94420 + Interview Questions in Chemistry in Class 12 Page 1746 InterviewSolution

87251.	A solution is
Answer» a MIXTURE of TWO compounds a HOMOGENEOUS mixture of two compounds a homogeneous mixture of two compounds all the above Solution :A/c to DEFINITION.

Discussion

87252.	A solution is 0.10 M in Na_2 SO_4 . When 50 mL of 0.1 M Ba(NO_3) , is added to 50 mL of this solution, what fraction of the sulphate ion is not precipitated? K_(sp) (BaSO_4) = 1.1 xx 10^(-10)
Answer» SOLUTION :`2.1 XX 10^(-1)`

Discussion

87253.	A solution is 0.10 M Co^(2+)and 0.10 M Hg^(2+). Calculate the range of pH in whichonly one of the metal sulphides precipitates when the solution is saturated in H_2S. K_(sp) (CoS) = 4 xx 10^(-21) and K_(sp)(HgS) = 1.6 xx 10^(-52)
Answer» SOLUTION :PH LESS than 0.8

Discussion

87254.	A solution has pH = 5, it is diluted 100 times, then it will become
Answer» Neutral Basic Unaffected More acidic Solution :pH = 5, MEANS `[H^(+)] = 10^(-5)M` After DILUTION `[H^(+)] = 10^(-5)//100 = 10^(-7)M` `[H^(+)]` from `H_(2)O` cannot be neglected. Total `[H^(+)] = 10^(-7) + 10^(-7) = 2 XX 10^(-7)` `pH = 7 - 0.3010 = 6.6990 = 7` (neutral).

Discussion

87255.	A solution has pH=3. If its hydrogen ion concentration is decreased 1000 times, the pH of the solution will be
Answer» 6 0 3 None of these Answer :A

Discussion

87256.	A solution has [H_(3)O^(+)] as 10^(-4). This solution is
Answer» Neutral Basic ACIDIC Amphoteric SOLUTION :pH is 4. It's acidic.

Discussion

87257.	A solution has an osmotic pressure of 8.314 pa at 300K. It's concentration would be:
Answer» 0.056 M 0.56 M 0.0034 M 0.034 M SOLUTION :`pi=CRT` (C = CONCENTRATION) `:. C=8.314/(8.314 XX 300)=0.0034M`

Discussion

87258.	A solution has a _____vapour pressure than the pure solution.
Answer» same HIGHER lower DEPENDS upon the SOLUTE and solvent Solution :lower

Discussion

87259.	A solution has an osmotic pressure of 0.821 atm at 300 K. Its concentration would be:
Answer» 0.033 M 0.66 M 0.033 M 0.33 M Answer :C

Discussion

87260.	A solutionhas 1:3ratioof cyclopentanetocyclohexane . Thevapourpressure of thepurecompoundsat 25^(@)Care 331 mmHg forcyclohexane. Whatis themolefractionofcyclopenyance inthevapourabovethesolution?
Answer» `0.42` `0.39` `0.49` `0.23` ANSWER :C

Discussion

87261.	A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20^@C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in vapour phase woule be
Answer» 0.786 0.478 0.549 `0.200` ANSWER :B

Discussion

87262.	A solutions 16 g of methanol and 90 g of water. The mole fraction of methanol in the solution is :
Answer» `0.200` `0.549` `0.786` `0.478` Solution :p (pentane) `=440` mm HG p (HEXANE)`=120` mm Hg `X` (pentane)`=(1)/(1+4)=0.2` `x` (hexane)`=(4)/(1+4)=0.8` p (pentane)`=0.2xx440=88`mm p (hexane)`=0.8xx120=96` mm total vapour pressure `=88+96=184` mm MOLE fraction of pentane in vapour PHASE y (pentane) `=(88)/(184)=0.478`

Discussion

87263.	A solution has 1:4 mole ratio of pentane to hexane . The vapour pressure of pure hydrocarbons at 20^@Care 440mmHgfor pentane and 120mmHg for hexane .The mole
Answer» 0.2 0.549 0.786 0.478 Solution :`(n_(C_(5)H_(12)))/(n_(C_(6)H_(14)))=1/4` `X_(C_(5)H_(12))=1/(1+4)=1/5` `X_(C_(6)H_(14))=4/(1+4)=4/5` `P_(C_(5)H_(12))^(@)=440mmHg,P_(C_(6)H_(14))^(@)=120 mm Hg` `P_(("TOTAL"))=P_(C_(5)H_(12))^(@)xxX_(C_(5)H_(12))+P_(C_(6)H_(14))^(@)xxX_(C_(6)H_(14))` `=440xx1//5+120xx4//5` `=88+96=184 mm Hg` Mole fraction `C_(5)H_(12)` in vapour phase `X_(C_(5)H_(12))=("Partoa V.P. of "C_(5)H_(12))/("Partoa V.P. of "C_(6)H_(14))=(88mm)/(184mm)=0.478`

