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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. |
Find the ratio of volumes of H_(2) and 0_(2) reactions and also to the water vapour formed under similar condition when 1000 ml of H_(2) reacts with 500 ml O_(2) to form 1000 ml of water vapour. |
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| 2. |
Find the ratio of energy of a photon of 2000 Å wavelength radiation to that of 4000 Å radiation. |
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| 3. |
Find the ratio of rates of diffusion of hydrogen and oxygen. |
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Answer» SOLUTION :Graham.s law for the diffusion of two GASES is GIVENAS`(r_1)/(r_2) = sqrt((M_2)/(M_1))` Substituting the molecular weights, 32 for 2 oxygen and 2 for HYDROGEN, we get the ratio of rates of diffusion of `H_2 : O_2 = sqrt(32//2) = 4:1`. |
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| 4. |
Find the quantu number 'n' corresponding to the excited state of He^(+) ion if on transition to the ground state that ion emits two photons in succession with wavelength 108.5 and 30.4 nm respectively. |
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Answer» Solution :Suppose the electron in the excited state is present in the shell `n_(2)`. First it falls from `n_(1) " to " n_(1)` and then from `n_(1)` to ground state (for which n =1). Thus, the two transitions INVOLVED and the corresponding wavelengths emitted are (i) `n_(2) rarr n_(1), lamda_(1) = 108.5 nm = 108.5 xx 10^(-7)cm` (ii) `n_(1) rarr 1, lamda_(2) = 30.4 nm = 30.4 xx 10^(-7) cm` Applying Rydberg's formula first to case (ii), we get `bar(V) = (1)/(lamda) = RZ^(2) ((1)/(1^(2)) - (1)/(n_(1)^(2))), " i.e., " (1)/(30.4 xx 10^(-7)) = 109677 xx 2^(2) xx ((1)/(1^(2)) - (1)/(n_(1)^(2)))` or `(1)/(n_(1)^(2)) = 1 - (1)/(30.4 xx 10^(-7) xx 4 xx 109677) = 1 - 0.75 = 0.25 or n_(1)^(2) = (1)/(0.25) = 4 or n_(1) = 2` Applying Rydberg's formula now to case (i), we get `(1)/(108.5 xx 10^(-7)) = RZ^(2) ((1)/(2^(2)) - (1)/(n_(2)^(2))) = 109677 xx 2^(2) xx ((1)/(4) - (1)/(n_(2)^(2)))` or `(1)/(n_(2)^(2)) = (1)/(4) - (1)/(108.5 xx 10^(-7) xx 4 xx 109677) = 0.25 - 0.21 = 0.04` or `n_(2)^(2) = (1)/(0.04) = 25 or n_(2)=5` |
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| 5. |
Find the Q value of the reaction H_(2) (g) + I_(2) (g) hArr 2 HI (g) at an instant where concentration of H_(2) , I_(2) and HI are found to be 0.2 mol L^(-1) , 0.2 mol L^(-1) and 0.6 mol L^(-1) respectively . |
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Answer» 48 |
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| 6. |
Find theradiusof firstsecondthirdorbit ofhydrogen |
| Answer» SOLUTION :`0.0529 nm0.2116 NM 0.476 nm` | |
| 7. |
Find the pressure of neon gas having density 0.9 gm L^(-1) at 350 K temperature.(R = 8.314xx10^(-2)" bar L K"^(-1)mol^(-1)) |
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| 8. |
Find the pressure of a CO_(2) gas when 6.022xx10^(22) molecules are placed in 2 L vesel at 27^(@)C temperature. [Molar volume = 22.4 L] |
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| 9. |
Find the pressure of 5 mole Cl_(2) gas filed in a 2L vessel t 27^(@)C temperature. |
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| 10. |
Find the pressure of 4g of O_2 and 2g of H_2 confinedin a bulb of 1 L at 0^(@)C. |
| Answer» SOLUTION :`25.215 ATM` | |
| 11. |
Find the precentage degree of dissociation of 0.05 M NH_3 at 25^(@)C in a solutionof pH = 11 |
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Answer» `rArr [OH^(-)] =10 ^(-3)= Calpha ` ` rArr alpha = (10^(-3))/(5 xx 10 ^(_2) ) = 2 XX1 0 ^(_2)=2%` |
