94420 + Interview Questions in Chemistry in Class 12 Page 1745 InterviewSolution

87201.	A solution of a metal ion when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. the metal ion is
Answer» `Pb^(2+)` `HG^(2+)` `Cu^(2+)` `Co^(2+)` Solution :`Hg^(2+)+2KItoHgI_(2)darr+underset("red PPT")(2K^(+))` `HgI_(2)+2KItoK_(2)HgI_(4)` `Hg^(2+)+Co^(2+)+underset("deep BLUE crystalline")(4SCN^(-)toCoHg(SCN)_(4)darr"")`

Discussion

87202.	A solution of a metal ion when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution. Morever, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is :
Answer» `Pb^(2+) ` `Hg^(2+) ` `CU^(2+)` `Co^(2+)` SOLUTION :`Hg^(2+) (aq) + 2KI (aq) to underset("(Scarlet RED)")(HgI_2 (s)) + 2K^(+) (aq)` `HgI_2(s) + 2KI (aq) to K_2[HgI_4](aq)` `Co^(2+)(aq) + Hg^(2+) (aq) + 4SCN^(-) (aq) to underset("DEEP blue ")(Co[Hg(SCN)_4])(s)`

Discussion

87203.	A solution of a metal ion when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution.Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate.The metal ion is:
Answer» `Pb^(2+)` `Hg^(2+)` `Cu^(2+)` `Co^(2+)` SOLUTION :`Hg^(2+)+2I^(-)to HgI_(2)DARR` (scarlet/red) `HgI_(2)+2I^(-)("excess")to[HgI_(4)]^(2-)`(soluble complex) `Hg^(2+) +Co(SCN)_(2) to Co[Hg(SCN)_(4)]darr` (DEEP blue)

Discussion

87204.	A solution of A is mixed with an equal volume of a solution of B containing the same number of moles, and the reaction A + B = C occurs. At the end of 1h, A is 75% reacted. How much of A will be left unreacted at the end of 2 h if the reaction is (a) first order in A and zero order in B, (b) first order in both A and B , and (c) zero order in both A and B ?
Answer» SOLUTION :(a) 6.25, (B) 14.3, (C )0%

Discussion

87205.	A solution of A and B with 30 moles present of A is in equilibrium with its vapour which contain 60 mole percent of A. Assuming that the solution and the vapour behave ideally, calculate the ratio of the vapour pressures of pure A and pure B.
Answer» Solution :`{:("In solution",x_(A)=0.30, therefore, x_(B)=0.70),("In the vapour phase,",y_(A)=0.60,therefore,y_(B)=0.40):}` But`y_(A)=(p_(A))/("TOTAL V.P.")=(p_(A))/(p_(A)+p_(B))=(x_(A)p_(A)^(@))/(x_(A)p_(A)^(@)+x_(B)p_(B)^(@))=(0.30p_(A)^(@))/(0.30p_(A)^(@)+0.70p_(B)^(@))=0.60"...(i)"` `y_(B)=(p_(B))/(p_(A)+p_(B))=(x_(B)p_(B)^(@))/(x_(A)p_(A)^(@)+x_(B)p_(B)^(@))=(0.70p_(B)^(@))/(0.30p_(A)^(@)+0.70p_(B)^(@))=0.40"...(II)"` Dividing EQN. (i) by eqn. (ii), `(0.30p_(A)^(@))/(0.70p_(B)^(@))=(0.60)/(0.40) or (p_(A)^(@))/(p_(B)^(@))=(0.60)/(0.40)xx(0.70)/(0.30)=3.5`

Discussion

87206.	A solution of a 0.4 g sample of H_(2)O_(2) reactedwith 0.632g of KMnO_(4)in the pressure of sulphuricacid.Calculatedthe percentage purity of the sample of H_(2)O_(2) .
Answer» Solution :`{:(2KMnO_(4),+3H_(2)SO_(4),+5H_(2)O_(2) ,to K_(2)SO_(4),+2MnSO_(4),+8H_(2)O,+5O_(2)),("+7",,,,"+2",,):}` Eq . Wt . Of `KMnO_(4) = (" mol.wt")/("change in ON per mole ") = 158/5 = 31/6` Again from the above reaction we see that 2 moles of `KMnO_(4)` combine with 5 moles of `H_(2)O_(2)` Again from the above reaction we see that 2 moles of `KMnO_(4)` combine with 5 moles of `H_(2)O_(2)` or 1 equivalentof `KMnO_(4)` combines with `1/2 ` molesof `H_(2)O_(2)` ` :. ` equivalentwt. of `H_(2)O_(2) = (" mol .wt")/2 = 34/2 = 17 ` Now , m.e of `H_(2)O_(2)` = m.e of `KMnO_(4)` oreq. of `H_(2)O_(2)`= eq. of `KMnO_(4)` `x/17=(0.632)/(31.6) "" ` ....(EQN . 4I) x being the amount of `H_(2)O_(2)` is GRAMS , x = 0.34 g Percentage of `H_(2)O_(2)` in the sample = `(0.34)/(0.4) xx 100 = 85 % `