Discussion

87264.	A solution contians Na_(2)CO_(3) and NaHCO_(3), 20cm^(3) of this solution requires 5.0 cm^(3) of 0.1 M H_(2)SO_(4) Solution for neutralization using phenolphthalein as the indicator. Methylorange is then added when a further 5.0cm^(3) of 0.2 M H_(2)SO_(4) was required. Calculate the masses of Na_(2)CO_(3) and NaHCO_(3) is 1L of this solution.
Answer» Solution :`Na_(2)CO_(3) + ( 1)/( 2) H_(2)SO_(4) rarr NaHCO_(3) + ( 1)/( 2) Na_(2) SO_(4)`phenolphthalein would change the colour after this reaction. Meq. of `H_(2)SO_(4)` used for 5ml mixture using phenolpthalein as indicator. `= 2 xx 0.1 xx 5 = 1` ` :. ( 1)/( 2) ` Meq. of `Na_(2) CO_(3) = 1 ` Now METHYL orange is ADDED in this solution after 1 end point. Meq. of `H_(2)SO_(4)` used for solution after 1 end point using methyl orange as Indicator `= 5 xx 0.2 xx 2 = 2 ` `:. ( 1)/( 2) ` Meq. of `Na_(2) O_(3) +` Meq. of `NaHCO_(3) = 2` Meq. of `NaHCO_(3) = 2-1= 1 ` `( W)/( 84) xx 1000 =1:. W = 0.084 g ` `:.` Weight of `NaHCO_(3)` in one litre `= ( 0.084 xx 1000)/( 20) = 4.2 g ` `:.` Meq. of `Na_(2) CO_(3) = 2 :. ( W)/( 53) xx 1000 = 2 ` or `W = ( 106)/( 1000) = 0.106 ` `:. ` Weight of `NaHCO_(3)` in litre

Discussion

87265.	A solution contians a mixutre of Na_(2) CO_(3) and NaOH. Using phenolphthalein as indicator 25ml of mixturerequired 19.5 ml of 0.995 N HCl for the end point. With methyl organge. 25 ml of solution required 25ml, of the same . HCl for the end point. Calculate grams per litre of each substance in the mixture.
Answer» Solution :meq of `Na_(2) CO_(3) =2 xx `moles`Na_(2) CO_(3) = a ` meq of NaOH =`b=1xx` moles of NaOH with PHENOL PHTHALEIN `Na_(2)CO_(3) + HCl rarr NaHCO_(3) + NACL`nf = 1 `NaOH + HCl rarr NaCl + H_(2) O`nf = 1 `:. a//2+b =` meq of HCl = `9.5 xx 0.995 = 19.4 `with methyl orange `Na_(2) CO_(3) + HCl rarr NAcL +H_(2) CO_(3)`nf = 2 `NaOH + HCl rarr NaCl + H_(2) O`nf = 1 `a+b =` meq of HCl `= 25 xx .995 = 24.875` `a//2=5.475 rArr a = 10.95 .b = 13.925` wt of `Na_(2) CO_(3) //` lit `= ( 10.95)/( 25) xx 10^(-3) xx ( 106)/( 2) xx 1000 = 23.2 gm//lit ` wt ofNaOH `//` lit `= b xx 10^(-3) xx 84 // 25 xx1000 = 22.8 gm //lit `

Discussion

87266.	A solutioncontaining 1.8 g of a compound (empirical formula CH_(2)O) in 40 g of water is observed to freeze at -0.465^(@)C. The molecules formuleaof the compound is (K_(f) of water =1.86kg Kmol^(-1)):
Answer» `C_(2)H_(4)O_(2)` `C_(3)H_(6)O_(3)` `C_(5)H_(10)O_(5)` `C_(6)H_(12)O_(6)` ANSWER :D

Discussion

87267.	A solution contains Na_(2)CO_(3) and NaHCO_(3) 20 cm^(3) of this solution requires 5.0 cm3 of 0.1 M H_(2) SO_(4) solution for neutralization using phenolphthalein as the indicator. Methylorange is then added when a further 5.0 cm^(3) of 0.2 M H_(2)SO_(4) was required. Calculate the masses of Na_(2)CO_(3) and NaHCO_(3) in 1 L of this solution.
Answer» SOLUTION :5.3g,4.2g

Discussion

87268.	A solution contains Na_(2)CO_(3) and NaHCO_(3) . 20 ml of this solution required 4 ml of 1 N HCl for titration with Ph indicator. The titration was repeated with the same volume of the solution but with MeOH. 10.5 ml of 1 N HCl was required this time. Calculate the amount of Na_(2)CO_(3) & NaHCO_(3).
Answer» ANSWER :0.424 GM; 0.21 gm

