Saved Bookmarks
This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In a 7.0 Levacuated chamber, 0.50 mol H_(2) and 0.50 mol I_(2) react at 427^(@) C. H_(2) (g) + I_(2) (g) hArr HI (g)+ Heat , At the given temperature, K_(c) = 49for the reaction.The K_(c)for above reaction at 500^(@) C is |
|
Answer» LESS than 49 |
|
| 2. |
In a 500 mL flask, the degree of dissociation of PCI_(5) at equilibrium is 40% and the initial amount is 5 moles. The value of equilibrium constant in mol L^(-1) for the decomposition of PCI_(5) is |
|
Answer» 1.33  `K_(C)=([PCl_(3)](Cl_(2)])/([PCl_(5)])=(((2)/(0.5))((2)/(0.5)))/(((3)/(0.5)))=(8)/(30)=2.66` |
|
| 3. |
In a 20 litre vessel initially each have 1-1 mole. CO,H_(2)OCO_(2) is present, then for the equilibrium of CO+H_(2)OhArrCO_(2)+H_(2) following is true: |
|
Answer» `H_(2)`, more then `1` mole `{:(t=0,1,1,1,0),(t=teq,1-x,1-x,1+x,x):}` at EQUILIBRIUM, only `CO_(2)` has `(1+x)` moles. |
|
| 4. |
In 6^(th) period if n = 6 and I = 0, 1, 2, 3, 4 there also why only 32 elements are there in this period ? |
Answer» Solution : Total 16 sub orbitals present so it CONTAIN 32 elements. Energy LEVEL ` = ns lt (n - 2) F lt (n - 1) d lt NP ` NOTE : `l = 4`, 3g orbital present it is not possible. |
|
| 5. |
In 2M H_(2)SO_(4)solution, what is % w/w and mole fraction of H_(2)SO_(4) respectively ? |
|
Answer» 16.39, 0.035 There are 2 mole `H_(2)SO_(4)` in 1000 gm WATER. 2 mole `H_(2)SO_(4) = 2 xx 98 = 196 gm H_(2)SO_(4)` % w/w `=(100xx196)/(1196)=16.387` Mole of `H_(2)SO_(4) = 2` Mole of water `=(1000)/(18) = 55.55` Molefraction of `H_(2)SO_(4) = (2)/(57.55) = 0.03475` `=0.035` |
|
| 6. |
In 3D representation of organic molecule, which model is indicate atomic size? |
| Answer» SOLUTION :The space filling model emphosises the RELATIVE size of each atom based on its van DER Waals RADIUS. It converys the volume occupied by each atom in the MOLECULE | |
| 7. |
In 20 mL of a solution of HCl, 3 g of CaCO_(3) were dissolved, 0.5 g of CaCO_(3) being left undissolved. Find out the strength of this solution in terms of (i) normality and (ii) g//L. Find the volume of this acid which would be required to make 1 litre of normal solution of this acid. |
|
Answer» |
|
| 8. |
In 10 L vessel SO_3, SO_2 and O_2 gases and definite temperature of K_c = 100. So reaction 2SO_(2(g)) + O_(2(g))hArr 2S0_((3)g) at equilibrium if SO_3 and SO_2 are same in mol than find the moles of O_2. If SO_3 is double than SO_2 than what is the mol of O_2 ?? |
| Answer» SOLUTION :0.1 and 0.4 MOL | |
| 9. |
In 1 L saturated solution of AgCl [K_(sp)(AgCl)=1.6xx10^(-10)], 0.1 " mol of " CuCl [ K_(sp)(CuCl)=1.0xx10^(-6)] is added. The resultant concentration of Ag^(+) in the solution is 1.6xx10^(-x). The value of "x" is |
|
Answer» If `[Ag^(+)]=y "mol" L^(-1)` , then `K_(sp) (AgCl) = [Ag^(+)][Cl^(-)]` i.e., `1.6xx10^(-10) = y xx 10^(-3) or y = 1.6 xx 10^(-7)` . HENCE , x = 7 . |
|
| 10. |
In 0.1 M HA the [H^+]= 3.16xx10^(-5) M, then what is [HA] and [A^-] ? |
