Discussion

86603.	A volume of 12.53 ml of 0.05093 M SeO_(2) reacted with exactly 25.52 ml of 0.1 M CrSO_(4) . In the reaction, Cr^(2+) was oxidized to Cr^(3+). To what oxidation state was selenium converted by the reaction
Answer» ANSWER :ZERO

Discussion

86604.	A voltic cell is set up at 25^@C with the half cells Ag^(+) (0.001M) Ag and Cu^(2+)(0.10M) . What should be its cell potential.[Ecu=+0.34V,E∘Ag=+0.8V]
Answer» Solution :`Cu+2AG^(+) to Cu^+ +2Ag` Half cell REACTIONS: Cathade (reduction ) `2Ag^+ (0.001M)+2e^(-) to 2Ag(s) ` Anode (oxidation): `Cu_((s))to Cu^(2+)(0.10M)+2e^-` `therefore n=2` `E_(cell)^@=0.46V` `E_(cell)=E_(cell)^@-0.059/2log""([Cu^(2+)])/([Ag^+]^2)` `=0.46-0.059/2log""((0.1))/(0.001)^2` `=0.46-0.059/2log 10^5` `=0.46 -0.059/2 times (5)` `=0.46-0.1475` =0.3125V

Discussion

86605.	A voltaic cell stops working after some time, because the electrode potential of both the electrodes
Answer» increases decreases become equal connected by SALT bridge Answer :C

Discussion

86606.	A Voltaic cell is set up at 25^(@)C with the following half cells ?
Answer» SOLUTION :`K = (1)/(R) xx ((1)/(A))` `^^_(m) = (K xx 1000)/(M)` (`{:("at anode": , Al(s) to Al^(3+) (aq) + 3e^(-)"]" xx 2), ("at CATHODE" , Ni^(2+)(aq) + 2e^(-) to Ni(s) xx 3):}/(2AL(s) + 3Ni^(2+)(aq) to Al^(3+) (aq)+ Ni(s)))/` `E_("cell") = E_("cell")^(0) - (0.0591)/(n) "log" ([Al^(3+)]^(2))/([Ni^(2+)]^(3))` `n = 6 , [Al]^(3+) = 0.001 M = 1 xx 10^(-5) M` `[Ni^(2+)] = 0.5 M E^(@) Ni^(2+)//Ni - E^(@) Al^(3+)//Al = -0.25 V - (-1.66V)` `E_("cell")^(@) = 1.41 V` `E_("cell")^(@) = 1.41 - (0.0591)/(6) "log" ((10^(-3))^(2))/((0.5)^(3)) = 1.41 - (0.0591)/(6)"log" (10^(-6))/(0.125)` `= 1.41 - (0.0591)/(6) "log" (10^(-6) xx 8) = 1.41 - (0.0591)/(6) ("log" 10^(-6)+ "log" 2^(3))` `= 1.41 -(0.0591)/(6) (-6 "log" + 3 "log" 2)` `= 1.41 - (0.0591)/(6) (-6 + 3 xx 0.3010) = 1.41 - (0.0591)/(6) (-5.097)` `= 1.41 + (0.3012)/(6) = 1.46V`

Discussion

86607.	A voltaic cell can be changed into an electrolytic cell by
Answer» PROVIDING HIGHER potential from outside changing the electrolytes in TWO half cells reversing the electrodes changing the conc. Of the electrolytes Solution :Reversible cell condition.

Discussion

86608.	A volatilesolution AB is dissolved in water , in which mole fraction of water is 5//6 . If 50% of molecules of AB get dissociated into A and B and remaning 50% get dimerised into (AB)_(2). Then relative lowering of vapour pressure will
Answer» Increase by `20%` Increase by `25%` Remain same because mass of solute is unchanged Remain same because no. of solute particles will not CHANGE. Answer :B

Discussion

86609.	(A) Vitamins are important in diet. (R) Deficiency of vitamins causes diseases.
Answer» Both A & R are TRUE, R is the correct explanation of A Both A & R are true, R is not correct explanation of A . A is true, R is FALSE A is false, R is true Answer :A

Discussion

86610.	A vitamin which plays a vital role in the coagulating property of blood is :
Answer» VITAMIN A Vitamin D Vitamin E Vitamin K Answer :D

Discussion

86611.	A vitamin that contains both N and P is
Answer» <P>VITAMIN C vitamin K vitamin `B_(12)` vitamin D Solution : Vitamin `B_(12)` contains both N and P

Discussion

86612.	A vitamin that contains both N and P is:
Answer» <P>Vitamin C Vitamin K Vitamin `B_(12)` Vitamin D Solution :Vitamin `B_(12)` CONTAINS both N and P .

