94420 + Interview Questions in Chemistry in Class 12 Page 1819 InterviewSolution

90901.	A certain quantity of electricity is passed through an aqueous solution of AgNO_(3) and cupric salt solution connected in series . The amount of Ag deposited is 1.08 g. then the amount of copper deposited is (At. Mass Cu = 63.5 , Ag = 108) .
Answer» 0.6454 G 6.354 g 0.3177 g 3.177 g Solution :`("Wt. of Cu")/("Wt. of AG") = ("Eq. of MASS of Cu")/("Eq. mass of Ag")` `("Wt. of Cu")/(1.08) = (63.5//2)/(108)` or Wt. of Cu = `0.3177` g

Discussion

90902.	A certain organic compound A decomposes by two parallel first order mechanisms. If k_(1) : k_(2)=1 : 9 and k_(1)=1.3xx10^(-5)s^(-1). Calculate the concentration ratio of C to A if an experiment is started with only A and allowed to run for one hour.
Answer» ANSWER :`1.17`

Discussion

90903.	A certain oxide of iodine has been found to contain iodine and oxygen. The ratio iodine : oxygen is 254: 112. On being dissolved in water this oxide can produce
Answer» `HIO_(2)` `HIO_(3)` `HIO_(4)` `H_(5)IO_(6)` Solution :Mole ratio of iodine : oxygen `= 254/127 : 112/16 = 2:7` The oxide is `I_(2)O_(7)`. When dissolved in water it can PRODUCE `HIO_(4)` or `H_(5)IO_(6)` `I_(2)O_(7) + H_(2)O to 2HIO_(4)` `I_(2)O_(7) + 5H_(2)O to 2H_(5)IO_(6)`

Discussion

90904.	A certain nuclide has a half-life period of 30 minutes. If a sample containing 600 atoms is allowed to decay for 90 minutes, how many atoms will remain
Answer» 200 ATOMS 450 atoms 75 atoms 500 atoms Solution :`n = (90)/(30) = 3 RARR N = (600)/(2^(3)) = 75` atoms

Discussion

90905.	A certain nuclide has a half life period of 30 minutes. If a sample containing 600 atoms is allowed to decay for 90 minutes, how many atoms will remain
Answer» 450 atoms 150 atoms 75 atoms 200 atoms Answer :C

Discussion

90906.	A certain nucleide has half life period of 30 min. If a sample containing 6xx10^(10) atoms is allowed to decay for 90 min, how many atoms will remain ?
Answer» `2XX10^(10)` ATOMS `4.5xx10^(9)` atoms `7.5xx10^(9)` atoms `1.5xx10^(9)` atoms. Answer :C

Discussion

90907.	A certain mixture of HCl and CH_3-COOH is 0.1 M in each of the acids.20 ml of this solution is titrated against 0.1M NaOH.By how many units does the pH changes form the start to the stage when the HCl is almost completely neutralised ? K_a for acetic acid=1.8xx10^(-5).
Answer» 2.03 0.775 2.325 3.175 Solution :INITIAL `[H^+]`=0.1 (`H^+` from weak acid can neglect ) After NEUTRALISATION of HCl volume would double, HENCE concentration of `CH_3COOH=(0.1xx20)/40=0.05` `[H^+]=sqrt(K_aC)` `[H^+]=sqrt(1.8xx10^(-5)xx0.05)=9.48xx10^(-4)` pH=3.03 change in pH unit =3.03-1=2.03

Discussion

90908.	A certain metal M occurs in four compounds namely A, B, C and D. A has 20% of M, B has 68% of M, C has 73% of M and D has 60% of M. If metal M is extracted from A, B, C and D, it cost RS 35 per kg, RS 40 per kg, RS 100 per kg and RS 45 per kg respectively. Which mineral can be considered as an effective ore of M>
Answer» A B C D Answer :B

Discussion

90909.	A certain metal M ocas in four compounds namely A,B,C and D. A has 20% of M, B has 68% of M, C has 73% of M and D has 60% of M. If metal M is extracted from A,B,C and D, it costs Rs 35 per kg, Rs 40 per kg, Rs 100 per kg and Rs 45 per kg respectively. which mineral can be considered as an effective ore of M?
Answer» A B C D Answer :B

