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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 50751. |
Assertion: According to Mendeleeff, periodic properties of elements is a function of their atomic masses. Reason : Atomic number is equal to number of protons |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 50752. |
(A): According to Mendeleeff atomic weight of 'Be' is 9.1 but experimentally it is '7' (R): In the atomic weight of Be valency was taken as '3' by Mendeleeff. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 50753. |
(A) : A solution which contains one gram equivalent of solute pre litre of the solution is called a normal solution (R) : A normal solution means a solution in which the solute does not associate or dissociate. |
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Answer» If both (A) and (R) are correct and (R) is the correct EXPLANATION for (A) |
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| 50754. |
(a). A transition metal cation x^(3+) has magnetic momen sqrt(35) BM. What is the atomic number of x^(3+)? (b). Select the coloured ion and the ion having maximum Magnetic moment (i). Fe^(2+) (ii). Cu^(+) (iii). Sc^(3+) and (iv) Mn^(2+). |
Answer» Solution : (a). 26, `._(26)Feto3d^(6)4s^(2)` `Fe^(3+)to3d^(5)4s^(0)` `mu=SQRT(N(n+2))=sqrt(5xx7)=sqrt(35)` (b). Both these ions will be coloured and magnetic moment of `MN^(2+)` will be greater. |
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| 50755. |
(A) : A spectral line will be seen for 2p_(x) to 2p_(y) transition. (R) : Energy is released in the form of wave of light when electron drops from 2p_(x) to 2p_(y) orbital. |
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Answer» Both (A) and (R) are true and (R) is the correct explanation of (A) |
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| 50756. |
(a) A solution containing methoxide ions, CH_(3)O^(-) ions(as NaOCH_(3)), in methanol can be prepared by adding sodium hybride (NaH) to methanol (CH_(3)OH). A flammable gas is the other product. Write the acidbase reaction that takes place. (b) Write the nucleophilic substitution that takes place when CH_(3)I is added and the resulting solution is heated. |
Answer» Solution :(a) We recall from SECTION 3.15 that sodium hybride consists of `Na^(+)` ions and hybride ions(H ions) and that the hybride ion is a very strong base. [It is the conjugate base of `H_(2)`, a very weak acid `(pK_(a) = 35)` The acid-base reaction that TAKES place is  (b) the methoxide ion REACTS with the alkyl hallide `(CH_(3)L)` in a NUCLEOPHILIC substitution. 
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| 50757. |
(a) A powdered substance (A) on treatment with fusion mixturegives a green coloured compound (B). (b) The solution of (B) on acidification with hot dilute H_(2)SO_(4) gives a pink coloured compound (C). (c) The aqueous solution of (A) on treatement with excess of NaOH and bromine water gives a compound (D). (d) A solution of (D) in conc. HNO_(3) with lead dioxide at boiling temperature produces a compound (E), which has of the same colour as that of (C). (e) A solution of (A) in dilute HCI on treatment with a solution of barium chloride gave a white precipitate of compound (F), which was insoluble in conc. HNO_(3) and conc. HCI. What is (D)? |
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Answer» `MnO_(2)` (D) `MnO_(2)` (E) `HMnO_(4)` (F) `BaSO_(4)` |
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| 50758. |
(a) A powdered substance (A) on treatment with fusion mixturegives a green coloured compound (B). (b) The solution of (B) on acidification with hot dilute H_(2)SO_(4) gives a pink coloured compound (C). (c) The aqueous solution of (A) on treatement with excess of NaOH and bromine water gives a compound (D). (d) A solution of (D) in conc. HNO_(3) with lead dioxide at boiling temperature produces a compound (E), which has of the same colour as that of (C). (e) A solution of (A) in dilute HCI on treatment with a solution of barium chloride gave a white precipitate of compound (F), which was insoluble in conc. HNO_(3) and conc. HCI. Compound (C) and (E) are coloured due to: |
