52671 + Interview Questions in Chemistry in Class 11

1.	In chromium (III) chloride, CrCl_3, chloride ions have cubic close packed arrangement and Cr (III) ions are present in the octahedral holes. What fraction of the octahedral holes is occupied ? What fraction of total number of holes is occupied ?
Answer» Solution :Suppose number of atoms Q=N Then number of tetrahedral sites =2 n `therefore` Number of atoms P=2n `therefore` Ratio P:Q=2n:n =2 :1 , i.e., FORMULA is `P_2Q`

Discussion

2.	In ClF_(3), NF_(3)" and " BF_(3) molecules the chlorine, nitrogen and boron atoms are
Answer» `sp^(3)` hybridised `sp^(3) ,sp^(3) and sp^(2)` RESPECTIVELY `sp^(2)` hybridised `sp^(3) d, sp^(3) and sp^(2)` hybridised respectively Solution :`CIF_(3) - sp^(3)` d HYBRIDISATION `NF_(3)- sp^(3)` hybridisation `BF_(3) - sp^(2)` hybridisation

Discussion

3.	In chromium (III) Chloride, CrCl_(3) , Chloride ions have cubic close packed arrangement and Cr (III) ions are presnent in the octahedral holes. What fraction of the octahedral holes is occupied ? What fraction of totalnumber of holes is occupied ?
Answer» <P> ANSWER :`Q_(4) P_(3)`

Discussion

4.	In chromatography, which of the following statement is Incorrect for R_f ?
Answer» `R_f` VALUE depends on the TYPE of chromatography. The value of `R_f` cannot be more than one. HIGHER `R_f` value means higher adsorption. `R_f` value is dependent on the mobile phase. Answer :C

Discussion

5.	In chromatography, if the stationary phase is solid, the basis is ............
Answer» SOLUTION :ADSORPTION

Discussion

6.	In chlorophyll pigment how many number of nitrogen atoms surrounds the central 'Mg' atom ?
Answer» Solution :Mg. ATOM is surrounded by 4 nitrogen ATOMS

Discussion

7.	In chlorobenzene, chlorine is ________ but _____ directing.
Answer» ANSWER :DEACTIVATING, o, p-directing

Discussion

8.	in chlorine.^(35)CIand .^(37) CI isin3:1ratiocalculatechargeatomicmass.
Answer» Solution :three `.^(35) CI` and one `.^(37)CI` totalfour CI massof fourCI= 3 (35)+ 1 (37) 142 AMU atomicmass ofCI` =(142)/( 4) = 35.5`amu

Discussion

9.	In chemistry, .mole.is an essential tool for the chemical calculations. It is a basic S.I. unit adopted by the 14^(th) general conference on weights and measurements in 1971. A mole contains as many elementary particles as the number of atoms present in 12 g of C. 1 mole of a gas at STP occupies 22.4 litre volume. Molar volume of solids and liquids is not definite. Molar mass of a substance is also called gram atomic mass or gram molar mass. The virtual meaning of mole is plenty, heap or the collection of large numbers. 1 mole of a substance contains 6.023xx10^(23) elementary particles like atom or molecule. Atomic mass unit (amu) is the unit of atomic mass, e.g., atomic mass of single carbon is 12 amu. x L N_(2) gas at STP contains 3xx10^(22) molecules. The number of molecules in x L ozone at STP will be
Answer» `3XX10^(22)` `4xx10^(23)` `6.02xx10^(23)` `3xx10^(2)` Solution :`ValphanalphaN` `THEREFORE` same VOLUME implies same MOLECULES

Discussion

10.	In chemistry, .mole.is an essential tool for the chemical calculations. It is a basic S.I. unit adopted by the 14^(th) general conference on weights and measurements in 1971. A mole contains as many elementary particles as the number of atoms present in 12 g of C. 1 mole of a gas at STP occupies 22.4 litre volume. Molar volume of solids and liquids is not definite. Molar mass of a substance is also called gram atomic mass or gram molar mass. The virtual meaning of mole is plenty, heap or the collection of large numbers. 1 mole of a substance contains 6.023xx10^(23) elementary particles like atom or molecule. Atomic mass unit (amu) is the unit of atomic mass, e.g., atomic mass of single carbon is 12 amu. How many atoms are present in 49 g of H_(2)SO_(4)?
Answer» `7xx6.023xx10^(23)` `5xx6.023xx10^(23)` `6xx6.012xx10^(23)` `7xx3.02xx10^(23)` Solution :`49g=(1)/(2)` mole `(1)/(2)` mole implies `(N_(0))/(2)` MOLECULES = `(N_(0))/(2)xx7` ATOMS

