94420 + Interview Questions in Chemistry in Class 12 Page 1741 InterviewSolution

87001.	A steel sample is to be analysed for Cr and Mn simultaneously . By suitable treatement Cr is oxidized as Cr_(2)O_(7)^(2-) and the Mn to MnO_(4)^(-) Cr rarr Cr_(2) O_(7)^(2-) Mn rarr MnO_(4)^(-) A 10 gm sample of steel is used to produce 250.0 mL of a solution containing Cr_(2)O_(7)^(2-) and MnO_(4)^(-). A 10mL portion of this solution is addedto a BaCl_(2) solution and by proper adjustment of the acidicty , the chromum is completely precipitated as BaCrO_(4) 0.0549 g is obtained. Cr_(2) O_(7)^(2-) overset( H^(o+))(rarr)BaCrO_(4) A second 10mL portion of this solution requires exactly 15.95mL of 0.0750 M standard Fe^(3+) solution for its titration ( in acid solution ) . Equivalent of Fe^(2+) required for reduction of MnO_(4)^(-) is
Answer» `5.44 XX 10^(-4)` `0.544 xx 10^(-2)` `1.196 xx 10^(-3)` `11.96 xx 10^(-4)` ANSWER :A

Discussion

87002.	A steel sample is to be analysed for Cr and Mn simultaneously . By suitable treatement Cr is oxidized as Cr_(2)O_(7)^(2-) and the Mn to MnO_(4)^(-) Cr rarr Cr_(2) O_(7)^(2-) Mn rarr MnO_(4)^(-) A 10 gm sample of steel is used to produce 250.0 mL of a solution containing Cr_(2)O_(7)^(2-) and MnO_(4)^(-). A 10mL portion of this solution is addedto a BaCl_(2) solution and by proper adjustment of the acidicty , the chromum is completely precipitated as BaCrO_(4) 0.0549 g is obtained. Cr_(2) O_(7)^(2-) overset( H^(o+))(rarr)BaCrO_(4) A second 10mL portion of this solution requires exactly 15.95mL of 0.0750 M standard Fe^(3+) solution for its titration ( in acid solution ) .% of chromium in the steel sample
Answer» 1.496 2.82 1.96 5 Answer :B

Discussion

87003.	(A) Steel is an alloy of iron with a metal. (R) Same steels contain carbon as a constituent
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is FALSE Both (A) and (R) are false Solution :Both (A) and (R) are true and (R) is not the correct explanation of (A)

Discussion

87004.	A steel ball of density 8.0 xx 10^(3) kg//m^(3)and radius 2 mm is observed to fall with a terminal velocity 1.0 xx 10^(-2) m//sin a liquid of density 1.8 xx 10^(3) kg//m^(3) . Calculate the viscosity of the liquid.
Answer» SOLUTION :We have, `ETA = (2R^(2)(d-d_(0))g)/(9u)` `=(2(2 xx 10^(-3))^(2) (8.0 xx 10^(3) - 1.8 xx 10^(3)) xx 9.8)/(9 xx 1.0 xx 10^(-2))` `=5.4 Nm^(-2) s`

Discussion

87005.	A steam boiler made up of steel weights 900 kg. The boiler contains 400 kg of water. Assuming 70% of the heat is delivered to boiler and water. Heat required to raise the temperature of the whole from 10^(@)C to 100^(@)C is (Heat capacity of steel is 0.11 kcal/kg-K and heat capacity of water is 1 kcal /kg-K)
Answer» 63550kcal 64200k cal 64157 K cal 65250 k cal Solution :`DeltaH` required for heating `=underset(("boiler"))(mSDeltaT)+underset(("water"))(mSDeltaT)` `=0.11xx900xx90+1xx400xx90=44910` k cal Since only 70% of HEAT GIVEN is used up to do so. THUS actual heat required `=(44910xx100)/70=64157` KCAL

Discussion

87006.	A steady current passing through a solution of AgNO_(3) solution deposits 0.50 g of Ag in 1 h. Calculate the number of coulombs. What volume of hydrogen at 27^(@)C and 750 mm pressure would the same current liberate in one hour ?
Answer» SOLUTION :`446.7` COULOMBS, 57.7 ML

