52671 + Interview Questions in Chemistry in Class 11

1.	At the suggestion of Earnest Rutherford, hans Geiger and ernest Marsden bombarded a thin gold foil by alpha-particles from a polonium source. It was expected that alpha-particles would go right through the foil with hardly any deflection. Although, most of the alpha particles indeed were not deviated by much, a few were scattered through veryi large angles. Some were even scattered in the backward direction. The nly way to explain the results, rutherford found, was to picture an atom as being compoed of a tiny nucleus in which its positive charge and nearly all its mass are concentrated. Scattering of alpha-particles is proportional to target thickness and is inversely proportional to the fourth power of sin((theta)/(2)), where, theta is scattering angle. Distance of closest approach may be calculated as: r_("min")=(Z_(1)Z_(2)e^(2))/(4piepsi_(0)K) where, K=kinetic energy of alpha-particles. Q. Alpha particles that come closer to the nuclei:
Answer» are DEFLECTED more are deflected LESS MAKE more collision are SLOWED down more. Answer :A

Discussion

2.	At the suggestion of Earnest Rutherford, hans Geiger and ernest Marsden bombarded a thin gold foil by alpha-particles from a polonium source. It was expected that alpha-particles would go right through the foil with hardly any deflection. Although, most of the alpha particles indeed were not deviated by much, a few were scattered through veryi large angles. Some were even scattered in the backward direction. The nly way to explain the results, rutherford found, was to picture an atom as being compoed of a tiny nucleus in which its positive charge and nearly all its mass are concentrated. Scattering of alpha-particles is proportional to target thickness and is inversely proportional to the fourth power of sin((theta)/(2)), where, theta is scattering angle. Distance of closest approach may be calculated as: r_("min")=(Z_(1)Z_(2)e^(2))/(4piepsi_(0)K) where, K=kinetic energy of alpha-particles. Q. Rutherford's scattering formula fails for vary small scattering angles because:
Answer» The gold foil is very thin the kinetic energy of `alpha`-PARTICLE is very high The full nuclear charge of the TARGET atom is PARTIALLY screened by its electron There is strong repulsive force between the `alpha`-particles and NUCLEUS of the target. Answer :C::D

Discussion

3.	At the suggestion of Earnest Rutherford, hans Geiger and ernest Marsden bombarded a thin gold foil by alpha-particles from a polonium source. It was expected that alpha-particles would go right through the foil with hardly any deflection. Although, most of the alpha particles indeed were not deviated by much, a few were scattered through veryi large angles. Some were even scattered in the backward direction. The nly way to explain the results, rutherford found, was to picture an atom as being compoed of a tiny nucleus in which its positive charge and nearly all its mass are concentrated. Scattering of alpha-particles is proportional to target thickness and is inversely proportional to the fourth power of sin((theta)/(2)), where, theta is scattering angle. Distance of closest approach may be calculated as: r_("min")=(Z_(1)Z_(2)e^(2))/(4piepsi_(0)K) where, K=kinetic energy of alpha-particles. Q. From the alpha-particle scattering experiemnt, rutherfod concluded that:
Answer» `alpha`-particle can approach within a DISTANCE of the order of `10^(-14)m` of the NUCLEUS. the radius of the nucleus is less than `10^(-14)`m scattering follows COULOMB's law The positively charged parts of the atom move with extremely HIGH velocities. Answer :A::B::C

Discussion

4.	At the sublimation temperature , for the process CO_(2)(s) rarr CO_(2)(g),
Answer» `DeltaH , DeltaS` and`DeltaG` are all POSITIVE `DeltaHgt 0,DeltaSgt 0` and `DeltaG LT 0` `DeltaH lt 0, DeltaS gt 0` and `DeltaG lt 0` `DeltaH GT0, DeltaSgt 0` and`DeltaG = 0` Solution :As the PROCESS is in equilibrium,`DeltaG =0`. But `DeltaG=0`only when the TWO factors `DeltaH`and `T DeltaS` oppose each other and equal in magnitude `( DeltaG =DeltaH - T DeltaS)`. Thus, we should have either `DeltaH` and `DeltaS`both `+ve` or both -ve, i.e., `DeltaH gt 0` and`DeltaS gt 0` or `DeltaH lt 0` and `DeltaS lt0`

Discussion

5.	At the same temperature carbon monoxide molecules have the same most probable velocity as the molecules of
Answer» Nitrogen DIOXIDE Nitrogen NITROUS Oxide Oxygen Solution :Same VELOCITY = same M.

