52671 + Interview Questions in Chemistry in Class 11 Page 1035 InterviewSolution

51701.	100 ml aqueous 0.1 molar M(CN)_(2), (80% ionized) solution is mixed with 100 ml of 0.05 molar H_(2),SO_(2) solution (80% ionized). (K_(6), of CN" = 10^(-6))
Answer» `6` `8` `9` `7` Solution :`M(CN)_2 hArr M^(+2)+ 2CN^(-) ` `10m " moles" ` ` {:( 10(1- alpha ) , 10 alpha =8 , 2 xx 10 alpha = 16 ),( H_2SO_4 hArr, 2H^(+) , SO_4^(2-)),( 5 " moles" ,,),(5(1- beta ), 2 xx 5 beta =8, 5 beta =4 ), (H^(+) + , CN^(-) hArr, HCN),( 8,16,):}` `8,8 ` i.e., BUFFER of HCN & `CN^(-) ` ` therefore PH =pKa +LOG ""( S)/(A)=pKa =14 -6=8`

Discussion

51702.	100 ml, 1.5M Fe(NO_(3))_(3). 100 ml, 2M FeCl_(3), 100 ml 2M Mg(NO_(3))_(2) are mixed and 700 ml water is added. The molar concentration of total ions in the solution in ______M
Answer» Solution :Total MOLES of ions from `Fe(NO_(3))_(3)` `=0.1xx1.5xx4=0.6` From `FeCl_(3)=0.1xx2xx4=0.8` From `MG(NO_(3))_(2)=0.1xx2xx3=0.6` `therefore` Total moles of ions = 2 `therefore M=(2)/(1)=2M`

Discussion

51703.	100 ml of 0.1 M NaCl and 100 ml of 0.2 MNaOH are mixed. What is the change in pH of NaCI solution ?
Answer» Some `KMnO_4`is LEFT unreacted 0.02 M `MN^(2+)` ions are present in the solutions `K_2SO_3` is PARTIALLY left unreacted Newly formed `SO_4^(2-)` has 0.05 M cocentration Solution :`100xx0.2 =20` meq `O_3^(2-)` `100xx0.5 =50`meq `MnO_4^(-)` `:. SO_3^(2-)` in limiting REAGENT `MnO_4^(-)` left = 30 meq `:.[Mn^(+2)]=(20//5)/200=0.05M` `[SO_4^(-2)]=(20//2)/200 = 0.05M`

Discussion

51704.	100 ml 0.1 M urea solution is diluted upto 200 ml than the molarity is ...............M.
Answer» 0.2 0.1 0.05 0.025 Solution :`M_(1)V_(1) = M_(2)V_(2)` `0.1xx100=M_(2)xx200 "" :.M_(2)=0.05M`

Discussion

51705.	100 ml, 0.1 M K_(2)SO_(4) is mixed with 100 ml, 0.1M Al_(2)(SO_(4))_(3) solution. The resultant solution is
Answer» `0.1NK^(+)` ions `0.4NK^(+)` ions `0.2MSO_(4)^(-2)` ions `0.1MAl^(+3)` ions SOLUTION :`[K^(+)]=(100xx0.1xx2)/(200)=0.1M=0.1N` `[SO_(4)^(2-)]=(100xx0.1xx1+100xx0.1xx3)/(200)` `=0.2M=0.4N` `[AL^(+3)]=(100xx0.1xx2)/(200)=0.1M=0.3N`

Discussion

51706.	100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOHsolution. The pH of the reaction mixture keeps increasing with addition of NaOH. The successive dissociation constant of H_3PO_4 " are "10^(-3) , 10 ^(-6) and 10^(-14)respectively.What is the pH of the solution after adding 150 ml of NaOH solution?
Answer» `3` `6` `6. 7` `5.3` SOLUTION :`PH = P^(Ka_2)= 6`