|
Answer» Solution :`HA HARR H^(+)+ A^(-)` THUS, `[A^-]=[H^+]=3.16xx10^(-5)` But [HA] =0.1 M , because it is weak ACID. |
|
| 11. |
In 0.1M solution of salt NaA, the amion is 8% hydrolysed. Calculate the dissociation constant of acid, HA |
| Answer» SOLUTION :`1.43 XX 10^(-11)` | |
| 12. |
Impure Napthalene is purified by |
|
Answer» FRACTIONAL crystallisation |
|
| 13. |
Naphthalene is a volatile solid. It is purified by |
|
Answer» FRACTIONAL cysallization |
|
| 14. |
Impure glycerol decomposes at a temperature below its boling point. How will you purify it ? |
| Answer» SOLUTION :By DISTILLATION under REDUCED PRESSURE. | |
| 15. |
Impure glycerine is purified by |
|
Answer» STEAM DISTILLATION |
|
| 16. |
Impure glycerine can be purified by |
|
Answer» steam DISTILLATION |
|
| 17. |
Important contribution of Mendeleev's periodic table. |
|
Answer» SOLUTION :Mendeleev.s PERIODIC table was one of the greatest achievements in the development of chemistry. Some of the important contributions of his periodic table are: (A) Systematic study of the elements : The Mandeleev.s periodic table simplified the study of chemistry of elements. KNOWING the properties of one element in a group the properties of other element in the group can be easily guessed. Thus, if became very useful in studying and remembering, the properties at a large number of elements. (B) Correction of Atomic mass : The mandeleev.s periodic table helped in correcting the atomic massed of some elements BASED on their positions in the table. For example, atomic mass of beryllium was corrected from 13.5 to 9. Similarly, with the help of this table atomic masses of indium, gold, platinum etc. were corrected. (C) Prediction of new elements : At the time of Mendeleev, only 56 elements were known. While arranging these element, he left some gaps. These gaps represented the undiscovered elements. Mendeleev predicted the properties of these undiscovered elements on the BASIS of their positions. For example he predicted the properties of these undiscovered later. The observed properties of these elements were found to be similar to those predicted by Mendeleev.  (D) Empty space for undiscovered elements or empty space for EKa Aluminum and Eka silicon. His Primary aim of a arranging the elements of similar properties in the same group. He proposed that some of the elements were still undiscovered therefore left several gaps in the table. (For Gallium & Silicon) Example-1 : Galium was not discovered when Mendeleeve published periodic table so he placed empty space and named Eka Aluminium. Example-2 : Germanium was not discovered when Mendeleev published periodic table so he placed empty space under the silicon and named Eka Silicon. Where later Germanium is found. Mendeleev.s Predictions for the Elements Eka-aluminium (Gallium) and Eka-sillicon (Germanium)  (E) Medelee.s periodic table published in 1905 is shown at back page in figure . 
