94420 + Interview Questions in Chemistry in Class 12

1.	How much gm of Ag is obtained on cathode by passing 9650 Coulomb electricity through aqueous solutionof AgNO_(3) using inert electrode ? (Ag=108" gm "mol^(-1))
Answer» 108 gm 10.8 gm 1.08 gm 32.4 gm Solution :`9650` COULOMB=`(9650)/(96500)=0.1` FARADAY `Ag_((aq))^(+)+E^(-) to Ag_((S))`. . . Cathodic REACTION 1 faraday gives 1 mol Ag=108 gm Ag So, 0.1 Faraday gives 0.1 mol Ag=10.8 gm Ag

Discussion

2.	Howmuch gas (in litres) will be produced at 0^(@)Cand 760 m m of pressure when 10 g of oxalic acid was heated withconcentratedsulphuricacid ?
Answer» SOLUTION :`(CO OH)_(2)OVERSET(H_(2)SO_(4))toCO + CO_(2)+ H_(2)O` Apply POAC for C atoms.

Discussion

3.	How much faster would a reaction proceed at 25^@C than at 0^@C if the activation energy is 65 kJ:
Answer» 2 times 16 times 11 times 6 times Answer :C

Discussion

4.	How much faraday current is required for reduction of 1.5 mol Cr_(2)O_(7)^(-2) to Cr^(3+) ?
Answer» 15F 9F 6F 3F Solution :`Cr_(2)O_(7(aq))^(2-)+14H_((aq))^(+)+6e^(-) to 2Cr_((aq))^(3+)+7H_(2)O_((L))` So, for 1 mol `Cr_(2)O_(7)^(-2)6F` current is REQUIRED And for 1.5 mol `Cr_(2)O_(7)^(-2)` required `(6xx1.5)=9.0F`

Discussion

5.	How much ethly alcohol much be added ot 1.0 litre of water so that the solution may treeze at -10^(@)C. (K_(f) "for water =1.88 K kg" mol^(-1), "density of water=1 g "mL^(-1)).
Answer» SOLUTION :Step I. CALCULATION of molality of the solution. `DeltaT_(f)=K_(f)xxm` `DeltaT_(f)=0^(@)C-(-10^(@)C)=10^(@)C=10 K, K_(f)=1.86" K kg mol"^(-1)` `m=(DeltaT_(f))/K_(f)=((10K))/((1.86" K kg mol"^(-1)))=5.376" mol kg"^(-1)=5.376 m.` Step II. Calculation of amount of ethyl alcohol to be added. `"Molality (m)"=("Mass of ethyl alcohol / Molar mass")/("Mass of water / 1000")` `5.376" mol kg"^(-1)=(W_(B)//(46"G mol"^(-1)))/(1" kg")` `W_(B)=(5.376" mol kg"^(-1))xx(46" g mol"^(-1))xx(1 kg)=247.3 g.`

Discussion

6.	How much ethanol is present in Azeotropic mixture ?
Answer» `5%` `95%` `15%` `100%` SOLUTION :`95%`

Discussion

7.	How much energy is released when 6 mole of octane is burnt in air?Given DeltaH_(f)^(@) " for " CO_(2)(g), H_(2)O(g) and C_(8)H_(18)(l) respectively are - 490, -240 and + 160 kJ/mol
Answer» `- 6.2 KJ` `- 37.4 kJ` `- 35.5 kJ` ` - 20.0 kJ` SOLUTION :`C+O_(2)rarrCO_(2) , DeltaH_(F)=-490 kJ//mol - (i)` `H_(2)+1//2O_(2)rarrH_(2)O, DeltaH_(F)=-240 kJ//mol - (ii)` `8C+9H_(2)rarrC_(8)H_(18), DeltaH_(F)=+160 kJ//mol - (iii)` applying (i)`xx8+(ii)xx9-(iii)` `C_(8)H_(18)+(25)/(2)O_(2)rarr8CO_(2)+9H_(2)O` `DELTAH^(@)=-3920-2160-160=-6240 kJ//mol` `=37440 kJ//mol =-37.4 kJ`.