Discussion

51707.	100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOHsolution. The pH of the reaction mixture keeps increasing with addition of NaOH. The successive dissociation constant of H_3PO_4 " are "10^(-3) , 10 ^(-6) and 10^(-14)respectively. How much volume of the given NaOH must be added such that a buffer H_2PO_4^(-) // HPO_4^(-2)of maximum capacity is formed.
Answer» 100 ml 150 ml 200 ml 250 ml Solution :` {:(H_3PO_4 +, NAOH to, NaH_2PO_4, +H_2O ),(10 m " moles " , 10 m " moles " ,0,),(-,-, ,10 m " moles " ):}` ` {:(NaH_2PO_4 +, NaOH to , Na_2HPO_4+ H_2O ),(10 m " moles " , 5 m " moles " ,"-" ),( 5 m " moles " , , ):}` To get BUFFER with maximum buffer capacity HALF of the `NaH_2PO_4` should be consumed. In 1st STEP, To get 10 m moles NaOH, volume should be 100 ml. In 2nd, step , To get 5 m moles NaOH , volume should be 50 ml Total vol = 150 ml

Discussion

51708.	100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOHsolution. The pH of the reaction mixture keeps increasing with addition of NaOH. The successive dissociation constant of H_3PO_4 " are "10^(-3) , 10 ^(-6) and 10^(-14)respectively. What is the pH of the reaction mixture after adding 50 ml of NaOH ?
Answer» `3` `4` `6` `5.3` Solution :After ADDING 50 ml NaOH, buffer ` (H_3PO_4 + NaH_2 PO_4) ` with MAXIMUM capacity is obtained. ` therefore pH =P^(Ka_2) =3`

Discussion

51709.	100 mL of 0.01M KMnO_4 oxidises 100mL of acidified H_2O_2 . The volume of Oxygen in (MnO_4^(-)) changes to Mn^(2+)in acidic medium and to MnO_4 in alkaline medium).
Answer» 26 ML 260 mL `2.6 mL` 26 Lt Solution :`(N_(1)V_(1))_(KMnO_(4))=(N_(2)V_(2))_(H_(2)O_(2))` `N=0.05 rArr 11.2` VOL of `H_(2)O_(2)=(11.2xx0.05)/(2)` Vol of `O_(2)=(11.2xx0.05)/(2)xx100 mL`

Discussion

51710.	100 gm oleum sample (labelled as 107.8%) is mixed with 7.8 gm water and requires, 1.1 L of x molar aq. Solution of NaOH for complete neutratization. The value of x is:
Answer» ANSWER :2

Discussion

51711.	112% labelled oleum is diluted with sufficient water. The solution on mixing with 5.3xx102gm Na_(2)CO_(3) liberates CO_(2). The volume of Co_(2) given out at 1 atm at 273 K will be
Answer» 1.12 lit 1.23 lit 2.2 lit 25.5 lit Solution :Final weight of `H_(2)SO_(4)=112` gms `Na_(2)O_(3)+H_(2)SO_(4)rarrNa_(2)SO_(4)+H_(2)O+CO_(2)` 530 GM 112 gm = 5 MOLES 1.14 moles IMPLIES VOLUME of `CO_(2)=1.14xx22.4=25.5` lit.

Discussion

51712.	10.0 g sample of Cu_(2)O is disoolved dissolved in dil. H_(2)SO_(4) where it undergoes disproportionation quantitatively. The solution is filtered off and 8.3 g pure KI cyrstals are added to clear filtrate in order to precipitate CuI with the liberation of I_(2). The solution is again filtered and boiled till all the I_(2) is removes. Now excess of an oxidising agent is added to the rfiltrate which liberates I_(2) again. The liberated I_(2) now percentrage by mass of Cu_(2)O in the sample.
Answer» Solution :`Cu_(2)OtoCu^(2+)+Cu^(0)` The solution after dissolution of `Cu_(2)O` in dil `H_(2)SO_(4)` CONTAINS `Cu^(2+)` IONS and `Cu`. `Cu^(2+)` ions react with KI to give `CuI_(2)` which is converted to CuI (or `CU_(2)I_(2))` and `I_(2)`. `Cu^(2+)+2I^(ɵ)toCuI_(2)to(Cu_(2)I_(2) or CuI)+(1)/(2)I_(2)` Millimoles of KI taken`=(8.3)/(166)xx10^(3)=50(mE of KI=166)` Now, KI left unused reacts with oxidising agent to liberate `I_(2)` again. `2I^(ɵ)overset("oxidising agent")toI_(2)overset(2S_(2)O_(3)^(2-))toS_(4)O_(6)^(2-)+I^(ɵ)` Millimoles of KI left`=` millimoles of `S_(2)O_(3)^(2-)` used `(n=(2)/(2)=1)=(n=(2)/(2)=1)` `=10xx1.0xx1` (n factor)`=10` Therefore, millimoes of KI used for `Cu^(2+)=50-10=40` Therefore, MILLI moles of `Cu_(2)O=20` `{:(Coverset(+1)(u_(2))to2overset(+2)(Cu)),(mol ratio of overset(+1)(Cu_(2)):overset(+2)(Cu)=1:2):}` `(Weight)/(Mw)xx10^(3)=20` (Mw of `Cu_(2)O=143)` `(weight)/(143)xx10^(3)=20` `thereforeW_(Cu_(2)O)=2.86` `% of Cu_(2)O=(2.86xx100)/(10.0)=28.6%`