Discussion

87007.	A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO_(4) and ZnSO_(4) until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow?Calculate the mass of Zn deposited at the cathode of cell Y. (Molar mass : Fe=56 g"mol"^(-1)Zn=65.3"g mol"^(-1), F=96500"C mol"^(-1))
Answer» SOLUTION :`m=z I t` `2.8g=(56xx2xx t)/(2xx96500)` `t=4825s` `(m_(1))/(m_(2))=(E_(1))/(E_(2))` `(2.8)/(MZN)=(56)/(2)xx(2)/(65.3)` `mZn=3.265g`

Discussion

87008.	(a) State two advantages of H_2 - O_2 fuel cell over ordinary cell. (b) Silver is electron deposited on a metallic vessl of total surface area 900 cm^2 by passing a current of 0.5 amp for two hours. Calcuate the thickness of silver metal deposited, [Given : Density of silver = 10.5 g cm^(-3) Atomic mass of silver = 108 u, 1F = 96500 C mol^(-1)].
Answer» Solution :(a) (i) It is highly efficient and do not produce pollution. (ii) The `H_2O` so produced can be USED by astronauts for drinking purpose. (B) `m = Z xx I xx t` ` m = (108)/(96500) xx 0.5 xx 2 xx 60 xx 60` `= (108 xx 5)/(965 xx 10) xx 2 xx 6 xx 6 = 4.03 g` `4.03 g = V xx d` `4.03 g = V xx 10.5 g cm^(-3)` `V = "Area" xx "THICKNESS"` `V = (4.03)/(10.5)` `(4.03)/(10.5) = 900 cm^2 xx "thickness"` `"Thickness"= (0.338 cm^3)/(900 cm^2)` `= 4.26 xx 10^(-4) cm`.

Discussion

87009.	(a) State the hybridisation and magnetic behaviour of [Cr(CO)_(6)]. (b) What are the various factors affecting crystal field splitting energy. (c ) Which of the two is more stable and why ? K_(4)[Fe(CN)_(6)] or K_(3)[Fe(CN)_(6)]
Answer» Solution :(a) `d^(2)sp^(3)`, diamagnetic (b) (i) Nature of LIGAND (ii) CHARGEON metal ion (c ) `K_(4)Fe(CN)_(6)`

Discussion

87010.	(a) State the law which helps to determine the limiting molar conductivity of weak electrolytes. (b) Calculate limiting molar conductivity of CaSO^4 (limiting molar conductivity of calcium and sulphate ions are 119.0 and 106.0 S cm^2 mol^(-1)respectively).
Answer» Solution :(a) The law which helps to determine the limiting molar conductivity of weak electrolytes is Kohlaursch Law. It states : The limiting molar conductivity of an electrolyte can be represented as the sum of the individual CONTRIBUTION of the anion and the cation of the electrolyte. (b) Limiting molar conducitivity of `CaSO_(4)` can be DETERMINED as under: `Lambda_(m)^(@) CaSO_(4) = lambda_(CA^(2+))^(@) + lambda_(SO_(4)^(2-))^(@) = 119.0 S cm^(2) mol^(-1) + 106.0 S cm^(2) mol^(-1) = 225.0 S cm^(2) mol^(-1)`.

