Explore topic-wise InterviewSolutions in Current Affairs.

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.

Heat of neutralisation of CH_3COOH_((aq)) with NaOH_((aq)) is - 55.2kJ. What is heat of ionisation of CH_3COOH ?

Answer»


ANSWER :`+2.1KJ MOL^(-1)`
2.

Heat of neutralisation of a strong acid is ____ than that of a weak acid.

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SOLUTION :`DeltaH_("ionisation")^(sigma)=+"42.77 KJ MOL"^(-1)`.
3.

Heat of neutralisation of a weak acid HA by NaOH is -"12.13 kJ mol"^(-1). Calculate the enthalpy of ionization of HA.

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SOLUTION :`DeltaH_("ionisation")^(SIGMA)=+"42.77 kJ mol"^(-1)`.
4.

Heat of neutralisation of a polybasic acid by strong base is -54.8 kcal mol^(-1). What is basicity of the acid?

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SOLUTION :`-54.8= -13.7 X RARR x= 4`
5.

Heat of neutralisation for the reaction NaOH + HCl rarr NaCl + H_(2)O " is " -57.1 kJ "mole"^(-1). The heat relased when 0.25 moles of NaOH is treated with 0.25 moles of HCl is

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22.5 K.J/mole 
57.1 K.J/mole
14.3 K.J/mole 
28.6 K.J/mole 

ANSWER :C
6.

Heat of Hydration of Na^(+) (size 102 Pm) = - 397J Kj mol^(-1) whereas Ca^(2+) (size 100 Pm) = - 1650 kJ mol^(-1). Explain the difference.

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Solution :`Ca^(2+)` is smaller in size then `Na^(+)` and ALSO it has HIGHER charge therefore, its HYDRATION ENERGY is more than that of `Na^(+)`
7.

Heat of formation of water and heats of combustion of ethylene and acetylene are respectively 286, -1410 and -1299 kJ mol^(-1), Calculate the heat of the reaction, C_2H_(2(g)) + H_(2(g)) to C_(2)H_(4(g)).

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ANSWER :`-175 KJ MOL^(-1)`
8.

The reaction which shows standard heat of formation of water correctly is

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`-251.9KJ`
`241.9KJ`
`-241.9KJ`
`301.9KJ`

ANSWER :C
9.

Heat of formation of CO and CO_2 are -26.4, -94.0 kcal/mole respectively. What is the heat of combustion of CO in kcals?

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`+26.4`
`120.6`
`-67.6`
`135.2`

ANSWER :C
10.

Heat of formation of 2 moles of NH_(3)(g) " is" -90 kJl bond energies of H-H and N-H bonds are 435 kJ and 390 kJ mol^(-1) respectively. The value of the bond energy of N -= N is (1000- (x^(2) + x +25)) kJ/mol What is x ?

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Solution :From the data we can say that heat if formation of `NH_(3)` is -45 kJ
So, `(1)/(2) N_(2(G)) + (3)/(2) H_(2(g)) rarr NH_(2(g)), Delta H = - 45kJ`
So, `-45 = (3)/(2) (H - H) + (1)/(2) (N -= N) - 3 (N - H)`
`(3)/(2) (435) + (1)/(2) (N -= N) - 3(390)`
`N -= N = 945` From the given CONDITION
`945 = 1000 - (x^(2) + x + 25)`
So, `x^(2) - x + 25= 55`, On solving x = 6S
11.

Heat of dissociationof aceticacidis 0.30 kcalmol^(-1) . Hence,enthalpy change when 1 molofCa(OH)_(2) is completelyneutralized by aceticacid would be

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`-13.4 kcal`
`-27.1 kcal`
`-26.8 kcal`
`-27.4 kcal`

Solution :`Ca(OH)_(2) rarr Ca^(2+) + 2OH^(-) , i.e., 1` MOLE of`Ca(OH)_(2)` gives two moles of `OH^(-)` ions. HENCE, when `Ca(OH)_(2)` is COMPLETELY neutralized,heat produced `= 2 xx 13.7 kcal = 27.4 kcal`
2 moles of `OH^(-)` ions require 2 moles of `H^(+)` ions for complete neutralization.
Heat of dissociation of acetic acidto produce 2 moles of`H^(+)` ions `= 2 xx 0.30 kcal = 0.6 kcal`.
Hence, net heat produced `=27.4 - 0.6 =26.8 kcal, i.e., DeltaH =- 26.8 kcal`
12.