Discussion

87269.	A solution contains Na_(2)CO_(3) and NaHCO_(3) . 10 mL of the solution requires 2.5 mL of 0.1 " M " H_(2)SO_(4) for neutralisation using phenlphthalein as an indicator . Methyl orange is added when a further 2.5 mL of 0.2 " M " H_(2)SO_(4) was required . calculate the amountof Na_(2)CO_(3) and NaHCO_(3)in one litre of the solution .
Answer» SOLUTION :The NEUTRALISATION reactions are ` 2Na_(2)CO_(3) +H_(2)SO_(4) to 2NaHCO_(3) +na_(2)SO_(4)` `2NaHCO_(3) +H_(2)SO_(4) to Na_(2)SO_(4) +H_(2)O +CO_(2)` The volume of `H_(2)SO_(4)` (2.5 mL ) , used while using phenophthalein correspondsto the volume required for conversion of `NaHCO_(3) " to " Na_(2)HC_(3)` .THUS at the end pointwith phenolphthalein , we have , m.eof `2.5 " mL of " 0.1 M (i.e , 0.2 N) H_(2)SO_(4) = m.e of Na_(2)CO_(3)` or m.eof `Na_(2)CO_(3) = 0.5` Equivalent of `na_(2)CO_(3) =(0.5)/(1000)` Wtof `Na_(2)CO_(3)//10 m L = (0.5)/1000 XX 106 = 0.053 g ` equivalent wt . of `Na_(2)CO_(3)`is 106 according to given reaction ) ` :. ` wtof `Na_(2)CO_(3) =(0.5)/(1000)` Wt . of `Na_(2)CO_(3)//10 mL = (0.5)/1000 xx 106 = 0.053 g ` (equivalent wt . of `Na_(2)CO_(3)` is 106 according to given reaction ) ` :. ` wt of `Na_(2)CO_(3)` per litre = 5.3 g . Againwith methyl ORANGE , we have , m.e of 2.5 mL of `0.2 ` M (i.e., 0.4 N ) `H_(2)SO_(4)` solution = m.e of `NaHCO_(3)` produced from `na_(2)CO_(3)of NaHCO_(3)` originally present . ` :. ` m.e of `NaHCO_(3)` originally present = `1- 0.5 = 0.5` ` :. ` equivalentof `NaHCO_(3) = (0.5)/1000` ` :. ` wt . of `NaHCO_(3) ` per mL = `(0.5)/1000 xx 84 = 0.042 g ` ( eq. wt of `NaHCO_(3) = 84` according to given reaction Wt. of`NaHCO_(3)` per mL = `(0.5)/1000 xx 84 = 0.042 g ` (eq.wt f `NaHCO_(3) =84 `according to given reaction ) Wt. of `NaHCO_(3) ` per litre = 4.2 g .

Discussion

87270.	A solution contains mixture of H_(2)SO_(4)" and " H_(2)C_(2)O_(4).25 ml of this solution requires 35.5 ml of N/10 for neutralization and 23.45ml of N/10 KMnO_(4) for oxidation, calculate (i) Normality of H_(2)C_(2)O_(4) " and " H_(2)SO_(4) (ii) Strength of H_(2)C_(2)O_(4)" and " H_(2)SO_(4)
Answer» SOLUTION :N/A

Discussion

87271.	A solution contains Fe^(2+),Fe^(3+) and I^(-) ions. This solution was treated with iodine at 35^(@)C. E^(@) for Fe^(3+)//Fe^(2+) is +0.77V and E^(@) for I_(2)//2I^(-)=0.536V. The favourable redox reaction is
Answer» `I^(-)` will oxidised to `I_(2)` `Fe^(2+)` will be oxidised to `Fe^(3+)` `I_(2)` will be reduced to `I^(-)` There will be no redox reaction Solution :`2(e^(-)+Fe^(+3)toFe^(+2)""E^(@)=0.77V` `underline(2I^(-)toI_(2)+2E^(-)""E^(@)=0.536V"")` `2FE^(+3)+2e^(-)to2Fe^(+2)+I_(2)""E^(o)=E_(o x)^(o)+E_(RED)^(o)` `=0.77=0.536=0.164V` So, Reaction will TAKEN place.

Discussion

87272.	A solution contains Fe^(2+),Fe^(3+) and I^(-) ions. This Fe^(3+)//Fe^(2+) is +0.77 V and E^(@) for I_(2)//2I^(-) is 0.536 V. the favourable redox reaction is
Answer» `I_(2)` will be reduced to `I^(-)` There will be no redox reaction `I^(-)` will be oxidized to `I_(2)` `FE^(2+)` will be oxidised to `Fe^(3+)` SOLUTION :The reaction will take place for which `E_(cell)^(@)` is positive. (a) `I_(2)+2e^(-)to2I^(-),E_(red)^(@)=0.536V` `Fe^(2+)toFe^(3+)+e^(-),E_(o x )^(@)=-0.770V` `thereforeE_(cell)^(@)` will be -ve. (b) Redox reaction will occur. (c) `2I^(-)toI_(2)+2e^(-),E_(o x)^(@)=0.536V` `underline(""Fe^(3+)+e^(-)toFe^(2+),E_(red)^(@)=+0.770V"")` `2FE^(3+)+2I^(-)to2Fe^(2+)+I_(2),E_(cell)^(@)=+0.164V` Thus, `E_(cell)^(@)` is +ve. hence, this reaction will occur. (d). also will not occur.

Discussion

87273.	A solution contains Fe^(2+), Fe^(3+) and I^(–) ions. This solution was treated with iodine at 25^(@)C. E^(@) for Fe^(3+)// Fe^(2+) is + 0.77 V and E^(o) for I2//2I– = 0.536 V. The favourable reaction is
Answer» `I _(2)` will be REDUCED to` I^(–)` There will be no REDOX reaction `I^(–)` will be oxidised to `I_(2)` `Fe^(2+ )` will be oxidised to `Fe^(3+)` . ANSWER :B

