Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

89001.

A hydrocarbon contains 10.5 gm carbon and 1 gm hydrogen. Its 2.4 gm has 1 L volume at 1 atm and 127^(@)C, hydrocarbon is

Answer»

`C_(6)H_(7)`
`C_(6)H_(8)`
`C_(5)H_(6)`
None of these

Solution :`C = 10.5 GM = (10.5)/(12) mol = 0.87 mol`
`H = 1 gm = (1)/(1) = 1MOL`
`:. (C_(0.87)H_(1))_(7) = C_(6.09)H_(7) = C_(6)H_(7)`
`PV = nRT`, `PV = (w)/(m)RT`
`1 XX 1 = (2.4)/(m) xx 0.082 xx 400`
`m = 2.4 xx 0.082 xx 400 = 78.42 = 79`.
Discussion
89002.

A light bluish-green crystalline compound responds to the following tests: (i). Its aqueous solution gives a brown precipitate on reaction with alkaline K_2[HgI_4] solution. (ii). Its aqueous solution gives a blue colour with K_3[Fe(CN)_6] solution. (iii). Its solution in hydrochloric acid gives a white precipitate with BaCl_2 solution. Identify the ions present and suggest the formula of the compound.

Answer»

Solution :Its aqueous solution gives a brown precipitate or COLOURATION with alkaline `K_2[HgI_4]` solution. This suggests ammonium IONS.
Its aqueous SOLUTIONS gives a blue colour with `K_2[Fe(CN)_6]` solution. This suggests ferrous ions. Its solution in hydrochloric acid gives a white precipitate with `BaCl_2` solution. This suggest sulphate ions. The formula of the compound can be `FeSO_4(NH_4)_2SO_4.6H_2O` (MOHR's SALT.)
Discussion
89003.

A hydrocarbon containing 2 carbon atoms gives Sabatier and Senderens reaction but does not give precipitate with ammoniacal silver nitrate solution. The hydrocarbon in the question is

Answer»

Ethane
Acetylene
ETHYLENE
None of these

Solution :`CH_2=CH_2+H_2underset(300^@C)OVERSET(Ni)to CH_3-CH_3`
Ethylene does not give precipitate with AMMONICAL SILVER nitrate solution because it does not have acidic hydrogen
Discussion
89004.

A light bluish-green crystaline compound responds to thefollowingtests i. Its aqueous solutiongivesa brownprecipitate orcolourtion with alkline K_(2)[HgI_(4)] solution ii Its aqueous solutiongivesa blue colourwith K_(3)[Fe(CN)_(6)] solution iii Its solution in hydrochloricacidgives a whiteprecipitatewith BaCI_(2) solution Identifythe ions present and suggest the formula ofthecompound

Answer»

Solution :The ions are `NH_(4)^(o+) ,Fe^(2+),SO_(4)^(2-)`
FORMULA is `(NH_(4))_(2),SO_(4).FeSO_(4).6H_(2)O` (MOHR's SALT)
Discussion
89005.

A light bluish green crystalline compound responds to the following tests (i) Its aqueous solution gives a brown precipitate or colouration with K_(2)[HgI_(4)] (ii) Its aqueous solution gives a blue colur with K_(3)[Fe(CN)_(6)]. (iii) Its solution in hydrochloric acid gives white precipitate with BaCl_(2). Identify the ions present and suggest the formula of the compound.

Answer»
Discussion
89006.

A hydrocarbon containing 2 carbon atoms give Sabatier and Senderen's reaction but does not give precipitate with ammoniacal silver nitrate solution. The hydrocarbon in question is:

Answer»

ETHANE
Acetylene
Ethylene
None of these

Answer :B
Discussion
89007.

A light green coloured salt soluble in water gives black precipitate on passing H_(2)S which dissolves readily in HCl. The metal ion present is :

Answer»

`Co^(2+)`
`FE^(2+)`
`Ni^(2+)`
`MN^(2+)`

ANSWER :B
Discussion
89008.

A hydrocarbon C_(5)H_(12) gives only one monochlorination product. The hydrocarbon is

Answer»

2-Methyl butane
2, 2-Dimethyl propane
1, 2-Dimethyl propane
2, 3-Dimethy butane

ANSWER :B
Discussion
89009.