|
|
| 18. |
Important allotrpic forms of phosphorus are white phosphrous, red phosphorus and black phosphorus. Among these which allotropic form is more reactive? Why? |
|
Answer» Solution :White phosphours is the most reactive because of the FOLLOWING two REASONS : (i) White phosphorus consists of `P_(4)` molecules which are held together by weak van der Waals forces of attraction. (ii) The `P_(4)` molecules have considerable angle STRAIN because in them PPP angle is just `60^(@)`. Therefore, due to considerable angle strain and WEAKER forces of attraction, white phosphorus is most reactive. |
|
| 19. |
Importance of hydrocarbons. |
|
Answer» SOLUTION :Fuel used in our daily life : LPG, CNG, LNG and kerosene etc. are mixed hydrocarbons. * Hydrocarbons are also used for the manufacture of polymers LIKE polyethene, polypropene, polystyrene etc. * Higher hydrocarbons are used as solvents for PAINTS. Higher hydrocarbons like TURPENTINE kerosene etc. are used in paints. * They are also used as the starting materials for manufacture of many dyes and drugs. * "Thus hydrocarbons compounds are much impotant in our routine life". |
|
| 20. |
imagnine a universe in which: (A) Prinicipal quantum n canhavevalues from1,2,3,……….oo. (b)Azimuthal quantum no. l can have values from 1 to n+1correspoding to A,B,C,D,E,F,.................. (c ) magntriic quantumno m can have interal values from-(l)/(2) "to " +(l)/(2)(including zero if possible ). (d) spinquantum no s canhave6 possiblevalues . All rules of fillingremainsintact .what will be thelast shell foran elemetshaving Z=117 ? |
|
Answer» N=3 `{:(n,l,n+1),(1,1","2(A","B),2","3),(2,1","2","3","(A","B","C),3","4","5),(3,1","2","3","4,4","5","6","7),(4,1","2","3","4","5,5","6","7","8):}` `{:(l,m,),(1,-(1)/(2)","(1)/(2),implies3),(2,-1","0","+1,impliesB(3)),(3,-(3)/(2)","-(1)/(2)","(1)/(2)","+(3)/(2),impliesC(4)),(4,-2","-1","0","+1","+2,impliesD(4)),(5,-(5)/(2)","-(3)/(2)","-(1)/(2)","+(1)/(2)","+(3)/(2)","+(5)/(2),impliesE(6)):}` |
|
| 21. |
imagnine a universe in which: (A) Prinicipal quantum n canhavevalues from1,2,3,……….oo. (b)Azimuthal quantum no. l can have values from 1 to n+1correspoding to A,B,C,D,E,F,.................. (c ) magntriic quantumno m can have interal values from-(l)/(2) "to " +(l)/(2)(including zero if possible ). (d) spinquantum no s canhave6 possiblevalues . All rules of fillingremainsintact .what will be the shell andsubshell for an elementwith Z=36 ? |
|
Answer» Solution :`1A lt1Blt2A lt2B lt3Alt2C` 12" "18" "6` |
|
| 22. |
imagnine a universe in which: (A) Prinicipal quantum n canhavevalues from1,2,3,……….oo. (b)Azimuthal quantum no. l can have values from 1 to n+1correspoding to A,B,C,D,E,F,.................. (c ) magntriic quantumno m can have interal values from-(l)/(2) "to " +(l)/(2)(including zero if possible ). (d) spinquantum no s canhave6 possiblevalues . All rules of fillingremainsintact .what is themaximum e^(-) capcity of a shell for whichn=4? |
|
Answer» 32 `n+1=2" "3" "4" "4" "6` ` "total orbital "=19 ` `19xx6=114` |
|
| 23. |