Discussion

87207.	A solution of 7 g of a non-volatile solute in 250 g of water boils at 100.26^(@)C. Find the molecular mass of th solute K_(b) for water is 0..52 K kg mol^(-1) .
Answer» Solution :Elevation in BOILING point, `DeltaT_(b)=100.26-100=0.26^(@)C=0.26 K` `M_(B)=(K_(B)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.52" K kg mol"^(-1))XX(7g))/((0.26 K)xx(0.250" Kg"))=56" g mol"^(-1)`

Discussion

87208.	A solution of 6.2 g ethylene glycol is dissolved in 55 g of water (Kf=1.86-kg "mol2"^(-1)), is cooled to -3.72^(@)C. What is the weight (in grams) of the ince separated from the solution ?
Answer»

Discussion

87209.	A solution of 3.0xx10^(4) kg of comphor (C_(6)H_(16)O) in 25.3xx10^(-3) kg of cholorform boils at 61.3^(@)C if K_(b) for chloroform is 3.83 k kg "mol"^(-1). Calculate the boiling of chloroform. (b) Two liquids (A) and (B) form an idal solution. At 300 K thevapour pressure of solution containing 1 mole of A and 3 moles of B is 550 of Hg. At the same temperature if one moel of B is added to this solution. the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour of A and B.
Answer» Solution :(i) `60.86^(@)C` (ii) `P_(A)^(@)=400` MM and `P_(A)^(@)=600 mm`

Discussion

87210.	A solution of 500 ml of 0.2 M KOH and 500ml of 0.2 M HCl is mixed and stirred, the rise in temperature is T_(1). The experiment is repeated using 250ml each of solution, the temperature raised is T_(2). Which of the following is true
Answer» `T_(1)=T_(2)` `T_(1)=2T_(2)` `T_(1)=4T_(2)` `T_(2)=9T_(1)` Solution :Suppose HEAT evolved in `I^(st)` case is `Q_(1)` and that in the `II^(ND)` case it is `Q_(2)`. Then `Q_(2)=(1)/(2)Q_(1)` But `Q_(1)=1000T_(1) and Q_(2)=500T_(2)` `therefore500T_(2)=(1)/(2)xx1000 T_(1)i.e.T_(2)=T_(1)`.

Discussion

87211.	A solution of 1.73g of 'A' in 100 "cc" of water is found to be isotonic with a 3.42% (wt./vol.) solution of sucrose (C_(12)H_(22)O_(11)). Calculate molecular weight of A. (C_(12)H_(22)O_(11)=342)
Answer» SOLUTION :We know that the isotonic SOLUTIONS have the same molar CONCENTRATION (i.e., moles/lit) LET the molecular WEIGHT of A be M Moles of `A=(1.73)/(M)` `:.` molar conc.of A (moles/litre) `=(1.73)/(M)xx(1000)/(100)=(17.3)/(M)` Molar conc.of sucrose `=(3.42)/(342)xx(1000)/(100)=0.1` `:.(17.3)/(M)=0.1`, `M=173`

Discussion

87212.	A solution of 1.25 g of 'P' in 50 g of water lowers freezing point by 0.3^(@) C. Molar mass of 'P' is 94.K_("(water)") = 1.86 K kg mol^(-1). The degree of association of ‘P’in water is
Answer» 0.8 0.6 0.65 `75%` SOLUTION :1.25 g of P in 50 GO `H_2O implies ` 25 g in 1000 g of `H_2O` MOLALITY `=(25)/(94)` `Delta t = i xx K_(p) xx m` `0.3 = ixx1.86 xx (25)/(94)` `i=(94 xx 0.3)/(1.86 xx 25) = 0.6064` ` alpha = (i-1)/((1)/(n)-1) = (0.6064 -1)/((1)/(2)-1)=(-0.3936)/(-0.5) = 0.787=78.7% = 80%`