Discussion

6.	At the same temperature and pressure, the ratio of the masses having equal volumes of NH_3 and H_2S gas is
Answer» `1:1` `2:1` `4:3` `1:2` ANSWER :D

Discussion

7.	At the same T and P, which of the following gases will have the highest average kinetic energy per mole? (at. Wt. H=12,O=16,8=32,F=19)
Answer» `H_(2)` `O_(2)` `CH_(4)` `SF_(6)` ANSWER :E

Discussion

8.	At the same condition of temperature, pressure and volume the ration mass of O_(2), O_(3)and SO_(2)is ..........
Answer» `2:1.3:1` `2:3:4` `4:2:1` `1:2:2` SOLUTION :`(w_(1))/(M_(1)):(w_(2))/(M_(2)):(w_(3))/(M_(3))` `(w_(1))/(32) : (w_(2))/(48) : (w_(3))/(64)` `(64)/(32) : (64)/(48) : (64)/(64) = 2 : 1.3:1`

Discussion

9.	At the one bar pressure the volume of gas is 0.6 litre. If the gas receives 122 Joule of heat at 1 atmosphere pressure, the volume become . 2 liter the calculate its internal energy. (1 litre bar = 101.32 Joule)
Answer» SOLUTION :`-19.85` JOULE

Discussion

10.	At the equilibrium of the reaction ,2X (g) + Y (g) to X_(2)Y (g)" the number of moles of " X_(2)Yat equilibrium is affected by the
Answer» temperature and pressure temperature only pressure only temperature , pressure and catalyst used Solution :The given reaction is ccompanied by decrease of randomness, i.e., randompanied factor MUST favour opposes the process . Hence , the energy factor must favour , u.e ` Delta H = -ve` which MEANS the reaction is exothermic . Alternatively , backward is dissociation reaction and hence endothermic. Also `n_(p) != n_(R) ` Hence , both temperature and pressure will affect the equilibrium .

Discussion

11.	At the critical micelle concentration, the surfactant molecules
Answer» decompose Dissoiciate Associate Become completely soluble Solution :At critical micellization CONCENTRATION (CMC), the SURFACTANT molecules associate to form micelles.for SOAP, CMC is `10^(-3)` mol/litre.

Discussion

12.	At T kelvin and a pressure of a P atm, certain gas is present in a vessel. If the vessel is divided into two equal compartments by a partition, the pressure in each compartment is equal to
Answer» <P>4P ATM P atm P/4 atm 2 P atm Answer :B

Discussion

13.	At STP, the volume of hydrogen is 22.72L mol^(-1) . Calculate the volume occupied by 10 gram of the same gas under similar conditions .
Answer» Solution :GMW of HYDROGEN = 2.016 G 20.16 g of hydrogen = 1 MOLE of hydrogen 10 g of hydrogen = ? `= (10)/(2.016) = 4.96` mole of hydrogen Volume occupied by 4.96 moles (10 g) of hydrogen at STP = `4.96 XX 22.72 = 112.7 L`.

Discussion

14.	At STP the order of root mean square velocity of molecules of H_2, N_2, O_2 and HBr is
Answer» `N_2 GT HBR gt O_2 gt H_2` `H_2 gt N_2gt O_2gt HBr` `HBr gt H_2gt O_2 gt N_2` `H_2 gt O_2 gtHBr gt N_2` ANSWER :B