Discussion

90910.	A certain metal fails to liberate H_2 gas from a moderately conc. HCI solution. However it desplaces Ag from AgNO_3 solution. Which among the following may it be :
Answer» Mg Fe Cu Cd Answer :C

Discussion

90911.	A certain metal (A) is boilded in dilute nitric acid to give a salt (B) an oxide of nitogen (C ). An aqueous solution of (B) with brine gives a precipitate(D) which is soluble in ammonium hydroxide . On adding aqueous solution of (B) to hypo solution, a white precipitate (E) is obtained. (E) on standing turns to a black compound (F). Identify (A) to (F).
Answer» SOLUTION :`{:(A=Ag,,,,B=AgNO_(-3)),(C=NO,,,,D=AgCl),(E=Ag_(2)S_(2)O_(3),,,,F=Ag_(2)S):}`

Discussion

90912.	A certain mass of a substance when dissolved in 100g C_(6)H_(6) lowers the freezing point by 1.28^(@)C. The same mass of solute dissolved in 100g of water lowers the freezing point by 1.40^(@)C. If substance has normal molecular mass in benzene and is completely dissociated in water, into how many ions does it dissociate in water, K_(f) for H_(2)O and C_(6)H_(6) are 1.86 and 5.12 K mol^(-1) kg respectively
Answer» Solution :`Delta T_(f)=(1000xxK'_(f)xxw)/(m xx W)` For the solution in BENZENE using the data given `1.28=(1000xx5.12xx w)/(m_(N)XX100)` (`m_(N)` = normal mol. mass) ……(i) For the solution in water in which solute dissociates `1.40=(1000xx1.86xx w)/(m_(esp.)xx100)`.......(ii) Dividing eq. (ii) by (i). `i=(m_(N))/(m_(EXP.))=(1.40)/(1.28)xx(5.12)/(1.86)=3.01~~3.0` Now, suppose that formula for solute is `{:(A_(x)B_(y),hArr,xA^(+),+,yB^(-)),(1,,0,,0),((1-alpha),,x alpha,,y alpha):}` `therefore i=1-alpha+x alpha+y alpha` `because` i=3 and `alpha = 1`(Given that `alpha = 1`) `therefore` No. of ions given (x+y)=3

Discussion

90913.	A certain mass of a substance when dissolved in 100 g C_(6)H_(6) lowers the freezing point by 1.28^(@). The same mass of the solute dissolved in 100 g of water lowers the freezing point by 1.40^(@). If the substance has normal molecular mass in benzene and is completely dissociated in water, into how many ions does it dissociate in water ? K_(f) for H_(2)O and C_(6)H_(6) are 1.86 and "5.12 K kg mol"^(-1) respectively.
Answer» Solution :`DeltaT=(1000xxK_(f)xxw_(2))/(w_(1)xxM_(2))` As substance has normal MOLECULAR mass in BENZENE, when dissolved in benzene, we have `1.28=(1000xx5.12xxw_(2))/(100xxM_("normal"))"…(i)"` As substance dissociates in water, observed molecular mass can be calculated as `1.40=(1000xx1.86xxw_(2))/(100xxM_("observed"))"...(ii)"` Dividing eqn. (ii) by eqn. (i) `(M_("normal"))/(M_("observed"))=(1.40)/(1.28)xx(5.12)/(1.86)=3.01""THEREFORE""i=3.01~=3.0` As the substance (solute) is `100%` ionized, `alpha=1.` Suppose the solute is `A_(x)B_(y)` `{:(,A_(x)B_(y),hArr,xA^(+),+,yB^(-),),("Initial moles ",1,,0,,0,),("After disso.",1-alpha,,xalpha,,yalpha",","Total "=1-alpha+xalpha+yalpha):}` `therefore""i=1-alpha+x alpha+yalpha` `therefore""3=1-1+x+y"" (because i=3, alpha=1)` `"or"x+y=3` `therefore"No. of ions produced on DISSOCIATION = 3"`

Discussion

90914.	A certain mass of a substance in 100 g of C_6H_6lowers the f.p. by 1.28^@C. The samemass of the substance in 100 g of water lowers the f.p. by 1.395^@C . If the substance has a normal molecular weight in C_6H_6and is completely dissociated in water, calculate the number of moles of ions produced by the dissociation of 1 mole of the substance in water. K_f for H_2O and C_6H_6are 1.86 and 5.00 respectively.
Answer» ANSWER :3