Discussion

8.	How much energy in joules must be needed to convert all the atoms of sodium to sodium ions present in 2.3 mg of sodium vapours ? Given : Ionisation energy of sodium is 495 kJ "mol"^(-1)
Answer» Solution :According to the definition of ionisation energy, ` Na(g) + IE to Na^(+) (g) +e^(-), I.E. = 495 kJ mol^(-1)` The amount of energy needed to ionise 1 mole of SODIUM vapours = `495 kJ mol^(-1)` Moles of sodium vapours present in given sample `=(2.3 xx 10^(-3))/23 = 1 xx 10^(-4)`mol L `therefore` Amount of energy needed to ionise `1 xx 10^(-4)` moles of sodium vapours `= 495 xx 10^(-4)` kJ/mol = 49.5 J/mol

Discussion

9.	How much electricity is required in coulomb for the oxidation of (i) 1 mol of H_(2)O to O_(2) ? (ii) 1 mol of FeO to Fe_(2)O_(3) ?
Answer» Solution :(i) Calculation for REQUIRED ELECTRICITY for production of `O_(2)` from the oxidation of 1 mol `H_(2)O` : ACCORDING to following reaction `1/2` mol of `O_(2)` is produced from 1 mol of `H_(2)O`. `H_(2)O_((l)) to (1)/(2)O_(2(g))+2H^(+)+2e^(-)` 1 mol `H_(2)O to 1/2` mol of `O_(2)`, in this 2 mole `2e^(-)` is used. So, to produce `O_(2)` from 1 mol `H_(2)O` 2F electricity is required. So, `2F=2xx96500=193000` coulomb electricity. (ii) Calculation for required electricity for production of `Fe_(2)O_(3)` from oxidation of 1 mol of `FeO`: 1 mol of FeO gives following reaction to obtain `Fe_(2)O_(3)`. `underset((+2))(Fe)O to (1)/(2) underset((+3))(Fe_(2))O_(3)+underset(1mol)(e^(-))+H^(+)` So, oxidation of1mol FeO gives `1/2` mole of `Fe_(2)O_(3)` and in this reaction 1 mol ELECTRON =1F=96500 coloumb electricity is used.

Discussion

10.	How much electricity in terms of Faraday is required to produce: (i) 20.0 g of Ca from molten CaCl_(2) ? (ii) 40.0 g of Al from molten Al_(2)O_(3) ?
Answer» Solution :(i) Required electricity in FARADAY to obtianed 20.0 g calcium from the MOLTEN `CaCl_(2)` Ionization reaction: `CaCl_(2) to Ca^(2+) +2Cl^(-)` Reduction reaction on cathode : `Ca^(2+)+2e^(-) to Ca` `therefore 2" mol "e^(-) to 1" mol "Ca` `therefore 2F to 40g` Ca `therefore`Faraday required to produce 20 g Ca `=(20gxx2F)/(40g)=1F` (ii) Required electricity in Faraday to obtained 40.0 g ALUMINIUM from the molten `Al_(2)O_(3)`. The following reaction occurs on cathode of molten `Al_(2)O_(3)`. `Al_(2)O_(3(l))to 2Al^(3+)+3O^(2-)` and reduction `Al^(3+)+3e^(-) to Al` So, we get 1 mol Al from 3 mole `e^(-)` `therefore` We get 27.0 g Al from 3F electricity. So, the electricity required to get 40.0 g Al `=(3Fxx40.0g)/(27.0g)=4.444F`

Discussion

11.	How much electricity is required in coulomb for the oxidation of (i) 1 mol of H_2Oto O_2 ? (ii) 1 mol of FeO to Fe_2O_3 ?
Answer» SOLUTION :(i) The electrode reaction for 1 mol of `H_(2)O` is REPRESENTED as: `H_(2)O to H_(2) + 1/2 O_(2)` i.e., `O^(2-) to 1/2O_(2) + 2e^(-)` `therefore` QUANTITY of ELECTRICITY required = `2F= 2 xx 96500 C = 193000 C` (ii) The electrode reaction for 1 mol of FeO is represented as: `FeO to 1/2O_(2)` i.e. `Fe^(2+) to Fe^(3+) +e^(-)` `therefore` Quantity of electricity required = 1F = 96500 C

Discussion

12.	How many coulombs are required for the oxidation of 1 mol of FeO to Fe_(2)O_(3) ? (Hint. Fe^(2+) to Fe^(3+)+e^(-))
Answer» Solution :`FE^(2+)RARRFE^(3+)+1e^(-)` `"So,1F"=1xx96500C=96500C`

Discussion

13.	How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl_(2) (ii) 40.0 g of Al from molten Al_(2)O_(3).
Answer» Solution :(i) `Ca^(2+)+2e^(-)TOCA` 1 MOLE of Ca i.e., 40 g require electricity = 2F 20 g of Ca will require electricity `=(2xx20)/(40)F` =1F (ii) `Al^(3+)+3E^(-)toAl` 1 mol of Al =27 g `:.` 27 g of Al require electricity =3F 40 g of Al will require electricity `=(3)/(27)xx40F` =4.44F.