A ligand should contain

Answer»

odd electrons
even NUMER of electrons
LONE pair of electrons to donate
VACANT orbital to accept the lone pair

Answer :C
Discussion
89010.

A hydrocarbon C_6H_12 on oxonolysis gives only one product which does not reduce Fehling solution. The hydrocarbon is :

Answer»

2 - HEXENE
3- Hexene
3-Methylpentene
2,3 - Dimethlpentene.

Answer :D
Discussion
89011.

A ligand having an unshared pair of electrons can be a

Answer»

neutral molecule
 POSITIVELY CHARGED ion
negatively charged ion
group CONTAINING a lone PAIR of electrons

Answer :A::B::C::D
Discussion
89012.

Ahydrocarbon ,C_5 H_(12) givesonlyonemonochlorinationproduct .Identifythathydrocarbon .

Answer»

Solution :SINCE the hydrocarbon gives only one monochlorination PRODUCT, it INDICATES that all the HYDROGENS are identical. The formula of the hydrocarbon SUGGESTS that it is a saturated aliphatic hydrocarbon. That hydrocarbon is dimethylpropane (neopentane).
Discussion
89013.

A ligand is

Answer»

LEWIS ACID
BRONSTED acid
EITHER a Lewis acid or a Lewis BASE
Lewis base.

Answer :D
Discussion
89014.

A hydrocarbon C_(5)H_(10) does not react with chlorine in dark but gives a single monochloro compound C_(5)H_(9)Cl in bright sunlight. Identify the hydrocarbon .

Answer»

SOLUTION :(i) The hydrocarbon with MOLECULAR formula`C_(5)H_(10)` can be either a cycloalkane or an alkene .
(ii) Since the hydrocarbon does not react with `Cl_(2)` in the DARK , it cannot be an alkene. Thus, it must be a cycloalkane.
(III) Since the cycloalkane reacts with `Cl_(2)` in the presence of bright sunlight , to give a single monochloro compound , `C_(5)H_(9)Cl`, all the hydrogen atoms of the cycloalkane must be EQUIVALENT. Thus, the cycloalkane is cyclopentane.
Discussion
89015.

A ligand can not be ( A) neutral, (B) anionic ,(C) anionic

Answer»

B only
C only
A and B
NONE

ANSWER :D
Discussion
89016.

A hydrocarbon C_(5)H_(10) does not react with chlorine in dark but give a single monochloro compound C_(5)H_(9)Cl in bright sunligh. Identify the hydrocarbon.

Answer»

SOLUTION :As the compound gives the single monochlorinatedproduct in bright SUNLIGHT with a chlorine, THEREFORE all the HYDROGENS of the compound are equivalent. Thus is possible only if a compound is a cyclic ALKANE. Thus, the hydrocarbon is cyclopentane.
Discussion
89017.

A : LiF and Csl both are less soluble in water . R : LiF is with high lattice energy and Csi is with smaller hydration energy.

Answer»

If both Assertion & Reason are true and the reason is the correct explanantion of the assertion, then MARK (1).
If both Assertion & Reason are true and the reason is the correct explanantion of the assertion, then mark (2).
If Assertion is true statement but Reason is false. Them mark (3).
If both Assertion and reason are false statement, then mark (4).

Answer :A
Discussion
89018.

A hydrocarbon C_(5)H_(10) does not react with chlorine but gives a single monochloro compound, C_(5)H_(9)Cl in bringht sunlight. Identify the hydrocarbon.

Answer»

SOLUTION :(i) The HYDROCARBON with M.F. `C_(5)H_(10)` can be either a cycloalkane or an alkene.
(ii) Since the hydrocarbon does not react with `Cl_(2)` in TH dark, it cannot be an alkene but must be a cycloalkane.
(iii) Since the cycloalkane reacts with `Cl_(2)` in the presence of bright sunlight, to GIVE a single monochloro compound, `C_(5)H_(9)Cl`, therefore, all the ten hydrogen atoms of the cycloalkane must be equivalent. thus,
Discussion
89019.