imagnine a universe in which: (A) Prinicipal quantum n canhavevalues from1,2,3,……….oo. (b)Azimuthal quantum no. l can have values from 1 to n+1correspoding to A,B,C,D,E,F,.................. (c ) magntriic quantumno m can have interal values from-(l)/(2) "to " +(l)/(2)(including zero if possible ). (d) spinquantum no s canhave6 possiblevalues . All rules of fillingremainsintact .what will be the atomicno of theelementin which lst e^(-) fills2 nd shell commpletlty ? |
|
Answer» 38 12" "18" "12" "18" "12" "24" "18` |
|
| 24. |
Imagine the molecular collisions to the gases were inelastic. What would have happened? |
|
Answer» Solution :If the molecular collisions were inelatic, there WOULD have been loss of energy, which ultimately slows down the molecules. FINALLY the pressure would gradually reduce to ZERO. |
|
| 25. |
IMNH_4 OHand JMHCl are mixedto make total volume of 300 mL. If pH of the mixture is 9.26andpK_a (NH_4 ^(+)) = 9.26then volume ratio of NH_4 and HClwill be : |
|
Answer» `1:1` then HCl = ` ( 300 -x )ml` pH of mixture= ` pKa of NH_4^(+) ` `rArr " buffer with "[S] =[B]` `1 xx ( 300 - x)= (1 xx x )- ( 300 -x ) ` ` rArr x = ( 600)/( 3 ) = 200 rArr " ratio " =(200)/(100)= (2)/(1) ` |
|
| 26. |
IM NaCl and IM HCI are present in an aqueous solution. The solution is |
|
Answer» Not a buffer SOLUTION with `P^(H) lt 7` |
|
| 27. |
Illustrate the formation of solid-vapour equilibrium with suitable example |
|
Answer» Solution :(i) CONSIDER a system in which the solid sublimes to vapour. e.g., `I_2` (or) camphor. (ii) When solid iodine is placed in a CLOSED transparent vessel, after sometime, the vessel gets filled up with VIOLET vapour due to sublimation of iodine. (III) Initially the intensity of the violet colour increases, after some time it decreases and finally it becomes constant as the following equilibrium is attained. `I_(2)(s) hArr I_(2)(g)` |
|
| 28. |
Illustrate rules for filling electrons in (n + l) orbital using an example. |
|
Answer» Solution :According to (n + 1) rule: a) In NEUTRAL atoms a subshell with lower value of (n + 1) has lower energy. Ex: 4s orbital hasloer energy than 3d orbital. For 4s orbital, n = 4 and 1 = 0. ·Hence, (n +1)= 4 + 0 = 4. For 3d orbital, n = 3 and 1= 2. Hence, (n +1)= 3 + 2 = 5. b) If two subshells have equal value of(n + 1), the subshell with lower value of n has lower energy. Ex:- consider 3 p and 4 s orbital. For 4 s orbital n = 4 and 1 = 0 Hence, (n + 1)= 4 + 0 = 4 For 3 p orbital n = 3 and L = 1 Hence (n + 1) = 3 1 = 4 Hence 3 p orbital has lower energy than 4 s because it has lower value of n. |
|
| 29. |
Illustrate how copper metal can give different products on reaction with HNO_(3). |
|
Answer» Solution :On heating with dil. `HNO_(3)`, copper gives copper nitrate and nitric oxide. `3Cu+8HNO_(3)(dil)OVERSET("Heat")to3Cu(NO_(3))_(2)+4H_(2)O+underset("Nitric oxide")(2NO)` With conc. `HNO_(3)`, instead of `NO,NO_(2)` is evolved. `Cu+4HNO_(3)(conc)overset("Heat")TOCU(NO_(3))_(2)+2H_(2)O+underset("NITROGEN dioxide")(2NO_(2))` |
|
| 30. |
Illustrate by taking examples of transition elements and non-transition elements that oxidation states of elements are largely based on electronic configuration |