Discussion

87011.	(a) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law. (b) Calculate Lambda_(m)^(@) , for acetic acid. Given that Lambda_(m)^(@) (HCl) = 426 S cm^(2) mol^(-1) Lambda_(m)^(@) (NaCl) = 126 S cm^(2) mol^(-1) Lambda_(m)^(@) (CH_(3)COONa) = 91 S cm^(2) mol^(-1)
Answer» Solution : Kohlrausch law states that limiting molar conductivity of an electrolyte can be represented as the SUM of the individual contributions of the anion and the cation of the electrolyte. In expression for the molar conductivity at INFINITE dilution. `Lambda_(m(NaCl))^(@) = lambda_(Na^(+))^(@) + lambda_(Cl^(-))^(@)` In general, if an electrolyte or dissociation gives n+ cations and n- anions, than its limiting molar conductivity is given by `Lambda_(m)^(@) =n_(+)lambda_(+)^(@) + n_(-)lambda_(-)^(@)` `Lambda_(m)^(@)(CH_(3)COOH) =Lambda_(m)^(@) (CH_(3)COONa) + Lambda_(m)^(@)(HCl) - Lambda_(m)^(@) (NaCl)` Given that: `Lambda_(m)^(@) (CH_(3)COONa) = 91 S cm^(2) mol^(-1)` `Lambda_(m)^(@) (HCl) = 426 S cm^(2) mol^(-1)` `Lambda_(m)^(@) (NaCl) = 126 S cm^(2) mol^(-1)` Substituting the VALUES in the equation above, we have: `Lambda_(m)^(@) (CH_(3)COOH) = 91 + 426 - 126 = 391S cm^(2) mol^(-1)`

Discussion

87012.	A standard hydrogen electrode has zero electrode potential is assumed to be
Answer» negative ZERO positive fractional SOLUTION :`E^(@)` for SHE has been ASSUMED to be zero.

Discussion

87013.	A standard hydrogen electrode has zero electrode potential because
Answer» Hydrogen is EASIER to oxidise This ELECTRODE potential is ASSUMED to be zero Hydrogen atom has only one electron Hydrogen is the lightest element. Solution :Standard hydrogen electrode is a REFERENCE electrode. Potential of any half cell is calculated with respect to standard hydrogen electrode (SHE). Hence, electrode potential of SHE has been arbitrarily taken to be zero.

Discussion

87014.	A standard electrochemical cell is made by dipping an Ag electrode into a 1.0 MAg^+solution and a Cd electrode into a 1.0 M Cd^(2+)solution.(a) What is the spontaneous chemical reaction and what is the maximum potential produced by the cell?(b) What would be the effect on the potential of this cell if Na_2Swere added to the Cd^(2+)half cell and CDS were precipitated? Why? (c) What would be the effect on the potential of the cell if the size of the silverelectrode was doubled?See E^@values from the table if required.
Answer» Solution :`(a) Cd(s) + 2AG^(+) = 2Ag(s) + Cd^(2+), 1.20V ` (b) It would increase (c) No EFFECT

Discussion

87015.	(a) Standard EMF of the cell, Cu\|Cu 2+\| Ag Ag\|is 0.46V at 25°C. Find the value of standard free energy change for the reaction that occurs in the cell.(b) Draw the neat labeled diagram of SHE and write its symbolic representation.
Answer» Solution :(a) `DeltaG^@ =-nFE^(@)_(cell) ` n = 2, F = 96, `500 C E^@ = 0.46V` `AG^@=-2xx96,500xx0.46` `AG^@ = -108080 joules` `AG^@ = -108.80 KJ` (b) SHE is a primary REFERENCE electrode because its SRP value (E) is zero V. Construction : It consists of a platinum foil fused to a platinum wire. It is enclosed in a narrow glass tube. It is surrounded by a cylindrical glass jacket having an inlet at the TOP and the platinum foil is dipped in 1 M HCI solution. Working : Pass pure and dry hydrogen gas continuously at 1 atm pressure into the HCI solution through an inlet. The hydrogen gas gets adsorbed on the surface of the Pt foil. The absorbed hydrogen is in contact with the H+ ions of 1 M HCI. The electrode reaction is as SHOWN below. `H_(2)(g) to 2H+ +2e^(-)` So hydrogen electrodes can be represented as Pt(s), `H_(2), ("g, Ibar") \| H+ (1M)`

Discussion

87016.	A stainless steel chamber contains Ar gas at a temperature T and pressure P. The total number of Ar atoms in the chamber is n. Now Ar gas in the chamber is rep,aced by CO_(2) gas and the total number of CO_(2) molecules in the chamber is n//2 at the same temperature T. The pressure in the chamber now is P'. Which one of the following relations relations holds true ? (Both the gases behave as ideal gases)
Answer» P'=P P'=2P `P'=P//2` `P' P//4` SOLUTION : The increasein PRESSUREIS causedby the numberofparticleshittingagainst the WALLSOF thecontainer.Sincenumber ofmoleculesisreducedtohalfhencethepressurewillalsobe decreased accordinglyas pertheformula PV = nRThence ` P=P//2`