Discussion

90915.	A certain mass of gas occupies a volume of 300 c.c. at 27^(@)C and 620 mm pressure. The volume of this gas at 47^(@)C and 640 mm pressure will be
Answer» 400 c.c. 510 c.c. 310 c.c. 350 c.c. Solution :Given INITIAL volume `(V_(1))= 300C c `,initial temperature `=(T _(1)0= 27^(@)C = 300 K `, initial pressure `(P_(1)) = 620 mm,` final temperature `T_(2)= 47^(@)C = 320 K` and final pressure`=(P_(2))= 640 mm`. We know from the general gas equation . `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))=(620 xx 300)/(300)=(640 xx V_(2))/(320)implies V_(2) = 310 c c`

Discussion

90916.	A certain ion B^- has an Arhenius constant of basic character (equ. Constant :2.8xx10^(-7)) What is the equilibrium constant for lowry bronstad character.
Answer» `2.8xx10^(-7)` `3.57xx10^(-8)` `3.57xx10^8` `2.8xx10^7` ANSWER :D

Discussion

90917.	A certain liquid lias a viscosity of 1.0 xx 10^4 poise and a density of 3.2 g/mL. How long will it take for a platinum ball with a 2.5-mm-radius to fall 1.0 cm through the liquid? The density of platinum is 21.4 g/cc.
Answer» ANSWER :40.5 s

Discussion

90918.	A certain inorganic compound (X) shows the following reactions:
Answer» Solution :Cation of X`Sn^(2+)` Reactions: (i) `Sn^(2+)+underset("BROWN ppt.")(H_(2)S)overset(H^(+))toSnSdarr` (ii) `Sn^(2+)+underset("White ppt.")(2NaOH)to2Na^(+)+Sn(OH)_(2)darr` `Sn(OH)_(2)+2NaOHunderset("Soluble")toNa_(2)SnO_(2)+2H_(2)O` (III) `Sn^(2+)+2Fe^(3+)toSn^(4+)+2Fe^(2+)`

Discussion

90919.	A certain hydrocarbon (A) was found to contain 85.7% of C and 14.3% of H. This compound consumes 1 molar equivalent of hydrogen to give a saturated hydrocarbon (B). 1g of hydrocarbon (A) just decolourised 38.05g of a 5% solution (by weight) of Br_(2) in C Cl_(4). Compound (A), on oxidation with concentrated KMnO_(4), gave compound (C ) (molecular formula C_(4)H_(8)O) and acetic acid. Compound (C ) could easily be prepared by the action of acidic aqueous mercuric sulphate on 2-butyne. Determine the molecular formula of (A) and deduce the structures of (A), (B) and ( C)
Answer» Solution :Moles of `C: H` in (A) `=(85.7)/(12): (14.7)/(1)= 7.14: 14.3= 1:2` `therefore` empirical formula of (A) is `CH_(2)` and empirical formula weight is 14. As (A) consumes I molar equivalent of hydrogen to give a saturated hydrocarbon (B), (A) should have one double bond. Reaction of (A) with `Br_(2)` will also be an addition reaction in which 1 MOLE of (A) will combine with 1 mole of `Br_(2)`. THUS, Moles of (A)= moles of `Br_(2)` or `(1)/(M)= ((5 xx 38.05)//100)/(160) " " {{:("M = mol . wt of (A)"),(Br_(2)=160):}}` `therefore M=84` `therefore n=("molecular formula wt.")/("empirical formula wt")=(84)/(14)=6` `therefore` molecular formula of (A) is `C_(6)H_(12)` Now from the given reaction sequece, we have, `{:(C_(6)H_(12)underset(KMnO_(4))overset((O))rarrC_(4)H_(8)O+ CH_(3)COOH),("(A)I(C)" ),(""IIuarr H^(+)//HgSO_(1)),(""CH_(3).C -=C.CH_(3)),("2-butyne"):}` We CONCLUDE that `C_(6)H_(12)` (A) must be represented as `C_(4)H_(8)=CH.CH_(3)` (step I) and `C_(4)H_(8)O` (C ) must be a KETONE `CH_(3).overset(overset(C )(\|\|))(C )CH_(2).CH_(3)` (step II) further, addition of `H_(2)` to compound (A) will produce `C_(6)H_(14)` (B) (an alkane). The structure of (A), (B) and (C ) MAY be represented as