Discussion

15.	At S.T.P. a mixture of 280 mL of CH_4 " and 140 mL of " H_2 iscompletely burnt. Calculate the required volume of oxygen and weight of water formed, assuming that whole of the steam condenses to water.
Answer» SOLUTION :For the combustion of methane : `{:(CH_4+2O_2to+2H_2O),(1 " mole " 2 " MOLES "2xx18=36g),(1"volume" 2" VOLUMES"):}` `:.1 "volume of " CH_4` requires for combustion, `O_2 = 2` volumes `:.280 " mL of "CH_4` will require for combustion, `O_2 = 2 xx 280 = 560 mL` Number of moles of `CH_4` in 280 mL at S.T.P. `=280/22400 =0.0125` `:. 1 " mole of " CH_4` gives on combustion, water = 36 G `:.0.0125 " mole of " CH_4` will give on combustion, water = `36/1 xx0.0125 = 0.45g` For the combustion of `H_2`: `{:(2H_2+O_2to2H_2O),(1 " mole " 2 " moles " 36g),(2 "volume " 1"volumes"):}` `:.2 " volumes of " H_2` require for combustion, oxygen = 1 volume `:. 140 mL H_2`will require for combustion, oxygen `=1/2 xx140 = 70 mL` Number of moles of `H_2` in 140 mL at S.T.P. `=140/22400 = 6.25 xx 10^(-3)` `:. 2 " moles of " H_2` produce water = 36 g `:. 6.25 xx 10^(-3) " moles of " H_2` will produce water `36/2xx6.25 xx 10^(-3)` `= 0.113 g` Hence, total volume of `O_2 " required " = 560 + 70 = 630 mL` and total mass of `H_2O " formed " = 0.45 +0.113 = 0.56 g`

Discussion

16.	At STP, the order of mean square velocity of molecuels of H_(2), N_(2),O_(2) and HBr is:
Answer» `H_(2)gtN_(2)gtO_(2)gtHBr` `HBrgtO_(2)gtN_(2)gtH_(2)` `HBrgtH_(2)gtO_(2)gtN_(2)` `N_(2)gtO_(2)gtH_(2)gtHBr` ANSWER :a

Discussion

17.	If 0.02 g of a volatile compound on heatingdisplaces 11.2 ml of dry air at STP, the molecular weight of the compound is
Answer» SOLUTION :112 CC at STP=0.3g 22400 cc at STP =? Molecular WEIGHT `=(22400xx0.3)/(112)=60`

Discussion

18.	At some temperature and under a pressure of 4 atm, PCl_(5) is 10% dissociated . Calculate the pressure at which PCl_(5) will be 20% dissociated , temperature remaining same.
Answer» Solution :`{:(,PCl_(5) ,hArr,PCl_(3),+,Cl_(2)), (" Atm eqm. " ,1-01=09"mole",,01 "mole",,01"mole"):}` Total no. of MOLES ` = 09+ 01 + 01= 1 1"mole " ` ` p_(PCl_(5))= (09)/(1 1) xx 4"atm " , p_(PCl_(3))= (01)/(11)xx 4 " atm " , p_(Cl_(2)) = (01)/(1.1) xx 4 " atm " ` ` :. K_(p) = (p_(PCl_(3))xx p_(Cl_(2)))/ (p_(PCl_(5))) = ((04)/(11) xx (04)/(11))/((09 xx 4)/(11) )=00404 ` 2nd caseWhen `PCl_(5)` is20% dissociated . Suppose total pressure= P atm . Then `{: (,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("INTIAL " ,1 "mole",,,,),("At equilibrium " ,1-02=08,,02,,02 "mole"),(,,,,," Total no. of moles"= 08 + 02 + 02 = 12 "moles"):}` ` p_(PCl_(5)) = (08)/(12) xxP atm , p_(PCl_(3)) = (02)/(12) xx P atm , p_(Cl_(2))= (02)/(12) xx P atm ` ` K_(p)= ((02 P)/(12) xx(02)/(12))/(( 08P)/(12))= (02)/(12) xx (02)/(08) P = 00404("calculated above")` which GIVES `P= 0 97 ` atm

Discussion

19.	At some temperature and under a pressure of 4 atm, PCI_(5) is 10% dissociated. Calculate the pressure at which PCI_(5) will be 20% dissociated, temperature remaining same
Answer» <P>9.7 atm 0.97 atm 97 atm 7.9 atm Solution :`UNDERSET(1-alpha)underset(1)(PCl_(5(g))) harr underset(alpha)underset(0)(PCl_(3(g)))+underset(alpha)underset(0)(Cl_(2(g)))` `KP=(alpha^(2))/((1-alpha))[(p)/(1+alpha)]^(1)` Since, Kp remains constant and therefore EQUATING for `alpha=0.1` at p=4atm and `alpha=0.2` at p=p atm `((0.1)^(2))/(0.9) xx (4)/(1.1)=((0.2)^(2) xx p)/(0.8 xx 1.2) implies p=0.97` atm

