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. |
The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is......... |
|
Answer» DISTILLATION |
|
| 2. |
The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapours in vapour phase. A suitable method for the extraction of these oils from the flowers is : |
|
Answer» Distillation |
|
| 3. |
The fractional distillation is used for separation of …….difference of boiling point containing liquid and simple distillation is used for separation of ……….difference of boiling point containing liquid |
| Answer» SOLUTION :LESS, HIGH | |
| 4. |
The fractional distillation: (i) The freezing of vapour of liquid with high boiling point take place first. (ii) In coloum volatile liquid is more in vapour at higher level. (iii) Condense liquid comes down and give heat to the vaporizing. (iv) The liquid with highest boiling point reaches at the top of coloum |
| Answer» SOLUTION :(i-T), (ii-F), (iii-T), (iv-F) | |
| 5. |
The fraction of the total volume occupied by the atoms present in a simple cube is |
|
Answer» `PI/4` ` = 8 xx 1/8 =1 ` Volume of the atom = ` 4/3 pi r^(3)` volume of the cube = ` a^(3) = ( 2r)^(3) = 8 r^(3)` Fraction occupied = `( 4/3 pi r^(3))/( 8 r^(3) ) = pi/6` |
|
| 6. |
The fraction of thetotal volume occupied by the atoms present in a simple cube is |
|
Answer» `pi//4` `=8times 1/8=1` `" VOLUME of the cube"=4//3pir^(3)` `" Volumeof the cube" = a^(3)=(2r)^(3) = 8 r^(3)(therefore a =2r)` `therefore "FRACTION occupied" = (4//3timespir^(3))/(8r^(3)) = pi/6` |
|
| 7. |
The fourth line of the Balmer series corresponds to the electronic transition between two orbits the H atom, Identify the orbits. |
|
Answer» 3 and 1 |
|
| 8. |
The fourth line of the Balmer series corres ponds to the electronic transition between two orbits of the H atom, Identify the orbits. |
|
Answer» `3 and 1 ` |
|
| 9. |
The fourteen types of space lattices are collectively called __________ |
|
Answer» |
|
| 10. |
The four tyres of a motor car are filled with CO_(2) , He, H_(2)and O_(2)respectively. The order in which they are to be reinflated is |
|
Answer» `CO_2, O_2, He, H_2` |
|
| 11. |
The four quantum number of the outermost orbital of K (atomic no. =19) are |
|
Answer» n=2,l=0, m=0, `s= + 1/2` for `4s^1` ELECTRONS. n=4, l=0, m=0 and `s=+1/2` |
|
| 12. |
thefour quantum numberof the outermostelectronof k (z=19) are ….. |
|
Answer» n=4 ,L = 0, m=0, s= `+(1)/(2)` |
|
| 13. |
The formulas of ammonium cynate and urea are …..and ……respectively |
| Answer» Solution :`NH_(4)CNO, NH_(2)CONH_(2)` | |
| 14. |
The formula which represents the composition of molecule is called ........ |
|
Answer» molecular FORMULA |
|
| 15. |
The formula weight of metal chloride is 136 and specific gravity of metal is 0.06 ""gcc""^(-1). How many grams of oxygen can combine with one gram of metal? |
|
Answer» |
|
| 16. |
The formula weight of ethanol (C_(2)H_(5)OH) is |
|
Answer» = 24 + 6+ 16 = 46 AMU |
|
| 17. |
The formula weight of an acid is 82. In a titration, 100 cm^(3) of a solution of this acid containing 39.0 g of the acid per litre were completely neutralised by 95.0 cm^(3) of aqueous NaOH containing 40.0 g of NaOH per litre. What is the basicity of the acid ? |
|
Answer» Normality of acid `=(1xx95)/(100)=0.95` LET the eq. MASS of the acid be E. `therefore (39)/(E )=0.95 " or " E=41, "" "Basicity"=(82)/(41)=2` |