Discussion

90920.	A certain inorganic compound (X) shows the following reactions : i) On passing H_(2)S through an acidified solution of (X) a brown precipitate is obtained. ii) The precipitate obtained at step (i) dissolves in excess of yellow ammonium sulphide. iii) On adding an aqueous solution of NaOH to a solution of (X), first a white prcipitate is obtained which dissolves in excess of NaOH. iv) The aqueous solution of (X) reduces ferric chloride. Identify the cation of (X) and give chemical equations for reactions sat steps (i), (iii) and (iv).
Answer» Solution :Acidified solution of (X) gives brown PRECIPITATE with `H_(2)S` , soluble in yellow ammonium sulphide, INDICATE that (X) should be a salt of `Sn^(2+)`. The given reactions of the CATION of X, i.e., `Sn^(2)` can be written as below. `i) Sn^(2+)+H_(2)S rarr SnS +2H^(+)` `iii) Sn^(2+)+2NAOH rarr Sn(OH)_(2)+2Na^(+)` `Sn(OH)_(2)+2NaOH rarr Na_(2)SnO_(2)+2H_(2)O` `iv) Sn^(2+)+2FeCl_(3)rarr Sn^(4+)+2FeCl_(2)`

Discussion

90921.	A certain homogenous gas-phase reaction is exothermic and proceeds with an increase in the number of moles of gas. Which of the following statements concerning this reaction is true?
Answer» This reaction is spontancous at lower TEMPERATURES but non-spontancous at higher temperatures. This reaction is is non-spontaneous over all possible temperatures This reaction is is spontancous over all possible temperatures There is a temperature at which this reaction can be at equilibrium where all species are PRESENT at STANDARD concentrations. Answer :C

Discussion

90922.	A certain gas takes 8 times as long to effuse out as compared to hydrogen gas for same volume at constant temperature and pressure. Its molecular mass will be
Answer» 16 u 32 u 64 u 128 u Answer :2

Discussion

90923.	A certain first order reaction is half completed in 46 min. Calculate the rate constant and also time for 75% completion of the reaction.
Answer» SOLUTION :DATA : `t_(1//2) = 46` min, ` k = ? ` (i) To Find value of k `k = (0.693)/(t_(1//2)) = (0.693)/(46) = 0.015 min^(-1)` (ii) To find t at 75% completion of reaction. ` k = (2.303)/(t_(75%)) XX log""([R]_(0))/([R])` `t_(75%) = (2.303)/(0.015) xxlog""(100)/(35)` `t_(75%) = 92.42`min

Discussion

90924.	A certain gas effuses out of two different vessels A and B.A has a circular orifice while B has a square orifice of length equal to the radius of the orifice of vessel A. The ratio of rate of diffusion of the gas from vessel A to that from vessel B is
Answer» `pi:1` `1:pi` `1:1` `3:2` Solution :Under DIFFERENT CONDITIONS of pressure and temperature: rate `alpha(PA)/SQRT(T M) "":. R_1/r_2 =P_1/P_2 A_1/A_2sqrt((T_2M_2)/(T_1M_1))`

Discussion

90925.	A certain gas diffuses from two different vessels A and B The vessel A has a circular orifice while vessel B has square orifice of length equal t the radius of the orifice of vessel A. The ratio of the rates of diffusion of the gas from vessel A to vessel B, assuming same temperature and pressure is :
Answer» `PI` 7:22 1:1 2:1 Answer :A