Discussion

86613.	(A) Vital Force Theory (i) Steam distillation (B) Wohler (ii) Rectified spirit (C) Azeotropic mixture (iii) Synthesis of urea (D) Kolbe (iv) Berzelius (E) Aniline (v) Synthesis of acetic acid
Answer» SOLUTION :NA

Discussion

86614.	(A) Visible light can not be used to study crystals (R) The wave length of visible light is much smaller than the atomic dimensions
Answer» Both (A) and (R) are true and (R) is the correct explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :C

Discussion

86615.	(A) Vitamin 'C' is ascorbic acid. (R) All acids are vitamins
Answer» Both A & R are true, R is the CORRECT EXPLANATION of A Both A & R are true, R is not correct explanation of A . A is true, R is FALSE A is false, R is true Answer :C

Discussion

86616.	A violet compound of manganese (P) decomposes on heating to liberate oxygen and compounds (Q) and (R ) of manganese are formed. Compound ( R) reacts with KOH in the presence of potassium nitrate to give compound (Q). On heating compound (R ) with conc. H_2SO_4 and NaCl, Chlorine gas is liberated and a compound (S) of manganese along with other products is formed. Compound P to S are
Answer» <P>`{:(" "P," "Q," "R," "S),(KMnO_4,K_2MnO_4,MnCl_2,MnO_2):}` `{:(" "P," "Q," "R," "S),(K_2MnO_4,MnO_2,KMnO_4,MnCl_2):}` `{:(" "P," "Q," "R," "S),(KMnO_4,K_2MnO_4,MnO_2,MnCl_2):}` `{:(" "P," "Q," "R," "S),(K_2MnO_4,KMnO_4,MnO_2,MnCl_2):}` Solution :`P=KMnO_4,Q=K_2MnO_4,R=MnO_4,S=MnCl_2` `UNDERSET((P))(2KMnO_4)OVERSET(DELTA)tounderset((Q))(K_2MnO_4)+underset((S))(MnO_2)+O_2` `2MnO_2+4KOH+O_2tounderset((S))(MnCl_2)+2NaHSO_4+2H_2O+Cl_2`

Discussion

86617.	A violet compound of manganese (A) decomposes on heating to liberate oxygen compounds (B) and (C ) of manganese are formed. Compound (C ) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C ) with conc. H_(2)SO_(4) and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.
Answer» SOLUTION :`A: KMnO_(4), B: K_(2)MnO_(4), C: MnO_(2), D: MnCl_(2)` `underset((A))(2KMnO_(4)) underset("decomposes")overset(Delta)rarr underset((B))(K_(2)MnO_(4))+ underset((C ))(MnO_(2)) + O_(2)` `underset(( C))(2MnO_(2)) + 4KOH + O_(2) overset(KNO_(3))rarr underset((B))(2K_(2) MnO_(4)) + 2H_(2)O` `underset((C ))(MnO_(2)) + 4NaCl + 4H_(2)SO_(4) overset(Delta)rarr underset((D))(MnCl_(2)) + 2NaHSO_(4) + 2H_(2)O + Cl_(2)`

Discussion

86618.	A violet compound of manganes (B). On heating compound (C) with conc. H_(2) SO_(4) and NaCl, C_(12) gas is liberated and compound (D) of manganese is formed. Identify A, B, C, D alongwith reactions involved.
Answer» SOLUTION :`A : KNnO_(3)B:K_(2)MnO_(4)C :MnO_(2)D:MnCI_(2)` `KMnO_(4)OVERSET(Delta )(to)K_(2) MnO_(4) +MnO_(2) + O_(2)` `MnO_(2) + KOH+ KNO_(3) toK_(2)MnO_(4)` `(c )"" ( B )` `MnO_(2)+ NaCI+ "conc"H_(2)SO_(4) toMnSI_(2)` `(C ) "" (D )`