A ligand can also be regarded as

Answer»

Lewis base
Bronsted acid
Lewis acid
Bronsted base

Solution :A LIGAND can be regarded as Lewis base because it is capable of donating its LONE pair to the central atom of the CO- ordination COMPOUND.
Discussion
89020.

A hydrocarbon A(molecular formula C_8H_10 ) on ozonolysis gives B(C_4H_6O_2)only. Compound C( C_3H_5Br )on treatment with magnesium in dry ether gives (D) which on with CO_2followed by acidification gives(B). Identify A, B, C and D.

Answer»

SOLUTION :
Discussion
89021.

(A): Levigation is used for the separation of oxide ores from impurities (R) : Ore particles are removed by washing in a current of water.

Answer»

Both A & R are TRUE, R is the correct explanation of A
Both A & R are true, R is not correct explanation of A
A is true, R is FALSE
A is false, R is true

Answer :C
Discussion
89022.

A hydrocarbon A(C_(10)H_(12)) has no chiral carbon. A gives a white precipitate with ammonical solution of silver nitrate. A on treatment with H_(2)//pt gives B(C_(10)H_(20)). A on ozonolysis gives C(C_(8)H_(12)O_(4)) as one product which on heating with soda lime gives D(C_(6)H_(12)). D on monochlorination with Cl_(2)//hv gives C_(6)H_(11)Cl as sole isomer. Identify A to D.

Answer»


SOLUTION :N//A
Discussion
89023.

(A) Levigation is used for the separation of oxide ores fromimpurities (R) Ore particles are removed by washing in a currentof water.

Answer»

Both (A) and (R) are true and (R) is the correct EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanationof (A)
(A) is true but (R) is false
Both (A) and (R) are false

ANSWER :C
Discussion
89024.

A hydrocarbon A(molecular formula C_(8)H_(10)) on ozonolysis gives B(C_(4)H_(6)O_(2)) only. Compound C(C_(3)H_(5)Br) on treatment with magnesium in dry ether gives (D) which on treatment with CO_(2) followed by acidification gives(B). Identify A, B, C and D.

Answer»

Solution :The compound with molecular formula `C_(8)H_(10)` is <BR>
Compound ( C) with molecular formula `C_(3)H_(5)Br` is
Discussion
89025.

A: Lesser the activation energy, greater is the rate of reaction R: Activation energy of a reaction is independent of temperature

Answer»

Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A)
Both (A) and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is false
Both (A) and (R ) are false

Answer :B
Discussion
89026.

A hydrocarbon (A) (molecular formula C_5H_(10)), on catalytic hydrogenation, produces 2-methylbutane. The compound (A) combines with HBr according to Markownikoff's rule to form the compound (B) which reacts with silver hydroxide to produce an alcohol (C ) having the molecular formula C_5H_(12)O. On oxidation, the alcohol(C ) yields a ketone (D). idenify A,B,C,D and give the reactions involved.

Answer»

Solution :The given changes are as follows:

The molecular formula, `C_(5)H_(10)` is in AGREEMENT with the general formula of alkenes `(C_(n)H_(2n))` and on catalytic hydrogenation. (A) PRODUCES the SATURATED hydrocarbon, 2-methylbutane. THEREFORE, the compound (A) is an alkene. Now, the alcohol (C), on oxidation, produces the ketone (D). So, the alcohol must be a secondary `(2^(@))` alcohol. Again, this alcohol is obtained by the reaction of (B) with `AgOH`. So, (B) is a secondary bromide.
`underset("2-bromo-3-methylbutane(B)")(CH_(3) - overset(overset(CH_(3))(|))(C)H-overset(overset(Br)(|))(C)H - CH_(3))`
Now, (B) is produced by the reaction of (A) with HBr according to Markownikoff's. rule. So, the structure of (A) is:
`CH_(3) - overset(overset(CH_(3))(|))(C)H - CH = CH_(2)`[3-methylbut-1-ene(A)]
Discussion
89027.

