94420 + Interview Questions in Chemistry in Class 12 Page 1841 InterviewSolution

92001.	""_(92)^(235)U nucleus absorbs a neutron and disintegrates into ""_(54)Xe^(139),_(38)Sr^(94) x. What will be the product x ?
Answer» 3 neutrons 2 neutrons `ALPHA`-PARTICLE `BETA`-particle. Answer :A

Discussion

92002.	5.35 g of a salt Acl (of weak base AOH) is dissolved in 250 ml of solution. The pH of the resultant solution was found to be 4.827. Find the ionic radius of A^(+) and Cl^(-) if ACl forms CsCl type crystal having 2.2 g/cm3. Given K_(b(AOH))=1.8xx10^(-5), (r^(+))/(r^(-))=0.732 for this cell unit.
Answer» ANSWER :`1.715 Å`

Discussion

92003.	""_(92)^(235)U+nto""_(92)^(235)Uto fission product + Neutron 3.20xx10^(-11)J. The energy released when 1 g of ""_(92)^(235)U undergoes fission is
Answer» `12.75xx10^(8)kJ` `18.60xx10^(9)kJ` `8.21xx10^(7)kJ` `6.55xx10^(6)kJ` ANSWER :C

Discussion

92004.	5.35 g of a salt A Cl(of weak base AOH) is dissolved in 250 ml of solution.The pH of the resultant solution was found to be 4.85 .Find the ionic radius of A^+ & Cl^(-) if A Cl forms CsCl type crystals having density 2.2g/cc.Given K_b(AOH)=2xx10^(-5) r_(+)/r_(-)=0.731 for this unit cell.
Answer» Solution :`pH=1/2xx[pK_w-pK_b-log C]IMPLIES 4.85\1/2xx[14-4.7-log C]implies C=0.4` `:. C=0.4=(5.34)/("molar mass")xx4 implies` molar mass of the salt=53.5gm/mole Now, density`=(1xx53.5)/(6.022xx10^(23)xxa^3)=2.2 " " implies " " a=3.43xx10^(-8)` CM and for CsCl type of crystals, `sqrt3a=2(r_(+)+r_(-))` `implies " " r_(+)=1.26xx10^(-3)cm and r_(-)=1.71xx10^(-8) cm`

Discussion

92005.	._(92)^(235)U+_(0)^(1)n rarr _(56)^(139)Ba +_(36)^(94)Kr +3_(0)^(1)n+200MeV Total energy released (in MeV) after 5^(th) stage of fission is:
Answer» 48600 16200 24200 None of these Answer :C

Discussion

92006.	5.33 mg of salt [Cr(H_(2)O)_(5)Cl].Cl_(2).H_(2)O is treated with excess of AgNO_(3)(aq) then mass of AgCl precipitate obtained will be : Given : [Cr = 52, Cl = 35.5 ,Ag = 108 ]
Answer» ANSWER :`5.74` mg `[CR(H_(2)O)_(5)Cl].Cl_(2).H_(2)Ounderset(0.02"MOL")+2AgNO_(3)(a)to 2Ag Cldarr+[Cr(H_(2)O)_(5)Cl]UNDERSET(0.04"mol")(NO_(3))_(2)`

Discussion

92007.	92 g. of ethanol is dissolved in 108g. Of water. The mole fraction of water in the solution is
Answer» 0.25 0.75 0.5 0.35 Solution :`"Moles of ETHANOL"="MASS"/("Molar Mass")=92/46=2` and moles of WATER `= 108/18=6` mole FRACTION of ethanol `=2/(2+6)=0.25`

Discussion

92008.	5.3 % (w/v) Na_(2)CO_(3) solution and 6.3 % (w/v) H_(2)C_(2)O_(4).2H_(2)O solution have same
Answer» molality MOLARITY normality mole fraction. Solution :`M_(Na_2CO_3) = (5.3)/(106) xx (1000)/(100) = (1)/(2)=0.5` `N_(Na_2CO_3) = 0.5 xx 2 =1.0` `MH_(2)C_(2)O_(4).2H_(2)O = (6.3)/(126) xx (1000)/(100) = 1/2 =0.5` `NH_(2)C_(2)O_(4).2H_(2)O= 0.5 xx 2=1`Both solutions have same molartiy and normality .