Discussion

87017.	(a) Square planar complexes with coordination number four exhibitgeometrical isomerism whereas tetrahedral complexesdo not , why ? (b) Three geometrical isomers of the square planarcomplex [Pt(NH_(3))(H_(2)O)(py)(NO_(2))]^(+)are possible . What are they?
Answer» SOLUTION :(a) Tetrahedral complexes do not show GEOMETRICAL ISOMERISM because the relative positions of the ligands ATTACHED to the central metal atom are same with respect to each other . (b)

Discussion

87018.	A standard hydrogen electrode has zero electrode potential because :
Answer» Hydrogen is easiest to oxidise This electrode POTENTIAL is assumed to be zero Hydrogen ATOM has only ONE electron Hydrogen is the lightest element Answer :B

Discussion

87019.	A square planar complex is represented as:
Answer» GEOMETRICAL ISOMERS OPTICAL isomerism Linkage isomerism None Answer :A

Discussion

87020.	A square planar complex is formed by hybridisation of which atomic orbitals ?
Answer» `s, p_(X), p_(y), p_(yz)` `s, p_(x), p_(y), p_(x^(2)-y^(2))` `s, p_(x), p_(y), d_(z^(2))` `s, p_(x), p_(y), d_(XY)` ANSWER :B

Discussion

87021.	A square planar complex is formed by hybridisation of which atomic orbitals?
Answer» `s,p_(x),p_(y),d_(yz)` `s,p_(x),p_(y),d_(x^(2)-y^(2))` `s,p_(x),p_(y),d_(z^(2))` `s,p_(y),p_(z),d_(XY)` ANSWER :B

Discussion

87022.	A spoon used as a cathode is dipped in AgNO_(3) solution and a current of 0.2 amp is passed for one hour. Calculate how much silver plating has occurred ? (b) how many electrons were involved in the process ? (c) what amount of copper would have been plated under similar conditions ?
Answer» SOLUTION :`0.805 G, 4.5 XX 10^(21), 0.237 g`

Discussion

87023.	(A) Spontaneous reaction may be slow or fast. (R ) Spontaneous nature deals with feasibility of the reaction but not rate.
Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R ) are false Answer :A

Discussion

87024.	A spontaneous process is one in which the system suffers
Answer» No energy change A lowering of FREE energy A lowering of entropy An INCREASE in internal energy Solution :Negative VALUE of `DeltaG` indicates that the process is spontaneous. Spontaneous process leads to the lowering of free energy `(DeltaG)`.

Discussion

87025.	A spontaneous change is one in which the system suffers:
Answer» An INCREASE in INTERNAL energy A LOWERING of entropy A lowering of FREE energy No energy Answer :C

Discussion

87026.	A spherical balloon of volume 5 litre is to be filled up with H_2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0^@C.The no of balloons that can be filled up is :
Answer» SOLUTION :VOLUME of gas at NTP is , `P_1V_1=P_2V_2` `6xx 6 =1xxV_2` `V_2`=36 LITRES. Outcoming gas = 36-6 =30 litres No of balloon =`30/5=6`

Discussion

87027.	A spherical ballon of 21 cm diameter is to be filled up with hydrogen at NTP from a cylinder containing the gas at 20 atm and 27^(@)C. The cylinder can hold 2.82 litre of water at NTP Calculate the number of ballons that can be filled up.
Answer» SOLUTION :Volume of one balloon `=(4)/(3) pi r^(3)` `=(4)/(3)xx(22)/(7)xx((21)/(2))^(3)=4851 cm^(3)=4.851` litre Let n BALLOONS be filled, thus total volume of hydrogen USED in filling balloons `=4.851xxn` litre Total volume of hydrogen in the cylinder at NTP `V=(20xx2.82xx273)/(300xx1)=51.324` litre ACTUAL volume of `H_(2)` to be transferred to balloons `=51.324-2.82` `=48.504` litre No. of balloons filled `.n. =(48.504)/(4.851)=10`