Heat of combustion of methane is ……….. KJ/ mol .

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`-87.78`
`+87.78`
`-394.55`
`+394.55`

ANSWER :A
13.

Heat of combustion of carbon is -96 k, cals. When some quantity of carbon is burnt in oxygen, 48 k.cals of heat is liberated. What is the volume (in lits) of O_2 at STP reacted with this carbon

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48 
22.4 
11.2

ANSWER :C
14.

Heat of combustion of C_(2)H_(4) is -337 kCal. If 5.6 lit O_(2) is used at STP, in the combustion heat liberated is _____ kCal

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28.08 
14.04 
42.06 
56.16 

ANSWER :A
15.

Heat of combustion of benzoic acid (C_6H_5COOH) at constant volume at 25^@C is -3233KJ//"mole". When 0.5g of benzoic acid is burnt in bomb calorimeter, the temperature of calorimeter increased by 0.53^@C. Now in the same bomb calorimeter 1g of C_2H_6 burnt then temperature increased by 2.04^@C. DeltaHfor combustion of C_2H_6is

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`-1530KJ//"MOLE"`
`-1536.2KJ//"mole"`
`-1522.8KJ//"mole"`
`+1536.2KJ//"mole"`. 

ANSWER :B
16.

Heat of combustion of A(s) is -10 k"cal" mol^(-1) and that of B(s) is -15 Kcal mol^(-1). Mixture containing 3 moles A(s) and x moles B(s) on combustion released 105 kcal of heat. What is the value of x?

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SOLUTION :`A+ O_(2) RARR Delta H_(1) = -10` Kcal
`B + O_(2) rarr Delta H_(2) = -15` Kcal
`3A xx xB = 10^(5)`
`rArr 3(10) + x (15) = 10^(5) rARr x = 5`
17.

Heats of combustion of benzene, carbon and hydrogen are - 782, -94, -68 kcal mol^(-1) respectively. Calculate heat of formation of benzene.

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Solution :`C_(6)H_((s))+O_(2)(g)to 6CO_(2)(g)+3H_(2)O_((l)),DeltaH_(1)=-782k.cal`
`C_((s))+O_(2)(g)to CO_(2)(g),DeltaH_(2)=-94Kcal`
`H_(2)(g)+1/2O_(2)(g)to H_(2)O_((l)),DELTAH=-68Kcal `
The required EQUATION is `6C_((s))+3H_(2)(g)to C_(6)H_(6)(l),DeltaH=?`
Heat of formation of benzene =`DeltaH=-DeltaH_(1)+6DeltaH_(2)+3DeltaH_(2)=+14Kcal`.
18.

Heat of combustion is always

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positive
negative
ZERO
either positive or negative

ANSWER :B
19.

Heat of combustion is always ………….. .

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POSITIVE
Negative
Zero
Either positive or negative

Answer :B
20.

Heat of combustion is always ____

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POSITIVE
NEGATIVE
zero
either positive or negative

ANSWER :B
21.

Heats of atmoisation of ozone and oxygen are 25 units and 30 units respectively. What is the heat of ozonisation of one mole of oxygen is the same units?

Answer»


Solution :`O_(3) rarr 30, Delta H_(1) = 25`
`O_(2) rarr 20, Delta H_(2) = 30`
`O_(2) rarr (2)/(3) O_(3), Delta H`
`Delta H = +30 - (2)/(3) xx 25 = (15)/(3) = 5`
22.