Discussion

20.	At standard pressure and temperature conditions the density of a gas in g.lit^(-1), whose molecular weight is 45
Answer» 2 22.4 11. 2 1000 ANSWER :A

Discussion

21.	At some high temperature, K_(w) of water is 10^(-13) Then the P^(H) of the water at the same temperature is
Answer» `7.0` `6.5` `7.5` `7.23` Solution :` [H^(+)] [OH^(-) ] =10 ^(-13), [ H^(+) ] =10 ^(-6.5)rArr pH = 6.5 `

Discussion

22.	At same temperature , which pair of the following solutions are Isotonic?
Answer» `0.2 M BaCl _(2) and 0.2 M ` ures `0.1 M ` GLUCOSE and `0.2 ` M UREA `0.1 M NaCl and 0.1 M K _(2) SO _(4)` `0.1 M BA (NO _(3)) _(2) and 0.1 M Na _(2) SO _(4)` Solution :`0.1 xx 3 ` ION `[ Ba ^(2+) 2NO _(3) ^(-) ] 0.1 3` ion `[ 2Na^(+) , SO _(4) ^(-)]`

Discussion

23.	At same temperature, which pair of the following solutions are isotonic?
Answer» 0.2 M BaCl, and 0.2M UREA 0.1 M glucose and 0.2 M urea 0.1 M NACL and 0.1 `MK_2SO_4` `0.1 M Ba (NO_3)_2 and 0.1 M Na_2 SO_4` Solution :`0.1xx3` ion `[Ba^(2+), 2NO_(3)^(-)]`, `0.1xx3` ion `[2NA^(+) , SO_(4)^(-)]`

Discussion

24.	At same temperature and pressure 10cc of an organic compound in the gaseous state were sparked with an excess of O_(2). 20cc of CO_(2) and 5cc of N_(2) were obtained among the products. Which of the following molecular formulas would fit these data-
Answer» `C_(2)H_(7)N` `C_(2)H_(3)N` `C_(2)H_(6)N_(2)` `CH_(5)N` Answer :A::B

Discussion

25.	At room temperature , the eclipsed and staggered forms of ethane cannot be isolated because
Answer» they interconvert rapidly both the conformers are EQUALLY STABLE the energy difference between the conformers is large there is a large energy barrier of ROTATION about the `sigma`-bond Answer :A

Discussion

26.	At room temperature, the density of water is 1.0 g/mL and the density of ethanol is 0.789 g/mL. What volume of ethanol contains the same number of molecules as are present in 175 mL of water?
Answer» Solution :LET the volume of ethanol CONTAINING the same number of molecules as are present in 175 mL of `H_2O` be v mL. As given, Now, `("WT. of "C_(2)H_(5)OH)/("mol. Wt of" C_(2)H_(5)OH) = ("wt of" H_(2)O)/("mol. Wt of" H_(2)O)` or `(0.789 XX v)/46 = (1.0 xx 175)/18` `THEREFORE v= 566.82` mL

Discussion

27.	At room temperature, Polonium cryatsillises in Cubic primitive cell. If edge length is 3.0Å, calculate the theoretical density of Po. (Atomic wt of Po = 207g)
Answer» 25/3 AMU/`Å^(3)` 23/3 amu/`Å^(3)` 21/3 amu/`Å^(3)` 27/3 amu/`Å^(3)` ANSWER :B

Discussion

28.	At room temperature, Polonium (atomic weight 209 gm mol^(-1)) crystallises in a primitive cubic unit cell. If a = 3.36 A^(0), calculate the theoritical density of Polonium..
Answer» Solution : A primitive cubic UNIT cell contains atoms only at the 8 corners with each corner contributing`1//8^(TH)` of an atom . HENCE n = `8 xx (1//8) = 1` . VOLUME `V = a^(3) = (3.36 Å)^(3)` From Eq. 1 `rho = (n M m)/(N_(0) V)` `= ((1) 209 G mol^(-1)))/((6.022 xx 10^(23) mol^(-1)) (3.36 xx 10^(-8) cm)^(3))` `= 9.15 g cm^(-3)`