Discussion

92009.	5.3g of Na_2CO_3 added to 100 cc of 1NH_2SO_4. The final solution is :
Answer» ACIDIC alkaline neutral basic Answer :B

Discussion

92010.	52.5millimoles ofLiAlH_(4)was treated with15.6 g (210 millimoles ) of t- butyl alcohol . A total of 157.5 millimolesof hydrogen wasevolved or for the readtion Li AIH_(4) + 3 (CH_(3))_(3) COH to 3H_(2) + Li [(CH_(3))_(3) O]_(3) ALHThe addition of an excess of anotheralcohol, methanol to the above reactionmixture caused the fourth H atom of the LiAIH_(4)to be replace according tothe equation Li [(CH_(3))_(3) O]_(3) AlH + CH_(3) OH to H_(2) + Li [(CH_(3))_(3) O] _(3) [CH_(3) O]Al How mauch H_(2)was evolved due to theaddition of CH_(3) OH?
Answer» Solution :According to the given EQUATIONS, 1 MOLE of Li `[(CH_(3))_(3)O]_(3)` AIHproduces1 MOLEOF `H_(2)` and 1 mole of Li `[(CH_(3))_(3)O]_(3)` AIHis produced by 1 mole of `LiAIH_(4)` `:.` 1 mole of `LiAIH_(4)` shouldproduce 1 mole of `H_(2)""(byCH_(3)OH)` or 1 MILLIMOLE of `LiAIH_(4)` should produce 1 millimole of `H_(2)(by CH_(3) OH)` 52 . 2 millimolesof `LiAIH_(4)` should produce 52.5 MILLIMOLES of`H_(2) (by CH_(3) OH)`

Discussion

92011.	""_(90)Th, a member of III group on losing alpha-particle forms a new element belonging to
Answer» IV group II group III group I group Answer :B

Discussion

92012.	52 mLof a solution , containing 1 g each of Na_(2)CO_(3) ,NaHCO_(3) and NaOHwas titrated with N HCl . What will be the titre readings if methyl orange is addedafter the first end point with phenolphthalein ?
Answer» Solution :The reactions in this case are , `NaHCO_(3) (" PRODUCED ") +HCl to NaCl +H_(2)O +CO_(2)` and `NaHCO_(3) +m.e of v_(3)` mL (SAY) of N HCl `1/106 xx 1000 +1/84 xx1000 = 1xx v_(3)` ` :. "" v_(3) = 21/3 mL `

Discussion

92013.	90g of silver coin was dissolved in strong nitric acid, and excess of sodium chloride solution was added. The silver chloride precipitate was dried and weighed 71.75 g. Caculate the percentage of silver in the coin (Atomic mass of Ag = 108 ) Ag + 2HNO_(3) to AgNO_(3) + NO_(2) + H_(2)O AgNO(3) + NaCl to AgCl + NaNO_(3)
Answer» ANSWER :60