A lead storage cell is discharged which causes the H_(2)SO_(4) electrolyte to change from a concentration of 34.6% by weight (density 1.261 gmL^(-1) at 25^(@)C) to one of 27% by weight. The original volume of electrolyte is one litre. How many faraday have left the anode of battry ? Note the water is produced by the cell reaction as H_(2)SO_(4) is used up. Over all reaction is : Pb_(s) + PbO_(2) + 2H_(2)SO_(4(l)) rarr 2PbSO_(4(s)) + 2H_(2)O

Answer»


ANSWER :`1.255` FARADAY
Discussion
89028.

A hydrocarbon 'A', (C_(4)H_(8)) on reaction with HCl gives a compound 'B', (C_(4)H_(9)Cl), which on reaction with 1 ol of NH_3 gives compound 'C', (C_(4)H_(11)N). On racting with NaNO_(2) and HCl followed by treatment with water, compound 'C' yields an optically active alcohol, 'D'. Ozonolysis of 'A' gives 2 mols of acetaldehyde. identify compounds 'A' to 'D'. explain the reaction involved.

Answer»

Solution :(i) M.F. of compound `A(C_(4)H_(8))` corresponds to the general formula `C_(n)H_(2n)` where n=4, therefore, A is an alkene. Since ozonolysis of compound `A(C_(4)H_(8))` gives two moles of acetaldehyde, therefore, alkene (A) must be symmetrical, i.e., but-2-ene.
`underset("But-2-ene")(CH_(3)CH=CHCH_(3)) underset((ii)Zn//H_(2)O)overset((i)O_(3)//CH_(2)Cl_(2))to underset("Acetaldehyde (A)")(2CH_(3)CH=O)`
(ii) Since compounds (A), i.e., but-2-ene reacts with HCl to form compound (B), therefore, (B) must be 2-chlorobutane.
`underset("But-2-ene")(CH_(3)CH=CHCH_(3)) +HCl to underset("2-Chlorobutane (B)")(CH_(3)CH_(2)-CHCl-CH_(3))`
(III) Since compound (B), i.e., 2-chlorobutane reacts with one mole of `NH_(3)` to give compound (C), therefore, compound (C) must be a `1^(@)` amine, i.e., butan-2-amine.
`underset("2-Chlorobutane (B) (Optically active)")(CH_(3)CH_(2)-underset(CL)underset(|)overset(**)(C)H-CH_(3)) underset(-HCl)overset(NH_(3)" (1 mole)")to underset("Butan-2-amine (C)")(CH_(3)CH_(2)-underset(NH_(2))underset(|)overset(**)(C)H-CH_(3))`
(iv) Since compound (C) reacts with `NaNO_(2)//HCl`, to give an alcohol (D), therefore, compound (D) must be butan-2-ol. please note that 2-clorobutane (B), butan-2-amine (C) and butan-2-ol (D) contain a chiral CARBON, therefore, all these compounds are optically active.
`underset("Butan-2-amine (C) (Optically active)")(CH_(3)CH_(2)-underset(NH_(2))underset(|)overset(**)(C)H-CH_(3)) overset (NaNO_(2)//HCl)to underset("Butan-2-ol (D) (Optically active)")(CH_(3)CH_(2)-underset(OH)underset(|)overset(**)(C)H-CH_(3))`
Discussion
89029.

(A) Lead, tin and bismuth are purified by liquation method . (R) Lead, tin and bismuthhave low melting points as compared to impurities

Answer»

Both (A) and (R) are TRUE and (R) is the correct explanation of (A)
Both (A) and (R) are true and (R) is not the correct explanationof (A)
(A) is true but (R) is FALSE
Both (A) and (R) are false

Answer :A
Discussion
89030.

A lead storage battery is the most important type of secondary cell having a lead anode and a grid of lead packed with PbO_(2) as cathode. A 38% solution of sulphuric acid is used as electrolyte. (Density = 1.294 g mL^(-1)) The battery holds 35L of the acid. During the discharge of the battery, the density of H_(2)SO_(4) falls to 1.138 g mL^(-1). (20%H_(2)SO)_(4) by mass)write the product of electrolysis wheni dilute sulphuric acid is electrolysed using platinu electrodes .