Discussion

87213.	A solution of 1.25 g of P in 50 g of water lowers freezing point by 0.3^(@)C. Molar mass of P is 94 and K_(f) (water) = 1.86 "K kg mol"^(-1). The degree of association of P if it forms dimers in water is :
Answer» 0.8 0.6 0.65 0.75 Solution :`DeltaT_(f)=03^(@)C " " M_(B)=94"g mol"^(-1)` `K_(f)=186"K kg mol"^(-1)""W_(B)=125g` `a=?""W_(A)=50g` In order to calculate `alpha`, .`i`. must be known `DeltaT_(f)=(ixxK_(f)xxW_(B)xx1000)/(W_(A)xxM_(B))` `03=(ixx186xx125xx1000)/(50xx94)` or `i=(03xx50xx94)/(186xx125xx1000)=0606` It is GIVEN that P undergoes ASSOCIATION to form DIMER. `{:(,2P,HARR,P_(2),),("Initial",1,,0,),("after association",1-alpha,,alpha//2,):}` `:.` Total number of moles `=1-alpha+alpha//2` `=1-alpha/2` `i=("Observed number of moles")/("Normal number of moles")` `i=(1-alpha//2)/(1)=1- alpha/2` `0606=1- alpha/2` `0606=(2-alpha)/(2)` `1212=2alpha` `alpha=2-1212=0788` or `78*8%` `~~80%`

Discussion

87214.	A solution of 12.5 g of an unknown solute in 170 g water gave a boiling point elevation of 0.63 K. Calculate the molar mass of the solute. (K_(b)=0.53 K kg mol^(-1)).
Answer» Solution :`W_(B)=12.5g, W_(A)=0.170 kg, DeltaT_(b)=0.63 K, K_(b)=0.53" K kg MOL"^(-1), M_(B)=?` `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.53" K kg mol"^(-1))(12.5g))/((0.63 K)xx(0.170 kg))=61.85" g mol"^(-1)`.

Discussion

87215.	A solution of 1.25 g of a non-electrolyte in 20 g of water freezing at 271.94. If K = 1.86 K molality-then the molecular wt. of the solute is:
Answer» 207.8 g/mol 179.79 g/mol 209.6 g/mol 109.6 g/mol Answer :D

Discussion

87216.	A solution of 1.25 g of a certain non-electrolyte in 20.0 g of water freezes at 271.94 K. Calculate the molecular mass of the solute, K_(f)for water is 1.86 K/m.
Answer» Solution :`W_(B)=1.25g, W_(A)=0.02 KG, K_(f)-=1.86" K kg mol"^(-1),T_(f)^(@)=273 K` `T_(f)=271.94 K, DeltaT_(f)=273-271.94=1.06 K, M_(B)=?` `M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((1.86" K kg mol"^(-1))XX(1.25 G))/((1.06 K)xx(0.02 kg))=109.67" g mol"^(-1).`

Discussion

87217.	A solution of 10^(-3)M each in Mn^(2+), Fe^(2+), Zn^(2+) and Hg^(2+) is heated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and HgS are 10^(-15), 10^(-23), 10^(-20) nd 10^(-54) respectively which one will precipitate first ?
Answer» FeS MnS HGS ZnS Solution :HgS which has LARGER `K_(SP)` VALUE.

Discussion

87218.	A solution of 1.25 g of a non-electrolyte in 20 g of water freezing at 271.9K with its molal depression constant , then the molecular wt. of the solute is:
Answer» 207.8 g/mol 179.79 g/mol 209.6 g/mol 105.68 g/mol Answer :D

Discussion

87219.	A solution of 1 molal concentration of a solute will have maximum boiling point elevation when the solvent is
Answer» ETHYL ALCOHOL ACETONE BENZENE Chloroform Answer :C