Discussion

92014.	5.2 cc of a gaseous hydrocarbon was exploded with excess of oxygen and product cooled. A contraction of 7.8 cc was observed. A further contraction of 10.4 cc was noted on treatment with aqueous potash. Find the formula of the hydrocarbon and give I.U.P.A.C. name to it.
Answer» Solution :Experimental values : `"Volume of HYDROCARBON taken = 5.2 cc"` `"Contraction on explosion and cooling = 7.8 cc"` `"Contraction on introducing KOH"` `"i.e.Volume of "CO_(2)" formed = 10.4 cc"` Theoretical values : Combustion of hydrocarbon (say `C_(X)H_(y)`) can be represented as follows :- `{:(C_(x)H_(y),+,(x+y//4)O_(2),rarr,xCO_(2),+,""y//2H_(2)O),("1 vol.",,x+y//4"vol.",,"x vol.",,"(negligible volume)"),("1 cc",,x+y//4" cc",,"x cc",,),("5.2 cc",,5.2(x+y//4)" cc",,"5.2 x cc",,):}` Equating experimental values and the theoretical values of carbon dioxide formed, we have `5.2x=10.4"or"x=2` Theoretical value of contraction after explosion and coling for 1 cc. of the hydrocarbon `=(1+x+y//4)-x=1+y//4" cc"` Since for 5.2 cc of hydrocarbon, contraction on explosion and cooling = 7.8 cc `therefore"For 1 cc of hydrocarbon, contrction on explosion and colling "=(7.8)/(5.2)=(3)/(2)` Equating the two values, we have `1+(y)/(4)=(3)/(2)"or"(y)/(4)=(1)/(2)"or"y=2` Hence the formula of the hydrocarbon is `C_(2)H_(2)" (acetylene)"` I.U.P.A.C. name for this compound is Ethyne.

Discussion

92015.	51 mLof a solution , containing 1 g each of Na_(2)CO_(3) ,NaHCO_(3) and NaOHwas titrated with N HCl . What will be the titre readings if only methyl orange is used as indicator from the very beginning ?
Answer» Solution :The reactions in this case are , `Na_(2)CO_(3) +HCl to NaHCO_(3) +NaCl ` `NaHCO_(3) +HCl to NaCl +H_(2)O + CO_(2)` (produced) `NaHCO_(3) +HCl to NaCl +H_(2)O + CO_(2)` (originally present ) and `NaOH +HCl to NaCl +H_(2)O ` THUS , we have m.e of `Na_(2)CO_(3) + m.e of NaHCO_(3) + m.e of NaHCO_(3) + m.e of NaOH ` (produced )(originally present ) ` = " m.e of " v_(2)"mL (say ) of N HCl " ` `{:(1/106, xx 1000 +1/106 xx 1000 + 1/84 xx 1000 + 1/40 xx 1000 = 1 xx v_(2)),( :.,""v_(2)=55.8):}`.

Discussion

92016.	50mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is 36, the molecular mass of gas A will be :
Answer» 64 96 128.4 20.2 Solution :`(r_(1))/( r_(2)) = ( v_(1) //t_(1))/( v_(2) //t_(2)) = SQRT((M_(B))/( M_(A)))` `(t_(2))/( t_(1)) = sqrt((M_(B))/( M_(A))) ( :. v_(1) = v_(2))` `( 200)/( 150) = sqrt(( 36)/( M_(A)))` or `( 36)/( M_(A)) = (( 200)/( 150) )^(2)= 1.78` `:. M_(A) = ( 36)/( 1.78) = 20.23`

Discussion

92017.	50ml of a solution which is 0.05M in the acid HA (pK_(a) = 3.80) and 0.1M in HB(pK_(a) = 8.20) is titrated with 0.2 M-NaOH solution. Calculate the pH of solution at the first equivalence point. Given your answer after multiplying with 100.
Answer» ANSWER :585

Discussion

92018.	50mL of 0.05M Propane -1,2-diamine solution is titrated with 0.1M HCl solution at 25^(@)C. Detemine the pH of solution obtained by adding following volumes of HCl solution.
Answer» Solution :(a) HALF neutralisation :`pH=pK_(a_(1))=14-4.18=9.82` (B)`pH=((pK_(a_(1))+pK_(a_(2)))/(2))=(14-4.18+14-7.39)/(2)=8.215` (c ) BUFFER: `pH=14-7.39+log((2.5-0.5)/(0.5))=7.21`