| Answer» Solution :The outermost electronic configuration of nitrogen `(2S^(2)2p_(X)^(1)2p_(y)^(1)2p_(z)^(1))` is very STABLE because p-orbital is half filled. Addition of extra electron to any of the 2p orbital requires ENERGY Oxygen has 4 electrons in 2p orbitals and acquires stable configuration i.e., `2p^(3)` configuration after removing one electron. | |
| 31. |
Illustrate bytakingexamplesof transitionelements andnon- transition elementthat oxidationstatesof elementare largelybasedon electronicconfiguration . |
|
Answer» Solution :Oxidation state of an elementis the charge(+ veor - ve)with an atomof theelementhas on its ion orappears to havewhenpresentin thecombinedstatewith otheratoms.The actualsign and exactvalue ofoxidationstate onan elementhoweverdependsuponits electronicconfiguration. For non-transition (representationelement ), the VALUE of theoxidationstate is equalto the numberof electronspresent intheoutermostshellor eightminus thenumber of electrons present in theoutermost shells(valenceshell ) . Forexamples- blockelementhave eitherone (alkali metals ) ortwo (alkalineearthmetals)electrons in theoutermost shellwhich theycan loseveryeasilyto acquirethe stableelectronicconfigurationof the nearestnoblegas and hence show oxidationstatesof either+ 1( alkalimetals ) or + 2(alkalineearthmetals).p-Blockelements on theother handshowpositivenegativeand evenzerooxidationstates. Forexampleacquiregroup 13 ELEMENT(i.e.,B , A1 ETC ) . have three electrons in theoutermost shellwhichthey canloseeasily toacquirestableelectronicconfiguration of the nearestnoblegas andhenceshowa maximumoxidationstateof + 3 . Elementsof group 14(i.e.,C, Si )neither haveastrongtendencyto lose or gain fourelectrons andhenceshowa maximu oxidationstate of+ 4or minimumof -4 group15 elements (i.e.,N, P , etc ) have a somewhatmore tendencyto gainthreeelectron than tofive electronsand henceshow a minimumoxidationstate of-3 ormaximumof + 5. Group16elements(i.e.,O, S, etc )haveeven moretendency togain twoelectrons than tolose six electrons and hence showa minimu oxidation state of -2 or maximumof +6Group 17elements(i.e.,halogens , F, C1 etc ) have thehighesttendency to gainone thanoflosesevenelectronsand henceshow aminimumoxidationstate of-1 ormaximumof + 7 . The storyof group18 elements (i.e.,noblegases ) is however quite different . Theyhave theirrespective valenceshells complete(Hehas two , all other have eightelectrons) and henceshow notendecyto loseor gainelectrons. Thereforein general , they showan oxidationstateof zero. forexamplehelium, neonand argondo not at allenterinto chemicalcombination andhence alwaysshow an oxidationstate ofzero . Howeverthere aresomeexceptions . Forexamplekryptonshowssometendencytoenter intochemicalcombinationand shows amaximum oxidationstatesof + 8 sinceit haseightelectrons in thevalenceshell. Likesome p- blockelements, transitionelements( d- block) andinnertransitionelement(f - block )alsoshowvariableoxidationstates. But unlikep- blockelements, the variableoxidationstates of transitionandinnertransitionelements arisedue toinvolvement ofnot onlyhte valenceelectronsbut d-or f-electronsaswell. However theirmostcommonoxidationstatesare + 2and + 3. |
|
| 32. |
IIIustrate the significance of de Broglie equation with an iron ball and an electron(i)6.626kg iron ball moving with10ms^(-1) (ii)An electron moving at 72.73ms^(-1) |