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Answer» d-d transition (D) `MnO_(2)` (E) `HMnO_(4)` (F) `BaSO_(4)` |
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| 50759. |
(a) A powdered substance (A) on treatment with fusion mixturegives a green coloured compound (B). (b) The solution of (B) on acidification with hot dilute H_(2)SO_(4) gives a pink coloured compound (C). (c) The aqueous solution of (A) on treatement with excess of NaOH and bromine water gives a compound (D). (d) A solution of (D) in conc. HNO_(3) with lead dioxide at boiling temperature produces a compound (E), which has of the same colour as that of (C). (e) A solution of (A) in dilute HCI on treatment with a solution of barium chloride gave a white precipitate of compound (F), which was insoluble in conc. HNO_(3) and conc. HCI. What is (A)? |
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Answer» `FeSO_(4)` (D) `MnO_(2)` (E) `HMnO_(4)` (F) `BaSO_(4)` |
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| 50760. |
(A): A gas behaves as an ideal gas at high temperature and low pressure. (R): Helium behaves as an ideal gas under all conditions. |
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Answer» Both A and R are CORRECT and R is the correct EXPLANATION of A. |
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| 50761. |
(A): A drop of liquid acquires spherical shape. (R): Acquiring spherical shape by a liquid drop is give to its capillary action. |
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Answer» Both A and R are CORRECT and R is the correct explanation of A. |
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| 50762. |
(a) A cylinder of gas is assumed to contain 11.2 kg of butane. If a normal family needs 20,000 kJ of energyper dayfor cooking , how long will the cylinder last ? Given that the heat of combustion of butane is 2658 kJ mol^(-1) (b) If the air supply of the burner is sufficient (i.e., you havea yellow instead of a blue flame), a portion ofthe gas escapes without combustion. Assuming that 33% of the gas is wasted due to this inefficiency , how long would the cylinder last ? |
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Answer» No.of days which it will LAST` = 513268.9 // 20,000= 25.7 `days `= 26 ` days (b) After wastage, heat available `= ( 67)/( 100) xx 513268.9 = 343890J` No. of days for which it will last `= 343890 // 20000 = 17` days |
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| 50763. |
(a) A 1000 watt hour is kept on ina room whose dimensions are 5m xx 5m xx 3m. Howmuch temperature of the room will rise after half an hour? Given the following data : Heat capacityofair at room temperature and 1 atm= 0.71 Jg^(-1)K^(-1) Density ofair = 1.22xx 10^(-3) g mL^(-1) (b) How much temperature of the room willrise if 25% heat is lost by radiation ? |
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Answer» Solution :(a) 1 Watt `= 1 J s^(-1)` `:. `Heat evolved by 1000 W heaterin 30MIN. `= 100 xx 30 xx 60 J = 18 xx 10^(5) J` Volume of the room ` 5 xx4 xx 3m^(3) = 60 m^(3) = 60xx 10^(6) cm^(3) ` Density of air`=1.22 xx 10^(-3) g mL^(-1)` Massof air in the room`= ( 60xx 10^(6)) xx ( 1.22 xx 10^(-3)) g = 73.2 xx 10^(3)g` Heat evolved by the heater `=` Heat absorbed by WALLS, roof and air INSIDE. `18 xx10^(5) J =( 5 xx 10^(4) xx Delta T ) +( 73.2 xx 10^(3)) xx 0.71 xx Delta T` `= ( 50000+ 51972 )Delta T` `= 101972 Delta T = 1.01 972 xx 10^(5) DeltaT ` or`Delta T = ( 1.8 xx 10^(5))/( 1.01972 xx 10^(5)) = 17.65^(@) ` (b) If `25%` heat is lost by radiation, net heat available in the room ` = (75)/( 100) xx 18 xx 10^(5) J = 13.5 xx 10^(5) J` ,BRGT `13.5 xx 10^(5) J = ( 5 xx 10^(4) xx Delta T ) + ( 73.2 xx 10^(3)) xx 0.71xx DeltaT` or`Delta T =( 13.5 xx 10^(5))/( 1.01972xx 10^(5)) = 13.24^(@)` |
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| 50764. |
A 9.40 g sample of KBr is dissolved in 105 g of H_(2)O at 23.6^(@) C in a coffee cup. Find the final temperature of this system. Assume that no heat is transferred to the cup or the surroundings. |
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Answer» `20.0^(@)` C |
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| 50765. |