Discussion

86619.	A violet compound of manganes (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of KNO_(3) to give compound
Answer» Solution :`A : KNnO_(3)B:K_(2)MnO_(4)C :MnO_(2)D:MnCI_(2)` `KMnO_(4)overset(Delta )(to)K_(2) MnO_(4) +MnO_(2) + O_(2)` `MnO_(2) + KOH+ KNO_(3) toK_(2)MnO_(4)` `(c )"" ( B )` `MnO_(2)+ NaCI+ "conc"H_(2)SO_(4) toMnSI_(2)` `(C ) "" (D )`

Discussion

86620.	A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H_(2)SO_(4) and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds (A) to (D) and also explain the reactions involved.
Answer» Solution :The compounds A, B, C and D are GIVEN as under: `A=KMnO_(4)""B=K_(2)MnO_(4)""C=MnO_(2)""D=MnCl_(2)` The reactions are EXPLAINED as under: `UNDERSET((A))(KMnO_(4))overset(Delta)to underset((B))(K_(2)MnO_(4))+underset((C))(MnO_(2))+O_(2)` `MnO_(2)+KOH+O_(2)to2K_(2)MnO_(4)+2H_(2)O` `MnO_(2)+4NaCl+4H_(2)SO_(4)to underset((D))(MnCl_(2))+2NaHSO_(4)+2H_(2)O+Cl_(2)`

Discussion

86621.	A violetcomopund of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C )of manganese are formed. Compounds (C ) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound(C ) with conc. H_(2) SO_(4) and NaCl, chlorine gas is liberated and a compound (D) of manganesealong with other products is formed. Identify compounds A to Dand also explain the reactions involved.
Answer» SOLUTION :`A=KMnO_(4) , B=K_(2)MnO_(4), C = MnO_(2), D = MnCl_(2)` Reaction `: underset((A))(2KMnO_(4))overset(Delta)(RARR)underset((B))(K_(2)MnO_(4))+underset((C ))(MnO_(2))+O_(2)` `2MnO_(2)+4KOH +O_(2)rarr2K_(2)MnO_(4)+2H_(2)O` `MnO_(2)+ 4NaCl+ 4H_(2)SO_(4) rarr underset((D))(MnCl_(2))+2NaHSO_(4)+2H_(2)O+ Cl_(2)`

Discussion

86622.	A violet colour is obtained on adding Cl_(2) water in solution of potassium halide in presence of chloroform and on adding excess of Cl_(2) water,violet colour disappears and colourless solution appears. The test shows the presence of
Answer» IODIDE ion BROMIDE ion Chloride ion Iodide and bromide ion Answer :A

Discussion

86623.	A vinyl polymer is a life saving substance as blood plasma. Write the empirical and structural formula of its monomer.
Answer» Solution :The polymer which SAVES life as blood plasma is POLYVINYL pyrrolidone. Its monomer is N-vinyl pyrrolidone. The empirical formula of the monomer is `C_(6)H_(9)NO`. The structural formula is

Discussion

86624.	A violet colour compound formed in detection of S in a compound is
Answer» `Na_(4)[FE(CN)_(5)NOS]` `Na_(3)[Fe(CN)_(5)NOS]` `Na_(2)[Fe(CN)_(5)NOS]` `Na_(2)[Fe(CN)_(5)NO]` SOLUTION :`UNDERSET(" Sodium nitroprusside")(Na_(2)S+Na_(2)[Fe(CN)_(5)CO])rarrunderset(" Violet COLOUR")(Na_(4)[Fe(CN)_(5)NOS])`

Discussion

86625.	A vicinal diol has two hydroxy group on :
Answer» Same CARBON atom Different carbon atoms Adjactent carbon atoms None Answer :C

Discussion

86626.	A vessel is filled with mixture of carbon dioxide and nitrogen. At what ratio of partial pressure mass of gases will be identical:
Answer» <P>`6.36` `0.636` `0.157` `1.57` Solution :`PV =NRT RARR PV =(W)/(m) RT rArr P_(CO_(2))V=(W)/(44) RT"…(1)"` `P_(N_(2))V=(W)/(28) RT"….(2)` `(P_(CO_(2)))/(P_(N_(2)))=(28)/(44)=(7)/(11)=0.636` Hence, (B) is the correct answer.