Discussion

51713.	1.00 g of oxygen combine with 0.126 g of hydrogen to form H_2O .1.00 g of nitrogen combine with 0.216g of hydrogen to form NH_3. Predict the weight of oxygen required to combine with 1.00 g of nitrogen to form an oxide.
Answer» ANSWER : 1.67 G or its SIMPLE MULTIPLE

Discussion

51714.	1.00 g of EuCl_(2), is treated with excess of aqueous AgNO_(3) and all the chlorine is recovered as 1.29 g of AgCl. Calculate the atomic weight of Eu. (Cl = 35.5, Ag = 108)
Answer» Solution :`underset(1g)(EU)CI_(2)+AgNO_(3) rarr underset(1.29 g)(AGCI)` Since CI atoms are conserved applying POAC for CI atoms moles of CI in `EuCI_(2)` = moles of CI in AgCI `2xx ` moles of `EuCI_(2)=1XX` moles of AgCI `2xx(1)/((x+35.5xx2))=1 xx(1.29)/((108+35.5))` ( where x = atomic weight of Eu ) `:. x= 152.48`

Discussion

51715.	100 g of an ideal gas (m01 wt 40) is pressent in a cylinder at 27^(@)C and 2 atm pressure During transportation cylinder The valve attached to cylinder cannot keep the pressure greater than 2 atm and therefore 10g of gas leaked out through cylinder Calculate (i) the volme of cylinder before and after dent (ii) the pressure inside the cylinder .
Answer» ANSWER :`30.79 LITRE, 27.71 litre 2.22 ATM`

Discussion

51716.	100 g of a solution of hydrochloric acid (sp.gr. 1.18) contains 36.5 g of the acid. The normality of the solution is:
Answer» `1.18` `11,8` 118 10 Solution :Volume `=("MASS")/("DENSITY") = 100/1.18 = 84.7 mL` `THEREFORE W = (NEV)/1000`

Discussion

51717.	1.0 g of non - electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.4 K. Find the molar mass of the solute. [Given : Freezing point depression constant of benzene = 5.12 K. kg mol].
Answer» Solution :`DeltaT_(f)=0.40K` `K_(f)="5.12 kg MOL"^(-1)` `W_(2)=1g` `W_(1)="50.5 gm "=(50.5)/(1000)kg` `M_(2)=(K_(f))/(DeltaT_(f))xx(W_(2))/(W_(1))` `therefore""M_(2)=(5.12)/(0.40)xx(1)/(50.5)xx1000` `="2.56 g mol"^(-1)` Thus, the molecule mass of the solute `=" 256 g mol"^(-1)`.

Discussion

51718.	1.00 g of a non-electrolyte dissolved in 50.5g of benzene lowered its freezing point by 0.40K. The freezing point depression constant of benzene is "5.12K.kg mol"^(-1). Find the molecular mass of the solute.
Answer» Solution :`DeltaT_(f)=0.40K` `K_(f)="5.12 kg mol"^(-1)` `W_(2)=1G` `W_(1)="50.5 GM "=(50.5)/(1000)kg` `M_(2)=(K_(f))/(DeltaT_(f))xx(W_(2))/(W_(1))` `therefore""M_(2)=(5.12)/(0.40)xx(1)/(50.5)xx1000` `="2.56 g mol"^(-1)` Thus, the molecule mass of the solute `=" 256 g mol"^(-1)`.