Discussion

87274.	A solution contains Fe^(2+), Fe^(3+) and I^(-) ions. The solution was treated with iodine at 35^(@)C. E^(@) for Fe^(3+)//Fe^(2+) is +0.77" V " and E^(0) for I_(2)//2I^(-) is +0.536" V". The favourable redox reaction is :
Answer» `I_(2)` will be reduced to `I^(-)` These will be no redox REACTION `I_(2)` will be oxidised to `I_(2)` `Fe^(2+)` will be oxidised to `Fe^(3+)` Solution :(c ) SINCE the reduction POTENTIAL of `Fe^(2+)//Fe^(3+)` is more than that of `l_(2)//L^(-)`, `l^(-)` in the redox reaction. `l^(-)` ions will be oxidised to `l_(2)` at the cost of `Fe^(3+)` ions which will be reduced : `2Fe^(3+)+2L^(-) to 2Fe^(2+)+l_(2)`

Discussion

87275.	A solution contains a mixture of isotopes of X""^(A""_(1))(t""_(1//2)=14 days) and X""^(A""_(2))(t""_(1//2)=25days). Total activity is 1 curie at t=0. The activity reduces by 50% in 20 days. Find : a) The initial activities of X""^(A""_(1)) and X""^A""_(2)). b) The ratio of their initial number of nuclei.
Answer» Solution :Let activity of `X""_(A""_(1))ANDX""_(A""_(2))` are a and B curie respectively at t = 0 `therefore` a + b =1 curie …(1) Now rate `mu` Number of atoms `therefore`For `X""^(A""_(1))" "t=(2.303)/(k)log""_(10)"(N""_(0))/(N)=(2.303)/(k)log""_(10).(r""_(0))/(r)` `20=(2.303xx14)/(0693)log""_(10).(a)/(r""_(1))` `thereforer""_(1)=0.3716a` For `X""^(A""_(2))""t=(2.303)/(k)log""_(10).(N""_(0))/(N)=(2.303)/(k)log""_(10).(r""_(0))/(r)` `20=(2.303xx25)/(0.693)log""_(10).(b)/(r""_(2))` , `r""_(2)=0.5744b` GIVEN activity after 20 days `=(1)/(2)`cruie 03716a+ 0.5744b `=(1)/(2)` or `0.7432a + 1.1488b = 1""...(2)` Byequation (1) and (2) a = 0.3669 Ci = 0.3669 `xx 3.7xx10""^(10)` dps b = 0.6331 Ci = 0.6331 `xx3.7xx10""^(10)` dps Now rate = k.N`because` a = 03669 curie For 0.3669 `xx10""^(10)xx` 3.7 = `(0.693)/(14xx24xx60xx60)N""_(0)""^(A""_(1))` For 0.6331 `xx10""^(10)xx` 3.7 =`(0.693)/(25xx24xx60xx60)N""_(0)""^(A""_(2))` `therefore(N""_(0)""^(A""_(1)))/(N""_(0)""^(A""_(2)))=0.3245`

Discussion

87276.	A solution contains Cu(CH_3COO)_2.Cd(CH_3COO)_2 and Zn(CH_3COO)_2. On passing H_2S gas , there is precipitation of sulphide of :
Answer» `ZN^(2+) , CD^(2+) ` `CU^(2+), Cd^(2+) ` `Zn^(2+), Cu^(2+) ` `Zn^(2+), Cu^(2+) , Cd^(2+)` . ANSWER :D

Discussion

87277.	A solution contains a mixture of Na_(2)CO_(3) and NaOH. Using Ph as indicator 25 ml of mix required 19.5 ml of 1 N HCl for the end point. With MeOH, 25 ml of the solution required 25 ml of the same HCl for the end point. Calculate gms/L of each substance in the mixture.
Answer» ANSWER :23.32 GM , 22.4gm

Discussion

87278.	A solution contains both Ag^+ (0.30 M) and Ba^(2+)(0.05 M).(i) If solid Na_2 SO_4is added very slowly to this solution, which will precipitate first, Ag_2SO_4 or BaSO_4?(ii) The addition of Na_2SO_4is continued until the second cation just starts precipitating as sulphate. What is the concentration of the first cation at this point?K_(sp)(Ag_2SO_4) = 1.2 xx 10^(-5), K_(sp)(BaSO_4) = 1.5 xx 10^(-9) .
Answer» SOLUTION :`BaSO_4, 1.15 XX 10^(-5) M`

Discussion

87279.	A solution contains 90 g of H_(2)O, 6.4 g of methanol and 18.4 g of glycerol. What is the mole fraction of glycerol ? ("Glycerol "=CH_(2)OH-CHOH-CH_(2)OH)
Answer» SOLUTION :`"MOLE FRACTION of GLYCEROL "=(18.4//92)/(18.4//92+90//18+6.4//32)=(0.2)/(0.2+5+0.2)=(0.2)/(5.4)=0.037`

Discussion

87280.	A solution contains 2.675 g of CoCl_3.6NH_3( molar mass =267.5 g mol^(-) ) is passed through a cation exchanger . The chloride ions obtained in solution were treated with excess of AgNO_3 to give 4.78 g of AgCl. The formula of the complex is_________.
Answer» `[CoCl(NH_3)_5]Cl_2` `[CO(NH_3)_6]Cl_3` `[Co(NH_3)_4Cl_2]CL` `[Co(NH_3)_3Cl_3]` Answer :B