Discussion

87028.	A spherical balloon of 21 cm diameter is to be filled up hydrogen at NTP from a cylinder containing the gas at 20 atmosphere at 27^(@)C. If the cylinder can hold 2.82 litres of water, calculate the number of balloons that can be filled up.
Answer» Solution :Volume of the balloon `= (4)/(3) pi r^(3)` `= (4)/(3) xx (22)/(7) xx (10.5)^(3) c c ""("radius" = (21)/(2) = 10.5)` = 4851 cc. Balloon is at a temperature of 273 K and 1 atm pressure, i.e, pressure `= 1 xx 76 xx 3.6 xx 981 = 1.014 xx 10^(6)` dynes/`cm^(2)`. `therefore` no. of moles of `H_(2)`, the balloon can CONTAIN at NTP `=(pV)/(RT)` `= (1.014 xx 10^(6) xx 4851)/(8.314 xx 10^(7) xx 273) = 0.2167`. No. of moles in the cylinder `= ((20 xx 76 xx 13.6 xx 981) xx 2820)/(8314 xx 10^(7) xx 300)` = 2.2929. While filling the last balloon, when the pressure ofthe cylinder will drop to 1 atm, gas cannot be withdrawn. Now, no. of moles of `H_(2)` remaining in the cylinder unused `= (1.014 xx 10^(6) xx 2820)/(8.314 xx 10^(7) xx 300) = 0.1146` `therefore` no. of balloons that can be filled `= ("no. of moles of " H_(2) " in the cylinder that can be used")/("no. of moles of " H_(2) " one balloon can contain")` `= (2.2929 - 0.1146)/(0.2167) = 10`

Discussion

87029.	A spectral line obtained when an electron jumps from sixth energy level to first energy level in spectrum of hydrogen atom falls in:
Answer» VISIBLE region ULTRA VIOLET region infra red region None of these Solution :It BELONGS of Lyman series which FALLS in ultra violet region.

Discussion

87030.	A sparingly soluble salt MX is dissolved in water to prepare 1 L saturated solution.Now 10^(-6) mole NaX (assume 100% dissociation ) is added into this.Conductivity of this solution is 29xx10^(-6) S//m.If K_(sp) of MX is axx10^(-b),then find value of (a+b),a is a natural number & 1le a le 9. Given:lambda_((x)^(-))^(0)=4xx10^(-3) S m^(2) mol^(-) lambda_((Na)^(+))^(0)=5xx10^(-3) S m^(2) mol^(-) lambda_((M)^(+))^(0)=6xx10^(-3) S m^(2) mol^(-)
Answer» Solution :`MX hArrM^++X^(-)` `x " " x+10^(-6)` `[NA^+]=10^(-6) M` `K_("SOL")=K_(M^+)+K_(X^(-))+K_(Na^+)` `29xx10^(-6) = 10^3 [6xx10x +(4xx10^(-3)(x+10^(-6))+(5xx10^(-3)xx10^(-6))]` `x=2XX10^(-6)` `K_(sp)=2xx10^(-6) xx3xx10^(-6)=6XX10^(-12)`

Discussion

87031.	A species has 16 protons, 18 electrons and 16 neutrons. Find the species and its charge.
Answer» <P>`S^(-)` `SI^(2+)` `P^(3-)` `S^(2-)` Solution :The SPECIES is `S^(2-)` .

Discussion

87032.	A space rocket is propelled by
Answer» An automobile ENGINE Rocketpropellant Nuclear submarine STEAM engine Solution :Rocket propellant are USED as fuel for space rocket.

Discussion

87033.	(A) Space or crystal lattice differ in symmetry of the arrangement of points. (R) n lambda.=2 dsin thetais known as Bragg's equation
Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :B

Discussion

87034.	A sp^3 - hybrid orbital contains:
Answer» 1/4s-character 1/2s-character 2/3s-character 3/4 s-character Answer :A

Discussion

87035.	A source of olives acid is:
Answer» ANIMAL fat Corn OIL Linseed oil None Answer :B

Discussion

87036.	A source of maximum energy is
Answer» Carbohydrates FATS Proteins VITAMINS Solution :Fats give more energy than carbohydrates, proteins or vitamins per UNITS mass.