Discussion

51719.	1.00 g of a chloride of an element contains 0.835 g of chlorine. If the vapour density of the chloride is 85, find the atomic weight of the element and its valency.
Answer» Solution :LET the formula of the chloride be `MCI_(X)` where x is the valency of the ELEMENT M. `x XX ` moles of M =1 `xx` moles of CI or `x xx (1-0.835)/("at . Wt . Of M)= (0.835)/(35.5)` Now we know molecular weight `= 2xx ` vapour density `=2xx85 =170` For `MCI_(x)` mol . wt. of `MCI_(x)` = at . wt. of M +x `xx` at . wt. of CI or 170 =at .wt . of M + 35.5 x. `:. ` at . wt. of M = 170 - 35.5 x or `x = (170-"at .wt.of M")/(35.5)` From (1) and (2), we get, at. wt. of M = 28.05. Substituting at. wt. of M in the eqn. (1) we get, x = 4.

Discussion

51720.	100 cm^3 of a centimolar solution of an acid contain 0.098 g of the acid. Find the molecular mass of the acid.
Answer» ANSWER :`98 G MOL^(-1)`

Discussion

51721.	100cc of gas is collected over water at 15^@C and 750torr. If the gas occupies 91.9 mL in dry state at STP, find the aqueous tension at STP.
Answer» SOLUTION :Let aqueous tension at `15^(@)C`is .f. `{:("GIVEN conditions",,"STP conditions"),(P_(1) = (750-f)"mm",,P_(2) = 760"mm"),(V_(2) = 100 "mL",,V_(2) = 91.9"mL"),(T_(1) = 288"K",,T_(2) = 273 "K"):}` APPLYING equation of state at two conditions, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2)) = ((750-f) xx100)/(288) = (760 xx 91.9)/(273)` ` = 750-f = 736.8` Aqueous tension .f. at `15^(@)C = 750-736.8 = 13.2` mm

Discussion

51722.	10% W/V NaOH is same as
Answer» 2.5 M 2.5 N 1.5 m 10% W/W NaOH Solution :`M=(10V%)/("M.wt")=(10xx10)/(40)=2.5M=2.5N`

Discussion

51723.	When P_4O_10 is hydrolysed with water ,it forms :-
Answer»

Discussion

51724.	10% PCl_5 decompose at definite temperature and 4 atm pressure. At same temperature if 20% PCl_5 decompose than find pressure. (Temperature not change.)
Answer» SOLUTION :`K_p`= 0.04 ATM, 0.96 atm

Discussion

51725.	10 moles of SO_(2) and 4 moles of O_(2) are mixed in a closed vessel of volume 2 litres. The mixture is heated in the presence of Pt catalyst. Following reaction takes place: 2SO_(2)(g)+O_(2)(g)to2SO_(3)(g) Assuming the reaction proceeds to completion. Select the correct statement.
Answer» 10 4 8 14 Solution :`underset("10 moles")(2SO_(2(G)))+underset("reagent")underset("limiting")underset("4 moles")(O_(2(g)))rarr2SO_(3(g))` `SO_(2)` moles remaining = `10-8=2`

Discussion

51726.	10 moles of pure PCl_(5) gas is put into a closed container of volume 'V' and temerature 'T' and allowed to reach equilibrium, at an equilibrium pressure 20 atm. The pure PCl_(5) is found to be 50% dissociated at equilibrium. PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) The partial pressure of PCl_(5), if equilibrium is established at new equlibrium pressure 35 atm by changing volume is :
Answer» 10 ATM 15 atm 18 atm 12 atm Answer :B

Discussion

51727.	10 moles of pure PCl_(5) gas is put into a closed container of volume 'V' and temerature 'T' and allowed to reach equilibrium, at an equilibrium pressure 20 atm. The pure PCl_(5) is found to be 50% dissociated at equilibrium. PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) If 20 moles of PCl_(5)(g) is added to original equilibrium mixture keeping total pressure constant to 20 atm at same tempereture 'T', then ratio of new equilibriujm volume to the initial volume V will be :
Answer» 1.5 `2.0` 1.8 `3.0` ANSWER :D

Discussion

51728.	10 moles of pure PCl_(5) gas is put into a closed container of volume 'V' and temerature 'T' and allowed to reach equilibrium, at an equilibrium pressure 20 atm. The pure PCl_(5) is found to be 50% dissociated at equilibrium. PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) K_(p) for the above the reaction is :
Answer» `(20)/(3)` 100 `(10)/(3)` `(20)/(3)` ANSWER :A