Heat liberated when 100 ml of 1N NaOH is neutralized by 300 ml of 1N HCl ____

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SOLUTION :BASE =`V_1`= 100 ml
`N_1=1N`
Acid = `N_2`=300 ml
`N_2`=1N
Enthalpy of neutralization of 1000 ml =57.3 KJ
`therefore` Enthalpy of neutralization of 100 ml =`(57.3 kJ)/1000xx100`= 5.73 kJ
23.

Heat liberated in the neutralisation of 500 ml of 1N HCl and 500ml of 1N NH_(4)OH is -1.36 K.Cals. The heat of ionisation of NH_(4)OH is

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`10.98`K.Cals
`-12.34` K.Cals
`-10.98`K.Cals
`12.34` K.Cals

ANSWER :A
24.

Heat is considered as a path function, but in an endothermic process , the heat absorbed by the system undercertain specific conditions is independent of path. What are those conditions ?

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Solution :Heat BECOME INDEPENDENT of the path , at constant volume and constant pressure.
(i) At constant Volume :
By I law of THERMODYNAMICS, `DeltaU=Q+w` or `q = DeltaU-W`. Sine `W=-PDeltaV` therefore , `q=DeltaU+PDeltaV` . But as volume remains constant, `DeltaV=0`
`:.qv=DeltaU`
Since `DeltaU` is a state FUNCTION , therefore qv is also state function.
(ii) At constant pressure :
`qp=DeltaU+PDeltaV`
But `DeltaM=DeltaU+PDeltaV`
`DeltaU=DeltaH-PDeltaV`
`q_(p)=DeltaH-PcancelDeltaV+PcancelDeltaV`
`:'qv=DeltaU`
since `DeltaH` is a state function , therefore `q_(p)` is a state function.
25.

Heat has randomisin influence on a system and temperature is the measure of average chaotic motion of particles in the system . Write the mathematical relation which relates these three parameters.

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SOLUTION : `DELTAS =( Q"REV")/( T)`
26.

Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.

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Solution :Heat has randomising influence on a system and TEMPERATURE is the measure of average chaotic motion of particles in the system.
The mathematical relation which relates these three PARAMETERS is `Delta S = (q_("rev") )/( T)`
Here, `Delta S= ` Change in ENTROPY
`q_("rev") =` Heat of reversible REACTION
T = Temperature
27.

Heat capacity (C_(p)) is an extensive property but specific heat (c) is intensive property. What will be the relation between C_(p) " and " c for 1 mole of water ?

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Solution :For water, molar heat CAPACITY, `C_(P) = 18 xx` SPECIFIC heat, C
`C_(P) = 18 xx c` Specific heat
`c = 4.18 Jg^(-1) K^(-1)`
Heat capacity, `C_(P) = 18 xx 4.18 JK^(-1) MOL^(-1)`
`= 753 JK^(-1) mol^(-1)`
28.

Molar heat capacity (C_(p)) of water of constant pressure is 75JK^(-1) mol^(-4). The increase in temperature (in K)of 100g of water when I kJof heat is supplied to it is

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`1.8 xx 10^(-3) MOL`
`3.6 xx 10 ^(-5)` mol
`0.0288 G`
`0.288mg `

ANSWER :D
29.

The specific heat capacity of ___ is maximum.

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SOLUTION :`(DQ)/(DT)`
30.

Heat capacity (C_(p)) in an extensive propertybut specificheat (c )is an intensive property . What will be the relationbetween C_(p)and c for 1 mol of water ?

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Solution :For water, molarheat CAPACITY `(C_(p)) = 18 xx `specific heat
`:.C_(p)= 18 xx c`
But specificheat of water , `c=4.18 J G^(-1) K^(-1)`
`:. `HEATCAPACITY , `C_(p) =18 xx 4.18 JK^(-1) MOL^(-1) = 75.3 JK^(-1) mol^(-1)`
31.

Heat capacity (C_p) is an extensive property but specific heat (c) is an intensive property. What will be the relation between C_p and c for 1 mol of water?