Discussion

86627.	A vessel is filled with a mixture of oxygen and nitrogen . At what ratio of partial pressure will the mass of gases be identical ?
Answer» <P>`p( O_(2)) = 0.5 p( N_(2))` `p ( O_(2)) = p ( N_(2))` `p ( O_(2)) = 1.14 p ( N_(2))` `p ( O_(2)) = 0.875p ( N_(2))` Solution :`p(O_(2))= ((m)/( 32))/((m)/(32)+ ( m)/( 28))P` `p(N_(2)) = ( ( m)/( 28))/( ( m)/( 32)+(m)/(28)) P` `( p ( O_(2)))/(p(N_(2)))=(28)/( 32)` or `p( O_(2)) = 0.875p ( N_(2))`

Discussion

86628.	A vessel has nitrogen gas and water vapours at a total pressure of 1 atm. The partial pressure of water vapours is 0.3 atm. The contents of this vessel are transferred to another vessel havig one third of the capacity of original volume, completely at the same temperature, the total pressure of the system in the new vessel is :
Answer» 3.0 atm 1 atm 3.33 atm 2.4 atm Answer :D

Discussion

86629.	A vessel contains three gases A, B and C in the equilibriumA iff 2B + CAt equilibrium, the concentration of A was 3 M and that of B was 4 M. On doubling the volume of the vessel, the new equilibrium concentration of B was 3 M. Calculate K, and the initial equilibrium concentration of C.
Answer» SOLUTION :28.8, 5.4

Discussion

86630.	A vessel contains O_(2)andH_(2) in 2 : 1 molar ratio at 10 atm pressure then calculate ratio of their rate of diffusion.
Answer» 1:4 1:8 1:2 1:6 Solution :Ratio of pressure of `O_(2)` and `H_(2)` = molar ratio = 2:1 `r_(O_2)/(rH_2)=(P_(O_(2)))/(P_(H_2))SQRT((M_(H_2))/(M_(O_(2))))` `2/1 sqrt(2/32)` `= 2/1 xx 1/4` `r_(O_(2))/(r_(H_(2)))=1/2` `r_(O_(2)):r_(H_(2))=1:2`

Discussion

86631.	A vessel contains dimethyl ether at a pressure of 0.4 atm. Dimethyl ether decomposes as CH_(3)OCH_(3(g))rarr CH_(4(g))+CO_((g))+H_(2(g)) The rate constant of decomposition is 4.78xx10^(-3) min^(-1). Calculate the ratio of initial rate of diffusion to rate of diffusion after 4.5 hour of initiation of decomposition. Assume the composition of gas present and gas diffused to be same.
Answer» SOLUTION :N//A

Discussion

86632.	A vessel contains 4.4 g of CO_(2). It means that it contains
Answer» 0.1 MOL of `CO_(2)` `6.02xx10^(22)` MOLECULES of `CO_(2)` 8.8 G atoms of oxygen 1120 ML of `CO_(2)` at S.T.P. Solution :`4.4g CO_(2)=(4.4)/(44)mol=0.1mol` `=0.1xx6.02xx10^(23)"molecules"` `=6.02xx10^(22)"molecules."`

Discussion

86633.	A vessel contains 1.6 g of dioxygen at sTP (273.15 K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate (i) volume of the new vessel. (ii) number of molecules of dioxygen.
Answer» SOLUTION :(i) `"1.6 g O"_(2)=1.6//32 mol="1.05 mol,1 mol of O"_(2)" at STP has volume of = 22.4 L"` `therefore"0.05 mol of O"_(2)" at STP has volume "=22.4xx0.05=1.12 L` (Remember. 1 mol of gas = 22.4 L if STP conditions are 273.15 K and atm `"= 22.7 L if STP conditions are 273.15 K and 1 bar")` `"Now,"V_(1)=1.12L,""P_(1)="1 atm,"V_(2)=?,""P_(2)=(1)/(2)" atm = 0.5 atm"` As temperature REMAINS constant, APPLYING Boyle's law, `P_(1)V_(1)=P_(2)V_(2) or V_(2)=(P_(1)V_(1))/(P_(2))=("1 atm"xx1.12L)/("0.5 atm")=2.24 L` (ii) Number of MOLECULES in 1.6 g `O_(2)`, i.e., 0.05 mol `O_(2)=0.05xx6.022xx10^(23)=3.011xx10^(22)`.