Answer»


Solution : At ANODE ,`O_(2)` is PRODUCED
At CATHODE : `H_(2)` is produced
Discussion
89031.

A hydrocarbon (A) [C - 90.56%, vapour density - 43] was subjected to vigorous oxidation to give a dibasic acid (B). 0.1 g of (B) required 24.1 mL of 0.05 N NaOH for complete neutralization. Nitration of (B) gave a single mononitro derivative. When (B) was heated strongly withsoda lime, it gave benzene. Identify (A) and (B).

Answer»

SOLUTION :
Discussion
89032.

A lead storage battery is the most important type of secondary cell having a lead anode and a grid of lead packed with PbO_(2) as cathode. A 38% solution of sulphuric acid is used as electrolyte. (Density = 1.294 g mL^(-1)) The battery holds 35L of the acid. During the discharge of the battery, the density of H_(2)SO_(4) falls to 1.138 g mL^(-1). (20%H_(2)SO)_(4) by mass)Lead storage battery is considered a secondary cell. Why ? Write the product battery is considered a secondary cell. Why?

Answer»


SOLUTION : This is because it can be RECHARGED after USE.
Discussion
89033.

A hydrocarbon (A) C_nH_(2n-4) on ozonolysis gives(CH_3)_2CHCH_2CHO, 2OH"CC"H_2CH_2CHO and CH_3COCH_3The value of n is

Answer»


ANSWER :16
Discussion
89034.

A hydrocabron coantaing 80% carbon, then the hydrocarbon is :-

Answer»

`CH_(4)`
`C_(2)H_(4)-`
`C_(2)H_(2)`
`C_(3)H_(9)`

ANSWER :D
Discussion
89035.

A lead storage cell is discharged which causes H_(2)SO_(4) electrolyte to change from a concentration of 40% by weight ( density 1.25g mL^(-1)C to one of 30% by weight. The original volume of electrolyte is 1L. How many Faradays have left the anode of battery. Overall reaction of lead storage cell is : ltbr. Pb(s)+PbO_(2)+2H_(2)SO_(4)(l) rarr 2PbSO_(4)(s)+2H_(2)O

Answer»

Solution :Weight of solution `(W_(sol))=V_(sol)(i n Ml) xx d_(sol)`
`=1000mL xx 1.25gmL^(-1)`
`=1250g`
Weight of `H_(2)SO_(4)=(40xx1250)/(100)=500g`
Weight of `H_(2)O=1250-500=750g`
After electrolysis,
Now during reaction, weight of `H_(2)O` formed `=Xg`
Moles of `H_(2)SO_(4)=` mol of `H_(2)SO_(4)` formed `=(X)/(18)`
`(:' mol ` ratio of `H_(2)SO_(4):H_(2)O=1:1)`
Weight of `H_(2)SO_(4)` USED`=(98X)/(18)=5.44Xg`
`[Mw` of `H_(2)SO_(4)=98g mol^(-1)]`
Weight of `H_(2)SO_(4)` left `=(500-5.44X)g`
New weight of solution`=[{:(Weight of old solution),(+Weight of H_(2)O fo rmed),(-Weight of H_(2)SO_(4) LOST):}]`
`=1250+X-5.44X`
`%` by weight of new solution
`((500-5.44X))/((1250+X-5.44X))=(30)/(100)`
`:.X=30.43g=(30.43)/(18)mol` of `H_(2)O` are formed
`(Mol` of `H_(2)O=Eq` of `H_(2)O)""( :'2H_(2)O` CONSUMES `2e^(-))`
`1mol` of `H_(2)O` formed by passage of `1F`.
`(30.43 )/(18)mol` of `H_(2)O` fo rmed`=(30.43)/(18)F=1.69F`
Discussion
89036.

A hydride of nitrogen which is acidic is

Answer»

`N_(2)H_(2)`
`N_(2)H_(4)`
`NH_(3)`
`N_(3)H`

Solution :`N_(3)H` (hyrazoic acid) is the acidic HYDRIDE of nitrogen.
Discussion
89037.