Discussion

11.	In chemistry, .mole.is an essential tool for the chemical calculations. It is a basic S.I. unit adopted by the 14^(th) general conference on weights and measurements in 1971. A mole contains as many elementary particles as the number of atoms present in 12 g of C. 1 mole of a gas at STP occupies 22.4 litre volume. Molar volume of solids and liquids is not definite. Molar mass of a substance is also called gram atomic mass or gram molar mass. The virtual meaning of mole is plenty, heap or the collection of large numbers. 1 mole of a substance contains 6.023xx10^(23) elementary particles like atom or molecule. Atomic mass unit (amu) is the unit of atomic mass, e.g., atomic mass of single carbon is 12 amu. The mass of one molecule of water is approximately
Answer» 1 g 0.5 g `1.66xx10^(-24)g` `3XX10^(-23)g` SOLUTION :1molecule = 18 AMU `=18xx1.66xx10^(-24)g=3xx10^(-23)g`

Discussion

12.	In chemical reaction, catayst:
Answer» alters the amount of the products LOWERS the activation ENERGY decrease the `DELTAH` of forward reaction increase the `DeltaH` of forward reaction Solution :Catalyst lowers the activation energy

Discussion

13.	In CH_(4),NH_(3) and H_(2)O, the central atom undergoes sp^(3) hybridisation - yet their bond angles are different. Why ?
Answer» Solution :(i) In `CH_(4),NH_(3)` and `H_(2)O` the central atom undergoes `sp^(3)` hybridisation. But their bond angles are different DUE to the presence of lone pair of electrons. It can be explained by VSEPR theory. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs are not same. (iii) Bond pair - Bond `lt` Bond pair - Lone pair `lt` Lone pair - Lone pair so due to the varying repulsive force the bond pairs and lone pairs are distored from regular geometry and organise themselves in such a way that repultion will be minimum and stabily will be MAXIMUM. (iv) In case of `CH_(4)`, there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry. i.e., tetrahedral with bond angle `=190^(@)28.` `H_(2)O` has 2 bond pairs and 2 lone pairs. There is large repsultion between lp-lp. Again repultion between lp-bp is more than that of 2 bond pairs. So 2 bond are more restricted to form inverted V shape (or) bent shape molecule with a bond angle of `104^(@)35.` (vi) `NH_(3)` has 3 bond pairs and 1 lone pair. There is repulsion between lp-bp. So 3 bonds are more retricted to form pyramidal shape with bond angle equal to `107^(@)18.`

Discussion

14.	In CH_(4),NH_(3) and H_(2)O the central atom undergoes sp hybridization - yet their bond angles are different. Why?
Answer» SOLUTION :(i) In `CH_(4),NH_(3)` and `H_(2)O` the central atom undergoes `sp^(3)` hybridisation. But their bond angles different due to the presence of lone pair of electrons. (ii) It can be EXPLAINED by VSEPR theory. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs are not same. (iii) Bond pair-Bond pair < Bond pair - Lone pair < Lone pair - Lone pair So due to the varying repulsive force the bond pairs and lone pairs are distorted from regular geometry and organise themselves in such a way that repulsion will be minimum and STABILITY will be maximum. (iv) In case of `CH_(4)` there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry. i.e., tetrahedral with bond ANGLE = `109^(@)28`’. (v) H.Ohas 2 bond pairs and 2 lone pairs. There is large repulsion between Ip-ip. Again repulsion between Ip-bp is more than that of 2 bond pairs. So 2 bonds are more restricted to form inverted V shape (or) bent shape MOLECULE with a bond angle of `104^(@)35`.. (vi) NH, has 3 bond pairs and I lone pair. There is repulsion between Ip-bp. So 3 bonds are more restricted to form pyramidal shape with bond angle equal to 107°