Discussion

87281.	A solution contains 25% water, 25% ethanol and 505 aceticacid by mass. Calculate the mole fraction of each component.
Answer» SOLUTION :Let us suppose that the total mass of the solution is 100 g. Then AMOUNT of water = 25 g , `""` Amount of ethanol = 25 g `"25 g of water "=("25 g ")/("18 g MOL"^(-1))="1.388 moles"[because"Malar mass of water "="18 g mol"^(-1)]` `"50 g of ethanol "=("25 g")/("46 g mol"^(-1))="0.543 moles"[because "Molar mass of ethanol "(C_(2)H_(5)OH)="46 g mol"^(-1)]` `"50 g of ACETIC acid "=("50 g")/("60 g mol"^(-1))="0.833 moles"` `""[because"Molar mass of acetic acid "(CH_(3)COOH)="60 g mol"^(-1)]` `therefore"Mole fraction of water "=(1.388)/(1.388+0.543+0.833)=(1.388)/(2.764)=0.503` Mole fraction of ethanol `=(0.543)/(2.764)=0.196"and mole fraction of acetic acid "=(0.833)/(2.764)=0.301.`

Discussion

87282.	A solution contains 25% water, 25% ethanol (C_(2)H_(5)OH ) and 50% acetic acid (CH_(3)COOH) by mass. The mole fraction of
Answer» water =0.502 ETHANOL = 0.302 acetic ACID = 0.196 ethanol + acetic acid = 0.497 Solution :MOLES of water `=25/18 = 1.388` Moles of ethanol `=25/46 = 0.543` Moles of acetic acid `=50/60 = 0.833` TOTAL moles `=1.388 + 0.543 + 0.833 = 2.764` `x_("water") = 1.388/2.764 = 0.502` `e_("ethanol") = 0.196` `e_("acetic acid") = 0.301` `x_("ethanol") + x_("ACETI acid") = 0.497`

Discussion

87283.	A solution contains 25% H_(2)O, 25% C_(2)H_(5)OH and 50%CH_(3)COOH by mass. The mole fraction of H_(2)O would be
Answer» `0.25` `2.5` `0.503` `5.03` Solution :`X_(H_(2)O)=(n_(H_(2)O))/(n_(H_(2)O)+n_(C_(2)H_(5)OH)+n_(CH_(3)COOH))`

Discussion

87284.	A solution contains 1.2046xx10^(24) hydrochloric acid molecules in one dm^(3) of the solution. The strength of the solution is
Answer» 6N 2N 4N 8N SOLUTION :STRENGTH of the solution =`(1.2046xx10^(24))/(6.023xx10^(23))=2N`

Discussion

87285.	A solution contains 1 mCi of L-phenylalanine C^(14) labelled in 2.0 mL solution. The specific activity of labelled sample is given as 150 mCi "mmol"^(-1). Calculate (a) The concentration of the sample in the solution in "mol" L^(-1) (b). The activity of solution in terms of counting per minute per mL at counting of 80%
Answer» SOLUTION :a. 1 mmol ` = 150 mCi` `1 mCi = (1)/(150)` mmol `= (1)/(150 xx 2) = 3.33 xx 10^(-2) M` b. `1 Ci = 3.7 xx 10^(10) dps` `= 3.7 xx 10^(10) xx 60 dpm` `= 3.7 xx 10^(10) xx 60 xx (80)/(100)` counts `"min"^(-1)` `= (177.6 xx 10^(10))/(2 ML) = 88.8 xx 10^(10)` counts `"min"^(-1) mL^(-1)` ` mCi = 88.8 xx 10^(7)` count `"min"^(-1) mL^(-1)`

Discussion

87286.	A solution contains 1 g mol each of p-toluenediazonium chloride and p-nitrophenyldiazonium chloride. To this 1 g mol of alkali solution of phenol is added. Predict the major product product. Explain your answer.
Answer» Solution :In ALKALINE MEDIUM, PHENOL forms PHENOXIDE ion which is more electron rich and hence more reactive than phenol. Coupling reactions between diazonium salts is an example of electrophilic aromatic substitution. In this reaction, diazonium salt acts as the electrophile and phenoxide ion as the nucleophile. evidently stronger the electrophile faster is the reaction. Now due to electrowithdrawing effect of the `-NO_(2)` group, p-nitrophenyldiazonium cation is a stronger electrophile than p-toluenediazonium cation (+I-effect of `CH_(3)` grou reduces its electrophilicity) and hence couples preferentially with phenol to FORM 4-hydroxy-4'-nitroazobenzene as the major product.

Discussion

87287.	A solution contains 0.1MH_(2)S and 0.3MHCl .Calculate the conc.of S^(2-) and HS^(-) ions is solutions.Given K_(a_(1)) and K_(a_(2)) are 10^(-7)and 1.3xx10^(-13) respectively.
Answer» Solution :`H_(2)ShArr H^(+)+HS^(-),K_(a_(1))=10^(-7)` `HS^(-)hArrH^(+)+S^(2-),K_(a_(2))=1.3xx10^(-13)` `HClrarr H^(+)+Cl^(-)` Due to coomon ion effect EXERTED by `H^(+)`of `HCl` ,the dissocation of `H_(2)S` are supposed and the `[H^(+)]` in solution is mainly due to `HCl`. `K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])` `10^(-7)=([0.3][HS^(-)])/([0.1])[:.[H^(+)]` from `HCl=0.3` would have DISSOCIATED NEGLIGIBLY] `[HS^(-)]=(10^(-7)xx0.1)/(0.3)=3.3xx10^(-8)M` Further `K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)])rArr 1.3xx10^(-13)([0.3][s^(2-)])/(3.3xx10^(-8))` `[S^(2=)]=(1.3xx10^(-13)xx3.3xx10^(-8))/(0.03)=1.43xx10^(-19)M`