Discussion

87220.	A solution of 0.2g of a compoundcontaining Cu^(2+) and C_(2)O_(4)^(2-) ions on titration with 0.02" M " KMnO_(4) in presence of H_(2)SO_(4) consumes 22.6 mL of theoxidant .The resultant solution is neutralised with Na_(2)CO_(2) acidfied with dilute acetic acid and treated with excess KI . The liberated I_(2) required 11.3 mL of0.05Na_(2)S_(2)O_(3) solution for complete reduction . Findout the mole ratio ofCu^(2+) " to " C_(2)O_(4)^(2-)in the compound. Writedown the balanced redoxreactions involvedin the above titrations .
Answer» Solution :At the first stage , `C_(2)O_(4)^(2-) ` is oxidisedto `CO_(2)` by the OXIDANT `KMnO_(4)` `{:(5C_(2)O_(4)^(2-)+2MnO_(4)^(-)+16H^(+)to,10CO_(2)+2Mn^(2+)+3H_(2)O),("+7","+2"):}` ` :. ` normalityof `KMnO_(4)` solution= `0.02 xx (7-2) = 0.1 N ` At the second stage `Cu^(2+)`liberates `I_(2)` from KI and this LIBERATED `I_(2)` requires 11.3 mL of `0.05 " M " Na_(2)S_(2)O_(3)` solution for complete reaction . `{:(2Cu^(2+)+2I^(-)to,2Cu^(+)+I_(2)),(2S_(2)O_(3)^(2-)+I_(2) to ,S_(4)O_(6)^(2-)+2I^(-)),(" +4"," +5"):}` Normalityof `Na_(2)S_(2)O_(3)` solution ` = 0.05 (5-4) = 0.05 N ` Now , `(" m.e of "Cu^(2+))/( "m.e of " C_(2)O_(4)^(2-))= (" m.e of"Na_(2)S_(2)O_(3))/("m.e of"KMnO_(4)) = (0.05 xx 11.3 )/(0.1 xx 22.3) = 1/4 ` As `{:(Cu^(2+)to,Cu^(+)and,C_(2)O_(4)^(2-) to,2CO_(2)),(+2,+1,+6,+8):}` ` :. ("mmol of "Cu^(2+)xx1)/("mmol of"C_(2)O_(4)^(2-)xx2) =1/4` or `("mole of "Cu^(2+))/(" mole of " C_(2)O_(4)^(2-)) = 1/2 `

Discussion

87221.	A solution of 0.2 g of a compound containing Cu^(2+) "and" C_(2)O_(4)^(2-) ions on titration with 0.02 M KMnO_(4) in presence of H_(2)SO_(4) consumes 22.6 mL of the oxidant. The resultant solution is neutralized with Na_(2)CO_(3) acidified with dil. acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.5 M Na_(2)S_(2)O_(3) solution for complete reduction. Find out the mole ratio of Cu^(2+) "to" C_(2)O_(4)^(2-)in the compound. Write down the balanced redox reactions involved in the above titrations.
Answer» Solution :Let 'a' moles of `Cu^(2+)` and 'b' moles of `C_(2)O_(4)^(2-)` be PRESENT in solution (i) The solution is oxidised by `KMnO_(4)` with only `C_(2)O_(4)^(2-)` `Mn^(7+) + 5 E^(-) rarr Mn^(2+)` `C_(2)^(6+) rarr 2C^(4+) + 2e^(-)` Meq. of `C_(2)O_(4)^(2-)` = Meq. of `KMnO_(4)` `b XX 2 xx 1000 = 0.02 xx 5 xx 2.26 ` `b = 1.13 xx 10^(-3)` (ii) After oxidation of `C_(2)O_(4)^(2-)` the resulting solution is neutralized by `Na_(2)CO_(3)` acidified with dilute `CH_(3)COOH`and then treated with excess of KI. The liberated `I_(2)` REQUIRED `Na_(2)S_(2)O_(3)` for itsneutralization, i.e. `Cu^(2+) overset("KI")to I_(2) overset(Na_(2)S_(2)O_(3))to Na_(2)S_(4)O_(6) + I^(-)` Meq. of `Cu^(2+)` = Meq. of `I_(2)` liberated = Meq. of `Na_(2)S_(2)O_(3)` used Meq. of `Cu^(2+)` = Meq. of `Na_(2)S_(2)O_(3)` used `a xx 1 xx 1000 = 11.3 xx 0.5 xx 1` `a = 5.65 xx 10^(-3)` Molar ratio of `Cu^(2+)//C_(2)O_(4)^(2-) = a/bimplies 5.65 xx 10^(-3)/(1.13 xx 10^(-3)) = 5/1 ` Hence, molar ratio of `Cu^(2+) , C_(2)O_(4)^(2-) = 5 : 1` Involved reactions are : `2 MnO_(4)^(-) + 5 C_(2)O_(4)^(2-) + 16H^(+) rarr 2MN^(2+) + 10 CO_(2) + 8H_(2)O` `2Cu^(2+) + 4I^(-) rarr Cu_(2)I_(2) + I_(2)` `I_(2) + 2S_(2)O_(3)^(2-) rarr 2I^(-) + S_(4)O_(6)^(2-)`

Discussion

87222.	A solution of 0.20 g of a non-electrolyte in 2 g of water freezes at 271.14 K . If K_(f) = 1.86 K "molality"^(-1)then the molar mass of the solute is :
Answer» 207.8 g//mol 179.79 g//mol 200.8 g//mol 100.0 g//mol Solution :`DELTA T_(F)=(1000_(f) XX W_(2))/(M_(2) W_(1))` or `1.86=(1000 xx 1.86 xx 0.20)/(M_(2) xx 20)` `:. M_(2) = (1000 xx 1.86 xx 0.20)/(1.86 xx 20 )=100.0`