Discussion

15.	In CH_(4),NH_(3) " and " H_(2)O the central atomundergoes sp^(3) hybridlsation - yet their bond angles are different.
Answer» Solution :(i) In `CH_(4), NH_(3) and H_(2)0` the central atom undergoes `sp^(3)` hybridisation. But their bond angles are different due to the presence of lone pair of electrons. (ii)It can be explained by VSEPR THEORY. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs are not same. (iii) Bond pair-Bond pair `lt` Bond pair - Lone pair `lt` Lone pair - Lone pair So due to the varying repulsive force the bond pairs and lone pairs are distorted from REGULAR geometry and organise themselves in such a way that repulsion will be minimum and stability will be maximum. (IV) In case of `CH_(4),`there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry. i.e., tetrahedral with bond angle `= 109^(@) 28..` (v) `H_(2)O` has 2 bond is pairs and 2 lone pairs. There is large repulsion between lp-lp. Again repulsion between lp-bp is more than that of 2 hond pairs. So 2 bonds are more restricted Om nverted V shape (or) bent shape molecule with a bond angle of `104^(@) 35..` (vi) `NH_(3)` has 3bond pairs and I lone pair, There is repulsion between lp-bp. So 3 bonds are more restricted to form pyramidal shape with bond angle EQUAL to `107^(@) 18..`

Discussion

16.	In CH_(3)CH_(2)OH the bond that undergoes heterolytic cleavage most readily is:
Answer» `C-C` `C-O` `C-H` `O-H` SOLUTION :Electronegativties of C, H and O are `2.5, 2.1` and `3.5` respectively. The greatest electronagetivity difference is for `O-H` bond and as such it READILY UNDERGOES heterolytic CLEAVAGE most easily.

Discussion

17.	InCH_(3) - underset(CH_(3))underset(\|)(CH) - CH_(3) most stable radicals/ions formed on homolysis is/a.
Answer» `CH_(3)-UNDERSET(CH_(3)) underset(\|)(CH) - CH_(2) ` `CH_(3) - underset(CH_(3))underset(\|)C -CH_(3) ` and H `CH_(3) - underset(CH_(3))underset(\|)OVERSET(+)C -CH_(3) ` and H `CH_(3) - underset(CH_(3))underset(\|)overset(-)C -CH_(3) ` and H Solution :N/A

Discussion

18.	In CH_2 Cl_2the oxidation number of C is
Answer» `-4` `+2` 0 `+4` SOLUTION :`UL(C)H_(2)Cl_(2),x+2-2=0 impliesx=0`

Discussion

19.	In certain respects, lithium differs from other alkali metals. The main reason for this is
Answer» Small size of LITHIUM atom and `LI^(+)` ion Extremely HIGH ELECTRO positivity of `Li` GREATER hardness of `Li` Hydration of `Li^(+)` ion Answer :A

Discussion

20.	In certain cases,the rate of reaction increase with time. This phenomenon is known as
Answer» INDUCED catalysis catalytic inhibition autocatalysis catalytic promotion Answer :C

Discussion

21.	In cell fluid, the most abundant cation is :
Answer» `Na^(+)` ION `K^(+)` ion `MG^(+2)` ion `CA^(+2)` ion Answer :B

Discussion

22.	In catalytic oxidation of benzene vapours in presence of V_2O_5 to give maleic anhydride no ofatomic/nasent oxygens required for 1 mole of benzene is _______
Answer» SOLUTION :

Discussion

23.	In Castner -Kellnerusedfor preparationofsodiumcarbonate,______actsas ananode .
Answer» Nickelrod GRAPHITE ROD copperrod zincrod Answer :B

Discussion

24.	In Castner - Kellner process, the cathode in the outer chamber is
Answer» Hg Graphite Pt Fe Solution :Hgacts as CATHODE in OUTER compartements in CASTNER kellener PROCESS

Discussion

25.	In Castner Kellner cell, reaction at mercury cathode is
Answer» `2H_(2)O+2e^(-)rarr H_(2)+2OH^(-)` `2Cl^(-)rarrCl_(2)+2e^(-)` `Na^(+)+OH^(-)rarrNaOH` `Na^(+)+e^(-)+Hg rarr NAHG` ANSWER :D

Discussion

26.	In case the difference in solubilty ofhte two substances in the solvent is not very marked, ………….., involving a series of repeated vrystallizations, is carried out.
Answer» fractional crystallization vacuume crystallization ULTRA crystallization multiple crystallization Solution :In this process, crystallization is repeated a number of TIMES.