A lead storage battery is the most important type of secondary cell having a lead anode and a grid of lead packed with PbO_(2) as cathode. A 38% solution of sulphuric acid is used as electrolyte. (Density = 1.294 g mL^(-1)) The battery holds 35L of the acid. During the discharge of the battery, the density of H_(2)SO_(4) falls to 1.138 g mL^(-1). (20%H_(2)SO)_(4) by mass)What is the molarity of sulphuric acid before discharge ?

Answer»


Solution : MOLARITY `=(38xx1.294)/(98)xx(1000)/(100)=5.02M`
Discussion
89038.

A hydro-carbon that shows acidic behaviour towards sodium in liquid NH_3is :

Answer»

`CH_3 -CH_2- C = C-H`
`CH_3 =C = CH -CH_3`
`CH_3-CH = CH - CH_3`
`CH_3-CH_2 -CH = CH_2 `

ANSWER :A
Discussion
89039.

A lead storage battery is the most important type of secondary cell having a lead anode and a grid of lead packed with PbO_(2) as cathode. A 38% solution of sulphuric acid is used as electrolyte. (Density = 1.294 g mL^(-1)) The battery holds 35L of the acid. During the discharge of the battery, the density of H_(2)SO_(4) falls to 1.138 g mL^(-1). (20%H_(2)SO)_(4) by mass) How much electricity in terms of faraday is required to carry out the reduction of one mole of PbO_(2) ?

Answer»


SOLUTION : ELECTRICITY in faradays to carry out the REDUCTION of one mole of `pbO_(2)=2` Faradyas
Discussion
89040.

A hydride of nitrogen having lowest oxidation number of N:

Answer»

`H_3N`
`N_3H`
`H_4N_2`
`H_2N_2`

ANSWER :A
Discussion
89041.

A lead storage battery is the most important type of secondary cell having a lead anode and a grid of lead packed with PbO_(2) as cathode. A 38% solution of sulphuric acid is used as electrolyte. (Density = 1.294 g mL^(-1)) The battery holds 35L of the acid. During the discharge of the battery, the density of H_(2)SO_(4) falls to 1.138 g mL^(-1). (20%H_(2)SO)_(4) by mass) Wrtie the reaction taking place at the cathide when the battery is in use.

Answer»


Solution : The cell reaction at the cathode when the battery is in use are GIVEN below :
Cathode : `PbO_(2)(s) +SO_(4)^(2-)(AQ)+4H^(+)(aq)+2E^(-)rightarrowPbSo_(4)+2H_(2)O(l)`
Discussion
89042.

A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducing agent (C) Identify A, B andC

Answer»

SOLUTION :(i) A HYDRIDE of 2nd period ALKALI metal (A) is lithium hydride (LiH).
(ii) Lithium hydrite (A) reacts with diborane (B) to give lithiumborohydride (C) which is ACT as reducing AGENT.
Discussion
89043.

A lead storage battery has initially 200 g of lead and 200 g of PbO_(2), plus excess H_(2)SO_(4). How long could this cell deliver a current of 10 amp, without recharging, if it was possible to operate it so that the reaction goes to completion ?

Answer»


ANSWER :4.48 HOUR
Discussion
89044.

(A) Hydration energy of iodide is highest among halides (R ) Greater the size of halide , higher is the hydration energy

Answer»

Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A)
Both (A) and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is false
Both (A) and (R ) are false

Answer :D
Discussion
89045.

A lead storage battery has initially 200 g of holf and 200 g of PbO_(2), plus excess H_(2)SO_(4). Theoretically, how long could this cell deliver a current of 10 amp, without reacharging, if it were possible to operate it so that the reaction goes to completion.

Answer»

Solution :Discharging of battery takes place through the reaction
`{:(Pb,+,PbO_(2),+,2H_(2)SO_(4),=, 2PbSO_(4),+,2H_(2)O),((200)/(207) "mole",,(200)/(239) "mole",,"(excess)",,,,):}`
As mole of `PbO_(2)` is less than that of Pb, `PbO_(2)` is the limiting REACTANT which shall be totally CONSUMED.
No. of FARADAY delivered by the battery
= no. of eq. of `PbO_(2)` lost
`= (200)/(239) xx 2`
`therefore` charge `= (400)/(239) xx 96500` coulombs
`therefore` time of discharge `= (400 xx 96500)/(239) xx (1)/(10)` seconds
`= (400 xx 96500)/(239 xx 10) xx (1)/(60 xx 60) h`
= 4.486 h.
Discussion
89046.