Discussion

87223.	A solution of 0.10 M of a weak electrolyte is found to be dissociated to the extent of 1.20% at 25^@C . Find the dissociation constant of the acid.
Answer» SOLUTION :GIVEN that `a=1.20%=1.20/100=1.2 TIMES 10^-2` `K_a=a^2c=(1.2 times 10^-2)^2 (0.1)` `=1.44 times 10^-4 times 10^-1=1.44 times 10^-5`

Discussion

87224.	A solution of 0.1 M KMnO_(4) is used for the reaction S_(2)O_(3)^(2-) + 2MnO_(4)^(-) + H_(2)O to MnO_(2)+SO_(4)^(2-) + OH^(-) What volume of solution in ml will be required to react with 0.158 gm of Na_(2)S_(2)O_(3)?
Answer» SOLUTION :N/A

Discussion

87225.	A solution of 0.01M Cd^(2+) contains 0.01IM NH_(4),OH. What conc. of NH _(4)^(+) form NH_(4)Cl is necessry to prevent precipitation of Cd(OH)_(2) ? K_(sp), of Cd(OH)_(2), = 2.0 xx 10^(-14) , K_(sp)of NH_(4)OH 1.8 xx 10^(-5) if answer is 1.272xx 10^(-x) mol/ltr then x =__________ ?
Answer» ANSWER :1

Discussion

87226.	A solution of (+) -2-chloro-2-phenylethane in toluene racemises slowly in the presence of small amount of SbCl_5 , due to the formation of
Answer» Carbanion Carbene Free-radical CARBOCATION Solution :`SbCl_5` is USED for the FORMATION of carbocation

Discussion

87227.	A solution of (+)-2-chloro-2-phenylethane in toluene racemises slowly in the presence of small amounts of SbCl_5, due to the formation of:
Answer» Carbanion Carbene Free radical Carbocation Answer :D

Discussion

87228.	A solution of (-) - 1 - chloro - 1 pheylethane in toluene raceimess slowly in the presence of a small amount of SbCI_(5), due of the formation of-
Answer» FREEE radical carbanion carbene carbocation Answer :D

Discussion

87229.	A solution of (-)-1- chloro -1- phenylethane in toluene racemises slowly in the presence of a small amount of SbCl_(5), due to the formation of
Answer» FREE radical carbene carbocation carbanion Answer :C

Discussion

87230.	A solution of (-)-1-chloro-1-penylethane in toluene racemises slowly in presence of a small amount of SbCl_(5) due to the formation of
Answer» FREE radical carbanion carbene CARBOCATION Solution :In presence of `SbCl_(5)`, (-)-1-chloro-1-phenylethane forms a carbocation as shown: `CH_(3)-CHCl-C_(6)H_(5)+SbCl_(5)to[CH_(3)-overset(+)(C)H-C_(6)H_(5)]SbCl_(6)^(-)` Since the carbocation is a planar species, therefore, it can be attacked by `SbCl_(6)^(-)` EITHER from the top or the BOTTON face with equal EASE. as a result, a 50:50 mixture of two enantiomers of 1-chloro-1-phenylethane are formed, i.e., (-)-1-chloro-1-phenylethane undergoes racemization due to the formation of a carbocation intermediate.

Discussion

87231.	A solution of (-)-1-chloro-1-phenylenthane is toluene racemises slowly in the presence of small amount of SbCl_(5), due to the formation of
Answer» CARBANION CARBENE carbocation free radical ANSWER :C

Discussion

87232.	A solutionmay contain any of the following ions: Fe^(2+),Ni^(2+), Cr^(3+),Zn^(2+),Mn^(2+) Based on the experiment and result therein , which of the ions would be present? Indicate any wrong infornation if any….. a. The original solution is treated with with (NH_(4)),S(a subestitate is obtain b. The ppt for (a) dissolves in regain c. The fitrate after sepration ppt in (a) is treated with NaOH and H_(2)O_(2) A dark ppt , is separate filtrate is colourless. e. The solution from (d) is turned with aq NH_(3) A dark ppt forms f. The ppt from (e) is solution in HCI (aq) and solutiondevelops an latense red colour whentreated with SCN^(Theta) (aq)
Answer» Solution :Precipitate can be of `NiS` b. `NiS` dissolves in aqua regain c.Precipitatecan be of `Fe(OH)_(3)` d.`Fe(OH)_(3)` is SOLUBLE in `HCI`(aq) giving (yellow) coloured soluble of `FeCI_(3)` e.This step is INCORRECT since filtrate of (c ) isalreadly colourlessincdicatingabsence of `Cr^(2+)` f. This confims `Fe^(3+)` by forming intense redcolour `Fe^(3+) + SCN^(THETA) rarr [Fe(SCN)]^(2+)` Thus, `{:(Fe^(3+),-"confirmed"),(Ni^(2+),-"probable"),(Zn^(2+)Mn^(2+)Cr^(3+),-"ABSENT"):}`

