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30201.

Theheat liberated when 1.89 g of benzoic acid burnt in a bomb calorimeter at 25^(@)C increases the temperature of 18.94 kg of water by 0.632^(@)C. If the specific heat of water at 25^(@)C is 0.998cal/g-deg, the value of the heat combustion of benzoic acid is

Answer»

771.1 KCAL
871.2 kcal
881.1 kcal
981.1 kcal

Solution :`Q=(18.94xx0.632xx0.998xx122)/(1.89)=771.1 kcal`.
30202.

The heat liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25^@C increase the temperature of 18.94 kg of water by 0.632^@C. If the specific heat of water at 25^@C is 0.998 cal/g-deg, the value of the heat of combustion of benzoic acid is

Answer»

881.1 kcal
771.124 kcal
981 kcal
871.2 kcal

Answer :B
30203.

The heat evolved in the combustion of benzene is given by: C_(6)H_(6)(l) + 7(1/2)O_(2)(g) to 3H_(2)(l) + 6CO_(2)(g) , DeltaH = -781.0 kcal//"mol"^(-1), When 156 g of C_(6)H_(6) is burnt in an open container, the amount of heat energy released will be:

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`150.2 kcal "MOL"^(-1)`
`1562.0 kcal "mol"^(-1)`
`6528.2 KJ "mol"^(-1)`
`2448.5 kJ "mol"^(-1)`

Answer :B::C
30204.

The heat evolved in the combustion of benzene is given by C_(6)H_(6)+7(1)/(2)O_(2)rarr6CO_(2)(g)+3H_(2)O(l),DeltaH=-3264.6 kJWhich of the following quantities of heat energy will be evolved when 39_(g)C_(6)H_(6) are burnt

Answer»

816.15 KJ
1632.3 kJ
6528.2 kJ
2448.45 kJ

Solution :78g of benzene on COMBUSTION PRODUCES
HEAT =-3264.6 kJ
`therefore" 39 g will produce = "(-3264.6)/(2)=-1632.3 kJ`.
30205.

The heat evolved in the combustion of benzene is given by C_(6)H_(6)(l) + 7 (1/2)O_(2)(g) to 3H_(2)O(l) + 6CO_(2)(g), DeltaH = -781.0 kcal//"mol"^(-1) When 156 g of C_(6)H_(6) is burnt in an open container, the amount of heat energy released will be

Answer»

150.2 kcal `"MOL"^(-1)`
1562.0 kcal `"mol"^(-1)`
`6528.2 KJ"mol"^(-1)`
`2448.5 kJ "mol"^(-1)`

Answer :B::C
30206.

The heat evolved in the combustion of112 litre of water gas (mixture of equal volume of H_(2) and CO) is H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g), DeltaH=-241.8 kJ CO(g) + 1/2O_(2)(g) to CO_(2)(g), DeltaH=-283 kJ

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`241.8` KJ
283 kJ
1312 kJ
1586 kJ

ANSWER :C
30207.

The heat evolved during the combustion of 112 litre of water gas (mixture of equal volume of H_2 and CO) is :GivenH2​ (g)+1/2O2​ (g)=H2​ O(g);ΔH=−241.8kJCO(g)+1/2O2​ (g)=CO2​ (g);ΔH=−283kJ

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241.8 kJ
283 kJ
1312 kJ
1586 kJ

Answer :C
30208.

The heat evolved during the combination of 24 g C and 128 g S following the change isC+S_(2)rarrCS_(2),DeltaH=22.0 kcal

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11 KCAL
32 kcal
44 kcal
22 kcal

Solution :`22.0xx2=44 kcal`.
30209.

The heat energy required to ionise the following molecules is given as follows N_(2)(g) overset(DeltaH_(1))to H_(2)^(o+)(g),O_(2)(g) overset(DeltaH_(2))to O_(2)^(+)(g) Li_(2)(g) overset(DeltaH_(3)) to Li_(2)^(o+)(g),C_(2)(g) overset(DeltaH_(4))to C_(2)^(o+)(g) The correct decreasing order of energy in terms of heat

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`DeltaH_(1) gt DeltaH_(3) gtDeltaH_(2) gt DeltaH_(4)`
`DeltaH_(2) gt DeltaH_(3) gt DeltaH_(1) gt DeltaH_(4)`
`DeltaH_(3) gt DeltaH_(4)gt DeltaH_(1) gt DeltaH_(2)`
`DeltaH_(3) gt DeltaH_(1) gt DeltaH_(4) gt DeltaH_(2)`

SOLUTION :Enthalpy of IONISATION depends upon the size of molecules. LARGER the size lesser will be ionisation energy.
30210.

The heat evolved during neutralisation is maximum in the reaction of:

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`NH_4OH and CH_3COOH`
`NH_4OH` and HCL
`NAOH and CH_3COOH`
`NaOH and HCl`

ANSWER :D
30211.