Discussion

14.	How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl_2 ? (ii) 40.0 g of Al from molten Al_2O_3 ?
Answer» Solution :In general for the REACTION, `M^(n+) + "ne"^(-) to M` n faraday of electricity is required for reduction of one mole of `M^(n+)` to give M. (i) `Ca^(2+) + 2e^(-) to Ca` Thus, 1 mol of Ca, i.e, 40 g of Ca require electricity = 2F `therefore 20 g` of Ca will require electricity = 1F (ii) `Al^(3+) + 3E^(-) to Al` Thus, 1 mol of Al, i.e. 27 g of Al require electricity = 3F `therefore 40 g` of Al will require electricity `=3/27 XX 40 = 4.44 F`

Discussion

15.	How much electricity in terms of faraday is required to produce (i) 20.0 g of Ca from CaCl_(2) (ii) 40.0 g of Al from molten Al_(2)O_(3)?
Answer» SOLUTION :(i) `Cu^(2+)+2e^(-)TOCA` Thus, 1 mol of Ca, i.e., 40 g of Ca require electricity=2F`therefore20` g of Ca will require electricity=1F (ii) `AL^(3+)+3e^(-)toAl` . Thus 1 mol of Al, i.e., 27G of Al required electricity=3F `therefore40g` of Al will require electricity=`(3)/(27)xx40=4.44E`

Discussion

16.	How much electricity in terms of Faraday is required to produce 20 g of Ca from molten CaCl_(2) ? (Atomic mass of Ca =40 )
Answer» Solution :`CaCl_(2)hArrCa^(2+)+2Cl^(-)` `UNDERSET("1 mole")(Ca^(2+))+underset(F)(2e)toCa` 40 g of Ca requires ELECTRICITY =2F 20 g of Ca will require electricity `=(2F)/(40)xx20` =1F

Discussion

17.	How much current in columb is required to decrease 1 mol of Cr_(2)O_(7)^(-2) from the following reaction ? Cr_(2)O_(7)^(-2)+14H^(+) +6e^(-) to 2Cr^(+3)+7H_(2)O
Answer» 5,80,000 COLUMB 5,79,500 columb 5,80,500 columb 5,79,000 columb Solution :`Cr_(2)O_(7)^(-2)+14H^(+)+6e^(-) to 2Cr^(+3)+7H_(2)O` `6xx96500C=5,79,000` columb current is required.

Discussion

18.	How much electric charge is requird for complete oxidation of 0.1 mole of MnO_(4)^(2-) ?
Answer» 96500 C `2xx96500C` `9650C` `96.50C` Solution :`underset(+6)(MnO_(4))^(2+) to underset(+7)(MnO_(4)^(-))` `0.1` MOLE=`96500xx0.1=9650` COULOMB

Discussion

19.	How much electricity in terms of Faraday is required to produce 40.0 g of Al from molter Al_(2)O_(3)?
Answer» SOLUTION :`Al^(3+)+3e^(-)rarrAl` 27 gram of Al require ELECTRICITY = 3F 40 gram of Al require electricity `=(3F)/(27)xx40=4.44F`

Discussion

20.	How much copper is deposited on the cathode of an electrolytic cell if a current of 5 ampere is passed through a solution of copper sulphate for 45 minutes? [Molar mass of Cu = 63.5 g mol^(-1), 1F = 96,500 Cmol^(-1)]
Answer» SOLUTION :`Cu^(2-)(aq) + 2e^(-) to Cu(s)` `m = Z xx I xx t` `(63.5)/(2 xx 96500) xx 5 AMP. xx 45 xx 60 = (857250)/(193000) = 4.44 g`.