Discussion

27.	In case of hydrogen spectrum wave number is given by barv = R_H [(1)/(n_2^2) - 1/(n_2^2)] where n_1 lt n_2
Answer» SOLUTION :Lyman `bar(UPSILON) = R. [1/(1^2) - 1/(n^2)]` Balmer `bar(upsilon) =R.[1/(2^2) - 1/(n^2)]`

Discussion

28.	In case of hydrogen , molar volume at STP is "_______" 22.414L.
Answer» more than LESS than equal to DOUBLE of ANSWER :A

Discussion

29.	In case of homologous series of alkanes, which one of the following statements is incorrect?
Answer» The numbers of the series have the general formula `C_nH_(2n+2)`, when N is an integar The difference between any two SUCCESSIVE member of the series corresponds to 14 UNITS of relative atomic mass The member of the series are isomers of each other The members of the series have similar chemical products Answer :C

Discussion

30.	In case of heteronuclear diatomics of the type AB, where A is more electronegative than B, bonding molecular orbital resembles the character of A more than that of B. The statement
Answer» is false is true cannot be EVALUATED SINCE DATA is not sufficient is true only for certain systems Solution : As A is more electronegative, there is less enegy difference I between atomic ORBITAL of A and bonding M.O. HENCE, bonding M.O. resembles A more closely.

Discussion

31.	In case of H_(2)O_(2) in solid state the angle between the planes containing H-atoms is
Answer» `100^(@)` `90^(@)` `109^(@)28^(1)` `180^(@)` ANSWER :B

Discussion

32.	In case of carbohydrates D, L indicates the
Answer» Configuration of - OH at last CHIRAL CARBON Configuration of - OH at first chiral carbon Configuration of - OH at SECOND chiral carbon Configuration of - OH at at all chiral carbon Solution :Configuration of -OH at last chiral carbon

Discussion

33.	In case of alkali metals , the covalent character decreases in the order :
Answer» `MI gt MBr gt MCl gt MF` `MCl gt MI gt MBr gt MF` `MF gt MCl gt MBr gt MI` `MF gt MCl gt MI gt MBr` Solution :ACCORDING to Fajan rules , as the size of the ANION , i.e., HALOGEN increases , the covalent CHARACTER increases and the ionic character decreases , i.e, option (a) is CORRECT .

Discussion

34.	Incase of alkali metal halides, the ionic character increases in the order________
Answer» MF < MCl < MBR < MI MI < MBr < MCl < MF MI < MBr < MF < MCl none of these Answer :B

Discussion

35.	In case, nitrogen sulphur both are present in an organic compound, sodium thiocynate is formed. If sodium fusion is carried out with excess of sodium, sodium thocyanate decomposes. Which of the following compounds is/are present in the extract after decomposition?
Answer» `NACN` `Na_(2)S` Both None of these Answer :C

Discussion

36.	In Carius tube, the compound ClCH_(2)COOAg was heated with fuming HNO_(3) & AgNO_(3). After filtration and washing, a white precipitate was formed. The precipitate is
Answer» `Ag_(2)SO_(4)` `AgNO_(3)` `AgCl` `ClCH_(2)COOAg` Solution :In Carius method, Cl is converted into AgCl.

Discussion

37.	In Carius method, the sulphur in an organic compound is oxidised to ............. .
Answer» SOLUTION :`H_(2)SO_(4)`

Discussion

38.	In carius method of estimation of halogens, 250 mg of organic compound gave 141 mg of AgBr. Th percentage of bromine in the compound is (at mass Ag = 108 , Br = 80)
Answer» 48<BR>60 24 36 Solution :`%Br = (80)/(188) xx (0.141)/(0.250) xx 100 = 24`

Discussion

39.	In Carius method of estimation of sulphur 0.466g of organic compound gave 0.233 g of BaSO_(4).The percentage of sulphur is
Answer» 6.9 13.8 0.69 2.3 Answer :A