Discussion

86634.	A vessel at 1000 K contains CO_2with a pressure of 0.5 atm. Some of the CO_2isconverted to CO on addition of graphite. Calculate the value of K, if the total pressure at equilibrium is 0.8 atm.
Answer» <P> SOLUTION :For the eqb , `CO_2 (g) + C(s) iff 2CO(g)` `0.5 + p + 2p = 0.8` 1.8 atm

Discussion

86635.	A vessel at 1000 K contains CO_(2) with a pressure of 0.5 atm. Some of the CO_(2) is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :
Answer» 0.3 atm 0.18 atm 1.8 atm 3 atm Solution :`{:(CO_(2)(g)+C(s),hArr,2CO(g)),(0.5,,0),(0.5-x,,2x):}` Now, `0.5-x+2x=0.8`atm x = 0.3 atm `P_(CO)=2xx0.3=0.6`atm `P_(CO_(2))=0.5-0.3=0.2` atm `K=((p_(CO))^(2))/(p_(CO_(2)))=(0.6xx0.6)/(0.2)=1.8` atm

Discussion

86636.	A very small amount of radioactive isotope of .^(213)Pb was mixed with a non-radioactive lead salt containing 0.01 g of Pb (atomic mass 207). The whole lead was brought into solution and lead chromate was precipitated by addition of a soluble chromate. Evaporation of 10 cm^(3) of the supernature liquid gave a residue having a radioactivity 1/24000 of that of the original quantity of .^(213)Pb. If the solubility of lead chromate is x xx 10^(-y) mol dm^(-3), then value of x is
Answer» SOLUTION :Since the activity of solution is `(1)/(24000)` times to the original mixture, THEREFORE the fraction of the radioactive lead obtained after evaporation of the supernatant liquid will be `(1)/(24000)`. Because almost whole amount lead is precipitated in the form of `Pb CrO_(4)` an insoluble salt, `:.` Fraction of non-radioactive lead MAY also be taken `= (1)/(24000)` The mass of non-radioactive lead obtained from `100c.c. = (0.01)/(24000)` `:.` Moles of non-radioactive lead obtained from 1 L i.e., solution `= (0.01)/(207) xx (1)/(24000) xx (1000)/(10) = 2.0 xx 10^(-7) mol. dm^(-3)` `:. X = 2`

Discussion

86637.	A very large swimming pool filled with water of temperture equal to 20^(@)C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties. Is the process reversible ? (Y/N)
Answer» SOLUTION :N//A

Discussion

86638.	A very small amount of a nonvolatile solute (that does not dissociate) is dissolved in 56.8cm^(3) of benzene (density 0.889gcm^(-3)). At room temperature, vapour pressure of this solution is 98.88mmHg while that of benzene is 100mmHg. Find the molality of this solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene what is the value of molal freezing point depression constant of benzene ?
Answer» <P> SOLUTION :We have , `(p^(0)-p)/(p^(0))=(N)/(N)` `(100-98.88)/(100)=(n)/(0.6474)` `{N=(56.8xx0.889)/(78)=0.6474}` `n=0.00725` Molality `=(0.00725xx1000)/(56.8xx0.889)=0.1435m` Further, `K_(f)=(DeltaT_(f))/(m)=(0.73)/(0.1435)=5.087`

Discussion

86639.	A very large swimming pool filled with water of temperture equal to 20^(@)C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties. Is the change of entropy of the system positive , negative or zero? (i) DeltaS_("total") gt 0 "" (ii) DeltaS_("total")=0 "" (iii) DeltaS_("total") lt 0
Answer» SOLUTION :N//A