A lead storage battery has been used for one month (30 days) at the rate of one hour per day by drawing a constant currentof 2 amperes. H_(2)SO_(4) consumed by the battery is:-

Answer»

1.12 mole
2.24 mole
3.36 mole
4.48 mole

Solution :The reactions occurring during discharge are:
Anode: `Pb(s)+SO_(4)^(2-)(AQ)toPbSO_(4)(s)+2E^(-)`
`underset("At cathode:"PbO_(2)(s)+4H^(+)(aq)+SO_(4)^(2-)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(l))`
Overall reaction reaction: `Pb(s)+PbO_(2)(s)+2H_(2)SO_(4)to2PbSO_(4)(s)+2H_(2)O(l)`
Quantity of electricity consumed
`=(2A)(30xx3600s)=216000C`
`2xx96500` coulombs consume `H_(2)SO_(4)=2` MOLES
`therefore216000` coulombs will consume `H_(2)SO_(4)`
`=(1)/(96500)xx216000=2.24` moles.
Discussion
89047.

A hydrideof 2^(nd) periodalkalimetal(A )on reactionwithcompoundof Boron (B )to givea reducingagent (c )I dentifyA, Band C.

Answer»

Solution :(i) A hydride of 2nd period alkali metal (A) is LiH.
(ii) Compound of BORON (B) is `B_(2)H_(6)` diborane.
(iii) Reducing AGENT ( C ) is `LiBH_(4)`.
(iv) `2LiH+B_(2)H_(6)overset("Ether")to 2LiBH_(4)`
(a strong reducing agent)
(v) A is LiH. B is `B_(2)H_(6)`. C is `LiBH_(4)`.
Discussion
89048.

A lead storage battery containing 5.01. of (IN)H_(2)SO_(4) solution is opcrated for 9.65 xx10^(5) s with a stcady current jof 100 mA. Assuming volume of the soution remaining constant, normality of H_(2) SO_(4) will

Answer»

remain tmchanged
increases by 0.20
increase by unity
DECREASE by 0.40

Solution :`({:(Pb+SO_(4)^(2-)to PbSO_(4)+2e^(-)),(PbO_(2)+ 4H^(+)+SO_(4)^(2-) +2e^(-) to PbSO_(4)+2H_(2)O):})/(Pb+PbO_(2)+2H_(2) SO_(4) to 2P bSO_(4)+2H_(2)O)`
The REACTIO INDICATES that 2 moles of `H_(2)SO_(4)`
corresponds ot `2xx96500` C and 2 moles `H_(2)SO_(4) -=4` equive. of `H_(2) SO_(4).`
`2xx96500` C consumed 4 equive. of `H_(2)SO_(4)`
and `100xx10^(-3) xx9.65 xx10^(5)C` consumed
`=(4xx100xx10^(-3) xx9.95 xx10^(5))/(2xx96500)=2`EQUIV. `H_(2) SO_(4)`
`therefore ` Decrease in normality `=2/5=0.40`
Discussion
89049.

A hydrated salt of iron sulphate weighing 2g, contains 0.9065g of water of crystallisation. Find the formula of the hydrated salt.

Answer»

SOLUTION :`(FeSO_(4).7H_(2)O)`
Discussion
89050.

A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% H_(2)SO_(4) by mass having a density of 1.294 g mL^(-1). The battery holds 3.5 L of the acid. During the discharge of the battery, the density H_(2)SO_(4) falls from 1.294 g mL^(-1) to 1.139 g mL^(-1) which is 20% H_(2)SO_(4) by mass Q. The number of ampere-hour for which the battery must have been used is

Answer»

2650.5
265.05
26.505
2.6505

Solution :Ampere-hour`=("COULOMBS")/(3600s)=(964178)/(3600)=262.05`
Discussion