Discussion

51729.	10 moles of an ideal gas expand isothermally and reversibly from a pressure of5 atm to 1 atm at300 K . What is thelargest mass that can be lefted through a height of 1 metre by this expansion?
Answer» Solution :`w_(exp) = - 2.303 nRT log.(P_(1))/(P_(2))=-2.303 ( 10 mol) xx( 8.314 JK^(-1) mol^(-1)) ( 300 K ) log. ( 5)/(1) = - 40.15 xx 10^(3)J` If M is the mass that can be lifted by this work ,through a height of 1metre, then work done `=` Mgh ` 40.15xx 10^(3)J= Mxx9.81 m s^(-2) xx 1m` or ` M =( 40.15 xx10^(3) kg m^(2) s^(-2))/( 9.18 ms^(-2)xx 1 m ) (J = kg m^(2) s^(-2))` `= 4092 .76 kg`

Discussion

51730.	10 moles of each N_(2) and H_(2) are made to react in a closed chamber. At equilibrium 40% of H_(2) was left. The total moles in the chamber are :
Answer» 16 12 8 10 Solution : TOTAL MOLES in CHAMBER at EQUILIBRIUM = 8+4+4=16

Discussion

51731.	1.0 mole of ethyl alcohol and 1.0 mole of acctic acid are mixed. At equilibrium, 0.666 mole of ester is formed. The value of equilibrium constant is
Answer» `1//4` `1//2` 4 3 Solution :`C_(2)H_(5)OH+CH_(3)COOH harr CH_(3)COOC_(2)H_(5)+H_(2)O` `K_(c)=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([C_(2)H_(5)OH][CH_(3)COOH])` `=(0.666 xx 0.666)/(0.333 xx 0.333)=4`

Discussion

51732.	1.0 mole of pure dinitrogen gas at SATP conditions was put into a vessel of volume 0.025 cm^(3), maintained at the temperature of 50^(@)C, what is the preussre of the gas in the vessel ?
Answer» <P> Solution : Volume of `1.0` mol of a gas at SATP `(V_(1))=248XX10^(-3)m^(3)` intial condition (SATP) `""` Final condition `V_(1)=248xx10^(-3)m^(3)"" V_(2)=01 m^(3)` `P_(1)1 "bar""" P_(2) ?` `T_(2)29815 k"" T_(2)50^(@)C32315K` Acoording to general gas equation. `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) or P_(2) =(P_(1)V_(1)T_(2))/(T_(1)V_(2))` `:. P_(2)=(10xx248xx10^(-3)xx32315)/(29815xx025)=1076` bar `:. P_(2) =1076` bar

Discussion

51733.	1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in figure. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.
Answer» Solution :The given diagram represent that the PRCESS is CARRIED out in infinite steps, HENCE it is isothermal REVERSIBLE expansion of the ideal gas from pressure 2.0 atm to 1.0 atm 298 K. `W = - 2.303 nRT log. (p_(1))/(p_(2))` `W = - 2.303 xx 1 mol xx 8.314 JK^(-1) xx 298 K log 2 "" ( :. (p_(1))/(p_(2)) = (2)/(1))` `W = - 2.303 xx 1 xx 8.314 xx 298 xx 0.3010 J` `W = - 1717.46 J`

Discussion

51734.	10 mL of water requires 1.47 mg ofK_(2) Cr_(2)O_(7)(M.wt. =294) fo oxidation of dissolved organic matter . C.O.D is
Answer» 2.44 ppm 24 ppm 32 ppm 1.6 ppm Answer :B

Discussion

51735.	10 ml of H_2O_2solution on decomposition liberated 200ml ofO_2at STP. Then the weight/volume percentage of that H_2O_2solution is
Answer» 3.03 6.12 9.1 3.4 Answer :B

Discussion

51736.	10 mL of H_(2)O_(2) solution is treated with KI and titration of liberated I_(2) required 10 mL of 1 N hypo . Thus H_(2)O_(2) is
Answer» 1N 5.6 VOLUME `17 gL^(-1)` all are correct Solution :Let the normality of `H_(2)O_(2)=N_(1)` `therefore N_(1)xx10(H_(2)O_(2))=1xx10(Na_(2)S_(2)O_(3))` `therefore ` Normality of `H_(2)O_(2)` solution =1N `1NH_(2)O_(2)=5.6 vol. H_(2)O_(2)=17 gL^(-1)` Therefore , all statement are correct.