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| 12. |
Find the precentage degree of dissociation of 0.05 M NH_3 at 25^(@)C in a solutionof pG = 11 |
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Answer» `rArr [OH^(-)] =10 ^(-3)= CALPHA ` ` rArr alpha = (10^(-3))/(5 XX 10 ^(_2) ) = 2 xx1 0 ^(_2)=2%` |
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| 13. |
Find the partial pressures of each gas in a mixture containing 4.4 gms. of CO_2 and 5.6 gms of N_2 present at a pressure of 1.5atm. at a given temperature. ) |
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Answer» <P> |
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| 14. |
Find the oxidation state of sulphur in the following compounds: H_2S_,H_2SO_4,S_2O_4^(2-),S_2O_8^(2-) |
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Answer» Solution :In `H_2S`, +2+x=0, x=-2 In `H_2SO_4`, +2+x-8=0,x=+6 In `S_2O_4^(2-)`, 2x-8=-2, 2x=6, x=+3 In `S_2O_4^(2-)`, let ONE of S=x, then 2x+4(-8)=2 `therefore x=+3` In `S_2O_8^(2-)` , there is a peroxide LINKAGE, therefore oxidation state of S is 6 because S has six VALENCE electrons and it can from 6 COVALENT bonds. |
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| 15. |
Find the oxidation state of sodium in Na_(2)O_(2). |
| Answer» SOLUTION :`Na_(2)O_(2)` is known as sodium PEROXIDE. Oxidation state of OXYGEN in peroxy compound is -1, so oxidation state of sodium is +1 | |
| 16. |
Find the oxidation state of sodium in Na_2O_2 |
| Answer» SOLUTION :In PEROXIDE links, OXYGEN is -.So oxidation state of Na is +1 | |
| 17. |
Find the oxidation number of the underlined atoms. underlineCIO_3^(-) |
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Answer» Solution :`{:( overline ClO_3^(-), underline (CO) , O_3^(-)), ( Let ON is, x,-2):}` `therefore i.e., x+(-2 times3)=-1 implies x-6=-1 implies x=-1+6 implies x=+5` OXIDATION number of CL in `CIO_3^(-)` is +5 |
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| 18. |
Find the oxidation number of the underlined atoms. underlineCH_4 |
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Answer» Solution :`{:(UNDERLINE CH_4 , underline C, H_4), (LET ON is, x , +1):}` `therefore i.e., (x)+(+1times4) =0 impliesx+4=0 implies x=-4` Oxidation number of C in `CH_4` is -4. |
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| 19. |
Find the oxidation number of the underlined atoms. underlineC Cl_4 |
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Answer» Solution :`{:(underline C Cl_4 , underline C, Cl_4), (Let ON is, x , -1):}` `therefore i.e., (x times 1)+ (-1 times 4)=0 impliesx+(-1 times 4) =0 IMPLIES x-4=0 implies x=+4` Oxidation number of |
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| 20. |
Find the oxidation number of the underlined atoms. underlineBrF_3 |
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Answer» Solution :`{:( underlineBrF_3, UNDERLINE(BR), F_3), (Let ON is, x , (-1)):}` `THEREFORE i.e., (x times1) +(-1 TIMES3)=0 implies x-3=0 implies x=+3` Oxidation number of Br in `BrF_3` is +3. |
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| 21. |
Find the oxidation number of the underlined atoms. HunderlineCIO_3 |
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Answer» SOLUTION :`{:(H underlineCIO_3, H, Cl, O_3), (Let ON is,+1,X,-2):}` `THEREFORE i.e., (+1 times1) +(x times1) +(-2 times3) =0 implies1+x-6=0 implies x=6-1 impliesx=+5` Oxidation NUMBER of Cl in `HCIO_3` is +5. |
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| 22. |
Find the oxidation number of the underlined atoms. Na_2underlineB_4O_7 |