|
Answer» Solution :(i)`gamma_(iron ball)=H/(mv)` `=(6.626xx10^(-34)Kgm^(2)s^(-1))/(6.626Kgxx10ms^(-1))` `1xx10^(-35)m` (II)`gamma_("electron")=h/(mv)` `(6.626xx10^(-34)Kgm^(2)s^(-1))/(9.11xx10^(-31)Kgxx72.73ms^(-1))` `(6.626)/(662.6)xx10^(-3)m=1xx10^(5)m` For an electron,the de Broglie wavelength is signification and measurable while for iron ball it is too small to MEASURE,hence it becomes insignificant. |
|
| 34. |
IIIA group element which forms only convalent compounds either in anhydrous state or in aqueous state is |
|
Answer» Al |
|
| 35. |
(iii) K_(2)Cr_(2)O_(7)+KCI+H_(2)SO_(4)to KHSO_(4)+CrO_(2)CI_(2)+H_(2)O. |
| Answer» SOLUTION :It is not a REDOX REACTION . | |
| 36. |
The best and latest technique for isolation, purification and separation of organic compounds is:a) Crystallisationb) Distillationc) Sublimationd) Chromatography |
|
Answer» |
|
| 37. |
(iii) S_(2)O_(3)^(2-)+I_(2)to S_(2)O_(4)^(2-)+I^(-). |
Answer» SOLUTION : `S_(2)O_(3)^(2-)+I_(2)to S_(2)O_(4)^(2-)+I^(-)` `S_(2)O_(3)^(2-)+I_(2)to S_(2)O_(4)^(2-)+2I^(-)` `S_(2)O_(3)^(2-)+I_(2)+H_(2)Oto S_(2)O_(4)^(2-)+2I^(-)` `S_(2)O_(3)^(2-)+I_(2)+H_(2)Oto S_(2)O_(4)^(2-)+2I^(-)+2H^(+)`. |
|
| 38. |
III A group element with highest density is |
|
Answer» B |
|
| 39. |
IIA group metal oxides are less basic than IA group metal oxides because |
|
Answer» II GROUP metals are LESS electropositive than I group metals |
|
| 40. |
(ii) Why type of mesomeric effect is observed in phenol ? Explain. |
Answer» Solution :(ii) RESONANCE is useful in explaining properties such as acidity of PHENOL. The phenoxide ion is more stabilised than phenol by resonance effect (+M effcet)and hence resonance favours ionisation of phenol to from `H^(+)` and shows acidity.  The above STRUCTURES shows that there is a charge separation in the resonance structure of phenol which needs energy, where as there is no such HYBRID structurers in the case of phenoxideion. This increased stability accounts for the ACIDIC character of phenol. |
|
| 41. |
The most stable carbocation among (CH_3)_2overset+CH, (CH_3)_3overset+C, CH_3overset+, CH_3CH_2overset+ is ….. |
|
Answer» |
|
| 42. |
(ii) Orbits are also called as stationary states. Say whether the above statement is true or false. Justify you answer. |
| Answer» Solution :(b) (ii) The statement is TRUE. ACCORDING to Bohr, as long as an electron REMAINS in a particular orbit, it does not lose or gain energy. This means that enegry of an electron in a particular path remains CONSTANT. Therefore, these orbit are also called stationary states. | |
| 43. |
(ii) KMnO_(4)+Na_(2)SO_(3)to MnO_(2)+Na_(2)SO_(4)+KOH (Alkaline medium). |
Answer» Solution : Equalise the increase / decrease in O N by multiplaying Mn species by 2 and S species by 3 `2KMnO_(4)+3Na_(2)SO_(3)to 2MnO_(2)+3Na_(2)SO_(4)+KOH` Balance all other atoms except H and O `2KMnO_4 +3Na_(2)SO_3 to 2MnO_(2) +3Na_(2)SO_(4)+2KOH` Balance O atoms by adding `H_2O` molecules on the SIDE falling SHORT of oxygen atom . `2KMnO_(4)+3NaSO_(3)+H_(2)Oto 2MnO_(2)+3Na_(2)SO_(4)+2KOH` . |
|
| 44. |