Discussion

87288.	A solution contains 0.1 M H_2S and 0.3 M HCl. Calculate the conc. of S^(2-) and HS^(-) ions in solution. Given K_(a_(1)) and K_(a_(2))" for "H_2S" are "10^(-7) and 1.3 xx 10^(-13)" respectively".
Answer» Solution :`H_(2)S LEFTRIGHTARROW H^(+)+HS^(-)""K_(a_(1))=10^(-7)` `HS^(-) Leftrightarrow H^(+)+S^(2-) K_(a_(2))=1.3 xx 10^(-13)` Due to COMMON ion effect the dissociation of `H_2S` is suppressed and the `[H^+]` in solution is due to HCl `therefore K_(a_(1))=([H^(+)][HS^(-1)])/([H_(2)S])` `10^(-7)= ([0.3][HS^(-1)])/([0.1]) [? [H^(+)]" fromHCl "=0.3]` `therefore [HS^(-1)]=(10^(-7) xx 0.1)/(0.3)` `=3.3 xx 106^(-8)M` Further `K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)]) and K_(a_(1))=([H^(+)][HS^(-1)])/([H_(2)S]` `therefore K_(a_(1)) xx K_(a_(2))=([H^(+)][S^(2-)])/([H_(2)S])` `10^(-7) xx 1.3 xx 10^(-13)=([0.3]^(3)[S^(2-)])/([0.1])` `therefore [S^(2-)]=(1.3 xx 10^(-20)xx 0.1)/(0.09)` `=1.44 xx 10^(-20)M`

Discussion

87289.	A solution contains 0.18 g/mL of a substance X, whose molecular weight isapproximately 68000. It is found that 0.27 mL of oxygen at 760 mmHg and 30^@Cwill combine with the amount of X contained in 1.0 mL of the solution. How many molecules of oxygen will combine with one molecule of X?
Answer» ANSWER :4

Discussion

87290.	A solution contains 0.1 M H_2 Sand 0.3 M HCl. Calculate the concentration of S2- and HS ions in the solution. For H_2S K_(a_1) = 1 xx 10^(-7) , K_(a_2) = 1.3 xx 10^(-13)
Answer» SOLUTION :`3.34 XX 10^(-8) M, 1.447 xx 10^(-20) M`

Discussion

87291.	A solutioncontains 0.1 MCl^(-)and 0.001MCrO_(4)^(2-). If solidAgNO_(3)is graduallyaddedto thissolutionwhich willprecipitatefirst, AgClor Ag_(2)CrO_(4) ? Assumethat the additioncauses no . Changein volume . GivenK_(ap(AgCl)) = 1.7 xx 10^(10) M^(2)and K_(ap(Ag_(2)CrO_(4))) = 1.79 xx10^(-12) M^(3). What %ofCl^(-)remainsin solution when CrO_(4)^(2-)startsprecipitating?
Answer» SOLUTION :`4XX10^(-3) %`

Discussion

87292.	A solution contains 0.06 M of Cu^(2+)ions and Ag^+of unknown concentration. Find the concentration of Ag * ions so that both the metals can be codeposited.E_(Cu^(2+) Cu)^@ = +0.33 V and E_(Ag^(+) , Ag = + 0.7991V
Answer» SOLUTION :`3.7 XX 10^(-9) M`

Discussion

87293.	A solution contains 0.01 M metal ion (Zn^(2+) and Cu^(2+)). It is saturated by passing H_(2)S gas in the M. The solubility products of ZnS and CuS are 3.0xx10^(-22)and8.0xx10^(-36)respectively.Which of the following will occur ?
Answer» ZNS will precipitate CUS will precipitate Both ZnS and CuS will precipitate Both `ZN^(2+)`and `Cu^(2+)`will remain in the solution. Answer :B

Discussion

87294.	A solution containing one mole per litre of each Cu(NO_(3))_(2),AgNO_(3),Hg_(2)(NO_(3))_(2) and Mg(NO_(3))_(2) is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are Ag//Ag^(+)=+0.80,2Hg//Hg_(2)^(2+)=+0.79,Cu//Cu^(2+)=0.34,Mg//Mg^(2+)=-2.37 with increasing voltage, the sequence of deposition of metals on the cathode will be
Answer» AG,Hg,Cu,Mg Mg,Cu,Hg,Ag Ag,Hg,Cu Cu,Hg,Ag Solution :A cation havig highest REDUCTION potential will be reduced first and so on. However, `Mg^(2+)` in aqueous solution will not be reduced `(E_(Mg^(2+)//Mg)^(o)ltE_(H_(2)O//(1)/(2)H_(2)+OH^(-)))` INSTEAD water would be reduced in preference.