A 60mL mixture of nitrous oxide and nitric oxide is totally decomposed. The volume of the mixture after passing through alkaline pyogallol is 48mL. Calculate the weight percentage of the diatomic oxide in the |
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Answer» Solution :The balanced equations for the DECOMPOSITION of NITROUS oxide `(N_(2)O)` and nitric oxide (NO) are given as `N_(2)O to N_(2)+(1)/(2)O_(2),NOto(1)/(2)N_(2)+(1)/(2)O_(2)` The volume of mixture initially = 60mL Let the volume of `N_(2)O` in the initial mixture =X Volume of `N_(2)O` in the initial mixture =60-x The volume of nitrogen finally =48mL `60-x+(x)/(2)=48` The RATIO of MOLES of NO:NO = ratio of volumes = 24:36 = 2:3 Ratio of weights =2 x 30:3x44=5:11 Weight percentage of No`=(5)/(16)x100=31.25%` |
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| 50766. |
A 600W mercurylampemitsmonochromaticradiationof wavelength313 : 3 nm . Howmanyphtonsare emittedfromlamppersecond ? (h=6.626 xx 10^(34)Jsvelocityof light3 xx 10^(8)ms^(-1)) |
| Answer» Answer :C | |
| 50767. |
A 600 W mercury lamp emits monochromatic radiation of wavelength 313.3 nm. How many photons are emitted from the lamp per second ? (h = 6.626 xx 10^(-34) Js, velocity of light = 3xx 10^(8) ms^(-1)) |
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Answer» `1 xx 10^(19)` i.e., Energy EMITTED per SEC = 600 J Energy of one photon, `E = hv = h (C)/(lamda)` `= (6.626 xx 10^(-34) Js xx 3 xx 10^(8) MS^(-1))/(313.3 xx 10^(-9) m)` `= 6.344 xx 10^(-19) J` `:.` Photons emitted per sec `= (600)/(6.344 xx 10^(-19))` `~= 1 xx 10^(21)` |
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| 50768. |
A 600 ml vessel containing oxygen at 800 mm and a 400 ml vessel containing Nitrogen at 600 mm at the same temperature are kept in communication with each other through a stop-cock. Neglecting the volume of the stopcock, the final pressure of the mixture is |
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Answer» 1400 m.m. |
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| 50769. |
A 60 ml of parcentamol pediatric oral suspension contains 3g of paracetamol. The mass percentage of paracetamol is ………….. |
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Answer» Solution :`5%` Mass percentage of paracetamol `= ("Mass of paracetamol")/("Volume of solution in ML") XX 100` `= (3)/(60) xx 100 =5%` |
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| 50770. |
A 60 ml of paracetamol pediatric oral suspension contains 3% of paracetamol. The mass percentage of paracetamol is ……………... |
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Answer» Solution :MASS PERCENTAGE of paracetamol =`("Mass of paracetamol")/("Volume of solution in mol")xx100` `=3/(60)xx100 = 5%` |
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| 50771. |
(A) ._(56)^(133)Be+e^(-) rarr ._(55)^(133)Cs + "X-ray" It is a process of K-electron capture. (R) The atomic number decreases by one unit as a result of K-capture. |
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Answer» Both (R) and (A) are TRUE and REASON is the correct EXPLANATION of assertion. |
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| 50772. |
A 500 gm sample of water is reacted with an equimoplar amount of CaO (both at an initial temperature of 25^(@)C) . What I sthe final emperature of the product ? [Assume that the poduct absorbs all of the heat released in the reaction heat product per mol of Ca(OH)_(2) is 65.2 KJ and specific heat Ca(OH)_(2)is 1.2 J//g^(@)C |
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Answer» `~~735^(@)C` |
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| 50773. |
A 50 ml solution of pH = 1 mixed with 50 ml solution of pH = 2. What will be the pH of the mixture ? |
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Answer» `pH = - LOG (5.5xx10^(-2))=2 - 0.74 = 1.26` |
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| 50774. |
A 5.0 mL solution of H_2O_2libearates 0.508g of iodine from an acidified KI solution. Calculate the strength of H_2O_2solution in terms of volume strength at STP. |
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Answer» 2.24 .V. 5.6 xx NORMALITY = VOLUME strength `:.` Volume strength `= 5.6 xx 0.08= 4.48` .V. |