(ii) I.E increases as we move across the period but Ionisation enthalpies (I.E) of second period of elements in the order. Li lt B lt Be lt C lt O lt N lt F lt Ne Explain why ? (1) Be has higher I.E and B (2) O has lower I.E than N & F |
|
Answer» Solution :(ii) (1)` ._(4)Be-1s^(2)2S^(2),_(5)B-1s^(2)2s^(2)2p^(1)` The I.E. of Be is more than that of B though the nuclear charge of boron atom `(Z=5)` is greater than that of beryllium atom `(Z=4)` . This can be explained as follows : Boron atom `(Z=5,1s^(2)2s^(2), 2p_(x)^(1) 2p_(y)^(0)2p_(z)^(0))` is having one paired electrons in the 2p-subshell. Be-atom `(Z=4,1s^(2)2s^(2))` is having paired electrons in the 2s-subshell. As the fully filled 2s-subshell in Be-atom is more stable than B-atom due to symmetry, more energy would be needed to REMOVE an ELECTRON from Be-atom. Hence, Be has high I.E. Hence I.E of Be `gt` B. `._(7)N-1s^(2),2s^(2),2p_(x)^(1),2p_(y)^(1),2p_(z)^(1)` (2)`._(8)O-1s^(2),2s^(2),2p_(x)^(2),2p_(y)^(1),2p_(z)^(1)` `therefore` O has lower I.E than N. |
|
| 45. |
(ii) C_2O_(4)^(2-)+Cr_(2)O_(7)^(2-) to Cr^(3+)+CO_(2) . |
Answer» SOLUTION : `3C_2O_(4)^(2-)+Cr_(2)O_(7)^(2-)to 2Cr^(3+)+6CO_(2)+7H_(2)O` `3C_2O_(4)^(2-)+Cr_(2)O_(7)^(2-)+14H^(+)to 2Cr^(3+)+6CO_(2)+7H_2O`. |
|
| 46. |
(i)H_(2(g))+I_(2(g)) hArr 2HI_((g)) (ii)1/2H_(2(g)) + 1/2 I_(2(g)) hArr HI_((g)) (iii)nH_(2(g)) +nI_(2(g)) hArr 2HI_((g)) For these reactions the equilibrium constant is respectively K_c(1), K_c (2) and K_c (3) state their relation ? |
|
Answer» Solution :(i)`K_c(1)=[HI]^2/([H_2][I_2])` (II)`K_c(2)=([HI])/([H_2]^(1/2)[I_2]^(1/2))=([HI]^2/([H_2][I_2]))^(1/2)=(K_c(1))^(1/2)` (III)`K_c(3)=[HI]^(2N)/([H_2]^n[I_2]^n)=([HI]^2/([H_2][I_2]))^n=(K_c(1))^n` Here , `K_c` (2) = `(K_c(1))^(1/2)` and `K_c(3)=(K_c(1))^n` |
|
| 47. |
Inniting MnO_(2) in air converts it quantitctively to Mn_(3)O_(4). A sample of pyrolusite is of the following composition. MnO_(2)=80% and othe inert constituents =15% and rest bearing H_(2)O. The sample is ignited to constant weight. What is the % of Mn is the igrited sample. |
|
Answer» `59.4%` `THEREFORE` 1 mol of `MnO_(2)=1//3` mol of `Mn_(3)O_(4)` Let 100 gm of PYROLUSITE sample be taken. Then GRAMS of `MnO_(2)=80` gm implies moles of `MnO_(2)=80//87` `therefore` Moles of `Mn_(3)O_(4)=(1)/(3)((80)/(87))` Moles of `Mn=(80)/(87)=1` wt of `Mn=(80)/(87)xx55` = 50.57 gm wt of `Mn_(3)O_(4)=(1)/(3)((80)/(87))xx229=70.19` Total weight of ignited sample `=70.19+15=85.19` `%Mn=(50.57)/(85.19)+100=59.36%` |
|
| 48. |
(i)[Fe(CN)_(6)]^(x-)+H_(2)O_(2)+H^(+)to[Fe(CN)_(6)]^(y-)+H_(2)O (ii) [Fe(CN)_(6)]^(p-)+H_(2)O_(2)+OH^(-)to[Fe(CN)_(6)]^(q-)+H_(2)O Select the correct statements: |
|
Answer» X and y are 3 and 4 respectively |
|
| 49. |
If Z is the number of atoms in the unitcell that represents the closest packing sequence -ABC ABC-, the number of tetrahedral voids in the unit cell is equal to |
|
Answer» Z |
|
| 50. |
If Z is a compressibility factor, Ven der Waals equation at low pressure can be written as : |
|
Answer» `Z=1-(PB)/(RT)` (1 mole REAL GAS) van der Waal.s equation `(P+(a)/(V^(2))) (V-b)=RT` At lower pressure `V-b ~~V` `(P+(a)/(V^(2)))V=RT` `therefore PV+(a)/(V)=RT` `therefore PV=RT-(a)/(V)` `therefore (PV)/(RT)=1-(a)/(VRT) ""` (`because` Dividing by RT on both side) `therefore Z=1-(a)/(VRT)` |
|