Discussion

51737.	10 ml of H_(2)O_(2) could release 224 ml of O_(2) at 273K and 2 atm pressure. What is the molarity of that H_(2)O_(2)
Answer» Solution :`N=(PV)/(RT)= n(2xx224xx10^(-3))/(0.0821xx273)` `n=(2xx224xx10^(-3))/(22.4) rArr n=2xx10^(-2) "MOLES" of O_(2)` `2H_(2)O_(2) to 2H_(2)O+O_(2)` `4xx10^(-2)` moles `2xx10^(-2)` moles `M=(n)/(Vlit)=(4xx10^(-2)xx1000)/(10), M=4`

Discussion

51738.	10 mLof H,A (weak diprotic acid) solutions is titrated against 0.1M NaOH. pH of the solution is plotted against volume of strong base added and following observation is made. If pH of the solution at 1^(st) equivalence point is pH_(1) and at 2^(nd) equivalnee point is pH_(2), Cal the value (pH_(2),-pH_(1)) at 25^(@)C. Given for H_(2),A, p^(Kal) =4.6 &p^(Ka2) =8
Answer» Solution :` H_2A +NaOH to NAHA +H_2O " at " I^(ST) " END point,"` ` pH = (p^(K_(a_1))+p^(K_a_2))/(2)=( 8+ 4.6)/( 2) = 6.3` ` N_1V_1 =N_2V_2 ` ` N_1xx 10=0 .1 xx 20 rArr N_1 =0.2` at ` 2^(nd) `end point , `Na_2 A` salt will be formed `{:(NaHA+ ,NaOHto ,Na_2 A + H_2O), (2m " moles " ,2m " moles",0),(-,-, 2 m " moles "):}` ` [Na_2 A]=(2)/(50)=0.04 M`

Discussion

51739.	10 mL of H_(2) combine with 5 mL of O_(2) to form H_(2)O when 200 mL of H_(2) at S.T.P. is passed through heated CuO, latter loses 0.144 g of its weight. Does the above data correspond to the law of constant composition ?
Answer» Solution :In the first experiment : Ratio by volume of `H_(2) and O_(2)` COMBINING to form `H_(2)O=2 :1` In the second experiment : `CuO+H_(2)overset(heat)(rarr)Cu+H_(2)O` The loss weight of CuO is due to oxygen which has been removed. It is 0.44 g Now, 32 g of `O_(2)` at S.T.P occupy = 22400 L 0.144 g of `O_(2)` at S.T.P occupy `= (("22400 mL"))/(("32 g"))xx(0.144g)=100.8mL` Ratio by volume of `H_(2) and O_(2)` combining to form `H_(2)O = 200 : 100.8 = 2:1` Since the TWO ratios are the same, the data CORRESPONDS to Law of Constant Composition.

Discussion

51740.	10 " mL of " a solution of H_(2)O_(2) of 10 violume strength decolourises 100 " mL of " KMnO_(4) solution acidified with dil H_(2)SO_(4). The amount of KMnO_(4) in the given solution is K=39, Mn=55)
Answer» 0.282 g 0.564 g 1.128g 0.155 g Solution :VOLUME of `O_(2)` at STP`=10mLxx10V=100mL` `22400" mL of " O_(2) at STP =1 mol=4Eq` `100 " mL of " O_(2) STP=(4)/(22400)xx100=(1)/(56)EQ` " Eq of "`KMnO_(4)=" Eq of "O_(2)` `=(1)/(56)" Eq of "KMnO_(4)` `=(1)/(56)xx31.5g of KMnO_(4)` `=0.564g` `(Ew of KMnO_(4)` in acidic medium `=(MW)/(5)=31.5`)

Discussion

51741.	10 ml of a mixture of carbon monoxide, marsh gas and hydrogen exploded with excess of oxygen gave a contraction of 6.5cc. There was further contraction of 7cc when the residual gas was treated with caustic potash. The volume of marsh gas present in original mixture as
Answer» 5cc 2cc 3cc 4cc Solution :VOLUME of CO = xCC Volume of MARSH GAS = yCC Volume of nitrogen = `10-(x+y)C C` `{:("2 CO",+,"O2",rarr,2CO_(2)),("x CC",,(x)/(2)"CC",,"xCC"):}` `{:(CH_(4)(g)+2O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l)),(""darr),("YCC2yCCyCCvolume negligible"):}` Contraction in volume due to bruning of CO `=x+(x)/(2)-x-(x)/(2)C C` Contraction in volume due to burning of `CH_(4)` `=y+2y-y=2yC C` `(x)/(2)+2y=6.5C Corx+4y=13C C` volume of `CO_(2)` produced = `x+y` contraction in volume with caustic POTASH = 7CC `x+4y=13impliesx+y=7impliesy=2impliesx=5` `CO=5C C""CH_(4)=2C C""N_(2)=3C C`