Discussion

29.	At room temperature, pollonium crystallizes in a primitive cubic unit cell. If a = 3.36 Å, calculate the theoretical density of pollonium, its atomic mass is 209 g mol^(-1).
Answer» Solution : A PRIMITIVE CUBIC unit cell contains atoms only at the 8 corners with each corner contributing`1//8^(th)` of an ATOM . Hence n = `8 xx (1//8) = 1` . VOLUME `V = a^(3) = (3.36 Å)^(3)` From Eq. 1 `rho = (n M m)/(N_(0) V)` `= ((1) 209 G mol^(-1)))/((6.022 xx 10^(23) mol^(-1)) (3.36 xx 10^(-8) cm)^(3))` `= 9.15 g cm^(-3)`

Discussion

30.	At room temperature normal hydrogen consists of……….
Answer» 25% ortho form + 75% PARA form 50% ortho form + 50% para form 75% ortho form + 25% para form 60% ortho form + 40% para form Answer :c

Discussion

31.	At room temperature normal hydrogen consists of ...........
Answer» 25% ORTHO form + 75% para form 50% ortho form + 50% para form 75% ortho form + 25% para form 60% ortho form + 40% para form Answer :C

Discussion

32.	At room temperature, Hydrogen reacts very Slowly. Explain (HOTS)
Answer» Solution :In elementary state, HYDROGEN exist as a diatomic molecule. The bond between two hydrogen ATOMS H-H is covalent. The bond dissociation energy is very HIGH `(435.9 Kjmol^(-1))`. So bond CLEAVAGE is extremely DIFFICULT, so hydrogen is less reactive at room temperature.

Discussion

33.	At room temperature among the following intermolecular forces are strongest in
Answer» `H_O` `H_2S` `H_2Se` `H_2 TE` SOLUTION :DUE to H- BONDING.

Discussion

34.	At room temperature, ammonia gas at 1 atm pressure and hydrogen chloride gas at P atm pressure are allowed to effuse through identical pin holes from opposite ends of a glass tube of one metre length and of uniform cross section. Ammonium chloride is first formed at a distance of 60 cm from the end through which HCl gas is sent in. What is the value of P?
Answer» Solution :In the present case, For ammonia, `P_1 = 1 " ATM, " M_1 =17` for HCl gas, `P_2 = P = ?, M_2 = 36.5`. Ift is the TIME taken by the TWO gases to meet together and to form ammonium chloride, rate of diffusion of `NH_3 (r_1) = (100 - 60)/t =40/t` ( `:. " The distance travelled by " NH_3 " in the tube " = 100 - 60 cm)` and rate of diffusion of `HCI (r_2) =60/t` For two gases at different pressures, Graham.s LAW can be expressed as `r_1/r_1 = sqrt(M_2/M_1)* P_1/P_2` We have `(49//t)/(60//t)=sqrt((36.5)/17) xx1/P` `:. ""P=2.198` atm

Discussion

35.	At point P and Q , the real gas deviation with respect to ideal gas is respectively :
Answer» POSITIVE, NEGATIVE Positive, positive Negative, positive Negative, negative Answer :A

Discussion

36.	At P_(C ) pressure, 13.1^(@)C, 31.5^(@)C, 50^(@)C, and 30.98^(@)C. What phase obtained by CO_(2) gas ?
Answer» Solution :`{:("At",,"Phase of "CO_(2)),(30.98^(@)C,to,"Equilibrium between"),(13.1^(@)C,to,"LIQUID"),(31.5^(@)C,to,"Gas"),(50^(@)C,to,"Gas"):}` if temperature is more than `T_(C )` then HIGHER pressure apply gas can.t convert into liquid.