Discussion

21.	How much copper is deposited on the cathode if a current of 3A is passed through aqueous CuSO_(4) solution for 15 minutes?
Answer» Solution :Quantity of electricity passed = Current in amperes x TIME in SECONDS. `Q = 3 xx 15 xx 60 = 2700` C `Cu^(++) + 2e^(-) to Cu` Two mole electron or 2F charge can DEPOSIT 1. Mole copper i.e. 63.5 g and so 2700 C will deposit `=(2700 xx 63.5)/(2 xx 96500) = 0.889` g

Discussion

22.	How much copper can be obtained from 100 g of copper sulphate (CuSO_(4))? (Atomic mass of Cu = 63.5 amu)
Answer» Solution :1 mole of `CuSO_(4)` contain 1 mole (1 g atom) of CU `"Molar MASS of "CuSO_(4)=63.5+32+4xx16=159.5" g mol"^(-1)` Thus,Cu that can be obtained from 159.5 g of `CuSO_(4)=63.5g` `THEREFORE" Cu that can be obtained from 100 g of "CuSO_(4)=(63.5)/(159.5)xx100g=39.81g`

Discussion

23.	How much chlorine will be liberated on passing one ampere current for 30 minutes through aqueous NaCl solution?
Answer» `0.66 ` MOLE `0.33` mole `0.66` GM `0.33` gm Answer :C

Discussion

24.	How much chlorine is obtained when 0.5 Faraday current is pass through aqueous solution of NaCl ? (Atomic weight of Cl=35.5 gm/mol)
Answer» 71.0 GM 35.5 gm 142.0 gm 17.75 gm Solution :`NACL to Na^(+)+CL^(-)` `Cl^(-) to (1)/(2) Cl_(2)+E^(-)0` `(1)/(2)` mole gives 1 F So, 1 F gives `(1)/(2)` mole=35.5 gm `Cl_(2)` So, 0.5 F gives `(0.5xx35.5)/(1)=17.75` gm `Cl_(2)`.

Discussion

25.	How much charge is required for the following reductions: (i) 1 mol of Al^(3+) to Al (ii) 1 mol of Cu^(2+) to Cu (iii) 1 mol of MnO_(4)^(-) to Mn^(2+).
Answer» Solution :(i) Required electric charge for reduction of 1 mol `Al^(3+)` to Al : Reduction on Al electrode : `underset (1" mol")(Al^(3+))+underset(3" mol")(3e^(-)) to underset(1" mol")(Al)` According to this reaction, 1 mol of `Al^(3+)` to get reduced into Al it requires 3 mol of electrons. So, electric charge of 1 mol of electron, 1F=96500 coulomb So, electric charge of 3 mol of electron, 3F=3(96500)=289500 coulomb (II) Required electric charge for reduction of 1 mol `Cu^(2+) ` to Cu : Reduction on Al electrode : `underset(1" mol")(Cu^(2+)) + underset(2" mol")(2e^(-)) to underset(1" mol")(Cu)` According to this reaction, 1 mol of `Cu^(2+)` to get reduced into Cu it requires 2 mol of electrons So, 2 mol electron=2 feraday electric charge is required. `=2xx96500` `=193000` coulomb electric charge (iii) Electric charge required for reduction of 1 mol `MnO_(4)^(-)` to `Mn^(2+)` : The reaction occurred on electrode when 1 mol of `MnO_(4)^(-)` is reduced to `Mn^(2+)` `underset(1MOL)(MnO_(4)^(-)) +8H^(+) + underset(5mol)(5E^(-)) to Mn^(2+)+4H_(2)O` According to this reaction, for reduction of 1 mol `MnO_(4)^(-)` it requires 5 mol electron and So, electric charge of 5 mol of electron =5Ferady `(5xx96500)` =193000 Coulomb So, 5 Ferady=193000 Coulomb electric charge is required.

Discussion

26.	How much charge is required, when 1 mole of Cr_(2)O_(7)^(2-) reduceto form I mole of Cr ^(3+) ?
Answer» 6F 3F 1F 2F Solution :Total of 6 electrons are REQUIRED to form 2 moles of `CR ^(3+)` therefore to form 1 mole of `Cr^(3+)3F` of CHARGE is required.

Discussion

27.	How much charge is required for the following reaction? (i) 1mol of Al^(3+) to Al. (ii) 1 mol of Cu^(2+) to Cu. (iii) 10 mole of MnO_(4)^(-) to Mn^(2+)
Answer» Solution :(i) The electrode reaction is: `Al^(3+)+3e^(-)toAl` `THEREFORE` QUANTITY of CHARGE required for reduction of 1 MOLE of `Al^(3+)=3F=3xx96500C=289500C`. (ii) The electrode reaction is: `Cu^(2+)+2e^(-)toCu` `therefore`Quantity of charge required for reduction of 1 mol of `Cu^(2+)=2` faradays`=2xx96500C=193000C` ltBrgt (iii) The electrode reaction is `MnO_(4)^(-)toMn^(2+)`, i.e., `Mn^(7+)5e^(-)toMn^(2+)` `therefore`Quantity of charge required `=5F=5xx96500C=482500C`.