Discussion

92019.	50mL of 0.05M Na_(2)CO_(3) is titrated against 0.1 M HCl ,pH of the solution will be [Given :For H_(2)CO_(3) ,pK_(a)=6.35,pK_(a)=10.33]
Answer» `6.35` `6.526` `8.34` `6.173` Solution :`{:(,CO_(3)^(2-),+,H^(+),rarr,HCO_(3)^(-),),("INITIAL milli-moles",50xx0.05,,40xx0.1,,-,),("FINAL milli-moles",-,,1.5,,2.5,),(,HCO_(3)^(-),+,H^(+),rarr,H_(2)CO_(3),),("Initial milli-moles",2.5,,1.5,,-,),("Final milli-moles",-,,1,,1.5,):}` ` pH=pK_(a_(1))+log(([HCO_(3)^(-)])/([H_(2)CO_(3)]))=6.173`

Discussion

92020.	50mL of 0.05 M Na_(2)CO_(3) is titrated against 0.1 M HCl. On adding 40mL of HCl, pH of the solution will be [ Given : For H_(2) CO_(3), pK_(a_(1)) = 6.35, pK_(a_(2)) = 10.33, log 3 = 0.477, log 2 = 0.30 ]. ( Express answer is nearest integer ) .
Answer» ANSWER :6

Discussion

92021.	50g. Of a solute is dissolved in 0.95 kg. of the solvent. The mass percent of the solution is
Answer» 5 0.9 0.52 `0.090` Solution :Mass of solute = 50g mass of solvent = 0.95kg = 950 g. Mass of solution` = 50 + 950 = 1000g` :. 1000 g of solutionequiv 50g of solute :. 100g of solution EQUIV 5 g of solute

Discussion

92022.	50g of a good sample of CaOCl_2, is made to react with CO_2. The volume of Cl_2 liberated at S.T.P is
Answer» 5.6 lit 11.2 lit 22.4 lit 4.48 lit Answer :A

Discussion

92023.	500ml of NH_(3) contains 6.00 xx 10^(23) molecules at S.T.P. How many molecules are present in 100 ml of CO_(2) at S.T.P.
Answer» `6 xx 10^(23)` `1.5 xx 10^(23)` `1.2 xx 10^(23)` None of these Solution :`(PV_(1))/(n_(2)RT)=(PV_(2))/(n_(2)RT)` or , `(V_(1))/(n_(1))=(V_(2))/(n_(2))` or , `(500)/(1)=(100)/(n_(2))` or , `n_(2)=(1)/(5)` `[6.0 xx 10^(23)` molecules of `NH_(3)=1` mole ] 1 mole of `CO_(2)` contains `6.0 xx 10^(23)` molecules `:. (1)/(5)` mole of `CO_(2)` contains `1.2 xx 10^(23)` molecules

Discussion

92024.	500mL of a hydrocarbon gas burnt in excess of oxygen yielded 2500 mL of Co_2 and 3.0 liter of water vapour (all volumes measured at the same temperature and pressure). The formula of the hydrocarbon is :
Answer» `C_3H_6` `C_2H_4` `C_5H_12` `CH_4` ANSWER :C

Discussion

92025.	500mL of 10^(-5)MNaOH is mixed with 500mL of 2.5xx10^(-5)M of Ba(OH_(2)),To the resulting solution ,99L water is added ,calculate pH of final solution.Take log0.303=-0.52.
Answer» Solution :`[OH^(+)]_(f)=((500xx10^(-5))+(500xx2xx2.5xx10^(-5)))/(1000)=3XX10^(-5)M` `V_(1)=1L&V_(f)=100L` no of molesof `[OH^(+)]` in resulting solution =no. of moles of `[OH^(+)]` in FINAL `3xx10^(-5)=[OH^(-)]_(f)xx100` `:.[OH^(-)]_(f)=3xx10^(-7)M( lt 10^(-6)M)` so `OH^(-)`ions coming form `H_(2)O` should also be considered. `H_(2)OhArrunderset(x) H^(+)+underset(x+3xx10^(-7))(OH^(-))` `K_(w)=x(x+3xx10^(-7))=10^(-14)` `,.x=((sqrt(13)-3)/(2))xx10^(-7)M=[H^(+)]` So `pH=7-log0.303=7.52`