Discussion

40.	In Carius method of estimation of halogens, 250 g of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (at. Mass Ag = 108, Br = 80)
Answer» 24 36 48 60 Answer :A::B::D

Discussion

41.	In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12g of AgBr. Find out the percentage of bromine in the compound (Ag = 108, Br= 80)
Answer» Solution :molar MASS of AgBr `= 108 + 80 = 188g mol^(-1)` <BR> `underset(80g)(Br) + Ag rarr underset(188g)(AgBr)` 188g AgBr CONTAINS 80g Br `:.` 0.12g AgBr contains, mass of `Br = (0.12g xx 80)/(188)`

Discussion

42.	In carius method for the quantitative estimation of sulphur, it is estimated by
Answer» BaS `CaSO_(4)` `BaSO_(4)` `BaCl_(2)` Solution :Sulphur is quantitatively estimated y `BaSO_(4)`

Discussion

43.	In Carius method for the quantitative estimation of phosphorous, by using magnesia mixture, phosphorous is estimatied by :
Answer» `MgNH_(4).PO_(4)` `Mg_(2)P_(2)O_(7)` `(NH_(4))_(3)PO_(4).12MoO_(3)` All Solution :`Mg_(2)P_(2)O_(7)` is formed in estimation of P by caries method.

Discussion

44.	In Carius method 0.468 g of a compound afforded 0.668 g barium sulphate. Find the percentage of S in the compound.
Answer» ANSWER :`%S=(32)/233 XX (0.668)/0.468 xx 100 = 19.6%`

Discussion

45.	In Carium method of estimation of halgoen, 0.15 g of an organic compound gave 0.12 g of AgBr. What is the percentage of bromine in the compound?
Answer» 0.6808<BR>0.3504 `42.1%` 0.5 Solution :Molar mass of `AgBr=108+80=188" g "mol^(-1)` LTBR Percentage of `Br_(2)=(80)/(188)XX(0.12)/(0.15)xx100=34.04%`

Discussion

46.	In Carius method, 0.1890 g of an organic compound gave 0.2870 of silver chloride. Calculate the percentage of chlorine in the compound
Answer» SOLUTION :WEIGHT of SUBSTANCE = 0.1890g Weight of silver CHLORIDE = 0.2870g % of chlorine `= ("Weight of " AgCl xx 35.5 xx 100)/("weight of substance " xx 143.5) = (0.2870 xx 35.5 xx 100)/(0.1890 xx 143.5) = 37.8`

Discussion

47.	In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.
Answer» SOLUTION :`BR = 34.04%`

Discussion

48.	In carious method of estimation of halogen 0.15 g of an organic compound gave 0.12 g of AgBr. Find the percentage of bromine in the compound.
Answer» SOLUTION :`BR = 34.04%`

Discussion

49.	In carious method 0.12g AgBr obtained from 0.15g organic compound. Find out the percentage of AgBr in compound. (Ag= 108, Br =80)
Answer» Solution :`underset(80g)(BR) RARR underset(188g)(AGBR)` Thus, 188g AgBr CONTAIN 80gm BROMINE. So, 0.120g AgBr contain, Bromine `= (80 xx 0.12)/(188)` % `Br = (80)/(188) xx (0.12)/(0.15) xx 100= 34.04%`

Discussion

50.	In carbocation the carbon bearing the positive charge is
Answer» `SP^(2)` hybridised `sp^(3)` hybridised `DSP^(3)` hybridised None Answer :A

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

In chromium (III) chloride, CrCl_3, chloride ions have cubic close packed arrangement and Cr (III) ions are present in the octahedral holes. What fraction of the octahedral holes is occupied ? What fraction of total number of holes is occupied ?

In ClF_(3), NF_(3)" and " BF_(3) molecules the chlorine, nitrogen and boron atoms are

In chromium (III) Chloride, CrCl_(3) , Chloride ions have cubic close packed arrangement and Cr (III) ions are presnent in the octahedral holes. What fraction of the octahedral holes is occupied ? What fraction of totalnumber of holes is occupied ?

In chromatography, which of the following statement is Incorrect for R_f ?

In chromatography, if the stationary phase is solid, the basis is ............

In chlorophyll pigment how many number of nitrogen atoms surrounds the central 'Mg' atom ?