Discussion

87233.	A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given k_(f)("water")=1.86Kkg"mol"^(-1) k_(f)("ethanol")=2.0Kkg"mol"^(-1) k_(b)("water")=0.52Kkg"mol"^(-1) k_(b)("ethanol")=1.2Kkg"mol"^(-1) Standard f.p. of water = 273 K Standard f.p. of ethanol = 155.7 K Standard b.p. of water = 373 K Standard b.p. of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mmHg Vapour pressure of pure ethanol = 40 mmHg Molecular weight of water = 18 g "mol"^(-1) Molecular weight of ethanol = 46 g "mol"^(-1) consider the solutions to be ideal dilute solutions and solutes to be nonvolatile and non-dissociative. The vapour pressure of the solution M is
Answer» 39.3 mmHg 36.0 mmHg 29.5 mmHg 28.8 mmHg Answer :A

Discussion

87234.	A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given k_(f)("water")=1.86Kkg"mol"^(-1) k_(f)("ethanol")=2.0Kkg"mol"^(-1) k_(b)("water")=0.52Kkg"mol"^(-1) k_(b)("ethanol")=1.2Kkg"mol"^(-1) Standard f.p. of water = 273 K Standard f.p. of ethanol = 155.7 K Standard b.p. of water = 373 K Standard b.p. of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mmHg Vapour pressure of pure ethanol = 40 mmHg Molecular weight of water = 18 g "mol"^(-1) Molecular weight of ethanol = 46 g "mol"^(-1) consider the solutions to be ideal dilute solutions and solutes to be nonvolatile and non-dissociative. The f.p. of the solution M is
Answer» 268.7 K 268.5 K 234.2 K 150.9 K Solution :SOLUTE is WATER and SOLVENT is ETHANOL

Discussion

87235.	A solution made by dissolving 40 g NaOH 1000 g of H_(2)O is
Answer» 1 molar 1 normal 1 molal NONE of the above Solution :`M=(40)/(40xx1)=1` m SOL

Discussion

87236.	A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given k_(f)("water")=1.86Kkg"mol"^(-1) k_(f)("ethanol")=2.0Kkg"mol"^(-1) k_(b)("water")=0.52Kkg"mol"^(-1) k_(b)("ethanol")=1.2Kkg"mol"^(-1) Standard f.p. of water = 273 K Standard f.p. of ethanol = 155.7 K Standard b.p. of water = 373 K Standard b.p. of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mmHg Vapour pressure of pure ethanol = 40 mmHg Molecular weight of water = 18 g "mol"^(-1) Molecular weight of ethanol = 46 g "mol"^(-1) consider the solutions to be ideal dilute solutions and solutes to be nonvolatile and non-dissociative. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The b.p. of this solution is
Answer» 380.4 K 376.2 K 375.5 K 354.7 K Solution :SOLUTE is ETHANOL and SOLVENT is WATER

Discussion

87237.	A solution is preparedb mixind 8.5 of g CH_(2) Cl_(2) and 11.95 g of CHCl_(3). If vapour pressure of CH_(2)Cl_(2) and CHCl_(2) and CHCl_(3) at 298 K are 415 and 200 mmHg respectively. The mole fraction of CHCl_(3) in vapour form is: (Molar mass of Cl=35.5 g mol ^(-1))
Answer» `0.162` `0.675` `0.325` `0.486` Solution :Molar mass os `CHCl_(3) = 119.5` g/mol Molar mass of `CH_(2) Cl_(2) =85` g/mol. molcs of `HCCl_(3 ) = (11.95)/( 119.5) =0.1` mol. Moles of `CH_(2)Cl_(2)=(8.5)/(85)=0.1` mol Molefraction of `ChCl_(3)=(0.1)/(0.2) =0.5` mol. Mole fraction of `CH_(2) Cl_(2) =(0.1)/(0.2) = 0.5` mol. Given-Vapour opressure of `CHCl_(3)=200 mm Hg = 0.263atm.` Vapour prcssure of `CH_(2) Cl_(2) =415 m m Hg = 0.546` ATM.) (1 atm = 760 mm Hg) `therefore P _(("above solution"))` = Mole fractin of `CHCl_(4)xx` (Vapour PRESSURE of `CHCl_(3)`) + Mole fraction of `CH_(2) Cl_(2)xx ` (Vapour pressure of `CH_(2) Cl_(2)`) `=0.5 xx0.263+ 0.5xx 546 =04045` Mole fraction of `CHCl_(3)` in vapour from `= (0.1315)/(0.4045) =0.325`