Discussion

90926.	A certain endothermic reaction: Ato Product, DeltaH=+ve proceeds ina sequence of three elementary steps with the rate constant K_(1),K_(2) and K_(3) and each one having energy of activation E_(1),E_(2) and E_(3) respectively at 25^(@)C. The observed rate constannt for the reaction is equal to K_(3)sqrt((K_(1))/(K_(2))). A_(1),A_(2) and A_(3) are Arrhenius parameters respectively.For a reversible Aunderset(K_(2))overset(K_(1))(hArr)B,DeltaH=q if pre exponential factors are same. The correct relation is
Answer» `K_(eq)=e(-q//RT)` Rate of REACTION `=(-d[A])/(dt)=K_(1)[A]-K_(2)[B]` At equilibrium `K_(1)[A]=K_(2)[B]` Either of these Solution :`r_(1)=K_(1).[A]^(1),K_(1)=A.e^(-epsilon_(1)//RT),r_(2)=K_(2).[B]^(1),K_(2)=A.e^(-epsilon_(2)//RT)` net rate `R=r_(1)-r_(2)=K_(1)[A]^(1)-K_(2)[B]^(1)` at eq state`r_(1)=r_(2)` `K_(1)[A]^(1)=K_(2)[B]^(1),K_(eq)=([B]^(1))/([A]^(1))=(K_(1))/(K_(2))=(A.e^(-epsilon_(1)//RT))/(A.e^(-E_(2)//RT)=A^(0).e^(((-epsilon_(1)+epsilon_(2))/(RT))))=A^(0).e^(((-epsilon_(1)-epsilon_(2))/(RT)))` `K_(eq)=A^(0).e^(-epsilon_(a)//RT)`.So`epsilon_(a)=(epsilon_(1)-epsilon_(2))`

Discussion

90927.	A certain endothermic reaction: Ato Product, DeltaH=+ve proceeds ina sequence of three elementary steps with the rate constant K_(1),K_(2) and K_(3) and each one having energy of activation E_(1),E_(2) and E_(3) respectively at 25^(@)C. The observed rate constannt for the reaction is equal to K_(3)sqrt((K_(1))/(K_(2))). A_(1),A_(2) and A_(3) are Arrhenius parameters respectively.If temperature coefficient of the observed reactionis 2,the numerical value of E_(1)-E_(2)+2E_(3) is
Answer» 12.73 kcal 25.4 kcal 38.19 kcal cannot be calculated Solution :`log((K_(2))/(K_(1)))=(E_(a))/(2.303R)(1/(T_(1))-1/(T_(2))),log(2)=(epsilon_(a))/(2.3.3R)(1/298-1/308)` `epsilon_(a)=((0.3010xx2.303xx2xx298xx308)/10),epsilon_(a)=12.72KCal=((2epsilon_(3)+epsilon_(1)-epsilon_(2))/2)` `12.72=((2epsilon_(3)+epsilon_(1)-epsilon_(2))/2)(2epsilon_(3)+epsilon_(1)-epsilon_(2))=12.72xx2=25.4499,(epsilon_(1)-epsilon_(2)+2epsilon_(3))~=25.4` Kcal

Discussion

90928.	A certain endothermic reaction: Ato Product, DeltaH=+ve proceeds ina sequence of three elementary steps with the rate constant K_(1),K_(2) and K_(3) and each one having energy of activation E_(1),E_(2) and E_(3) respectively at 25^(@)C. The observed rate constannt for the reaction is equal to K_(3)sqrt((K_(1))/(K_(2))). A_(1),A_(2) and A_(3) are Arrhenius parameters respectively. Presence of a catalyst decreasesthe energy of activation of each path by half the value of E_(1). Assuming the other factors same, the observed energy of activation would be
Answer» `E_(3)-(E_(2))/2` `2E_(3)-E_(2)` `E_(2)-(E_(3))/2` `(E_(1))/2+(E_(2))/2+E_(3)` Solution :`E^(1)=[(2(epsilon_(3)-(epsilon_(1))/2)+(epsilon_(1)-(epsilon_(1))/2)-(epsilon_(2)-(epsilon_(1))/2))/2]=(2((2epsilon_(3)-epsilon_(1))/2)+((2epsilon_(1)-epsilon_(1))/1)-((2epsilon_(2)-epsilon_(1))/2))/2` `=(4epsilon_(3)-2epsilon_(1)+epsilon_(1)-2epsilon_(2)+epsilon_(1))/(2xx2),EPSILON^(1)=((4epsilon_(3)-2epsilon_(2)))/4,epsilon^(1)=(epsilon_(3)-(epsilon_(2))/2)`