Discussion

92026.	500cm^(3) of 0.250 M Na_(2)SO_(4) solution added to an aqueous solution of 15.00 g of BaCl_(2) resulted in the formation of a white precipitate of BaSO_(4). How many moles and how many grams of BaSO_(4) are formed ?
Answer» SOLUTION :`500cm^(3)` of `0.25M Na_(2)SO_(4)` solution CONTAIN `Na_(2)SO_(4)=(0.25)/(1000)xx500=0.125" mole"` `15g BaCl_(2)=(15)/(208)" mole"=0.072 " mole"` `Na_(2)SO_(4)+BaCl_(2) rarr BaSO_(4)+2NaCl` EVIDENTLY, `BaCl_(2)` will be the limiting reactant. `BaSO_(4)` formed = 0.072 mole = `0.072xx233g=16.776g`

Discussion

92027.	500cc of a hydrocarbon gas burnt in excess of oxygen yields 2500cc of CO_(2) and 3 litres of H_(2)O vapour, all the vapours being meausred at the same temperature and pressure. What si the formula of the hydrocarbon gas?
Answer» Solution :`{:(,"Hydrocarbon",overset("combustion")rarr,CO_(2),+,H_(2)O),(,0.5 lit,,2.5lit,,3lit),(,0.5"moles",,2.5 "moles",,3"moles"),(THEREFORE, 1"moles",,5 "moles",,6 "moles"):}` Moles of C in `CO_(2)=1 xx` moles of `CO_(2)` `=1 xx5 =5` Moles of H in `H_(2)O=2xx` moles of `H_(2)O` `=2 xx 6= 12` SINCE 1 mole of the compound contains 5 moles of C and 12 moles of H, the molecular formula of the hydrocarbon is `C_(5)H_(12)`

Discussion

92028.	500 mL of each three samples of H_(2)O_(2) labelled 10 vol, 15 vol and 20 vol are mixed and then diluted with equal volume of water. Calculate the volume strength of resultant H_(2)O_(2) solution .
Answer» SOLUTION :N/A

Discussion

92029.	500 ml of vessel contains 1.5 M each of A, B, C and D at equilibrium. If 0.5 M each of C and D are taken out, value of K_(c ) forA+B hArr C+Dwill be :
Answer» `1.0` `1//9` `4//9` `8//9` ANSWER :A

Discussion

92030.	500 ml of a H_(2)O_(2) solution on complete decomposition produces 2 moles of H_(2)O. Calculate the volume strength of H_(2)O_(2) solution ?
Answer» ANSWER :45.4V

Discussion

92031.	500 ml of "0.3 M AgNO"_(3) are added to 600 ml of "0.5 M KI" solution. The ions which will move towards the cathode and anode respectively are
Answer» `Agl//Ag^(+)and NO_(3)^(-)` `Agl//l^(-) and K^(+)` `K^(+) and K^(+)` `K^(+) and Agl//l^(-)` ANSWER :C