Discussion

87238.	A Solution is prepared by dissolvingcertain amountofsolute in 500 g water. The %by massof asolutionsis 2.38 . Calculate mass ofthe solute.
Answer» Solution :Given : `W_(1)`= Massof a SOLVENT(water) = 500 g `W_(2)` Massof a solute =? %by MASS= % `(W//W) = 2. 38` Let massof thesolute`=W_(2) =X g` `:.` Massof the solution`= W_(1) + W_(2)= (500 + x) g` Nowper centby massof solute `= ("massof solute")/("massof solution") xx 100` `= ("mass of solute")/("mass ofsolvent+ massof solute") xx 100` `:. 2.38= (X )/((500 + X)) xx 100` `:. 2. 38(500 + X) =100 x` `:. 2.38 xx 500 + 2.38 X =100 x` `:.1190 + 2.38 X = 100 x` `:.100 x -2.38 x= 1190` `:. (100 -2.38) x= 1190` `:. 97 .62 x = 1190` `:. x = (1190)/(96.62) =12.19`

Discussion

87239.	A solution is prepared by dissolving 26.3 g of CdSO_(4)in 1000 g of water. The depression in freezing point of the solution was forund to be 0.284 K. Calculate Van't Hoff factor. The cryoscopic constant for water is 1.86 K kg mol^(-1). (Given molar mass of CdSO_(4)=208 g mol^(-1).)
Answer» Solution :`DeltaT_(F)=ixxK_(f)xxm=iK_(f)xxW_(B)/(M_(B)xxW_(A))` `DeltaT_(f)=0.284 K, K_(f)=1.86" K kg mol"^(-1), W_(B)=26.3 g,` `M_(B)=208.4" g mol"^(-1),W_(A)=1g.` `i=(DeltaT_(f)xxM_(B)xxW_(B))/(K_(f)xxW_(B))=((0.28K)XX(208.4"g mol"^(-1))xx(1g))/((1.86" K kg mol"^(-1))xx(26.3g))=1.21.`

Discussion

87240.	A solution is prepared by dissolving 10g of nonvolatile solute in 180g of H_(2)O. If the relative lowering of vapour pressure is 0.005, find the mol.wt. of the solute.
Answer» SOLUTION :Suppose the mol.wt. of the solute is M. Moles of solute `=(10)/(M)` Mole of solvent `(H_(2)O)=(180)/(18)=10` Mole fraction of solute `=(10//M)/((10)/(M)+10)=(1)/(M+1)` We KNOW, RELATIVE lowering of VAP. Pressure `=` mole fraction of solute `0.005=(1)/(M+1)` `M=199`

Discussion

87241.	A solution is prepared by dissolving 10 g of non-volatile solute in 200 g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute. [ Vapour pressure of pure water at 308 K = 32 mm Hg]
Answer» Solution :APPLY the RELATION `(p_1^0 - p_1)/(p_1^0) = (w_2 XX M_1)/(M_2 xx w_1)` Substituting the values in the above relation, we have `(32-31.84)/(32) = (10 xx 18)/(M_2 xx 200)` `(0.16)/(32) = (180)/(200M_2)` `M_2 = (180 xx 32)/(0.16 xx 200) = 180`

Discussion

87242.	A solution is one molar in each of NaCl , CdCl_(2) , ZnCl_(2) and PbCl_(2) . To this , tin metal is added . Which of the following is true ? Given : {:(E_(Pb^(2+) \|P)^(@) = -0.126 V "," E_(Sn^(2+)\|Sn) = -0.136 V) , (E_(Cd^(2+)\|Cd)^(@) = -0.40 V "," E_(Zn^(2+)\|Zn)^(@)= -0.763 V), (E_(Na^(+) \|Na)^(@) = -2.71 V):}
Answer» Sn can REDUCE `NA^(+)` to Na Sn can reduce `ZN^(2+)` to Zn Sn can reduce `Cd^(2+) ` to Cd Sn can reduce `Pb^(2+)` to Pb . Solution :CHECK the spontaneity of each reaction by finding the E.M.F.