Discussion

51742.	10 mL of a given solution of H_(2)O_(2) contains 0.91 g of H_(2)O_(2) . Express its strength in volume.
Answer» Solution :68 g of `H_(2)O_(2)` produce `O_(2)=22400` ML at NTP `therefore 0.91 g ` of `H_(2)O_(2)` will produce `O_(2)=(22400xx0.91)/(68)~= 300` mL at NTP `therefore ` Volume STRENGTH `~=(300)/(10)~=30`

Discussion

51743.	10 ml of a gaseous hydrocarbon is mixed with excess of oxygen and burnt . The gases are then cooled back . The reduction in volume was 25 ml . When the gases are passed into caustic potash , there is a further reduction in volume of 30 ml . If all volumes are measured under the laboratory conditions , the hydrocarbon is
Answer» `C_(3) H_(8)` `C_(3) H_(4)` `C_(3) H_(6)` `C_(4) H_(8)` ANSWER :C

Discussion

51744.	10ml of an alkane on complete combustion gave 40ml of CO_(2) under the same conditions. The formula of the alkane is
Answer» `C_(4) H_(6)` `C_(6) H_(10)` `C_(4) H_(8)` `C_(4) H_(10)` Answer :D

Discussion

51745.	50 ml of 0.1 M solution of sodium acetate and 50 mlof 0.01 M acetic acid are mixed . The pKa of acetic acid is 4.76 TheP^(H)ofthe buffer solution is
Answer» Solution :NUMBER of milliequivalents of acid `= 20 xx 0.2 = 2` Number of milliequivalents of salt `= 40 xx0.1 = 4` ` PH =pK_a +log ""(["Salt"])/("[acid]")=4.8 + log = 4.8+ 0.3= 5.1`

Discussion

51746.	10 ml N_(2) is reacted with 20 ml H_(2) to form NH_(3). The correct statements is /are
Answer» 13.3 ML `NH_(3)` is formed 20 ml `NH_(3)` is formed 3.4 ml `N_(2)` is left after the completion of the reaction 16.7 ml `NH_(3)` of mixture is left after the completion of the reaction Solution :`{:(,N_(2(g)),+,3H_(2(g)),rarr,2NH_(3(g))),("STP","22400 ml",,3xx22400ml,,2xx22400ml),("At START","10 ml",,"20 ml",,),("After reaction",10-6.6,,-,,"13.3ml"),(,"= 3.4ml",,,,):}`

Discussion

51747.	10 litres of a gas at S.T.P. weigh 19.64 g. Calculate the molecular mass of the gas.
Answer» ANSWER :44

Discussion

51748.	1.0 litre of a solution contains 5.3 g of Na_(2)CO_(3) and 8 g of NaOH. 20 mL of this solution are taken and titrated against N//10 HCl using separately (a) methyl orange as an indicator and (b) phenolphthalein as an indicator. What will be the titre values in these two cases ?
Answer» SOLUTION :N//A

Discussion

51749.	10 L of water has 33 mg of oxygen dissolved. What is the ppm of dissolved oxygen?
Answer» ANSWER :3.3 PPM

Discussion

51750.	10 kg of cold water at 2^(@)C absorbs 50J of heatfrom the surrounding which were at a temperature of37^(@)C . What is the entropy change of (i) the system (ii) the surroundings (iii) the universe ? Assumethe change in temperature of the water and the surroundings to be negligible.
Answer» Solution :`(i) DeltaS_("system") ("water")= ( +50J)/( 275K) = 0.18 JK^(-1)``(ii) DeltaS_("SURROUNDINGS") =( -50J)/(310K) = - 0.16 JK^(-1)` `(III) DeltaS_("universe") = DELTAS _("system") + DeltaS_("surroundings") = 0.18 + ( -.16) = 0.02 JK^(-1)`

Discussion

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