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Solution :Molar heat capacity for water,
`C_p= 18 XX` SPECIFIC heat, c
`C_p= 18 xx` Specific heat
`c= 4.18 "J" g^(-1) "K"^(-1)` (for water)
Heat capacity, `C_(p) = 18 xx 4.18 "JK"^(-1) "MOL"^(-1)`
`= 753 "JK"^(-1) "mol"^(-1)`
32.

Temperature of 1 mole of an ideal gas is increased from 300 K to 310 K under isochoric process. Heat supplied to the gas in this process is Q = 25 R, where R = universal gas constant. What amount of work has to be done by the gas if temperature of the gas decreases from 310 K to 300 K adiabatically?

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16X
`6.25X`
32X
`12.5X`

ANSWER :D
33.

Heat balance on earth surface is maintained by circulation of air in

Answer»

TROPOSPHERE
HYDROSPHERE
LITHOSPHERE
BIOSPHERE

ANSWER :A
34.

He_2 cannot exist as stable molecule. Justify this statement on the basis of bond order.

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Solution :The electronic configuration of `He_2` is `sigma_(1s)^2 sigma^*_1s^2`
BOND order = `1/2(N_b-N_a)` = 1/2(2-2) = 0, hence cannot EXIST
35.

He_(2) does not exist. Explain in terms of LCAO.

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SOLUTION :the COMBINING CAPACITY is ZERO
36.

The size of a microscopic particle is 1 micron and its mass is 6 xx 10^(-13)g. If its position may be measured to within 0.1% of its size, the uncertainity in velocity (in cm^(-1)) is approximately

Answer»

`(10^(-7))/(4PI)`
`(10^(-5))/(4pi)`
`10^(-5)`
`10^(-8)`

Solution :`Delta X = (0.1)/(100) = 10^(-4) cm , Delta x . M Delta v ge (h)/(4pi)`
`Delta V = h/(4pim) xx1/(Deltax) = (10^(-7))/(4pi)`
37.

The ratio of the wavelength of the last line of Balmer series and the last line of Lyman series is :

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`2.5 :1`
`3.5 :1`
`4.5 :1`
`5.5 :1`

SOLUTION :`bar(upsilon) = R [1/(1^2) -1/(n^2)]`
n=2 for `16(ST)` line , n = 3 for `2^(nd)` line
n = 4 for `3^(rd)` line , on SOLVING ratio is `2.5:1`
38.

he product obtained as a result of a reaction of nitrogen with CaC_2 is ........

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`CA(CN)_3`
`CaN_2`
`Ca(CN)_2`
`Ca_3N_2`

Solution :`Ca(CN)_2`
`CaC_2 +N_2 overset(300- 350^@ C)to UNDERSET("Calcium cyanide")(Ca(CN)_2)`
39.

Order of intemolecular attractive forces

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WATER `LT` ALCOHOL `lt` Ether
Ether `lt` Alcohol `lt` Water
Alcohol`lt `Water`lt`Ether
Ether`lt` Water`lt` Alcohol

Answer :B
40.

he figure indicates the energy level diagram of an atom and the origin of six spectral lines in emission (e.g. line no. 5 arises from the transition from level B to A). Which of the following spectral lines will also occur in the absorption spectrum ?

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4,5,6
1,2,3,4,5,6
1,4,6
1,2,3

Solution :SIMILAR TRANSITIONS CN be CAUSED by similar PHOTONS
41.

He atom can be excited to 1s^(1) 2p^(1) by lambda=58.44nm. If lowest excited state for He lies 4857cm^(-1) below the above. Calculate the energy for the lower excitation state.

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ANSWER :`3.3xx10^(-18)J`
42.

HCOOH reacts with conc. H_(2)SO_(4) to produce

Answer»

CO
`CO_(2)`
`NO`
`NO_(2)`

SOLUTION :`HCOOH underset("DEHYDRATION")overset("conc."H_(2)SO_(4))RARR CO + H_(2)O`
43.