Discussion

87037.	Iron (II) oxide has a cubic structure and each unit cell has a size of 5 Å. If density of this oxide is 4 g cm^(-3), calculate the number of Fe^(2+) and O^(2-) ions present in each unit cell. (Atomic mass of Fe = 56, O = 16, N_(A) = 6.023 xx 10^(23) and 1 Å = 10^(-8) cm)
Answer» Solution :(a) When amorphous solid (glass) are heated and cooled SLOWLY they acquire crystallinity at same TEMPERATURE. Glass objects of ancient monuments, over a period of years, are exposed to sunlight and cooled during night times, which has resulted in the crystalization of the glass object which imparts milky colour to the ancient monuments. (b) Solve by using `d = (Z xx M)/(a^(3) xx N_(A) xx 10^(-30))` `Z = (4.0 g cm^(-3) xx (5 xx 10^(-8)cm)^(3)(6.023 xx 10^(23) mol^(-1)))/((56 + 16) g mol^(-1))` = 4 i.e., `4Fe^(2+) and 4O^(2-)` ions PRESENT in each unit cell.

Discussion

87038.	A solution containing 8.6 g urea in one litre was found to be isotonic with a 5% (wt./vol.) solution of an organic non-volatile solute. The molecular weight of latter is:
Answer» 348.9 34.89 3489 861.2 Answer :A

Discussion

87039.	(a) Solutions of two electrolytes 'A' and 'B' are diluted. The limiting molar conductivity of 'B' increases 1.5 times while that of 'A' increases 25 times. Which of the two is a strong electrolyte ? Justify your answer. (b) The products of electrolysis of aqueous NaCl at the respective electrodes are : Cathode : H_(2), Anode : Cl_(2) and not O_(2). Explain.
Answer» SOLUTION :(a) 'B' is a strong elctrolyte sinece it is already almost completely ionised (`alpha~~1`). DILUTION simply helps in its dissociation i.e., the ions get separated. Therefore, the increase in molar conductivity upon dilution is small. However, 'A' is a weak electrolyte because it is weakly ionised `alpha lt 1)`. Dilution helps in its ionisation as well as dissociation. Therefore, the increase in molar conductivity is quite large. For more details, CONSULT SECTION 19. (b) For explanation, consult section 15.

Discussion

87040.	A solutionn of urea in water has a boiling point of 100.18^(@)C. Calculate the freezing point of the same direction.Molal constants for water K_(f) and K_(b) are 1.86 and 0.512 respectively.
Answer» Solution :For a solution of molality m we have `m=(DeltaT_(F))/(K_(f))=(DeltaT_(b))/(K_(b))`…………..(Eqns 7 and 8 ) `:.DeltaT_(f)=DeltaT_(b).(K_(f))/(K_(b))`, `(DeltaT_(b)=100.18-100=0.18^(@))` `=0.18xx(1.86)/(0.512)=0.654^(@)` As the f.p. of PURE WATER is `0^(@)C`, the f.p. of the solution will be `-0.654^(@)C`

Discussion

87041.	A solution with lower osmotic pressure is called ………………with respect to a more concentrated solution which is called……………. .
Answer» SOLUTION :HYPOTONIC, HYPERTONIC

Discussion

87042.	A solution which isresistant to changes of pH on addition of small amounts of an acid or a base is known as
Answer» BUFFER solution true solution isohydric solution ideal solution Answer :A

Discussion

87043.	A solution which is resistant to change of pH upon the addition of an acid or a base is known as
Answer» A colloid A crystalloid A BUFFER An INDICATOR SOLUTION :Because buffer solution have a CONSTANT PH.

Discussion

87044.	A solution which is 10M each in Mn^(2+),Fe^(2+), Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and Hgs are 10^(-15),10^(-23),10^(-20) and 10^(-54) and respectively, which one will precipitate first?
Answer» FeS MgS HgS ZnS Solution :Hgs having the LOWEST `K_(SP)` among the lot will PRECIPITATE first. Hence, (C) is the correct answer.