Discussion

28.	How much charge is required for the following reductions : (i) 1 mol of Al^(3+) to Al? (ii) 1 mol of Cu^(2+) to Cu ? (iii) 1 mol of MnO_(4)^(-)to Mn^(2+)?
Answer» Solution :(i) The electrode reaction is REPRESENTED as: `Al^(3+) + 3e^(-) to Al` `therefore` Quantity fo charge required for reduction of 1 mol of `Al^(3+)` =`3 F = 3 xx 96500 C = 289500 C` (ii) The electrode reaction is represented by `Cu^(2+) + 2E^(-) to Cu` `therefore` Quantity of charge required for reduction of 1 mol of `Cu^(2+)` `=2F= 2 xx 96500 C = 193000 C` (iii) The electrode reaction is represented as: `=5F = 5 xx 96500 C = 482500 C`

Discussion

29.	How much charge in terms of Faraday is required to reduce one mol of MnO_4^(-) to Mn^(2+)?
Answer» SOLUTION :3 FARADAY

Discussion

30.	How much charge in Faradays is required for the reduction of 1 mol of Al^(3+)to Al ? [
Answer» Solution :THREE Faradays of CHARGE is REQUIRED for the CONVERSION.

Discussion

31.	How much ampere current should be passed through the CuSO_(4) solution having graphite electrode to obtain 250 milliliter O_(2) gas per minute at 1 bar pressure and 300 K temperature. [R=0.08314]
Answer» ANSWER :64.33 AMPERE

Discussion

32.	How much amount of KCl must be added to 1 kg of water so that the freezing point is depressed by 2 K? (K_f"for water" = 1.86 kg mol^(-1))
Answer» 40 g 20 g 10 g 60 g Solution :Since KCl undergoes complete dissociation. `KCl rarr K^(+) Cl^(-)` One mole of KCl will give 2 mole particles, therefore, the VALUE of i wil be equal to 2. `Delta T_f =i K_f m, K_f=1.86 K` kg `mol^(-1)`, `DeltaT_(f) =2 K, i=2` `therefore 2=2 xx 1.86 xx m` or `m = (2)/(2 xx 1.86) =0.5376` mol /kg Grams of KCl `=0.5376 xx 74.5 = 40.05g ` per kg `therefore` 40.05 g of KCl should be added to 1 kg of WATER.

Discussion

33.	How much amount of CuSO_(4).5H_(2)O required for liberation of 2,55gI_(2) when titrated with KI
Answer» 2.5 gm 4.99 gm 2.4 gm 1.2 gm Answer :B

Discussion

34.	How much AgClwill be formed byadding 1.70 " g of " AgNO_(3) in 200 mL of 5 N HCl solution ? (Ag = 108, N = 14 , O= 16)
Answer» Solution :Equivalent of `AgNO_(3) = (1.70)/(170) = 0.71 ` m.e of HCl solution = `5 xx 200 = 1000` ` :. ` equivalent of HCl solution ` = 1000/(1000) = 1 ` Since equivalent of `AgO_(3)`is less than the eq. of HCl , equivalent of AgCl = eq. of `AgNO_(3) = 0.01 ` ` :. ` wt. of AgCl = `0.01 xx 1435` ` = 1.435 G `

Discussion

35.	How much amount of a substance is deposited by 1 coulomb? What is it called?
Answer» SOLUTION :One COULOMB will deposit `"Eq.wt"/96500` of substance. It is CALLED the electrochemical equivalent.

Discussion

36.	How much AgBr would dissolve in 1 litre of 0.40 M NH_3 ? K_(sp) (AgBr) = 5 xx 10^(-13), K_d [Ag(NH_3)_2^+] = 1 xx 10^(-8)
Answer» SOLUTION :`2.83 XX 10^(-3) M`

Discussion

37.	How much Ag^+would remain in solution after mixing equal volumes of 0.08 M AgNO_3and 0.08 M HOCN? K_a (HOCN) = 3.3 xx 10^(-4), K_(sp) (AgOCN) = 2.3 xx 10^(-7) .
Answer» SOLUTION :`5 XX 10^(-3) M`