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| 50775. |
(a) 50 mL of 0.2 N KMnO_(4) is required for complete oxidation of 0.45 g of anhydrous oxalic acid. Calculate the normality of oxalic acid solution. (b) In the titration of Fe^(2+) ions with KMnO_(4) in acid medium, why is dilute H_(2)SO_(4) used and not dilute HCl ? |
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Answer» `10FeSO_(4)+2KMnO_(4)+3H_(2)SO_(4) to 5Fe_(2)(SO_(4))_(3)+K_(2)SO_(4)+2MnSO_(4)+3H_(2)O` If HCL is taken in place of `H_(2)SO_(4)`, then HCl will be oxidised to `Cl_(2)` |
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| 50776. |
A 50% increase of CO_2 level in atmosphere causes |
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Answer» The INCREASE of surface temperature by `3^(@)`C |
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| 50777. |
A 5.0 cm^(3) solution of H_(2)O_(2) liberates 0.508 g of iodine from an acidified KI solution. Calculate the strength of H_(2)O_(2) solution in terms of volume strength of STP. |
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Answer» From the above equation `H_(2)O_(2) -=I_(2)` or 34G of `I_(2)` `therefore 0.508 g` of `I_(2)` will be liberated from `H_(2)O_(2)` `=(34)/(254)xx0.508` `=0.068` g The decomposition of `H_(2)O_(2)` occurs as `underset(2xx34=68 g )(2H_(2)O_(2))to underset(22400 cm^(3) " at " NTP)(2H_(2)O) + O_(2)` `therefore 0.068 of `H_(2)O_(2)` upon decomposition will give `O_(2)=(22400)/(68)xx0.068` `=22.4 cm^(3)` Now `5.0cm^(3)` of `H_(2)O_(2)` solution GIVES `O_(2)` `=22.4 cm^(3)` at STP `therefore 1.0 cm^(3) ` of `H_(2)O_(2)` solution will give `O_(2)` `=(22.4)/(5)=4.48 cm^(8)` at STP |
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| 50778. |
A 5-litre cylinder contained 10 moles of oxygen gas a t27^(@) C.Dueto sudden leakage through the hole, all the gas escaped into the atmosphere and the cylinder got empty. If the atmospheric pressure is 1.0 atmosphere, calculate the work done by the gas.( 1 L atm = 101.3 J) |
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Answer» Solution :`V_("initial") = 5L, T = 27^(@)C = 27 + 273 K = 300 K` `V _("final") = (nRT)/(P) = (10 xx 0.0821 xx 300)/( 1.0) = 246.3L` `Delta V = V_("final") - V_("initial") = 246.3 -5= 241.3 L` `W _(exp) = - PDeltaV = - 1 xx 241.3 L atm= - 241.3 xx 101 . 3 J = - 24443.7 J` |
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| 50779. |
A 5 L vessel contains 2.8 g of N_(2)only, when heated to 1800 K 30% molecules are dissociated into atoms. |
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Answer» Total no, of moles N in the CONTAINER will be 0.12 |
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| 50780. |
A 5 L flask containing 1.0g of hydrogen is heated from 300K to 600K. Which of the following statements are correct |
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Answer» The PRESSURE of the gas increases |
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| 50781. |
A 5cc of solution of H_(2)O_(2) liberales 0.508 gm of I_(2) from acidified Ki solution. The volume strength of H_(2)O_(2) Solution at S.TP is |
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Answer» 2.2 V |
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| 50782. |
A 4:1 molar mixture of He and CH_(4) is is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially ? |
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Answer» Solution :Applying Grahan's law of effusion, `(r_(He))/(r_(CH_(4)))=sqrt((M_(CH_(4)))/(M_(He)))=sqrt((16)/(4))=sqrt(4)=2`,i.e., He diffuse TWOTIMES faster than `CH_(4)`. As initially, the mixture CONTAINS He and `CH_(4)` in the MOLAR RATIO of 4:1, therefore, molar ratio of `He: CH_(4)` effusing out initially =8:1. |
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| 50783. |
A 4:1 mixture of helium and methane is contained in a vessel at 10 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The composition of the mixture effusing out initially is |
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Answer» `8:1` Pressure of `CH_(4)=2` bar `(r_(He))/(r_(CH_(4)))=(P_(1))/(P_(2))sqrt((M_(CH_(4)))/(M_(He)))=(8)/(2)sqrt((16)/(2))=(8)/(1)=8:1` |
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| 50784. |