Discussion

87243.	A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mus. Calculate the moss per centage of the resulting solution.
Answer» Solution :MASS of solute in solution - I = `(25)/(100 ) XX 300 ` g = 75 g Mass of solute in solution - II = `(40)/(100) xx 400 ` g 160 g Total mass of solute (I + II) = 75 + 160 = 235 g Total mass of solution = 300 + 400 = 700 g mass percentage of solute in final solution = `(235)/(70) xx 100 = 33.57%` mass percentage of soluent in final solution 100 - 33.57 = 66.43%.

Discussion

87244.	A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass.Calculate the mass percentage of the resulting solution.
Answer» Solution : 300 g of 25% solution CONTAINS = 75 g SOLUTE 400 g of 40% solution contains = 160 g solute Total solute = 75 + 160 = 235 g Total solution = 300 + 400 = 700 g PERCENTAGE of solute in the solution = `235/700 xx 100 = 33.5%` Percentage of water in the solution = 100 - 33.5 = 66.5%.

Discussion

87245.	A solution is obtained by mixing 200 ml 1 M CaCl_(2), 300 ml 2 M Na_(2)SO_(4) and 500 ml 1 M BaCl_(2) then which of the following is correct for the resultant solution obtained.
Answer» `[CL^(-)] = 2.4 M` `[Ca^(2+)] = 0.2 M` `[SO_(4)^(2-)] = 0.1 M` `[BA^(2+)] = 0.5 M` Solution :`{:(Na_(2)SO_(4),+,BaCl_(2),rarr,2 NaCl,+,BaSO_(4),darr,,),(0.6 mol,,0.5 mol,,,,,,,),(0.1 mol,,-,,1 mol,,,,,):}` `[Cl^(-)] = (2 xx 0.2 + 2) = 2.4 M` `[Ca^(2+)] = 0.2 M` `[SO_(4)^(2-)] = 0.1 M` `[Ba^(2+)] = 0M`

Discussion

87246.	A solution is obtained by mixing 300 g of 25% and 400 g of 40% solution by mass. Calculate the mass percentage of the resultng solution.?
Answer» SOLUTION :`"300 g of 25% solution contains solute = 75 g,400 g of 40% solution contains solute = 160 g"` `"Total solute "=160+75 = 235g,"Total solution "=300+400=700g` `"Mass % of solute in the final solutieon"=(235)/(700)xx100=33.5%` `"Mass % of solute in the final solutieon"=100-33.55=66.5%`.

Discussion

87247.	A solution is obtained by mixing 100 ml of 1M NaCl, 100 ml of 1M MgCl_(2), 300 ml of 1M Mg (NO_(3))_(2) and finally diluted to 1000 ml. Which of the following is incorrect for final concentration of ions ?
Answer» `[Na^(+)] = 0.1 M` `Mg^(+2) = 0.4 M` `[Cl^(-)] = 0.2 M` `[NO_(3)^(-)] = 0.6 M` Solution :`[Cl^(-)] = 0.1 + 0.2 = 0.3` mole/litre

Discussion

87248.	A solution is obtained by dissolving 0.2 moles of urea in a litre water. Another solution is obtained by dissolving 0.4 moles of sugar in a litre of water at the same temperature. The lowering of vapour pressure to the first solution is.
Answer» Same as that of the second SOLUTION Half to that of the second solution Double to that of the second solution NONE Solution :LOWERING of VAPOUR pressure depends on numbers of MOLES of solute

Discussion

87249.	A solution is made by dissoving 40.5 g of C_(2)H_(5)OH until the total volume of the solution is 100 mL. What is the percentage volume of C_(2)H_(5)OH? The density of pure C_(2)H_(5)OH is 0.78 g/mL.
Answer» SOLUTION :`"Intial mass of "C_(2)H_(5)OH=40.5 g.` `"DENSITY of "C_(2)H_(5)OH=0.78 g//ML.` `"Intial volume of solution "="Mass"/"Density"=((40.5g))/((0.78g//mL))=51.92 mL` Final volueme of solution = 100 mL `"PRECENTAGE volume of "C_(2)H_(5)OH=("Intial volume")/("Final volume")XX100=((51.92mL))/((100mL))xx100=51.92%`

Discussion

87250.	A solution is 0.25% by mass. The weight of solvent containing 1.25g. Of solutes would be
Answer» 506g 498.75 G 580.25 g 581.25 g Solution :MASS of solution =100 g. Mass of solute = 0.25 g. :. Mass of SOLVENT =100-0.25 = 99.75 g. 0.25 of solute is PRESENT in 99.75g of solvent :. 1.25 g. of solute will be present in `(99.75 xx 1.25)/0.25 =498.75` g of solvent.

Discussion

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