Discussion

38.	How mnay of the following liberate coloured vapour/gas with concentrated H_(2)SO_(4) ? KCl (s) +K_(2)Cr_(2)O_(7) (s), KNO_(2) (s), Kl (s), KBr(s), KCl(s) KBr (s) + MnO_(2)(s) , KNO_(3), KCl(s) + MnO_(2), K_(2)SO_(3)
Answer» Solution :`(i) 4Cl^(-)+Cr_(2)O_(7)^(2-)+6H^(+) to 2CrO_(2)Cl_(2) uarr ("deep red")+3H_(2)O` `(ii) NO_(2^(0-))+H^(+) to HNO_(2), "" 3HNO_(2) to HNO_(3)+NO uarr +H_(2)O` `2NO uarr +O_(2)uarr to NO_(2) uarr`(brown) (III) `3l^(-)+2H_(2)SO_(4) to l_(3^(-)) uarr ("violet") +SO_(4)^(2-)+2H_(2)O+SO_(2)` `(iv)2KBr +2H_(2)SO_(4) to Br_(2) uarr ("reddish brown")+SO_(2) uarr+NO_(4)^(2-)+2K^(+)+2H_(2)O` (v)` Cl^(-)+H_(2)SO_(4) to HCl uarr ("colourless")+HSO_(4^(-))` `(vi) 2KBr+MnO_(2)+2H_(2)SO_(4) to Br_(2) uarr ("reddish-brown") +2K^(+) +Mn^(2+) +2SO_(4)^(2-)+2H_(2)O` (vii) `4NO_(3^(-))+2H_(2)SO_(4) to NO_(2) uarr ("brown")` (viii) `KCl+MnO(OH)_(2)+2H_(2)SO_(4) to Mn^(2+ +Cl_(2) uarr ("yellowish green") +2SO_(4)^(2-)+3H_(2)O` (ix) `SO_(3)^(2-) +2H^(+) to SO_(2) uarr ("colourless")+H_(2)O`

Discussion

39.	How might you separate the following pairs of ions by addition of singles reagent include formula for the major products of the reactions(a)Fe^(3+)andNa^(+)(b)Cr^(3+)andFe^(3+)(c )Fe^(2+)and Cu^(2+)
Answer» Solution :(a)`FE^(3+)`and Na Add KOH(excess),`Fe^(3+)`is precipitaed as `Fe(OH)_(3)` Na remains in solution `Fe^(3+)(aq)+3OH(aq)TOFE(OH)_(3)(s)darr` (b) `CR^(3+)` and `Fe^(3+)` Add KOH(excess),ofaq NaOH,`Fe^(3+)` is PRECIPITATED as Fe(Oh) while `Cr(OH)_(3)` First formed dissolves in excess of aq.`NaOHFe^(3+)` `(aq)+3OH(aq)toFe(OH)_(3)(s)overset(OH)^(-))to`does not dissolve

Discussion

40.	How mnay constitutional isomers (excluding ring chain isomers) of molecular formula C_(5)H_(8) are possible?
Answer» 5 6 7 9 Solution :

Discussion

41.	How molar conductivity varies with dilution?
Answer» SOLUTION :It INCREASES with DILUTION.

Discussion

42.	How metallic adsorbents be activated ?
Answer» SOLUTION :By MAKING their SURFACE ROUGH by mechanical RUBBING.

Discussion

43.	How may the conductivity of an intrinsic semiconductor be increased ?
Answer» SOLUTION :CONDUCTIVITY of an intrinsic SEMICONDUCTOR can be increased by raising the TEMPERATURE.

Discussion

44.	How may of the following molecules have a definite geometry? "CCl"_(4),NH_(3),CHCl_(3),BF_(2)Cl,SO_(3),SF_(6),OSF_(4),PCl_(5),ClF_(3)
Answer» Solution :`"CCL"_(4), SO_(3),SF_(6),PCl_(5)`

Discussion

45.	How may possible angle FSeF bond angles are present in SeF_(4) molecule ?
Answer» ANSWER :6

Discussion

46.	How may chiral compounds are possible on monochlorination of 2-methylbutane?
Answer» 8 2 4 6 Solution :There are two CHIRAL carbo atoms for the MOLECULAR formula. `overset(Cl)overset(\|)(C)H_(2)-UNDERSET(CH_(3))underset(\|)overset()(C)H-CH_(2)-CH_(3)" "CH_(3)-underset(CH_(3))underset(\|)(C)H-overset(Cl)overset(\|)overet()(C)H-CH_(3)` there ar in all four chiral compounds possible `(2^(N)=2^(2)=4)`.