Discussion

86640.	A very large swimming pool filled with water of temperture equal to 20^(@)C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties. Is the change of entropy of the water positive, negative or zero? (i) DeltaS_("pool") gt 0 "" (ii) DeltaS_("pool")=0""(iii) DeltaS_("pool") lt 0
Answer» SOLUTION :N//A

Discussion

86641.	A very large swimming pool filled with water of temperture equal to 20^(@)C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties. Is the change of entropy of the resistor positive , negative or zero? (i) DeltaS_("rev") gt 0 "" (ii) DeltaS_("rev")=0 ""(iii) DeltaS_("rev") lt 0
Answer» SOLUTION :N//A

Discussion

86642.	A veritcal cylinder closed at both ends is divided into two parts by a frictionless piston, each part containing one mole of air. At temperature 300 K, the volume of the upper part is four times than that of the lower part. At what temperature will the volume of the upper part be three times than that of the lower part ?
Answer» Solution :At 300 K, the position of the piston in the cylinder o volume, say V, is represented as : Let `p_(1)` and `p_(2)` be the pressure at the upper and lower part of the cylinder respectively. Let the pressure at the lower part due to the weight of the piston of the cylinder be `p_(0)`. `therefore p_(2) = p_(1) + p_(0)`. ....(1) In the two PARTS of the cylinder, the no. of moles of air are same. As the temperature is also same (300 K), We have `p_(1) xx (4V)/(5) = p_(2) xx (V)/(5)`, or `4p_(1) = p_(2) = p_(1) + p_(0), p_(1) = (p_(0))/(3)` ......(2) Now let the temperature be T at which the volume of the upper part will be three times than that of the lower part. Let the PRESSURES at the uppr and lower parts be `p._(1)` and `p._(2)` respectively. Thus, `p._(2) = p._(1) + p_(0)` ........(3) Again in both the parts, temperature and no. of moles are the same, we have, `p._(1) xx (3V)/(4) = p._(2) xx (V)/(4)`. or `3p._(1) = p._(2) = p._(1) + p_(0)` or `p._(1) = (p_(0))/(2)`. .........(4) From (2) and (4), we have , `p._(1) = (3)/(2)p_(1)` ......(5) Now, for the upper part of the cylinder at temperature T, We have, `p._(1)xx (3V)/(4) = RT ""(n = 1)` Substituting `p._(1)` from (5), we have, `(3)/(2)p_(1) xx (3V)/(4) = RT` `(9)/(8) p_(1) V = RT` .......(6) Further, for the upper part of the cylinder at 300 K, We have, `p_(1) xx (4V)/(5) = R(300)` ......(7) From (6) and (7), we get , T = 421.9 K. Second method At 300 K, `P_(1)xx (4V)/(5) = R xx 300` and `P_(2) xx (V)/(5) = R xx 300` At a temperature T (say), `P._(1) xx (3V)/(4) = RT` and `P._(2)xx (V)/(4) = RT` Substituting `P_(1), P_(2), P._(1)` and `P._(2)` from the above equations in `P_(2) - P_(1) = P._(2) - P._(1)` we get, T = 421.9 K.

Discussion

86643.	A varibaleoppositeexternal potential E_("ext") is appliedto thecell Zn\|Zn^(2+)(1M)\|Cu^(2+)(1M)\|Cu of potential1.1 VwhenE_("ext")lt 1.1 V andE_("ext") gt 1.1 Vrespecitively electrons flowfrom
Answer» cathodeto ANODE in both cases cathode to anode and anode to cathode anode to cathode and cathode to anode anode to cathode in both cases SOLUTION :Forthe daniell CELL `E_("cell") = 0.34 -(-0.76) = 1.10 V` when`E_("ext") LT 1.10 V`electronsflowformcathode to anodein externalcircuit when `E_("ext")gt 1.10 V` electron flowform cathode to anodein externalcircuit (reverse reaction )