Discussion

90929.	A certain endothermic reaction: Ato Product, DeltaH=+ve proceeds ina sequence of three elementary steps with the rate constant K_(1),K_(2) and K_(3) and each one having energy of activation E_(1),E_(2) and E_(3) respectively at 25^(@)C. The observed rate constannt for the reaction is equal to K_(3)sqrt((K_(1))/(K_(2))). A_(1),A_(2) and A_(3) are Arrhenius parameters respectively. The observed Arrhebius parameter for the reaction is
Answer» `(2A_(1)-A_(2)+2A_(3))/2` `sqrt((A_(1))/(A_(2))).A_(3)` `A_(1)A_(2)A_(3)` `(A_(1).A_(3))/(A_(2))` Solution :`impliesA=A_(3)sqrt((A_(1))/(A_(2)))` and also `impliese^(((-E)/(RT_(1))))=e^([(-E_(3)(E_(1))/2+(E_(2))/2)/(2RT_(1))])` `implies(-EPSILON)/(RT_(1))=(-2epsilon_(3)-epsilon_(1)+epsilon_(2))/(2RT_(1))-epsilon=(-2epsilon_(3)-epsilon_(1)+epsilon_(2))/2,epsilon=((2epsilon_(3)+epsilon_(1)-epsilon_(2))/2)`

Discussion

90930.	A certain endothermic reaction: Ato Product, DeltaH=+ve proceeds ina sequence of three elementary steps with the rate constant K_(1),K_(2) and K_(3) and each one having energy of activation E_(1),E_(2) and E_(3) respectively at 25^(@)C. The observed rate constannt for the reaction is equal to K_(3)sqrt((K_(1))/(K_(2))). A_(1),A_(2) and A_(3) are Arrhenius parameters respectively.Which represents the correct value if catalyst is not present:
Answer» `DeltaH` is definitely lower than`E_(1)` `DeltaH` is definitely lower than `E_(3)` `DeltaH` is definitely lower than `E_(1)-E_(2)` `DeltaH` is definitely lower than `(E_(1)-E_(2)+2E_(3))/2` Solution :`Aepsilon_(R)gtAepsilon_(R)(epsilon)gtDeltaH,((2epsilon_(3)+epsilon_(1)-epsilon_(2))/2)gtDeltaH` `DeltaH=(E_(P)-epsilon_(R))=+ve,DeltaH=(Aepsilon_(R)-Aepsilon_(P))=+ve, (A epsilon_(R)gtAepsilon_(P)` & `EPSILONR)ltepsilon_(P)`

Discussion

90931.	A certain endothermic reaction: Ato Product, DeltaH=+ve proceeds ina sequence of three elementary steps with the rate constant K_(1),K_(2) and K_(3) and each one having energy of activation E_(1),E_(2) and E_(3) respectively at 25^(@)C. The observed rate constannt for the reaction is equal to K_(3)sqrt((K_(1))/(K_(2))). A_(1),A_(2) and A_(3) are Arrhenius parameters respectively. The observed energy of activation for the reaction is
Answer» `(2E_(1)-E_(2)+2E_(3))/2` `(E_(2)-E_(1)-2E_(3))/2` `SQRT((E_(1)E_(3))/(E_(2)))` `(E_(1)-E_(2))/2+E_(3)` Solution :Here `K=K_(3)sqrt((K_(1))/(K_(2)))`……….1 `K_(1)=A_(1)E^(-E_(1)//RT),K_(2)=A_(2)e^(-E_(2)//RT),K_(3)=A_(3)e^(-E_(3)//RT)`…..2 from (1) (2)

Discussion

90932.	A certain endothermic reaction: Ararr Product, DeltaH=+ve proceeds in a sequence of three elementary steps with the rate constants K_(1), K_(2) and K_(3) and each one having energy of activation E_(a), E_(2) and E_(3) respectively at 25^(@)C. The observed rate constant for the reaction is equal to K_(3) sqrt(K_(1)/K_(2)). A_(1), A_(2) and A_(3) are Arrhenius parameters respectively. For the reaction, H^(.)+^(.)CH_(2)CH_(3)overset(K)(rarr) CH_(3).CH_(3), a chain termination step of the reaction: 2C_(2)H_(6)rarr CH_(2)=CH_(2)+H_(2), the activation energy and Arrhenius parameter can be given as:
Answer» (a) `0, K` (b) `0, E_(a)` (C ) `A, 1` (d) `1, K` Solution :For a FREE radical combination, `E_(a)=0`