Discussion

37.	At particular temperatureK_C= 4xx10^(-2) for the reaction, calculate K_C for each of the follow reaction. (i) 2H_2 S_((g)) hArr 2H_(2(g))+ S_(2(g))"" (ii) 3H_2 S_((g)) hArr 3H_(2(g)) + 3//2 S_(2(g))
Answer» SOLUTION :`K_C = 4XX10^(-2) ` for the reaction, `i. H_2 S(g) HARR H_2(g) + 1/2 S_2(g)` `K_C =([H_2] [S_2]^1/2)/([H_2S]) = 4xx10^(-2)` For the reaction, `2H_2 S (g) hArr 2H_2 (g) + S_2(g)` `K_C = ([H_2]^2 [S_2])/([H_2S]^2) = (4xx10^(-2))^2` `=16xx10^(-4)` II. For the reaction, `3H_2 S(g) hArr 3H_2 (g) + 3/2 S_2 (g)` `K_C = ([H_2]^3 [S_2]^(3/2))/([H_2S]^3)=(4xx10^(-2))^3` `= 64xx10^(-6)`

Discussion

38.	At particular temperature K_(C)=4xx10^(-2) for the reaction H_(2)S(g)hArrH_(2)(g)+1//2S_(2)(g) Calculate K_(C) for each of the following reaction. i) 2H_(2)S(g)hArr2H_(2)(g)+S_(2)(g) ii) 3H_(2)(g)hArr3H_(2)(g)+3/2S_(2)(g)
Answer» SOLUTION :`K_(C)=4XX10^(-2)` for the RACTION, `H_(2)S(g)hArrH_(2)(g)+1/2S_(2)(g)` `K_(C)=([H_(2)][S_(2)]^(1//2))/([H_(2)S])` `rArr4xx10^(-2)=([H_(2)][S_(2)]^(1//2))/([H_(2)S])` For the reaction, `2H_(2)S(g)hArr2H_(2)(g)+S_(2)(g)` `K_(C)=([H_(2)]^(2)[S_(2)])/([H_(2)S]^(2))=(4xx10^(-2))^(2)=16xx10^(-4)` For the reaction, `3H_(2)S(g)hArr3H_(2)(g)+3/2S_(2)(g)` `K_(C)=([H_(2)]^(3)[S_(2)]^(3//2))/([H_(2)S]^(3))=(4xx10^(-2))^(3)=64xx10^(-6)`

Discussion

39.	At particular temperature K_C= 4 xx 10^(-2) for the reaction H_2S(g) hArr H_2(g)+ 1/2 S_2(g) Calculate K_C for each of the following reaction (i) 2H_2S (g) hArr 2H_2(s)+ S_2(g) (ii) 3H_2S (g) hArr 3H_2(g)+ 3/2S_2(g)
Answer» SOLUTION :`K_C = 4 xx 10^(-2)` for the REACTION, (i) `H_2S(G) HARR H_2(g)+ 1/2 S_2(g)` `K_C = ([H_2][S_2]^(1/2))/([H_2S]^2)= 4xx10^(-2)` For the reaction, `2H_S(g)hArr2H_2(g)+S_2(g)` `K_C=([H_2]^2[S_2])/([H_2S]^2)=(4xx10^(-2))^2=16xx10^(-4)` For the reaction, `3H_2S(g)hArr3H_2(g)+3/2S_2(g)` `K_C=([H_2]^3[S_2]^(3/2))/([H_2S]^3)=(4xx10^(-2))^(3)=64xx10^(-6)`

Discussion

40.	At one time H_(2)O_(2) was obtained by electrolysis of 50% H_(2)SO_(4) The process of electrolysis involves following reaction : 2H_(2)SO_(4) rarr 2H^(+)+2HSO_(4)^(-) At cathode: 2H^(+)+2e^(-) rarr H_(2)uarr At anode: 2HSO_(4)^(-)rarr X+2e^(-) X+H_(2)O rarr Y+Z, Y+H_(2)Orarr Z+H_(2)O_(2) The number of -OH groups in X:
Answer» 3 2 4 zero Answer :B

Discussion

41.	At one time H_(2)O_(2) was obtained by electrolysis of 50% H_(2)SO_(4) The process of electrolysis involves following reaction : 2H_(2)SO_(4) rarr 2H^(+)+2HSO_(4)^(-) At cathode: 2H^(+)+2e^(-) rarr H_(2)uarr At anode: 2HSO_(4)^(-)rarr X+2e^(-) X+H_(2)O rarr Y+Z, Y+H_(2)Orarr Z+H_(2)O_(2) Among X, Y and Z which is an oxidising agent ?
Answer» only X Only Y X,Y and Z only Z Answer :C

Discussion

42.	At one time, hand pump water was considered to be pure and used freely for drinking villages but not now . Why ?
Answer» Solution :Hand pump water is GROUND water. It collects below the surface of the earth after passing through the pores of the earthy materials which ACT as a filter for it and is pure . Now due to DISPOSAL of domestic wastes and industiral effuents and use of fertilizers and pesticides, a NUMBER of harmful soluble substances dissolve into rain water and pass through the soil and enter into ground water resulting in POLLUTION.