In chlorobenzene, chlorine is ________ but _____ directing.

in chlorine.^(35)CIand .^(37) CI isin3:1ratiocalculatechargeatomicmass.

In chemical reaction, catayst:

In CH_(4),NH_(3) and H_(2)O, the central atom undergoes sp^(3) hybridisation - yet their bond angles are different. Why ?

In CH_(4),NH_(3) and H_(2)O the central atom undergoes sp hybridization - yet their bond angles are different. Why?

In CH_(4),NH_(3) " and " H_(2)O the central atomundergoes sp^(3) hybridlsation - yet their bond angles are different.

In CH_(3)CH_(2)OH the bond that undergoes heterolytic cleavage most readily is:

InCH_(3) - underset(CH_(3))underset(|)(CH) - CH_(3) most stable radicals/ions formed on homolysis is/a.

In CH_2 Cl_2the oxidation number of C is

In certain respects, lithium differs from other alkali metals. The main reason for this is

In certain cases,the rate of reaction increase with time. This phenomenon is known as

In cell fluid, the most abundant cation is :

In catalytic oxidation of benzene vapours in presence of V_2O_5 to give maleic anhydride no ofatomic/nasent oxygens required for 1 mole of benzene is _______

In Castner -Kellnerusedfor preparationofsodiumcarbonate,______actsas ananode .

In Castner - Kellner process, the cathode in the outer chamber is

In Castner Kellner cell, reaction at mercury cathode is

In case the difference in solubilty ofhte two substances in the solvent is not very marked, ………….., involving a series of repeated vrystallizations, is carried out.

In case of hydrogen spectrum wave number is given by barv = R_H [(1)/(n_2^2) - 1/(n_2^2)] where n_1 lt n_2

In case of hydrogen , molar volume at STP is "_______" 22.414L.

In case of homologous series of alkanes, which one of the following statements is incorrect?

In case of heteronuclear diatomics of the type AB, where A is more electronegative than B, bonding molecular orbital resembles the character of A more than that of B. The statement

In case of H_(2)O_(2) in solid state the angle between the planes containing H-atoms is

In case of carbohydrates D, L indicates the

In case of alkali metals , the covalent character decreases in the order :

Incase of alkali metal halides, the ionic character increases in the order________

In case, nitrogen sulphur both are present in an organic compound, sodium thiocynate is formed. If sodium fusion is carried out with excess of sodium, sodium thocyanate decomposes. Which of the following compounds is/are present in the extract after decomposition?

In Carius tube, the compound ClCH_(2)COOAg was heated with fuming HNO_(3) & AgNO_(3). After filtration and washing, a white precipitate was formed. The precipitate is

In Carius method, the sulphur in an organic compound is oxidised to ............. .

In carius method of estimation of halogens, 250 mg of organic compound gave 141 mg of AgBr. Th percentage of bromine in the compound is (at mass Ag = 108 , Br = 80)

In Carius method of estimation of sulphur 0.466g of organic compound gave 0.233 g of BaSO_(4).The percentage of sulphur is

In Carius method of estimation of halogens, 250 g of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (at. Mass Ag = 108, Br = 80)

In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12g of AgBr. Find out the percentage of bromine in the compound (Ag = 108, Br= 80)

In carius method for the quantitative estimation of sulphur, it is estimated by

In Carius method for the quantitative estimation of phosphorous, by using magnesia mixture, phosphorous is estimatied by :

In Carius method 0.468 g of a compound afforded 0.668 g barium sulphate. Find the percentage of S in the compound.

In Carium method of estimation of halgoen, 0.15 g of an organic compound gave 0.12 g of AgBr. What is the percentage of bromine in the compound?

In Carius method, 0.1890 g of an organic compound gave 0.2870 of silver chloride. Calculate the percentage of chlorine in the compound

In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.

In carious method of estimation of halogen 0.15 g of an organic compound gave 0.12 g of AgBr. Find the percentage of bromine in the compound.

In carious method 0.12g AgBr obtained from 0.15g organic compound. Find out the percentage of AgBr in compound. (Ag= 108, Br =80)

In carbocation the carbon bearing the positive charge is

In chlorobenzene, chlorine is ____ but _ directing.