A 40 mL of a mixture of H_(2) and O_(2) at 18^(@)C and 1 atm pressure was sparked so that the formation of water was complete. The remaining pure gas had a volume of 10mL at 18^(@)C and 1 atm pressure. If the remaining gas was H_(2). The mole fraction of H_(2) in the 40mL mixture is: |
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Answer» 0.75 |
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| 50785. |
A 4.0 g sample contained Fe_2O_3,Fe_3O_4, and inert material. It was treated with an excess of aq KI solution in acidic medium, which reduced all iron to Fe^(2+) ions. The resulting solution was diluted to 50 mL and a 10 mL sample of it was taken the iodine liberated in the small sample was titrated with 12.0 " mL of " 0.5 M Na_2S_2O_3 solution. The iodine from another 25 mL was extracted, after which the Fe^(2+) ions were titrated with 16 " mL of " 0.25 M MnO_4^(ɵ) ions in H_2SO_4 solution. Calculate the mass of two oxides in the original mixture. |
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Answer» Solution :Let `Fe_2O_3-=xmmol` `Fe_3O_4-=ymmol-=FeO+Fe_2O_3` `Fe^(3+)=2(x+y)mmol and Fe^(2+)=y m mol` `2Fe^(3+)+2I^(ɵ)toI_2+2Fe^(2+)` `2m" mol of "Fe^(3+)-=1 mmol I_2` `mmol I_2-=2m" mol of "S_2O_3^(2-)` (in 20 mL sample) `(x+y)(10)/(50)=(1)/(2)xx12xx0.5` `impliesx+y=15` NOTE that now m" mol of "`Fe^(2+)` in 50 mL sample is `2(x+y)+y=2x+3y` Also, `5Fe^(2+)+MnO_4^(ɵ)to5Fe^(3+)MN^(2+)` `5m" mol of "Fe^(2+)-=1 m" mol of "MnO_4^(ɵ)` `IMPLIES(25)/(50)(2x+3y)=5(16xx0.25)` `implies2x+3y=40` Solving, we get `x=5 and y=10` Mass of `Fe_2O_3=(5)/(1000)xx160=0.8g` Mass of `Fe_2O_4=(10)/(1000)xx232=2.32g` |
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| 50786. |
A 4.0 dm^(3) flask containing N_(2) at 4.0 bar was connected to a 6.0 dm^(3) flask containing helium at 6.0 bar, and the gases were allowed to mix isothermally, then the total pressure of the resulting mixture will be |
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Answer» 10.0 bar `P_(1)V_(1)+P_(2)V_(2)=P_(3)(V_(1)+V_(2))` `(4.0" bar")(4.0" dm"^(3))+(6.0" bar")(6.0" dm"^(3))` `=P_(3)(4.0+6.0" dm"^(3))` or `P_(3)=(16+36)/(10)=(52)/(10)=5.2" bar"` |
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| 50787. |
A 4: 1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially? |
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Answer» partial pressure of Help `He(P_(He)) = 4/5xx 20 = 16` bar and partial pressure of `CH4(P_(Ch_4)) = 1/5 xx 20 = 4` bar If t is the time taken in effusion, rate of diffusion of `He(r_(He)) = (""^nHe)/t` and rate of diffusion of `CH_4(r_(CH_4)) =(""^nCH_4)/t` where `n""^Heand n""^CH_4` are respectively the number of moles of He and `CH_4` effused initially. According to Graham.s law of diffusion, for two gases at ifferent pressures, we have `r_(He)/r_(CH_4)=sqrt(M_(CH_4)/M_(He))xxP_1/P_2` `:. ""(""^nHe//t)/(""^nCH_4//t)=sqrt(16/4)xx16/4` or`(""^nHe)/(""^nCH_4)=8/1` Hence, the initially effused mixture contains the moles of helium and methane in the ratio 8: 1 |
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| 50788. |
(a) 3,6,-Bis-(1,1-dimetylethyl) octane (b) Octa-4,7-dien-2-amine (c) 3-Hydroxymethyl cyclohexanecarboxylicacid (d) 2-Nitropent-3-ene If the given IUPAC name is correct the write 1 and if it is wrong then write 2 |
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Answer» |
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| 50789. |
A 36 mL mixture of an alkene and propane required 171 " mL of " O_2 for complete combustion and yielded 109 " mL of " CO_2 (all volume measured at same temperature and presasure). Calculate the molecular formula of olefin and composition of the mixture by volume. |
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Answer» Solution :`x " mL of " C_(n)H_(2N)` `(36-x)" mL of " C_3H_8` `C_(n)H_(2n)+(3n)/(2)O_2tonCO_2+nH_2O` `C_(3)H_(8)+5O_2to3CO_2+4H_2O` Volume of `CO_2=nx+3(36-x)=108` `impliesn=3 or x=0` (impossible) Volume of `O_2` used`=171` mL `thereforex((3n)/(2))+5(36-x)=171` Substituting `n=3,ximplies18mL` The HYDROCARBON is `C_3H_6` and the mixture is `50%` composition by volume because `x=18mL` |
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| 50791. |
