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.

Molarity of a given orthophosphoric acid solution is 3 M. Its normalityis

Answer»

9 N
`0.3` N
3 N
1 N

Solution :MOLARITY of orthophosphoric ACID = 3 M Now, basicity of orthophosphoric acid `(H_(3)PO_(4)) = 3`
`therefore` Normality of orthophosphoric acid = `3 xx 3N = 9N`
Discussion
2.

Molarity of 0.2 N H_(2)SO_(4) is

Answer»

`0.2`
`0.4`
`0.6`
`0.1`

SOLUTION :Normality of ACID = MOLARITY `XX` BASICITY
i.e., Molarity = 0.2/2 = 0.1
Discussion
3.

Molarity of 1m aqueous NaOH solution [density of the solution is 1.02 g/ml]

Answer»

1M
`1.02M`
`1.2M`
`0.98M`

ANSWER :D
Discussion
4.

Molarity of 0.2 N H_(2) SO_(4) is

Answer»

A) 0.2
B) 0.4
C) 0.6
D) 0.1

Solution :NORMALITY of ACID = molarity `XX` BASICITY i.e. 0.2 = molarity `xx` 2:. Molarity = 0.2/2 = 0.1
Discussion
5.

Molarity is related to molality (m) , density of the solution (d) and molar mass of the solute (M^(@)) by which of following relation.

Answer»

`M=(10MD)/(M^(@))`
`M=(1000md)/(1000+mM^(@))`
`M=(100MD)/(100+mM^(@))`
`M=(10md)/(100+mM^(@))`

SOLUTION :`M=(1000md)/((1000+mM^(@)))`
Discussion
6.

Molarity (M), molality (m), and mole (x) are some methods for expressing concentration of solutions. Which of these are temperature independent ?

Answer»

SOLUTION :MOLALITY, MOLE FRACTION
Discussion
7.

Molarity is expressed in gram/litre

Answer»

litre/mol
litre/mol
mol/litre
mol/kg.

Answer :D
Discussion
8.

Molarity is

Answer»

the NUMBER of moles of SOLUTE present in `1 dm^(3)` volume of solution
the number of moles of solute dissolved in 1 kg of solvent
the number of moles of solute dissoved in 1 kg of solution
the number of moles of solute dissolved in 100 `dm^(3)` volume of solution

Solution :Molarity is defined as the number of moles of solute dissolved in one LITRE or one `dm^(3)` volume of solution. It is expressed by the SYMBOL 'M'
`"Molarity (M)"=("Number of moles of solute")/("Volume of solution (in L ordm"^(3)")")`
Discussion
9.

Molar masses of salutes are best measured from osmeticpressure. Why?

Answer»

Solution : Compared to other COLLIGATIVE properties, magnitude of OSMOTIC pressure is large, even for very dilute solutions. Determination of molar masses is particularly useful for polymers as they have poor solubility and for BIOMOLECULES as they are unstable at HIGHER TEMPERATURES.
Discussion
10.

Molar ionic conductivities of a two-bivalent electrolytes x ^(2+) and y ^(2-)are 57 and 73 respectively. The molar conductivity of the solution formed by them will be

Answer»

130 S `cm^(2) MOL^(-1)`
`65 S cm^(2) mol^(-1)`
`260 S cm^(2) mol^(-1)`
`1875 S cm^(2) mol^(-1)`

SOLUTION :`^^_(m)^(oo)=57+73=130S cm^(2) mol ^(-1)`
Discussion
11.

Molar mas sof an ideal gas can be calculated from the relation.

Answer»

`M = ( pd)/( RT )`
`M = ( d RT )/( p ) `
`M = ( RT)/( pd)`
`M = ( DRT )/( pV)`

ANSWER :B
Discussion
12.

Molar ionic conductivities of a bivalent electrolyte are 57 and 73. The molar conductivity of the solution will be

Answer»

`130 S cm^(2) MOL^(-1)`
`65 S cm^(2) mol^(-1)`
`260 S cm^(2) mol^(-1)`
`187 S cm^(2) mol^(-1)`

Answer :A
Discussion
13.

Molar heat of vapourisation of a liquid is 6kJ"mol"^(1) . If the entopy change is 16 J"mol"^(-1)K^(-1), the boiling point of the liquid is :-

Answer»

`375^(@)C`
375 K
273 K
`102^(@)C`

Answer :2
Discussion
14.

Molar heat of vaporization of a liquid is 6kJ mol^(-1). If the entropy change is 16 J mol^(-1) K^(-1), the boiling point of the liquid is

Answer»

`375^(@)C`
375 K
273 K
`102^(@)C`

Solution :`DeltaS=16 J mol^(-1) K^(-1)`
`T_(b.p.)=(DeltaH_("vapour"))/(DeltaS_("vapour"))=(6xx1000)/(16)=375 K`.
Discussion
15.

Molar heat of vaporisation of a liquid is 6 kJ mol^(-1). If the entropy change is 16 J mol^(-1) K^(-1) the boiling point of the liquid is

Answer»

A) `375^(@)C`
B) `375 K`
C) `273 K`
D) `102^(@)C`

Solution :`/_\S = 16 J mol E^(-1) K^(-1)`
`T_(b.p.)=(/_\H_("vapour"))/(/_\S_("vapour")=((6xx1000)/(16))=375k`
Discussion
16.

Molar heat of neutralization of NaOH with HCl in comparison to that of KOH with HNO_(3) is

Answer»

Less
More
Equal
Depends on pressure

Solution :Because both are STRONG ACID and strong BASE.
Discussion
17.