Discussion

92032.	500 mL of 0.2M aqurous solution of aceticacid is mixed with 500 mLof 0.2M HClat25^(@)C (i) Calculatethe degreeof dissociation acidin the resulting solutionandpH ofthe solution . (b) If 6 g of NaOH is addedto the abovesolution , determinethe final pH . [ There is nochangein volume on mixing , Ka of aceticacid is1.75 xx 10^(-5) mol L^(-1)
Answer» Solution :The volumebeingdoubledby mixingtwo SOLUTIONS, the molarityof eachcomponentwill be halved i.e ` [CH_(3)COOH] = 0.1 M , [HCl] = 0.1 M ` HClbeing a strongacidwill remaincompletelyionised and hence`H^(+)`ion concentrationfurnishedby it will be `0.1 `M . This wouldexert commonion effecton the dissociation ofaceticacid ,a weak acid. `{:(CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),(C,,0,,0),(C(1-alpha),,Calpha,,Calpha +0.1):}` `K_(a)= (Calpha(Calpha+0.01))/(C(1-alpha)) = (Calpha^(2)+0.1alpha)/((1-alpha))` Neglecting `alpha `in comparisonto UNITY and `Calpha^(2)` i.e ` 0.1 alpha^(2)` `K_(a) = 0.1 alpha ` oe `alpha= (K_(a))/(0.1) = (1.75 xx10^(-5))/(0.1) = 1.75xx 10^(-4)` ` [H^(+)] TOTAL = 0.1 +Calpha ,Calpha ` is negligiableas compared to `0.1 ` ` :. [ H^(+)] _(Total) = 0.1 ` ` :.pH = 1 ` (ii) `6g NaOH = 6/40= 0.15 ` mol `0.1 ` moleof NaOH will beconsumedby`0.1 ` mole of HCL . Thus, `0.05` mole of NaOH will react with aceticacid(No. of molof `CH_(3)COOH = 1 xx 0.2 = 0.1 ` ) accordingto the equation `{:(CH_(3)COOH,+,NaOH,to,CH_(3)COONa,+,H_(2)O),(0.1" mol",,0.05 " mol",,0,,0),(0.05 " mol",,0 " mol" ,,0.05 " mol",,0.05 " mol"):}` Thus , solutionof aceticacidand sodium acetate will becomeacidicbuffer . So pH of the buffer will be `pH = pK_(a) + log.(["salt"])/(["acid"])= - log(1.75 xx 10^(-5)) + log1 = 4.75`

Discussion

92033.	To 500 mL of 0.150 M AgNO_(3) solution were added 500 mL of 1.09 M Fe^(2+) solution and the reaction is allowed to reach an equilibrium at 25^(@)C Ag^(+)(aq)+Fe^(2+)(aq) hArr Fe^(3+)(aq)+Ag(s) For 25 mL of the solution, 30 mL of 0.0832 M KMnO_(4) was required for oxidation. Calculate the equilibrium constant for the the reaction at 25^(@)C.
Answer» ANSWER :3

Discussion

92034.	500 mL of 0.1 N solution of AgNO_(3) is added to 500 mL of a 0.1 N KCl solution .the concentration of nitrate in the resulting mixture is
Answer» `0.1` N `0.05` N `0.01` N `0.2 `N ANSWER :B

Discussion

92035.	500 mL of 0.01 AgNO_3 is mixed with 250 ml each of NaBr and NaCl, each having molarity 0.02 M. Find equilibrium concentrtion of Br^(-) (moles/L). Given : K_(sp)(AgBr)= 5 xx 10^(-13), K_(sp)(AgCl) =10^(-10)).
Answer» ANSWER :`2.49 XX 10^(-5)`