The heat content of a system is called

Answer»

<P>INTERNAL energy
Entropy
Free energy
Enthalpy

Solution :`q_(p)=DELTAH`.
30212.

The heat change in a chemical reaction at constant pressure is:

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`TRIANGLEH`
`TRIANGLEU`
`triangleT`
`triangleV`

ANSWER :A
30213.

The heat change taking place during the reaction H_2O(l)rarrH_2O(g) is ?[ Given triangleH_1=H_2O(g)=-57 kal, TriangleH_f=H_2O(l)=-68.3kcal]:

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`+11.3 KCAL`
`-11.3 kcal`
`-115.3 kcal`
`+115.3 kcal`

ANSWER :A
30214.

The heat change for the reaction. CO + (1)/(2) O_(2) rarr CO_(2) is called

Answer»

HEAT of reaction
heat of formation
heat of neutralisation
heat of combustion.

Answer :D
30215.

The heat change for the following reaction at 298K and at constant pressure is +7.3 kcal A_(2)B(s) rarr 2A(s) + 1//2 B_(2)(g), DeltaH =+7.3 kcal The heat change at constant volume would be C(s)+2S(s)rarrCS_(2)(l) is known as

Answer»

HEAT of VAPORIZATION
Heat of solution
Heat of fusion
Heat of FORMATION

Solution :ACCORDING to the DEFINITION of heat of formation.
30216.

The heat change for the following reaction at 298K and at constant pressure is +7.3 kcal A_(2)B(s) rarr 2A(s) + 1//2 B_(2)(g), DeltaH =+7.3 kcal The heat change at constant volume would be

Answer»

7.3 KCAL
More than 7.3
Zero
None of these

Solution :`DeltaH=DeltaE+DeltanRT or DeltaE=DeltaH-DeltanRT`
`thereforeDeltaE=+7.3-(1)/(2)xx0.002xx298=7.3-0.298=7` kcal.
30217.

The heat change DeltaH for the reaction 2CO+O_(2)rarr2CO_(2),DeltaH=-135 kcal is called

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Heat of formation
Heat of reaction
Heat of combustion
Heat of solution

Solution :ONE mole of a substance is COMPLETELY OXIDIZED in air.
30218.

The heat change at constant volume (q_(v)) is equal to

Answer»

`DELTAU`
`DELTAH`
RT
`DELTAG`

ANSWER :A
30219.

The heat capacity of liquid water is 75.6 J/mol K, while the enthalpy of fusion of ice is 6.0 kJ/mol. What is the smallest number of ice cubes at 0^(@)C, each containing 9.0g of water, needed to cool 500 g of liquid water from 20^(@)C to 0^(@)C ?

Answer»

1
7
14
None of these

Answer :C
30220.

Theheat capcity of liquildwater is 75.6 J //mol .K, while theenthalpy of fusion of ice is6.0 kJ//mol, What is the smallestnumberof icecubes at0^(@) C,eachcontainig9.0g of water ,needed to cool500 gofliquildwaterfromj 20^(@)to 0^(@) C ?

Answer»

1
7
14
None of these

Solution :`DeltaU =Q + W`
` = 100 -[1 XX 1 + 1//2xx 1]"bar" m^(3)""("1 bar " m^(3) = 10^(5) J= 100 k)`
`= - 50 kJ`
30221.

The heat capacity at constant volume of an ideal gas consisting of monatomic molecules is 3//2 R ( where R is the gas constant ) . The heat capacity at constant pressure is :

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R
`( 5)/( 2) R `
` ( 1)/( 2) R `
`( 3)/( 2) R `

SOLUTION :New PRESSURE `= ( 40)/( 100)`
Since PV is constant, new volume should be `( 100)/( 40) ` times
i.e. `( 100)/( 40) xx 100 = 250` ml
Volume of bulb A= 100ml
Volume of bulb B = 250 - 100 = 150 ml
30222.

The hcp and ccp structure of a given element. (Given radius of element id same in both structures)

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have same density
have same distance between TWO CONSECUTIVE LAYERS (A&B)
have same co-ordination number
have same fraction of unoccupid space.

Answer :ABCD
30223.

The h.c.p. and c.c.p. structures for a given crystal system of equal - sized spheres are expected to have

Answer»

the same VOLUME, `1.33 pi r^(3)`
the same NUMBER of particles within the given UNIT cell
the same ratio, 0.74, of the total sphere volume to the unit - cell volume
All of the above

Answer :A::C
30224.

The HCC bond angle in acetylene is:

Answer»

`90^@`
`120^@`
`109^@ 28'`
`180^@`

ANSWER :D
30225.

The hardness of water is estimated by:

Answer»

CONDUCTIVITY method
EDTA method
Titrimetric method
Distillation method

Answer :B
30226.

The hardness of metal increases with increase in number of …… involved in metallic bonding.