|
| 18. |
The formula used to calculate the Bohr'r radius is………………… |
| Answer» SOLUTION :`r_n=((0.529)N^(2))/(Z)A` | |
| 19. |
The formula used to calculate the angular momentum is…………. |
| Answer» SOLUTION :`SQRT(L(l+1))H/(2PI)` | |
| 20. |
The Formula of white lead is |
|
Answer» `PB(OH)_(2) PbCO_(3)` |
|
| 22. |
The formula of the compound is A_(2)B_(3). The number of electrons in the outermost orbits of and B respectively are , |
|
Answer» 3 and 6 |
|
| 23. |
The formula of soda ash is ….. |
|
Answer» `Na_(2)CO_(3)*10H_(2)O` `Na_(2)CO_(3)*10H_(2)O OVERSET(gt 373K)toNa_(2)CO_(3)*10H_(2)O` |
|
| 24. |
The formula of pyrogallol is |
|
Answer» `C_6H_5(OH)_3` |
|
| 26. |
The density( in g mL^(-1) ) of 3.60sulphuric acid solution that is 29 % H_(2)SO_(4) (Molar mass = 98 g mol^(-1) )by ,mass will be |
|
Answer» 1.64 `:.` 3.6 mole `H_(2)SO_(4)` in 1000 ML solution `:.` 1000 mL solution of `H_(2)SO_(4)` contains `(3.6xx98) g ` of `H_(2)SO_(4)` The solution is 29% by MASS `1:.` 100 g solution contains 29 g of `H_(2)SO_(4)` `:.` If `H_(2)SO_(4) 3.6 xx 98 g` WEIGHT of solution `= (3.6xx 98xx 100)/(29)` g Density (d) `= ("wt of solution")/("volume of solution")` `= (3.6xx 98xx100)/(29 xx 1000) = (3.6xx98)/(290)` `= 1.2166` `= 1.22 g mL^(-1)` |
|
| 27. |
The densities of two gases are in the ratio of 1:16. The ratio of their rates of diffusion is |
|
Answer» `16:1` |
|
| 28. |
The densities of graphite and diamond at 298 K are 2.25 and 3.31 gcm^(-3) respectively. If the standard fre energydifference(DeltaG^(@)) is eqal to 1895 J mol^(-1), the pressureat which graphite will be transformed into diamond at298 K is |
|
Answer» <P>`9.92 xx 10^(8) Pa` `1895= P [ ((12)/( 2.25)- (12)/( 331))xx 10^(-6) m^(2)]` Calculate P, [From thermodynamics , `[ (del[DeltaG])/(delP)]_(T)=DeltaV` or `( DeltaG _(2)-DeltaG_(1))/( P_(2)-P_(1))= DeltaV` or`DeltaG_(2)- DeltaG_(1) = (P_(2)= P_(1))DELTAU` Taking `P_(1)=1` bar`= 10^(5)Pa`, calculate `P_(2)]` |
|
| 29. |
The densily of air is maximum in |
|
Answer» TROPOSPHERE |
|
| 30. |
The Delta_(i)H_1) andDelta_(i) H_(2) of Mg(g ) are740and 1450 kJ mol^(-1). Calculatethe percentageof Mg^(+) ( g) andMg^(2+) ( g) if 1 g ofMg ( g)absorbs 50kj of energy . |
|
Answer» Solution :No . Of moes of Mgvapourspresent in1 g =`1//24 = 0.0417` Energy absorbed inthe ionizationof 0.0417mole ofMg ( g) to `MG^(2+) ( g) =0.0417xx 740= 30. 83 KJ` Energyleftunused= 50- 30. 83= 19 . 17kj Now 19. 17 kjwill be usedto ionize `Mg^(+)( g) ` to `Mg^(2+)` ( g) `:. ` No. of moles of `Mg^(+)(g )`convertedinto `Mg^(2+) ( g) = 19. 17 // 1450= 0.0132` No. ofmoles ofmagnesium ionsleft as `Mg^(+) ( g) =0.0417-0.0417-0.0132= 0.0285` `:. ` %ageof `Mg^(+) (g)= (0.0285// 0.0417 ) xx 100 =68.35 %` and% ageof `Mg^(2+) ( g) =100-68 . 35 = 31 .65 %` |
|
| 31. |
The DeltaH^(@)for the mutarotation of glucose in aqueous solution , alpha"-D- glucose (aq)" to beta"-D-glucose (aq)" has been measured in a microcalorimeter and found to be -1.16 kj .mol^(-1) . The enthalpies of solution of the two forms of glucose have been determined to be alpha"-D- glucose (aq)" to alpha"-D-glucose (aq)" DeltaH^(@)=10.72KJmol^(-1) beta"-D- glucose (aq)" to beta-"D-glucose (aq)" DeltaH^(@)=4.68KJmol^(-1) Calculate DeltaH^(@)(in KJ/mol) for the mutarotation of soild alpha"-D-glucose"to soild" beta"-D-glucose:" |