Discussion

87295.	A solution containing one mole per litre each of Cu(NO_(3))_(2),AgNO_(3),Hg(NO_(3))_(2) and Mg(NO_(3))_(2) is being electrolysed by using inerty electrodes. The values of the standard oxidation potentials in vlts are Ag//Ag^(+)=-0.8V,Ag//Hg^(2+)=-79V,Cu//Cu^(2+)=-0.34V,Mg//Mg^(2+)=2.37V. The order in which metals will be formed at cathode will be-
Answer» `Ag,Cu,Ag,MG` `Ag,Hg,Cu,Mg` `Ag,Hg,Cu` `Cu,Hg,Ag` SOLUTION :Metal having more VALUE of SRP will deposit first, but Mg does not DEPOSITION AQUEOUS solution.

Discussion

87296.	A solution containing one mole per litre each of Cu(NO_(3))_(2),AgNO_(3),Hg_(2)(NO_(3))_(2),Mg(NO_(3))_(2) is being electrolyesd by using inert electrodes. The value of standard reduction potentials are Ag^(+)//Ag= +0.80V,Hg_(2)^(2+)//Hg=0.79V, Cu^(2+)// Cu=+0.34V,Mg^(2+)//Mg=-2.7V. with increasing voltage, the sequence of deposition of metals on the cathode will be:
Answer» AG,MGHG,CU Cu,HG,Ag,Mg Ag,Hg,Cu Mg,Cu,Hg,Ag Answer :C

Discussion

87297.	A solution containing one mole per litre each of Cu(NO_3)_2, AgNO_3, Hg_2(NO_3)_2 and Mg (NO_3)_2 is being electrolysed by using inert electrodes. The values of standard electrode potentials in volt (reduction potentials) are, Ag\|Ag^+ =0.80.2Hg\|Hg_2^(2+) =0.79,Cu\|Cu^(2+)=+0.34 and Mg^(2+)=-2.37 With increasign voltage, the sequence of deposition of metals on the cathode will be :
Answer» Ag, HG , Cu,Mg Mg, Cu, Hg, Ag Ag, Hg, Cu Answer :C

Discussion

87298.	A solution containing Na_(2)CO_(3) and NaOH requires 300ml of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to the above titrated solution when a further 25ml of 0.2 N HCl is required . The amount of NaOH present in solution is ( NaOH = 40, Na_(2)CO_(3) = 106 )
Answer» 0.6g 1.0 G 1.5 g 2.0 g ANSWER :B

Discussion

87299.	A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 N HCl using phonolphthalein as a indicator. Methyl orange is then added to above titrated solution when a further 25 ml of 0.2 N HCl is required. The amount of NaoH present in the original solution is
Answer» 0.5 g 1 g 2 g 4 g Answer :B

Discussion

87300.	A solution containing Fe^(3+) is titrated against a standard solution of Ti^(3+) using ammonium thiocyanate as indicator. The colour of the solution at end point will be
Answer» Red Colourless Blue `Fe^(3+)` is not OXIDIZED by `TI^(3+)` Solution :`Ti^(3+)` is used in volumetric analysis for the determination of `Fe^(3+)` and also organic nitro COMPOUNDS. In the iron titration the end point may be detected with ammonium thiocyanate which REMAINS red whilst any `Fe^(3+)` is present and becomes colourless at the end point, or by methylene blue which is reduced and decolorized as SOON as `Ti^(3+)` is in excess.

Discussion

Explore topic-wise InterviewSolutions in .

A solution is

A solution is 0.10 M in Na_2 SO_4 . When 50 mL of 0.1 M Ba(NO_3) , is added to 50 mL of this solution, what fraction of the sulphate ion is not precipitated? K_(sp) (BaSO_4) = 1.1 xx 10^(-10)

A solution is 0.10 M Co^(2+)and 0.10 M Hg^(2+). Calculate the range of pH in whichonly one of the metal sulphides precipitates when the solution is saturated in H_2S. K_(sp) (CoS) = 4 xx 10^(-21) and K_(sp)(HgS) = 1.6 xx 10^(-52)

A solution has pH = 5, it is diluted 100 times, then it will become

A solution has pH=3. If its hydrogen ion concentration is decreased 1000 times, the pH of the solution will be

A solution has [H_(3)O^(+)] as 10^(-4). This solution is

A solution has an osmotic pressure of 8.314 pa at 300K. It's concentration would be:

A solution has a _____vapour pressure than the pure solution.

A solution has an osmotic pressure of 0.821 atm at 300 K. Its concentration would be:

A solutionhas 1:3ratioof cyclopentanetocyclohexane . Thevapourpressure of thepurecompoundsat 25^(@)Care 331 mmHg forcyclohexane. Whatis themolefractionofcyclopenyance inthevapourabovethesolution?