Discussion

86644.	Vapour pressure of benzene is 200 mm of Hg. When 2 gram of a non-volatilesolute dissolved in 78 gram benzene, benzene has vapour pressure of 195 mm of Hg. Calculate the molar mass of the solute. [Molar mass of benzene is 78 g/mol^(-1)]
Answer» Solution :(a) `M_(2) = (W_(2) xx M_(1) xx P_(1)^(0))/((P_(1)^(0) - P_(a)) xx W_(1))` `M_(2) = (2 xx 78 xx 200)/((200 - 195) xx 78)` `M_(2) = 80 g//mol` (b) Binary liquid mixtures having the same COMPOSITION in liquid and vapour phase and boil at a constant temperature. or Constant boiling point liquid mixture [Any suitable example for solution minimum boiling azeotrope.] benzene and ACETONE, n-Hexane and ethanol, water and ethanol, acetone and `CS_(2), C Cl_(4)` and `CHCl_(3), C Cl_(4)` and Toluene, Acetone and ethanol. (a) Given * Vapour pressure of pure benzene `(P_(0))` = 200 mm of Hg. * Mass of non-volatile SOLUTE `(W_(B) = 2g` * Mass of benzene as solvent `(W_(A)) = 78 g` * Molar mass of benzene `(M_(A)) = 78 g mol^(-1)`. * Vapour pressure of solution (P) = 195 mm of Hg `because` No. of moles (n) `= ("W(mass)")/("(M(Molar mass)")` `:.` Moles of benzene (n) `= (W)/(M) = (78)/(78) = 1` and Moles of non-volatile solute `(n_(B)) = (W_(B))/(M_(B)) = (2)/(M_(B))` To find : Value of `M_(B)`. also `(DELTA p)/(p^(0)) = (n_(B))/(n_(A))` `(p^(0) - p)/(p^(0)) = (n_(B))/(n_(A))` `(Delta p)/(p^(0)) = (200 - 195)/(200) = ((2)/(M_(B)))/((1)/(1))` `:. (5)/(200) = (2)/(M_(B))` or `M_(B) = (2 xx 200)/(5) = 80` Molar mass of solute `= 80 g mol^(-1)`. (b) Azotropes : it is a mixture of two liquids which has a constant boiling point and composition through out distillation. It is of two types : The minimum boiling azeotrope and the maximum boiling azeotrope. A solution that shows (+) ve DEVIATION from Raoult.s law is called minimum boiling zeotrope while a solution that shows (-) ve deviation from Raoult.s law is called maximum boiling azeotrope. Mixture of ethanol and water, in which ethanol is approximately 95% by volume is an example of minimum boiling azeotrope.

Discussion

86645.	(A) Vapour density of sulphur vapour relative in oxygen is 2 because sulphur atom is twice as heavy as that of oxygen atom. (R) Vapour density depends upon the molecular state of the substance in vapour state.
Answer» If both (A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A). If both (A) and (R) are correct but (R) is not the correct explanation of (A). If (A) is correct but (R) is WRONG. If (A) is wrong but (R) is correct. Answer :D

Discussion

86646.	A vander Waals' gas may behave ideally when
Answer» the volume is very low the TEMPERATURE is very high the pressure is very low the temperature, pressure and volume all are very high. Solution :A vander Waals GAS MAY behave ideally when pressure is very low as compressibility factor (Z) APPROACHES 1. At high temperature `Zgt1`.

Discussion

86647.	(A) Van Arkel method is a vapourphase refining process (R) Vapoursof nickel are condensed in van Arkel method
Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanationof (A) (A) is true but (R) is FALSE Both (A) and (R) are false ANSWER :C

Discussion

86648.	(A) Valency of xenon in second excited state is two (R) Xenon forms XeF_2in its second excited electronic state
Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :D

Discussion

86649.	A useful derivative to identify carboxylic acids is :
Answer» Osazone Amide 2,-4 DIPHENYL hydrozone Oxime. Answer :B

Discussion

86650.	A: Urethane on reaction with anmonia gives urea.R: Formula of urethane is NH_(2)-overset(O)overset(\|\|)C-NH-OCH_(2)CH_(3)
Answer» If both Assertion & Reason are true and the reason is the CORRECT explanation of the assertion, then mark (1). If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2) If Assertion is true STATEMENT but Reason is FALSE, then mark (3) If both Assertion and Reason are false statements, then mark (4) Answer :3

Discussion

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