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Answer» SOLUTION :`{:( Na_2underline B_4O_7,Na_2,B_4,O_7), (LET ON is ,+1,x,-2):}` `thereforei.e., (+1 times2) +( x times 4) +(-2 times7)=0 implies2+4x-14=0 IMPLIES 4x=12 implies x=+3` OXIDATION number of B in `Na_2B_4O_7` is +3. |
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| 23. |
Find the oxidation number of the underlined atoms. HunderlinePO_2^(-) |
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Answer» Solution :`{:(HunderlinePO_2^(-) , H, P, O_2^(-) ), ( LET ON is,+1,x,-2):}` `THEREFORE i.e., (+1 times1) +x+(-2 TIMES2)=-1 IMPLIES 1+x-4=-1implies x=-1+4-1 implies x=+2` Oxidation number of P in `HPO_2^(-)` is +2. |
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| 24. |
Oxidation number of Fe in Fe_(3)O_(4) is : |
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Answer» Solution :`{:( F_3O_4,Fe_3,O_4), (Let ON be,x,-2):}` `therefore i.e., (3 times x)+(-2times4)=0 IMPLIES 3x+(-2 times4)=0 implies 3x=8 implies =8/3` OXIDATION number of Fe in `Fe_3O_4` is `+ 8/3` |
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| 25. |
Find the oxidation number of the element in bold in the following speices (i) SiH_(4), BH_(3),BF_(3),S_(2)O_(3)^(2-),BrO_(4)^(-) "and" HPO_(4)^(2-) (ii) PbSO_(4),U_(2)O_(7)^(2-),CrO_(4)^(2-), K_(2)MnO_(4) |
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Answer» Solution :(i)`Si=-4 in SiH_(4), B=-3 in BH_(3) , B=+3 in BF_(3),S=+2 in S_(2)O_(2)^(2-),Br=+7 in Br_(4)^(-) ND P=+5 in HPO_(4)^(2-)` `(II) S=+6 in Pb SO_(4), U=+5 in U_(2)O_(7)^(4-), B=+3 in B_(4)O_(7)^(2-)Cr =+ CrO_(4)^(2-)` and Mn=+6in `K_(2)MnO_(4)` |
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| 26. |
Find the oxidation number of the element underlined in the following species, PbunderlineSO_4,underlineU_2O_7^(2-),underlineB_4O_7^(2-),underline(Cr)O_4^(2-) |
| Answer» SOLUTION :S=+6 in `PbSO_4,U=+5` in `U_2O_7^4,B=+3` in `B_4O_7^(2-),Cr=+6` in `CrO_4^(2-)` | |
| 27. |
Find the oxidation number of the element underlined in the following species, underline(Si)H_4,underlineBH_3,BF_3,underlineS_2O_3^(2-),andunderlineHPO_4^(2-) |
| Answer» SOLUTION :Si=-4in `SiH_4`,B=-3 in `BH_3,B=+3` in `BF_3,S=+2` in `S_2O_3^(2-)`, andP=+5 in `HPO_4^(2-)` | |
| 28. |
Find the oxidation number of Mn in the product of alkaline oxidative fusion of MnO_2 |
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Answer» Let the oxidation no. of Mn in `K_(2)MnO_(4)`is a `:. 2 XX 1 + a + 4 xx(-21) = 0 IMPLIES a = 6` |
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| 29. |
Find the number of yellow coloured water insoluble compound. {:(PbI_(2),PbCrO_(4),KI,NaI),(Ag_(2)CrO_(4),BaCrO_(4),K_(2)CrO_(4),(NH_(4))_(3)PO_(4).12MoO_(3)),(HgO,AgI,AgCI,):} |
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| 30. |
Find the number of waves made by a Bohr electron in one complete revolution in the 3rd orbit. |
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Answer» Solution :The NUMBER of WAVES in any orbit `= ("CIRCUMFERENCE")/("wavelength")=(2pir)/(LAMBDA)=2pir xx(mv)/(h)` The angular momentum of 3rd orbit is `(3h)/(2pi)` Number of waves `=(2pi)/(h)xx(3h)/(2pi)=3` |
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| 31. |
Find the number of unpaired electrons present in phosphorus (atomic no.15), chromium (atomic no. 24) and copper (atomic no. 29) after writing their orbital electronic configurations. |
Answer» SOLUTION :
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| 32. |
Find the number of sulphide compounds which are not soluble in hot and dil. HNO_(3) but soluble in aqua regia Ag_(2)S,PbS,HgS,CdS |
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| 33. |
Find the number of structural isomers of fully saturated cycloalkane of molecular formulae C_(6)H_(12) which give three monochloro structural products. |