Discussion

47.	How may methyl bromide be preferentially converted to methyl cyanide and methylisocyanide ?
Answer» Solution :REACTION with KCN produces methyl cyanide and with `AGCN` produces methyl isocyanide. The reactions are given as under: <BR> `CH_(3)Br + KCN to underset("Methyl cyanide") (CH_(3)CN ) + KBr ` `CH_(3)Br + AgCN to underset("Methyl isocyanide") (CH_(3)NC ) + AgBr `

Discussion

48.	How may coulombs of electricity are required for the oxidation of one mol of water to dioxygen ?
Answer» `9.65 xx 10^(4)C` `1.93 xx 10^(4)C` `1.93 xx 10^(5) C` `19.3 xx 10^(5)C` ANSWER :C

Discussion

49.	How many years would it take to spend Avogadro's number of rupees at the rate of 10 lakh rupees per second?
Answer» Solution :`"Time taken"=(6.02xx10^(23))/(10,00,000)s=6.02xx10^(17s` `=(6.02xx10^(17))/(60xx60xx24xx365)"YEARS"=1.9089xx10^(1)"years"`

Discussion

50.	How many will undergo haloform reaction among the following ? CH_(3)COCH_(3), "" CH_(3)CH(OH)CH_(3),"" CH_(3)OH, "" C_(2)H_(5)OH CH_(3)CH_(2)CH_(2)OH
Answer» Solution :`CH_(3)COCH_(3)CH_(3)CH(OH)CH_(3),C_(2)H_(5)OH` (THREE)

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

How much gm of Ag is obtained on cathode by passing 9650 Coulomb electricity through aqueous solutionof AgNO_(3) using inert electrode ? (Ag=108" gm "mol^(-1))

Howmuch gas (in litres) will be produced at 0^(@)Cand 760 m m of pressure when 10 g of oxalic acid was heated withconcentratedsulphuricacid ?

How much faster would a reaction proceed at 25^@C than at 0^@C if the activation energy is 65 kJ:

How much faraday current is required for reduction of 1.5 mol Cr_(2)O_(7)^(-2) to Cr^(3+) ?

How much ethly alcohol much be added ot 1.0 litre of water so that the solution may treeze at -10^(@)C. (K_(f) "for water =1.88 K kg" mol^(-1), "density of water=1 g "mL^(-1)).

How much ethanol is present in Azeotropic mixture ?

How much energy is released when 6 mole of octane is burnt in air?Given DeltaH_(f)^(@) " for " CO_(2)(g), H_(2)O(g) and C_(8)H_(18)(l) respectively are - 490, -240 and + 160 kJ/mol

How much energy in joules must be needed to convert all the atoms of sodium to sodium ions present in 2.3 mg of sodium vapours ? Given : Ionisation energy of sodium is 495 kJ "mol"^(-1)

How much electricity is required in coulomb for the oxidation of (i) 1 mol of H_(2)O to O_(2) ? (ii) 1 mol of FeO to Fe_(2)O_(3) ?

How much electricity in terms of Faraday is required to produce: (i) 20.0 g of Ca from molten CaCl_(2) ? (ii) 40.0 g of Al from molten Al_(2)O_(3) ?

How much electricity is required in coulomb for the oxidation of (i) 1 mol of H_2Oto O_2 ? (ii) 1 mol of FeO to Fe_2O_3 ?

How many coulombs are required for the oxidation of 1 mol of FeO to Fe_(2)O_(3) ? (Hint. Fe^(2+) to Fe^(3+)+e^(-))

How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl_(2) (ii) 40.0 g of Al from molten Al_(2)O_(3).

How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl_2 ? (ii) 40.0 g of Al from molten Al_2O_3 ?

How much electricity in terms of faraday is required to produce (i) 20.0 g of Ca from CaCl_(2) (ii) 40.0 g of Al from molten Al_(2)O_(3)?

How much electricity in terms of Faraday is required to produce 20 g of Ca from molten CaCl_(2) ? (Atomic mass of Ca =40 )

How much current in columb is required to decrease 1 mol of Cr_(2)O_(7)^(-2) from the following reaction ? Cr_(2)O_(7)^(-2)+14H^(+) +6e^(-) to 2Cr^(+3)+7H_(2)O

How much electric charge is requird for complete oxidation of 0.1 mole of MnO_(4)^(2-) ?

How much electricity in terms of Faraday is required to produce 40.0 g of Al from molter Al_(2)O_(3)?