A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20^@C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in vapour phase woule be

A solutions 16 g of methanol and 90 g of water. The mole fraction of methanol in the solution is :

A solution has 1:4 mole ratio of pentane to hexane . The vapour pressure of pure hydrocarbons at 20^@Care 440mmHgfor pentane and 120mmHg for hexane .The mole

A solutioncontaining 1.8 g of a compound (empirical formula CH_(2)O) in 40 g of water is observed to freeze at -0.465^(@)C. The molecules formuleaof the compound is (K_(f) of water =1.86kg Kmol^(-1)):

A solution contains mixture of H_(2)SO_(4)" and " H_(2)C_(2)O_(4).25 ml of this solution requires 35.5 ml of N/10 for neutralization and 23.45ml of N/10 KMnO_(4) for oxidation, calculate (i) Normality of H_(2)C_(2)O_(4) " and " H_(2)SO_(4) (ii) Strength of H_(2)C_(2)O_(4)" and " H_(2)SO_(4)

A solution contains Fe^(2+),Fe^(3+) and I^(-) ions. This solution was treated with iodine at 35^(@)C. E^(@) for Fe^(3+)//Fe^(2+) is +0.77V and E^(@) for I_(2)//2I^(-)=0.536V. The favourable redox reaction is

A solution contains Fe^(2+),Fe^(3+) and I^(-) ions. This Fe^(3+)//Fe^(2+) is +0.77 V and E^(@) for I_(2)//2I^(-) is 0.536 V. the favourable redox reaction is

A solution contains Fe^(2+), Fe^(3+) and I^(–) ions. This solution was treated with iodine at 25^(@)C. E^(@) for Fe^(3+)// Fe^(2+) is + 0.77 V and E^(o) for I2//2I– = 0.536 V. The favourable reaction is

A solution contains Fe^(2+), Fe^(3+) and I^(-) ions. The solution was treated with iodine at 35^(@)C. E^(@) for Fe^(3+)//Fe^(2+) is +0.77" V " and E^(0) for I_(2)//2I^(-) is +0.536" V". The favourable redox reaction is :

A solution contains Cu(CH_3COO)_2.Cd(CH_3COO)_2 and Zn(CH_3COO)_2. On passing H_2S gas , there is precipitation of sulphide of :

A solution contains a mixture of Na_(2)CO_(3) and NaOH. Using Ph as indicator 25 ml of mix required 19.5 ml of 1 N HCl for the end point. With MeOH, 25 ml of the solution required 25 ml of the same HCl for the end point. Calculate gms/L of each substance in the mixture.

A solution contains 90 g of H_(2)O, 6.4 g of methanol and 18.4 g of glycerol. What is the mole fraction of glycerol ? ("Glycerol "=CH_(2)OH-CHOH-CH_(2)OH)

A solution contains 2.675 g of CoCl_3.6NH_3( molar mass =267.5 g mol^(-) ) is passed through a cation exchanger . The chloride ions obtained in solution were treated with excess of AgNO_3 to give 4.78 g of AgCl. The formula of the complex is_________.

A solution contains 25% water, 25% ethanol and 505 aceticacid by mass. Calculate the mole fraction of each component.

A solution contains 25% water, 25% ethanol (C_(2)H_(5)OH ) and 50% acetic acid (CH_(3)COOH) by mass. The mole fraction of

A solution contains 25% H_(2)O, 25% C_(2)H_(5)OH and 50%CH_(3)COOH by mass. The mole fraction of H_(2)O would be

A solution contains 1.2046xx10^(24) hydrochloric acid molecules in one dm^(3) of the solution. The strength of the solution is

A solution contains 1 g mol each of p-toluenediazonium chloride and p-nitrophenyldiazonium chloride. To this 1 g mol of alkali solution of phenol is added. Predict the major product product. Explain your answer.

A solution contains 0.1MH_(2)S and 0.3MHCl .Calculate the conc.of S^(2-) and HS^(-) ions is solutions.Given K_(a_(1)) and K_(a_(2)) are 10^(-7)and 1.3xx10^(-13) respectively.

A solution contains 0.1 M H_2S and 0.3 M HCl. Calculate the conc. of S^(2-) and HS^(-) ions in solution. Given K_(a_(1)) and K_(a_(2))" for "H_2S" are "10^(-7) and 1.3 xx 10^(-13)" respectively".

A solution contains 0.18 g/mL of a substance X, whose molecular weight isapproximately 68000. It is found that 0.27 mL of oxygen at 760 mmHg and 30^@Cwill combine with the amount of X contained in 1.0 mL of the solution. How many molecules of oxygen will combine with one molecule of X?

A solution contains 0.1 M H_2 Sand 0.3 M HCl. Calculate the concentration of S2- and HS ions in the solution. For H_2S K_(a_1) = 1 xx 10^(-7) , K_(a_2) = 1.3 xx 10^(-13)

A solution contains 0.06 M of Cu^(2+)ions and Ag^+of unknown concentration. Find the concentration of Ag * ions so that both the metals can be codeposited.E_(Cu^(2+) Cu)^@ = +0.33 V and E_(Ag^(+) , Ag = + 0.7991V

A solution contains 0.01 M metal ion (Zn^(2+) and Cu^(2+)). It is saturated by passing H_(2)S gas in the M. The solubility products of ZnS and CuS are 3.0xx10^(-22)and8.0xx10^(-36)respectively.Which of the following will occur ?

A solution containing Na_(2)CO_(3) and NaOH requires 300ml of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to the above titrated solution when a further 25ml of 0.2 N HCl is required . The amount of NaOH present in solution is ( NaOH = 40, Na_(2)CO_(3) = 106 )

A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 N HCl using phonolphthalein as a indicator. Methyl orange is then added to above titrated solution when a further 25 ml of 0.2 N HCl is required. The amount of NaoH present in the original solution is

A solution containing Fe^(3+) is titrated against a standard solution of Ti^(3+) using ammonium thiocyanate as indicator. The colour of the solution at end point will be