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| 35. |
Find the number of sp carbons of Hept 3, 4 diene 1,6 diyne. |
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| 36. |
Find the number of salt(s) give(s) white coloured precipitate when treated with KI. Bi(NO_(3))_(3),Pb(CH_(3)COO)_(2), AgNO_(3),, HgCI_(2), Hg_(2)(NO_(3))_(2) |
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| 37. |
Find the number of significant figures in the following physical quantities: (i) 5.506 cm(ii) 0.0509 kg(iii) 08.0075 s. |
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Answer» SOLUTION :(i) 5.506 cm has four significant figures because zero PLACED between two non-zero DIGITS is significant. (II) 0.0509 kg has three significant figures because zero placed to the left of first non-zero digit is not significant. (iii) 08.0075 s has five significant figures because zero placed before the first non-zero digit is insignificant while ZEROS placed between two non-zero digits are significant. |
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| 39. |
Find the number of significant figures in the following constants: (i) 6.626 xx 10^(-34) Js (Planck's constant) (ii) 6.02 xx 10^(23)(Avogadro's number) (iii) 1.097 xx 10^5 cm^(-1) (Rydberg's constant). (iv) 5.29 xx 10^(-9) cm (First Bohr's radius in H) |
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Answer» SOLUTION :(i) Planck.s constant (`6.626 xx 10^(-34)`) has four significant figures. 23 (ii) Avogadro.s number (`6.02 xx 10^(23)` ) has THREE significant figures. (iii) Rydberg.s constant (`1.097 xx 10^5`) has four significant figures. (iv) FIRST Bohr.s radius in H ATOM (`5.29 xx 10^(-9)`) has three significant figures. |
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| 40. |
Find the number of reagent(s) which oxidized HCO_(3)^(-) ion wolution. MnO_(4)^(-)//H^(+),Cr_(2)O_(7)^(2-)//H^(+),CI_(2)-water, Br_(2)-water, I_(2)-water |
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| 41. |
Find the number of reagent(s) which form white ppt. with CO_(3)^(2-) ion. BaCI_(2),CuSO_(4),HgCI_(2),Pb(CH_(3)COO)_(2),CaCI_(2) |
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| 42. |
Find the number of reagents in which Ag_(2)S is insoluble. hot conc.HNO_(3), (NH_(4))_(2)S, NH_(3),KCN, Na_(2)S_(2)O_(3) |
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| 43. |
Find the number of reagent (s) which oxidizes HCO_(3)^(-) ion solution: MnO_(4)^(-)//H^(+),Cr_(2)O_(7)^(2-)//H^(+), CI_(2) water, Br_(2) water, I_(2) water |
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| 44. |
Find the number of reagent (s) which form white ppt. with CO_(3)^(2-) ion. BaCI_(2),CuSO_(4),HgCI_(2),Pb(CH_(3)COO)_(2),CaCI_(2) |
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| 45. |
Find the number of radicals with cannot form white coloured precipitate with K_(4)[fe(CN)_(6)] Fe^(+3),Fe^(+2),Cu^(+2),Zn^(+2),Ca^(+2) |
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| 46. |
Find the number of protons, electrons and neutrons in (a) ._(13)^(27)Al^(3+) (b) ._(8)^(15)O^(2-) |
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Answer» For `._(8)^(15)O, Z = p = e^(-) = 8, A = 15:. n = A - Z = 15 - 8 = 7, e^(-) " in " O^(2-) = 8 + 2 = 10` |
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| 48. |
Find the number of monobromo derivatives in the given reaction. |
Answer»  four sites are AVAILABLE for BROMINATION REACTION .
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| 49. |
Find the number of moles of solute present in 500mL of 0.4M KMnO_(4) solution. |
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