Discussion

87045.	A solution which is 10^(-3)M each in Mn^(2+),Fe^(2+),Zn^(2+) and Hg^(2+) is treated with 10^(-16)M sulphide ion. If K_(SP) of MnS, FeS, ZnS and HgS are 10^(-15),10^(-23),10^(-20) and 10^(-54) respectively, which one will precipitate first
Answer» FeS MgS HGS ZnS Solution :HgS having the lowest `K_(SP)` AMONG the lot will PRECIPITATE first.

Discussion

87046.	A solution which is 10^(-3)M each in Mn^(2+), Fe^(2+), Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and HgS are 10^(-15), 10^(-23), 10^(-20) and 10^(-54) respectively, which one will precipitate first
Answer» FeS MgS HgS ZnS Answer :C

Discussion

87047.	A solution which is 10^(-3) M each in Mn^(2+), Fe^(2+), Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and HgS are, 10^(-13), 10^(-18), 10^(-24) and 10^(-53) respectively, which one will precipitate first :
Answer» FeS MgS HgS ZnS Answer :C

Discussion

87048.	A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) "of " MnS , FeS, ZnS and HgS " are " 10^(-13),10^(-18),10^(-24) and 10^(-53) respectively. Which one will precipitate first ?
Answer» FeS MgS HgS Zns Answer :C

Discussion

87049.	A solution which is 0.1 M in NaI and also 0.1 M in Na_2(SO_4)is treated with solid Pb(NO_3)_2 . Which compound, PbI_2 or PbSO_4 , will precipitate first? What is the concentration of anions of the least soluble compound when the more soluble one starts precipitating?
Answer» SOLUTION :`PbSO_4, 0.021M`

Discussion

87050.	A solution which is 0.001 M each in Mn^(2+), Fe^(2+), Zn^(2+)and Hg^(2+)is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and HgS are 10^(-15), 10^(-23), 10^(-20)and10^(-54) respectively, which one will precipitate first ?
Answer» FeS MgS HgS ZnS Answer :C

Discussion

Explore topic-wise InterviewSolutions in .

(A) Steel is an alloy of iron with a metal. (R) Same steels contain carbon as a constituent

A steel ball of density 8.0 xx 10^(3) kg//m^(3)and radius 2 mm is observed to fall with a terminal velocity 1.0 xx 10^(-2) m//sin a liquid of density 1.8 xx 10^(3) kg//m^(3) . Calculate the viscosity of the liquid.

A steady current passing through a solution of AgNO_(3) solution deposits 0.50 g of Ag in 1 h. Calculate the number of coulombs. What volume of hydrogen at 27^(@)C and 750 mm pressure would the same current liberate in one hour ?

(a) State the hybridisation and magnetic behaviour of [Cr(CO)_(6)]. (b) What are the various factors affecting crystal field splitting energy. (c ) Which of the two is more stable and why ? K_(4)[Fe(CN)_(6)] or K_(3)[Fe(CN)_(6)]

(a) State the law which helps to determine the limiting molar conductivity of weak electrolytes. (b) Calculate limiting molar conductivity of CaSO^4 (limiting molar conductivity of calcium and sulphate ions are 119.0 and 106.0 S cm^2 mol^(-1)respectively).

A standard hydrogen electrode has zero electrode potential is assumed to be

A standard hydrogen electrode has zero electrode potential because

(a) Standard EMF of the cell, Cu|Cu 2+| Ag Ag|is 0.46V at 25°C. Find the value of standard free energy change for the reaction that occurs in the cell.(b) Draw the neat labeled diagram of SHE and write its symbolic representation.

(a) Square planar complexes with coordination number four exhibitgeometrical isomerism whereas tetrahedral complexesdo not , why ? (b) Three geometrical isomers of the square planarcomplex [Pt(NH_(3))(H_(2)O)(py)(NO_(2))]^(+)are possible . What are they?

A standard hydrogen electrode has zero electrode potential because :

A square planar complex is represented as:

A square planar complex is formed by hybridisation of which atomic orbitals ?

A square planar complex is formed by hybridisation of which atomic orbitals?