A 3 lit flask contains 3g of nitrogen and 6.4g of oxygen at 30^@C. Calculate the following for the mixture : (a) partial pressure of oxygen and (b) total pressure of the gaseous mixture. |
| Answer» SOLUTION :(a) 1.66atm and (B) 2.56 ATM | |
| 50792. |
A 3 litre flask contains 2.2 gms of CO_2 and some mass of oxgyen. If the pressure of the mixture is 1.2 atm at 27^@C, Calculate the partial pressure and the mass of oxygen. |
| Answer» SOLUTION :0.79 ATM, 3.07 G | |
| 50793. |
A 2L flask contains 1.6 g of methane and 0.5g of hydrogen at 27^(@)C. Calculate the partial pressure of each gas the mixture and calculate the total pressure. |
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Answer» <P> SOLUTION :`P_(CH_4)=1.23 ATM, P_(H_2)=3.079 atm,P_(" TOTAL ")=4.31 atm` |
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| 50794. |
A 2.5gsample containing As_(2) O_(5) Na_(2) HAsO_(3) and inert substance is dissolved in water and the pH is adjusted to neutralwith excessof NaHCO_(3). The solution is titrated with 0.15M I_(2) solution, requiring11.3 mL to just reach the end point, then the solutionis acidified with HCl, KI is added and the liberated iodine requires41.2 mL fo 0.015M Na_(2)S_(2)O_(3) under basic conditionswhere it converts to SO_(4)^(2-). Calculate per cent compositon of mixture. |
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Answer» inert MATERIAL `= 84.9%` |
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| 50795. |
A 25mL of NO_(3)^(-) Solution was treated with excess Al. The ammonia gas is passed into 50mL of 0.15N HCl. Excess unreacted acid HCl is back titrated with 32.1mL 0.1N NaOH solution. Calculate the molarity of NO_(3)^(-) in the original solution. |
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Answer» |
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| 50796. |
A 25.0ml sample of 0.1 M HCl is titrated with 0.1 M NaOH . What is the pH of the solution at the points where 24.9 and 25.1 ml of NaOH have been added ? |
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Answer» `3.70,10.70 ` `N_aV_a =25xx 0.1 , N_b V_b =24.9xx 0.1` ` [H^(+) ]_("left")=(2.5 - 2.49)/(50.1 )=2 xx 10 ^(_4)` ` pOH =3.7 rArr pH=10.3` |
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| 50797. |
A 25.0 mLof 0.1 M weak acid HA is titrated with 0.1 M NaOH to the equivalence point. If the pH at the equivalence point is 8.28 Calculate K_(a) for this acid |
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Answer» `2.8 xx 10^(-4)` `pH = 7 + 1/(2)pKa + 1/2 "LOG" C` |
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| 50798. |
A 25watt bulbemitsmonochromaticyellowlightofwavelengthof 0.57 mu m.Calculate therate ofemissionof quanta per second. |
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Answer» SOLUTION :ENERGYOF bulb=25watt=25 `js^(-1)` energy(E ) E = HV= `(HC)/(LAMBDA)` `=(6.626 xx 3.0)/( 0.57)xx 10^(20)J` perphoton Quantum velocity`=("Emitenergybulb")/( "Emitenergyof photon")` `=(25 js^(-1)xx 0.57)/(6.626 xx 3.0 xx 10^(20)J)` `7.169 xx 10^(19)s^(-1)` `=7.17 xx 10^(19) s^(-1)` |
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| 50799. |
A 25 watt bulb emits monochromatic tellow light of wavelength of 0.57 mum. Calculate the rate of emission of quanta per second. |
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Answer» Solution :ENERGY emitted by the bulb `=25` watt `=25" J s"^(-1)""( :' 1" watt"=1" J s"^(-1))` Energy of one photon `(E)=hv=h c/lambda` Here, `lambda=0.57 MU m=0.57xx10^(-6) m""(1 mu m=10^(-6) m)` PUTTING `c=3xx10^(8)" m s"^(-1), h=6.62xx10^(-34)J` s, we GET `E=((6.62xx10^(-34) Js)(3xx10^(8) ms^(-1)))/(0.57xx10^(-6) m)=3.48xx10^(-19) J` `:.` No. of photons emitted per sec `= (25"J s"^(-1))/(3.48xx10^(-19) J)=7.18xx10^(19)`. |
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| 50800. |
A 2.24 litre cylinder of oxygen at NTP is found to develop a leakage. When the leakage was plugged the pressure dropped to 570 mm of Hg. The number of moles of gas that escaped will be: |
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Answer» 0.025 MOLES LEAKED= `n_1 - n_2` `(2.24)/(RT) ((760 - 570)/(760)) = 0.025`. |
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