|
Answer» `+4.88KJ//mol` |
|
| 32. |
The DeltaH_(f)^(0)N_(2)O_(5(g)) in KJ/mol on the basis of the following data is 2NO_((g))+O_(2(g))rarr2NO_(2(g))DeltaH_(f)^(0)=-114KJmol^(-1) NO_(2)+O_(2(g))rarr2N_(2)O_(5(g)) Delta_(f)H^(0)NO_((g))=90.2KJmol^(-1) |
|
Answer» SOLUTION :`(1)/(2)N_(2(g))+(1)/(2)O_(2(g))rarrNO_((g))Delta_(F)H^(0)=90.2` `N_((g))+O_(2(g))rarr2NO_((g))Delta_(f)H^(0)=90.2xx2rarr(1)` `NO_((g))+O_(2(g))rarr2NO_(2(g))Delta_(f)H^(0)=-14rarr(2)` `2NO_(2(g))+(1)/(2)O_(2(g))rarrN_(2)O_(5(g))` `Delta_(f)H^(0)=(-102.6)/(2)=-51.3rarr(3)` From equations `(1)+(2)+(3)` `N_(2(g))+(5)/(2)O_(2(g))rarrN_(2)O_(5(g))` `Delta_(f)H^(0)N_(2)O_(5(g))=15.1KJmol^(-1)` |
|
| 33. |
The DeltaH_(f)^(@) of MgO is -602KJxxmol^(-1) .when 20.15g MgO is decomposed at constant pressure accordingto the equation below , how much heat will be transferred? 2MgO(s)to 2MG(s)+O_(2)(g) |
|
Answer» `1.20xx10^(3)KJ"of HEAT is released "` |
|
| 34. |
The DeltaH^(@) for CO_(2)(g), CO(g) and H_(2)O(g) are -393.5, -110.5 and -241.8 kJ mol^(-1) respectively.The standard enthalpy change (in kJ) for the reaction : CO_(2)(g) + H_(2)(g) to CO(g) + H_(2)O(g) is |
|
Answer» 524.1 = `110.5 + (-241.8) -(-393.5) -0` `= 41.2 KJ` |
|
| 35. |
The Delta n_(g) value for the reaction 2 NO (g) + O_(2) (g) hArr 2 NO_(2) (g) is ………… |
|
Answer» <P> SOLUTION :`-1``DELTA n_(g) = n_(p) - n_(r) = 2 - 3 = -1` |
|
| 36. |
The delocalised pi -molecular orbital in benzene contains three ......... bicentric molecular orbitals and possesses full ...... symmetry. |
| Answer» SOLUTION :LOCALISED, HEXAGONAL | |
| 38. |
The delocalisation of sigma electrons with P-orbital is known as Hyperconjugation. It is also known as sigma-pi conjugation 'or' No bond resonance. Presence of atleast one 'alpha' hydrogen at saturated carbon in an alkene, carbocation and free radical involves hyper conjucation. More is the no. of hyper conjugative structures more stable is the alkene. Bond lengths, dipolemoments are also effected by hyperconjugation. No bond resonance is not possible in |
|
Answer»
|
|
| 39. |
The delocalisation of sigma electrons with P-orbital is known as Hyperconjugation. It is also known as sigma-pi conjugation 'or' No bond resonance. Presence of atleast one 'alpha' hydrogen at saturated carbon in an alkene, carbocation and free radical involves hyper conjucation. More is the no. of hyper conjugative structures more stable is the alkene. Bond lengths, dipolemoments are also effected by hyperconjugation. Hyper conjugation involves the |
|
Answer» `SIGMA` BOND |
|
| 40. |
The delocalisation of sigma electrons with P-orbital is known as Hyperconjugation. It is also known as sigma-pi conjugation 'or' No bond resonance. Presence of atleast one 'alpha' hydrogen at saturated carbon in an alkene, carbocation and free radical involves hyper conjucation. More is the no. of hyper conjugative structures more stable is the alkene. Bond lengths, dipolemoments are also effected by hyperconjugation. Which of the following molecule has longest C = C bond length |
|
Answer» `CH_(3)-UNDERSET(underset(CH_(3))(|))OVERSET(overset(CH_(3))(|))(C)-CH=CH_(2)` |
|
| 41. |
The dehydration reaction is represented by : |
|
Answer» (III),(ii),(R) |
|
| 42. |
The dehydration of butan-1-ol gives |
|