A spoon used as a cathode is dipped in AgNO_(3) solution and a current of 0.2 amp is passed for one hour. Calculate how much silver plating has occurred ? (b) how many electrons were involved in the process ? (c) what amount of copper would have been plated under similar conditions ?

(A) Spontaneous reaction may be slow or fast. (R ) Spontaneous nature deals with feasibility of the reaction but not rate.

A spontaneous process is one in which the system suffers

A spontaneous change is one in which the system suffers:

A spherical balloon of volume 5 litre is to be filled up with H_2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0^@C.The no of balloons that can be filled up is :

A spherical ballon of 21 cm diameter is to be filled up with hydrogen at NTP from a cylinder containing the gas at 20 atm and 27^(@)C. The cylinder can hold 2.82 litre of water at NTP Calculate the number of ballons that can be filled up.

A spherical balloon of 21 cm diameter is to be filled up hydrogen at NTP from a cylinder containing the gas at 20 atmosphere at 27^(@)C. If the cylinder can hold 2.82 litres of water, calculate the number of balloons that can be filled up.

A spectral line obtained when an electron jumps from sixth energy level to first energy level in spectrum of hydrogen atom falls in:

A species has 16 protons, 18 electrons and 16 neutrons. Find the species and its charge.

A space rocket is propelled by

(A) Space or crystal lattice differ in symmetry of the arrangement of points. (R) n lambda.=2 dsin thetais known as Bragg's equation

A sp^3 - hybrid orbital contains:

A source of olives acid is:

A source of maximum energy is

Iron (II) oxide has a cubic structure and each unit cell has a size of 5 Å. If density of this oxide is 4 g cm^(-3), calculate the number of Fe^(2+) and O^(2-) ions present in each unit cell. (Atomic mass of Fe = 56, O = 16, N_(A) = 6.023 xx 10^(23) and 1 Å = 10^(-8) cm)

A solution containing 8.6 g urea in one litre was found to be isotonic with a 5% (wt./vol.) solution of an organic non-volatile solute. The molecular weight of latter is:

A solutionn of urea in water has a boiling point of 100.18^(@)C. Calculate the freezing point of the same direction.Molal constants for water K_(f) and K_(b) are 1.86 and 0.512 respectively.

A solution with lower osmotic pressure is called ………………with respect to a more concentrated solution which is called……………. .

A solution which isresistant to changes of pH on addition of small amounts of an acid or a base is known as

A solution which is resistant to change of pH upon the addition of an acid or a base is known as

A solution which is 10M each in Mn^(2+),Fe^(2+), Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and Hgs are 10^(-15),10^(-23),10^(-20) and 10^(-54) and respectively, which one will precipitate first?

A solution which is 10^(-3)M each in Mn^(2+),Fe^(2+),Zn^(2+) and Hg^(2+) is treated with 10^(-16)M sulphide ion. If K_(SP) of MnS, FeS, ZnS and HgS are 10^(-15),10^(-23),10^(-20) and 10^(-54) respectively, which one will precipitate first

A solution which is 10^(-3)M each in Mn^(2+), Fe^(2+), Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and HgS are 10^(-15), 10^(-23), 10^(-20) and 10^(-54) respectively, which one will precipitate first

A solution which is 10^(-3) M each in Mn^(2+), Fe^(2+), Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and HgS are, 10^(-13), 10^(-18), 10^(-24) and 10^(-53) respectively, which one will precipitate first :

A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) "of " MnS , FeS, ZnS and HgS " are " 10^(-13),10^(-18),10^(-24) and 10^(-53) respectively. Which one will precipitate first ?

A solution which is 0.1 M in NaI and also 0.1 M in Na_2(SO_4)is treated with solid Pb(NO_3)_2 . Which compound, PbI_2 or PbSO_4 , will precipitate first? What is the concentration of anions of the least soluble compound when the more soluble one starts precipitating?

A solution which is 0.001 M each in Mn^(2+), Fe^(2+), Zn^(2+)and Hg^(2+)is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and HgS are 10^(-15), 10^(-23), 10^(-20)and10^(-54) respectively, which one will precipitate first ?