How much copper is deposited on the cathode of an electrolytic cell if a current of 5 ampere is passed through a solution of copper sulphate for 45 minutes? [Molar mass of Cu = 63.5 g mol^(-1), 1F = 96,500 Cmol^(-1)]

How much copper is deposited on the cathode if a current of 3A is passed through aqueous CuSO_(4) solution for 15 minutes?

How much copper can be obtained from 100 g of copper sulphate (CuSO_(4))? (Atomic mass of Cu = 63.5 amu)

How much chlorine will be liberated on passing one ampere current for 30 minutes through aqueous NaCl solution?

How much chlorine is obtained when 0.5 Faraday current is pass through aqueous solution of NaCl ? (Atomic weight of Cl=35.5 gm/mol)

How much charge is required for the following reductions: (i) 1 mol of Al^(3+) to Al (ii) 1 mol of Cu^(2+) to Cu (iii) 1 mol of MnO_(4)^(-) to Mn^(2+).

How much charge is required, when 1 mole of Cr_(2)O_(7)^(2-) reduceto form I mole of Cr ^(3+) ?

How much charge is required for the following reaction? (i) 1mol of Al^(3+) to Al. (ii) 1 mol of Cu^(2+) to Cu. (iii) 10 mole of MnO_(4)^(-) to Mn^(2+)

How much charge is required for the following reductions : (i) 1 mol of Al^(3+) to Al? (ii) 1 mol of Cu^(2+) to Cu ? (iii) 1 mol of MnO_(4)^(-)to Mn^(2+)?

How much charge in terms of Faraday is required to reduce one mol of MnO_4^(-) to Mn^(2+)?

How much charge in Faradays is required for the reduction of 1 mol of Al^(3+)to Al ? [

How much ampere current should be passed through the CuSO_(4) solution having graphite electrode to obtain 250 milliliter O_(2) gas per minute at 1 bar pressure and 300 K temperature. [R=0.08314]

How much amount of KCl must be added to 1 kg of water so that the freezing point is depressed by 2 K? (K_f"for water" = 1.86 kg mol^(-1))

How much amount of CuSO_(4).5H_(2)O required for liberation of 2,55gI_(2) when titrated with KI

How much AgClwill be formed byadding 1.70 " g of " AgNO_(3) in 200 mL of 5 N HCl solution ? (Ag = 108, N = 14 , O= 16)

How much amount of a substance is deposited by 1 coulomb? What is it called?

How much AgBr would dissolve in 1 litre of 0.40 M NH_3 ? K_(sp) (AgBr) = 5 xx 10^(-13), K_d [Ag(NH_3)_2^+] = 1 xx 10^(-8)

How much Ag^+would remain in solution after mixing equal volumes of 0.08 M AgNO_3and 0.08 M HOCN? K_a (HOCN) = 3.3 xx 10^(-4), K_(sp) (AgOCN) = 2.3 xx 10^(-7) .

How mnay of the following liberate coloured vapour/gas with concentrated H_(2)SO_(4) ? KCl (s) +K_(2)Cr_(2)O_(7) (s), KNO_(2) (s), Kl (s), KBr(s), KCl(s) KBr (s) + MnO_(2)(s) , KNO_(3), KCl(s) + MnO_(2), K_(2)SO_(3)

How might you separate the following pairs of ions by addition of singles reagent include formula for the major products of the reactions(a)Fe^(3+)andNa^(+)(b)Cr^(3+)andFe^(3+)(c )Fe^(2+)and Cu^(2+)

How mnay constitutional isomers (excluding ring chain isomers) of molecular formula C_(5)H_(8) are possible?

How molar conductivity varies with dilution?

How metallic adsorbents be activated ?

How may the conductivity of an intrinsic semiconductor be increased ?

How may of the following molecules have a definite geometry? "CCl"_(4),NH_(3),CHCl_(3),BF_(2)Cl,SO_(3),SF_(6),OSF_(4),PCl_(5),ClF_(3)

How may possible angle FSeF bond angles are present in SeF_(4) molecule ?

How may chiral compounds are possible on monochlorination of 2-methylbutane?

How may methyl bromide be preferentially converted to methyl cyanide and methylisocyanide ?

How may coulombs of electricity are required for the oxidation of one mol of water to dioxygen ?

How many years would it take to spend Avogadro's number of rupees at the rate of 10 lakh rupees per second?

How many will undergo haloform reaction among the following ? CH_(3)COCH_(3), "" CH_(3)CH(OH)CH_(3),"" CH_(3)OH, "" C_(2)H_(5)OH CH_(3)CH_(2)CH_(2)OH