Discussion

92036.	50.0 kg of N_(2)(g) and 10.0 kg of H_(2)(g) are mixed to produce NH_(3)(g). Calculate the NH_(3)(g) formed. Identify the limiting reagent in the production of NH_(3) in this situtation.
Answer» Solution :The BALANCED chemical equaiton for the reaction is `N_(2)(g)+3H_(2)(g)rarr 2NH_(3)(g)` STEP 1. To convert the GIVEN amounts into moles Mass of `N_(2)` taken `=50.0kg = 50,000g` `"Moles of "N_(2)=("Mass of N"_(2))/("Molar mass of N"_(2))=(5000g)/("28 g mol"^(-1))="1785.7 moles"` Mass of `H_(2)` taken = 10.0 kg= 10,000 g Step 2. To identify the limiting reagent From the above balanced equation, 1 mole of `N_(2)` reacts with 3 moles of `H_(2)` `therefore"1785.7 moles of N"_(2)" will react with "H_(2)=3xx1785.7" mol = 5355.9 moles"` But we have only 4960.3 moles of `H_(2)`. Hence, `N_(2)` is not the limiting reactant. Testing in the reverse manner, 3 moles of `H_(2)` react with 1 moles of `N_(2)` To calculate the amount of `NH_(3)` formed As the amount of product formed depends UPON the limiting reagent, hence, we calculate `NH_(3)` formed as follows: `"3 moles of "H_(2)" from "NH_(3)="2 moles"` `therefore"4960.3 moles of "H_(2)" will form "NH_(3)=(2)/(3)xx4960.3mol=" 3306.9 moles"` Molar mass of `NH_(3)=17gmol^(-1)` `therefore"Mass of "NH_(3)" formed "=3306.9xx17g=56217g=56.217kg`

Discussion

92037.	500 cc of a hydrocarbon gas burnt in excess of oxygen yielded 2500 cc of CO_(2) and 3 litres of H_(2)O vapour, measured under identical conditions of temperature and pressure. The formula of the hydrocarbon is
Answer» `C_(5)H_(12)` `C_(5)H_(10)` `C_(4)H_(10)` NONE of these Answer :A

Discussion

92038.	50 volume of H_(2) takes 25 minutes to diffuse out of vessel. How long will 40 volumes of oxygen take to diffuse out from the same vessel under similar conditions?
Answer» 20 minutes 80 minutes 100 minutes 200 minutes Answer :B

Discussion

92039.	50% of the reagent is used for dehydrohalogenation of 6.45 g CH_(3)CH_(2)Cl . What will be the weight of the main product obtained ? [Atomic mass of H , C and Cl are 1 ,12 and 35.5 g/mol respectively]
Answer» 0.7 g 1.4 g 2.8 g 5.6 g Answer :B

Discussion

92040.	50% neutralisation of a solution of formic acid (K_a=2xx106(-4)) with NaOH would result ina solution having a hydrogen ion concentration of :
Answer» `2XX10^(-4)` 3.7 2.7 1.85 Answer :A

Discussion

92041.	50% of a reaction is completed in 20 minute and 75% is completed in 30 minutes from the beginning,What is the order of reaction?
Answer» SOLUTION :`t_(1/2)ALPHA1/(a^(n-1)),((t_(1/2))_(1))/((t_(1/2))_(2))=((a_(2))/(a_(1)))^(n-1),20/10=(50/100)^(n-1),2/1=(1/2)^(n-1),(2/1)^(1)=(2/1)^(1-n)1=1-n,n=1-1,n=0`

Discussion

92042.	50% of the reagent is used for dehydro halogenation of 6.45 gm CH_(3)CH_(2)Cl. What will be the weight of the main product obtained ? [At. Mass of H, C and Cl are 1, 12 and 35.5 gm / "mole"^(-1) respectively]
Answer» 0.7 GM 1.4 gm 2.8 gm 5.6 gm ANSWER :B

Discussion

92043.	50 ml water is added to a 50 ml solution of Ba(OH)_(2) of strength 0.01 M. The pH value of the resulting solution will be
Answer» 8 10 12 6 Solution :`0.01 M Ba(OH)_(2) = 0.02 N (Ba(OH)_(2)` `N_(1)V_(1) = N_(2)V_(2)` `[0.02 N] xx [ 50 ml] = N_(2)xx 100 ml` `N_(2) = (0.02 xx 50)/(100) = 10^(-2) N, [OH^(-)] = 10^(-2) N` `pOH = 2` or PH = 12.