HCOOH reacts with conc. H_(2)SO_(4) to give :

Answer»

`CO_(2)`
`CO`
OXALIC acid
acetic acid

Solution :See FORMIC acid
44.

HCN_((aq))+NaOH_((aq))to NaCN_((aq)) + H_(2)O_((1)), Delta H=-12.13 KJ Calculate heat of ionisation of HCN .

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Solution :`H_((aq))^(+)+OH_((aq))^(-) to H_(2)O_((1)),DeltaH=-57.3KJ` ……(1)
`HCN_((aq))+OH_((aq))^(-)to H_(2)O_((1))`,…….(2)
EQUATION (2) -equation (1) GIVES,
`HCN_((aq))to H_((aq))^(-) + CN_((aq))^(-),Delta H=+45.17KJ`
HeatOf ionisation of `HCN=+45.17KJmol^(-1)`.
45.

HCN_((aq)) + NaOH_((aq)) to NaCN_((aq)) + H_2O_((l)) , DetlaH = -12.13 kJ. Calculate heat of ionisation of HCN.

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SOLUTION :`H_((aq))^(+) + OH_((aq))^(-) to H_2O_((l)) , Delta H = -57.3 KJ` ……(1)
`HCN_((aq)) + OH_((aq))^(-) to H_2O_((l)) + CN_((aq))^(-)`,
`DELTAH = -12.13 kJ`.....(2)
Equation (2) – equation (1) gives ,
`HCN_((aq0) to H_((aq))^(-) + CN_((aq))^(-) , DeltaH = +45.17 kJ`
Heat of ionisation of `HCN = +45.17 kJ mol^(-1)`.
46.

HCN _((aq)) + NaOH _((aq)) to NaCN _((aq)) + H _(2) O_((l)) , Delta H =-12.13 KJ.Calculate heat of ionisation of HCN.

Answer»

SOLUTION :`H _((aq)) ^(+) + OH _((aq) ) ^(-) to H _(2) O _((l)) ,DELTA H =-57.3KJ…..(1)`
`HCN_((aq)) + OH _((aq))^(+) to H _(2) O _((l )) + CN _((aq)), ^(-),Delta H =-12.13 KJ…..(2)`
Equation (2) - equation (1) GIVES,
`HCN _((aq)) to H _((aq)) ^(-) + CN _((aq)) ^(-) , Delta H = + 45.17KJ`
Heat of ionisation of `HCN = + 45.17kJ mol ^(-1).`
47.

HClO_4 is a poor conductor in

Answer»

Water
`dil. NH_3`
Acetic ACID
NAOH SOLUTION

Solution :In acetic acid it is a weakelectrolyte
48.

HCl is produced in the stomach which can be neutralised by Mg(OH)_2 in the form of milk of magnesia. How much Mg(OH)_2 is required to neutralise one mole of stomach acid?

Answer»

29.16 g
34.3 g
58.33 g
68.66g

Solution :`Mg(OH)_2+2HCl to MgCl_2 + 2H_2O`
No. of moles of `Mg(OH)_2` REQUIRED for 2 moles of HCL = 1
No. of moles of `Mg(OH)_2` required for 1 mole of HCl = 0.5
MASS of 0.5 mol of `Mg(OH)_2=58.33xx0.5=29.16g`
49.

_______HCl is bad conductor of electricity but _______ HCl is good conductor

Answer»

Soild, AQUEOUS
Aqueous, Solid
Anhydrous, Solid
Anhydrous, aqueous

SOLUTION :HCl IONISES only in squeous solution
50.

HCl gas is covalent and NaCl is an ionic compound. This is because

Answer»

sodium is highly electropositive
HYDROGEN is a non metal
HCl isa gas
ELECTRONEGATIVITY difference H and CL is less than 2.1 .

Solution :A COMPOUND is SAID to be covalent if the electronegativity difference between the combining atoms is less than 2:1.