Discussion

43.	At one time H_(2)O_(2) was obtained by electrolysis of 50% H_(2)SO_(4) The process of electrolysis involves following reaction : 2H_(2)SO_(4) rarr 2H^(+)+2HSO_(4)^(-) At cathode: 2H^(+)+2e^(-) rarr H_(2)uarr At anode: 2HSO_(4)^(-)rarr X+2e^(-) X+H_(2)O rarr Y+Z, Y+H_(2)Orarr Z+H_(2)O_(2) Which of the following statements are correct with respect to X,Y and Z ? (i) In all compounds the covalency of Sulphur is 6 (ii) Peroxy bond is present in both Y and Z (iii)Basicity of all acids is 2 In X there is no S-S linkage
Answer» II,iv ii, III, iv I,ii, iv I,iii,iv Answer :D

Discussion

44.	At one litre vessel contains oxygen gas saturated with water vapour at 760 torr and 25^(@)C. At constant temperature, if the gaseous mixture transferred into 500 ml vessel, find out the total pressure in the new vessel. (aqueous tension at 25^(@)C = 23.8 torr)
Answer» Solution :In ONE litre VESSEL `P_(O_(2)) + P_(H_(2)O) = 750` torr `P_(H_(2)O) = 23.8` torr , `P_(O_(2)) = 760-23.8 = 736.2` torr When gaseous MIXURE transferred into 500 ml flask, partial pressure of OXYGEN is doubled but aqueous tension remains constantbecause some WATER vapour condenses to water Pressure in 500 ml flask ` = P_(O_(2)) + P_(H_(2)O) = 736.2 xx 2 + 23.8 = 1472.4 + 23.8 = 1496.2` torr

Discussion

45.	At 1 atm and 273 K the density of gas, whose molecular weight is 45, is:
Answer» 44.8 GM `//`litre 11.4 gm `//` litre 2 gm` //` litre 3 gm `//` litre Answer :C

Discussion

46.	At NTP, 1L of O_2 reacts with 3L of carbon monoxide. What will be the volume of CO and CO_2 after the reaction?
Answer» 1L `CO_2`, 1L CO 2L `CO_2`, 2L CO 1L `CO_2`, 2L CO 2L `CO_2`, 1L CO Solution :`UNDERSET(2 Vol )(2CO)+underset(1vol)(O_2) to underset(2vol)(2CO_2)` 1 vol of `CO_2` reacts with 2 vol of CO 1 L of `CO_2` reacts with 2 L of CO CO left after reaction = 3-2=1L 1L of `O_2` PRODUCES 2L of `CO_2` Hence, after the reaction, CO =1L, `CO_2=2L`

Discussion

47.	At S.T.P., 22.4 L of a gas contain ........... molecules.
Answer» SOLUTION :`6.022xx10^(23)`

Discussion

48.	At normal temp. dihydrogen react violently with dioxygen.
Answer» ANSWER :TRUE

Discussion

49.	At moderate pressures, the compressibility factor of a gas is given by: Z=1+0.4P-(200P)/(T) (P in bars and T in Kelvin) The Boyle's temperature is given by:
Answer» 200K 500K `(2400)/(27)K` 400K Answer :B

Discussion

50.	At low temperature, Nitrogen dioxide, a reddish brown gas gets associated to form the colourless dinitrogen tetroxide as in the reaction 2NO_(2(g)) harr N_2O_(4(g)). Then al equilibrium
Answer» There would be an INCREASE in colour intensity The MIXTURE would become colourless There would be a DECREASE in colour intensity There would be no change in colour intensity Solution :No change in colour `implies` equilibrium

Discussion

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