Discussion

90933.	A certain element is a metalloid that forms an acidic oxide with the formula R_(2)O_(5). Identify the element.
Answer» Solution :Since the ELEMENT forms an oxide with formula, `R_(2)O_(5)`, it must be an element of group 15. Now group 15 has two metalloids, As and Sb. Since the ACIDIC character of pentoxides of group 15 ELEMENTS decreases while the basic character increases down the group, therefore, `As_(2)O_(5)` is acidic while `Sb_(2)O_(5)` is amphoteric. Thus, the methalloid of group 15 elements which forms and acidic pentoxide `(As_(2)O_(5))` is As.

Discussion

90934.	A certain element forms a solid oxide which when dissolved in water forms an acidic solution. The element is:
Answer» Neon Sodium Phosphorus Sulphur Answer :C

Discussion

90935.	A certain electric motor produced 15 kJ of energy each second as mechanical work and lost 2 kJ as heat to the surroundings. What is the change in the internal energy of the motor and its power supply each second ?
Answer» ANSWER :17 KJ

Discussion

90936.	A certain dye was prepared from a plant grown on a large scale in India. Name the dye
Answer» MALACHITE green Congo red INDIGO Turmeric Solution :Indigo is obtained in India by PLANTS.

Discussion

90937.	A certain dye absorbs light of lamda=4530Å and then fluorescence light o 5080Å. Assuming that under given conditioins 47% of the absorbed energy is re emitted out as fluorescence. Calculate the ratio of quanta emitted out to the number of quanta absorbed.
Answer» ANSWER :0.527

Discussion

90938.	A certain dye absorbs 4000 Å and fluoresces at 8000 Å. These being wavelengths of maximum absorption that under given conditions 50 % of the absorbed energy is emitted. Calculate the ratio of the no. of quanta emitted to the number absorbed.
Answer» ANSWER :1

Discussion

90939.	A certain current liberated 0.504 gm of hydrogen in 2hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?
Answer» `31.75` `15.8` `7.5` `63.5` ANSWER :B

Discussion

90940.	A certain current liberated 0.504 gm of hydrogen in 2 hours. How many gramsof copper can be liberated by the same current flowing for the same time in acopper sulphate solution
Answer» 31.75 15.8 7.5 63.5 Answer :B

Discussion

90941.	A certain compound (X) when treated with copper sulphate solution yields a brown precipitate. On adding hypo solution, the precipitate tums white. The compound is
Answer» `K_(2)CO_(3)` `KI` KBr `K_(3)PO_(4)` Solution :KI reacts with `CuSO_(4)` solution to produce CUPROUS iodide (white precipitate) and `I_(2)` (which GIVES brown colour). Iodine reacts with hypo `(Na_(2)S_(2)O.5H_(2)O)` solution. Decolourisaition of solution shows the APPEARANCE of white precipitate.

Discussion

90942.	A certain compound Y has a formula C_(3)H_(6)O. It combines with hydroxylamine to form two compounds which are geometrical isomers of each other. Y is
Answer» `CH_3 CHO` ` CH_3 CH_2 CHO` ` CH_3 COCH_3` ` CH_2= CHCH_2OH` ANSWER :B

Discussion

90943.	A certain compound (x ) shows the following reactions. (i) When KI is added to an aqueous suspension of X containing aceticacid, iodine is liberated. (ii) When CO_(2)is passed through an aqueous suspension of 'X', the turbidity transforms to a ppt. (iii) When a paste of (X ) in water is heated with ethylalcohol, a product of anaesthetic use is obtained.The anaesthetic obtained in the above, treated with silver powder to form the following.
Answer» `C_(2) H_(2)` `C_(2)H_(6)` `CH_(4)` None of these Solution :`CaOCl_(2) + H_(2)O rarr Ca(OH)_(2) + Cl_(2)` `{:(CH_(2)),(CH_(2)):} :)CO +3Clrarr {:(C Cl_(3)),(CH_(3)):}:) CO + 3HCL ` `2 {:(C C l_(3)),(CH_(3)):} :)CO+Ca(OH)_(2)rarr {:(CH_(3)COO),(CH_(3)COO):}:) Ca+2CHCl_(3)` `UNDERSET("Amoesthetic")(2CHCl_(3)) + 6Ag overset( DELTA)(rarr) underset("gas")(C_(2)H_(2)) + 6AgCl`