Answer»

Atoms
Molecules
Electrons
All

Answer :C
30227.

The hardness Cr is due to ….And metallic lusture is due to

Answer»

Covalent BOND, METALLIC bond
Covalent bond, hydrogen bond
Metallic bond, covalent bond
Metallic bond, hydrogen bond

Solution :Covalent bond is CONSTITUTED by electrons of d-orbitals and luster is due to FREE electrons of s-orbital in metallic bond.
30228.

The hardest substance amongst the following is

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`Be_(2)C`
titanium
SiC
`B_(4)C`

SOLUTION :`B_(4)C` is the hardest SUBSTANCE after diamond and BN.
30229.

The hardest substance amonts the following is

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`Be_(2)C`
Graphite
Titanium
`B_(4)C`

Solution :`B_(4)C` is the hardest SUBSTANCE ALONG with DIAMOND.
30230.

The hard plastic covers of telephones are made of polymers of

Answer»

ACRYLONITRILE
STYRENE
Fluoromethane
PHENOL formaldehyde

ANSWER :D
30231.

The halogens, which is attacked by conc. HNO_(3) are :

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`F_(2)`
`Cl_(2)`
`Br_(2)`
`I_(2)`

Solution :CONCENTRATED nitric ACID OXIDISES `I_(2)` to `HIO_(3) : 10 HNO_(3) + I_(2) rarr 2HIO_(3) + 10 NO_(2)`
30232.

The hardest compound of boron is :

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BORON NITRIDE
Boron HYDROXIDE
Boron carbide
Boron oxide.

Answer :C
30233.

Which halogen has highest value of electron gain enthalpy?

Answer»

SOLUTION :CHLORINES
30234.

The halogen with the highest value of electron gain enthalpy is..................

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SOLUTION :CHLORINES
30235.

The halogen with the highest heat of vaporisation is :

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`F_(2)`<BR>`Cl_(2)`
`Br_(2)`
`I_(2)`.

ANSWER :D
30236.

The halogen with highest polarisibility is.........

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FLUORINE
CHLORINE
BROMINE
Iodine

Solution :Iodine is LARGEST in size among fluorine, chlorine and bromine.
30237.

The halogen with the highest electron affinity is :

Answer»

`F`<BR>`CL`
`Br`
`I.`

ANSWER :B
30238.

The halogen with highest ionisation potential is

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F
Cl
Br
I

Answer :A
30239.

The halogen which normally does not form oxoacid is .............. .

Answer»

SOLUTION :FLUORINE
30240.

The halogen which is solid at room temperature:

Answer»

F
Cl
Br
I

Answer :D
30241.

The halogen which is most easily reduced is :

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`F_(2)`
`Cl_(2)`
`Br_(2)`
`I_(2)`

Solution :`F_(2)` is most easily reduced because HF is the most stable ACID AMONG the HALOGEN acid.
30242.

The halogen which is most reactive in the halogenation of alkenes under sunlight is

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Chlorine
Bromine
Iodine
All equal

Solution :`C_2H_4+Cl_2underset"hv"overset(C Cl_4)to UNDERSET"Ethelene CHLORIDE"(Cl-CH_2-CH_2-Cl)`
30243.

The Halogen which has the highest electron affinity :

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Fluorine
Chlorine
Bromine
Iodine

Answer :B
30244.

The halogen which gives O_2 with hot conc. NaOH is

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`I_2`
`Br_2`
`Cl_2`
`F_2`

SOLUTION :Only `F_2` GIVES `O_2` with HOT CONC. NAOH
`4NaOH + 2F_2 to 2NaF + 2H_2O + O_2`
30245.

The halogen whichforms maximum number of interhalogen compounds is

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`F_(2)`<BR>`Cl_(2)`
`Br_(2)`
`I_(2)`

Solution :FLUORINE has more electronegtive hence it cane form no. of INTER HALOGEN compounds
30246.

The halogen that undergoes sublimation is

Answer»

`F_(2)` <BR>`Cl_(2)`
`Br_(2)`
`I_(2)`

ANSWER :D
30247.

Which shows same oxidation state in all its compounds with other elements :

Answer»

Hydrogen
Fluorine
Carbon
Oxygen

Answer :B
30248.

The halogen that oxidises water

Answer»

`F_(2)`
`Cl_(2)`
`Br_(2) `
`I_(2)`

Solution :`F_(2) + 3H_(2) O rarr 2H_(3) O_(2)^(+) + 2F^(-) + ( 1)/( 2) O _(2) , F_(2)^(0) rarr F^(-1)`
30249.

The halogen that occurs in liquid state at 25^(@)C

Answer»

`Cl_(2)`<BR>`Br_(2)`
`F_(2)`
`I_(2)`

SOLUTION :`Br_(2)` exhibitliquid state.
30250.

The halogen that is most readily reduced is:

Answer»

Chlorine
Bromine
Iodine
Fluorine

Answer :D