Discussion

92044.	50 mLsolution of BaCl_(2) (20.8%w//v) and 100 mL solution of H_(2)SO_(4)(9.8%w//v) are mixed (Ba = 137, Cl = 35.5, S = 32). CaSO_(4) formed will be
Answer» 11.65 g 23.3 g 29.8 g 46.6 g Solution :Molecular MASS of `BaCl_(2)=137+2xx35.5=208` Amount of `BaCl_(2)` present in 50 mL solution `=(20.8)/(2)=10.4g=(10.4)/(208)"mole"="0.05 mole"` Amount of `H_(2)SO_(4)` present in 100 mL solution `=9.8g=(9.8)/(98)"mole"="0.1 mole"` `BaCl_(2)+H_(2)SO_(4)rarrBaSO_(4)+2HCl` Thus, `BaCl_(2)` will be limiting REAGENT. 0.05 mole of `BaCl_(2)` will be produce `BaSO_(4)` = 0.05 mole `=0.05xx233 g` (mol. mass of `BaSO_(4)=137+32+64=233`) = 11.65 g

Discussion

92045.	50 ml of water sample, containing temporary hardness only, required 0.1 ml of M/50 HCl for complete neutralisation. Calculate the temporary hardness of water in ppm.
Answer» Solution :`M_1V_1=M_2V_2` `50xxM_1=1/50xx0.1` `M_1=(HCO_2^(-))=1/(25xx10^3)` mmoles of `HCO_3=1/(25xx10^3)xx50=1/500` mmoles of `CA^(2+)=1/1000` mmoles of `CaCO_3=1/1000` WEIGHT of `CaCO_3=1/1000xx10^(-3) XX 100=10^(-4)` hardness=[weight of `CaCO_3`/weight of water]x`10^6=10^(-4)/50xx10^6=2` ppm.

Discussion

92046.	50 ml of solution of pH = 1 is mixed with 50 ml of solution of pH = 2. The pH of the mixture is nearly :
Answer» 2 `1.74` `1.26` 0 Answer :C

Discussion

92047.	50 mL of hydrogen gas was collected over at 23^(@)C, 740 mm Hg baromatric pressure H_(2) was produced by the electrolysis of water. The voltage was constant at 2.1 volts the current averaged 0.50 amp for 12 minutes and 20 seconds. Calculate Avogadro constant.
Answer» SOLUTION :`5.85 XX 10^(23)`

Discussion

92048.	50 ml of hydrogen diffuses through a small hole from a vessel in 20 minutes. Time taken for 40 ml of oxygen to diffuse under similar conditions will be
Answer» 12 min. 64 min. 8 min 32 min. Solution :`( R ( H_(2)))/( r( O_(2))) = ( 50 // 20)/( 40 //t ) = SQRT( ( 32)/( 2))` or ` ( 50)/( 20) XX ( t )/( 40) = 4 ` or ` t = ( 4 xx 20 xx 40)/( 50) = 64` min

Discussion

92049.	50 ml of hydrogen diffuses out through a small hole from a vessel in 20 minutes. The time needed for 40 ml of oxygento diffuse out is
Answer» 12 min 64 min 8 min 32 min Solution :If 40 ml `O_(2)` will diffuse in t min. then . `r_(O_(2))=(40)/(t)` `r_(H)=(50)/(20)implies r_(O)=r_(H)sqrt((M_(H_(2)))/(M_(O_(2))))=(50)/(20)sqrt((2)/(32))=(50)/(20).(1)/(4)` `(40)/(t)=(50)/(80):. t= (40xx 80)/(50)=64` MINUTES

Discussion

92050.	50 ml of given H_(2)O_(2) solution is added to excess KI solution in acidic medium. The liberated I_(2) requires 20 ml of 0.04 M standard Hypo solution. The strength of given 250 m H_(2)O_(2) solution is
Answer» (a) 0.0272 g/100 CC (B)0.136 g/100cc (C)0.544g/100cc (d)NONE of these Answer :A

Discussion

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