Discussion

90944.	A certain compound (x ) shows the following reactions. (i) When KI is added to an aqueous suspension of X containing aceticacid, iodine is liberated. (ii) When CO_(2)is passed through an aqueous suspension of 'X', the turbidity transforms to a ppt. (iii) When a paste of (X ) in water is heated with ethylalcohol, a product of anaesthetic use is obtained.When X is palced in moist air, the smell of which as is evolved
Answer» `Br_(2)` and `Cl_(2)` `H_(2)S` `Cl_(2)` NONE of these Solution :X is placed in MOIST AIR, the smell of `Cl_(2)` gas is evolved.

Discussion

90945.	A certain compound (X) shows the following reactions : (i) When K_3 is added to an aqueous suspension of (X) containing acetic acid, iodine is liberated. (ii) When CO_2 is passed through an aqueous suspension of (X), the turbidity transforms to a precipitate. (iii)When (X) is heated with ethyl alcohol, a product of anesthetic use is obtained. The (X) is ...........
Answer» `CaCl_(2)` `CaOCl_(2)` `Cl_(2)` `CaCO_(3)` SOLUTION :(i) `CaOCl_(2) + 2Kl + 2CH_(3)COOH to Ca(CH_(3)COO)_(2) + 2KCl + H_(2)O + I_(2)` (ii) `CaOCl_(2) + CO_(2) to underset("White ppt")(CaCO_(3) + Cl_(2))` (iii) `CaOCl_(2) + H_(2)O to Ca(OH)_(2) + Cl_(2) +underset("excess")(Cl_(2)) + C_(2)H_(5)OH to C Cl_(3)CHO overset(Ca(OH)_(2))to CHCl_(3) + (HCOO)_(2)Ca`

Discussion

90946.	A certain compound (x ) shows the following reactions. (i) When KI is added to an aqueous suspension of X containing aceticacid, iodine is liberated. (ii) When CO_(2)is passed through an aqueous suspension of 'X', the turbidity transforms to a ppt. (iii) When a paste of (X ) in water is heated with ethylalcohol, a product of anaesthetic use is obtained.As per the above observation. X is
Answer» `CaCl_(2)` `CaOCl_(2)` `Ca(ClO_(3))_(2)` None of these Solution :X is `CaCOCl_(2)`

Discussion

90947.	A certain compound (X) shows the following re- actions : When a paste of (X) in water is heated with ethyl alcohol a product of anaesthetic use is obtained. Identify (X) and write down chemical equations for reactions involved in steps (i), (ii) and (iii)
Answer» SOLUTION :`X-CaOCl_(2)` `CaOCl_(2)+H_(2)O rarr Ca(OH)_(2)+Cl_(2)` `C_(2)H_(5)OH+Cl_(2) rarr CH_(3)CHO+2HCI` `CH_(3)CHO +3 Cl_(2) rarr C Cl_(3) CHO overset(Ca(OH)_(2))(rarr)UNDERSET("Anaesthetic")(CHCl_(3))`

Discussion

90948.	A certain compound when burnt gave three oxided. The first turned lime water milky, the second turned cobalt chloride paper pink and the thired formed an aqueous solution of pH_3 nearly. The elements present in the compound are:
Answer» C,S,O C,H,Na C,H,S C,H,Ca Answer :C

Discussion

90949.	A certain compound was known to have a formula which would be represented as [PdC_(x)H_(y)N_(2)] (ClO_(4))_(2). Analysis showed that the compound contained 30.15%C and 5.06%H. When converted to the corresponding thiocyanate, [PdC_(x)H_(y)N_(z)] (SCN)_(2), the analysis was 40.412%C and 5.94%H. Find x,y and z.
Answer» ANSWER :14,28,4

Discussion

90950.	A certain compound is a viscous, high boiling point liquid, miscible with water. The compound is most likely to be:
Answer» `CH_3CH_2OH` `CH_3CH-2CH_2OH` `CH_3CHOHCH_3` `CH_2OHCHOHCH_2OH` ANSWER :D

Discussion

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