Answer» 1-butene as the main PRODUCT `underset("Butan-1-ol")(CH_(3)CH_(2)CH_(2)CH_(2)OH) underset(-H_(2)O)overset(H^(+))rarrCH_(3)CH_(2)-CH_(2)underset((1^(@)))(-CH_(2)^(+)) overset("H-shift")rarr underset((2^(@)))(CH_(3)CH_(2))-overset(o+)(CH)-CH_(3) underset(-H^(+))rarr underset("2-Butene")(CH_(3)-CH=CH-CH_(3))` |
|
| 43. |
The dehydration of an amide with phosphorus pentoxide yields |
|
Answer» <P>ammonia |
|
| 44. |
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_(a) bromoacetic acid. |
|
Answer» Solution :`{:(,CH_(2)(Br)CO OH ,HARR,CH_(2)(Br)CO O^(-) ,+,H^(+),,,),("Initial conc.",C,,0,,0,,,),("Conc. at EQM.",C-C alpha ,,C alpha ,,C alpha ,,,),(,,,,,,,,):}` `K_(a) = (C alpha . C alpha)/(C ( 1-alpha))=(C alpha^(2))/(1-alpha)~=C alpha^(2) = 0.1 xx (0.132)^(2) = 1.74xx10^(-3)` `pK_(a) = - log (1.74xx10^(-3))=3-0.2405=2.76` `[H^(+)]=C alpha = 0.1 xx 0.32 = 1.32 xx 10^(-2) M` `pH = - log (1.32xx10^(-2))=2-0.1206=1.88` |
|
| 45. |
The degree of unsaturation of fat or oil is measured by |
| Answer» Solution :Iodine number | |
| 46. |
The degree of ionizationof a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid. |
|
Answer» Solution :Calculation of `[H^+]` of 0.1 M bromoacetic acid `(CH_2BrCOOH)`: Suppose the ionization degree of this acid = `ALPHA` `THEREFORE` Ionised acid = `Calpha = 0.1 alpha` So, at equilibrium acid = `(0.1-0.1alpha)=0.1 (1-alpha)` But VALUE of `alpha` is very LESS so, `0.1 (1-alpha) =0.1 (1) = 0.1` At equilibrium , `[H^+] = [CH_2BrCOO^-]=C alpha= 0.1 alpha` The ionic equilibrium is under : `{:(,CH_2BrCOOH_((aq)) + H_2O HARR, H_3O_((aq))^(+) + , CH_2BrCOO_((aq))^(-)),("Initial M:",0.1 , 0.0 M, 0.0 M),("Change in equilibrium :",-0.1 alpha ,+0.1 alpha,+0.1 alpha),("Concentration at M equilibrium",overset(0.1 (1-alpha)M)(approx0.1),0.1alpha,0.1 alpha):}` `K_a=([H_3O^+][CH_2BrCOO^-])/([CH_2BrCOOH])=0.132` `therefore ((0.1alpha)(0.1alpha))/0.1 = 0.132` `therefore alpha^2= 0.132/0.1 = 0.0132` `therefore alpha` = 0.1149 So, `[H^+]= Calpha = 0.1xx0.1149`=0.01149 M Calculation of pH: pH=-log `[H^+]`=-log (0.01149)=1.9397 = 1.94 Calculation of `pK_a` : `pK_a=-log K_a=-log (0.132)-0.8794` |
|
| 47. |
The degree of ionisation of 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid. |
|
Answer» SOLUTION :`K_a=calpha^2=0.1xx0.132xx0.132=1.74xx10^-3` `THEREFORE pKa=-logK_a=-log(1.74xx10^-3)=3-0.2405=2.76` `[H^+]=calpha=0.1xx0.132=1.32xx10^-2M` `therefore pH=-log(1.32xx10^-2)=2-0.1205=1.88` |
|
| 48. |
The degree of hardness of a given sample of hard water is 40 ppm. If the entire hardness in due to MgSO_(4) , how much of MgSO_(4) is present per kg of water ? |
|
Answer» Solution :Degree of HARDNESS of `H_(2)O` is 40 ppm, i.e., `10^(6)` g of water contain `CaCO_(3)`=40 g Since 1 MOLE of `CaCO_(3)-=1 `mole of `MgSO_(4)` `therefore 100 ` g of `CaCO_(3)-=120` g of `MgSO_(4)` `therefore10^(6)` g of water contain `MgSO_(4)=(40xx120)/(100)=48 g` or `10^(3)` g of water contain `MgSO_(4)=(48xx10^(3))/(10^(6))xx10^(3) ` MG or 1 kg of water will contain `MgSO_(4)=48` mg |
|
| 49. |
The degree of dissociation of 0.1 M weak acid is 10^(-2) and the degree of dissociation of the same acid in 0.025M concentration is '*', then find 100x. __________ ? |
|
Answer» ` X= sqrt( (K_a)/(C)) =sqrt((10 ^(_5))/(25 xx 10 ^(-3))) rArr x = (10^(_1))/(5)` ` therefore 100 x =(0.1)/(5) xx 100 =2` |
|