Molar heat capacity of water in equilibrium with the ice at constant pressure is:

Answer»

ZERO
`INFTY`
`40.45 KJ K^-1 mol^-1`
`75.48 JK^-1 mol^-1`

ANSWER :B
Discussion
18.

Molar heat capacity of water in equilibrium with ice at costant pressure is

Answer»

ZERO
`oo`
`40.45kJ K^(-1) MOL^(-1)`
`75.48 JK^(-1)mol^(-1)`

Answer :B
Discussion
19.

Molar heat capacityof water inequilibriumwith ice at canstantpressure is

Answer»

zero
`OO`
`40. 45 kJ K^(-1) "MOL"^(-1)`
`75.48 JK^(-1) "mol"^(-1)`

Solution :`C_(P) = (Q)/(nDeltaT) RARR DeltaT = 0 rArr C_(P) = oo`
Discussion
20.

Molar heat capacity of water at equilibrium with ice at constant pressure is :

Answer»

ZERO
infinity
`40.45 KJ mol^(-1)`
`75.48 kJ mol^(-1)`

Answer :B
Discussion
21.

Molar heat capacity of ethenol is 110.4 JK^(-1) mol^(-1). Its specific heat capacity is

Answer»

2.4
55.2
5.078 kJ
110.4

Solution :Specific heat capacity = `("MOLAR heat capacity")/("Molar mass")`
`= (110.4 JK)/(46)`
`= 2.4 JK^(-1) mol^(-1)`
Discussion
22.

Molar heat capacity of CD_(2)O (deuterated form of formaldehyde) at constant pressure in 9 cal mol^(-1) K^(-1) at 1000K. Calculate the entropy change associated with cooling of 3.2 g of CD_(2)O vapour from 1000 to 900K.

Answer»

`-0.1 "cal"K^(-1)`
`-0.25 "cal"K^(-1)`
`-0.4"cal"K^(-1)`
`-0.15"cal"K^(-1)`

Solution :`DeltaH` for cooling
`CD_(2)O=mSDeltaT=3.2/32xx14xx100=140` cal
`:.` Cooling is exothermic `Dletah=-140` cal
Now `DELTAS=(DeltaH)/T=-140/900=-0.15"cal"K^(-1)`
Discussion
23.

Molar heat capacity of aluminium is 25 JK^(-1)mol^(-1). The heat necessary to raise the temperature of 54 g of aluminium (Atomic mass 27gmol^(-1)) from 30^(@)C to 50^(@)C is

Answer»

1.5 KJ
0.5 kJ
1.0 kJ
2.5 kJ

Solution :`Q=nCDeltaT, n=(54)/(27)=2`
`=2xx25xx20=1000 J`
Q=1.0 kJ.
Discussion
24.

molar heat capacity of Aluminium is 25 JK^(-1)mol^(-1)the heat necessary to raise the temperature of 54 gram of aluminium (atomic mass 27 g mol^(-1))from 30°C to 50°C is

Answer»

1.5kJ
0.5kJ
1.0kJ
2.5kJ

Answer :C
Discussion
25.

Molar heat capacity of aluminium is 25 JK^(-1) mol^(-1). The heat necessary to raise the temperature of 54 g of aluminium (atomic mass 27 g mol^(-1)) from 30^(@)C to 50^(@)C is

Answer»

1.5 kJ
0.5 kJ
1.0 kJ
2.5 kJ

Solution :`Q = nC Delta T`
`N = (54)/(27) = 2 MOL`,
`Delta T = 50 - 30 = 20^(@)`
`Q = 2 mol xx 25 JK^(-1) mol^(-1) xx 20 K`
= 1000 J = 1.0 kJ
Discussion
26.

Molar enthalpy change for vaporisation of 1 mole of water at 1 bar and 100^(@)C is 41 kJ mol^(-1) (If water vapour is assumed to be perfect gas). Find out of the internal energy change. If 1 mole of water is vaporised at 1 bar pressure and 100^(@)C.

Answer»

`+37.904kJmol^(-1)`
`-37.904kJmol^(-1)`
`44.096kJmol^(-1)`
`-44.096kJmol^(-1)`

SOLUTION :`H_(2)O(l)toH_(2)O(G)`
`DeltaH=DeltaU+Deltan_(g)RT`
`DeltaU=41.00" KJ "mol^(-1)-8.3JK^(-1)mol^(-1)xx373K`
`=41.00-3.096" kJ "mol^(-1)=37.904" kJ "mol^(-1)`
Discussion
27.

Molar heat capacity is given by:

Answer»

dQ/dT
dqxdT
sumQ 1/dT
None

Answer :A
Discussion
28.

Molar heat capacity of a gas at constant temperature and pressure is:

Answer»

(3/2) R
(5/2) R
`infty`
DEPENDS upon atomicity of gas

Answer :C
Discussion
29.

Molar heat capacity at constant volume can be given a

Answer»

`C_(V)=(DE)/(DT)`
`C_(V)=(dH)/(dT)`
`C_(p)=(dE)/(dT)`
`C_(p)=(dH)/(dT)`

Solution :MOLAR HEAT capacity at constant VOLUME, `C_(V)=dE//dT`.
Discussion
30.

Molar heat capacity at constant P for substance is equal to:

Answer»

<P>`(deltaE//deltaT)_v`
`(deltaH//deltaT)_v`
`(deltaE//deltaT)_p`
`(deltaH//deltaT)_p`

ANSWER :D
Discussion
31.

Molar conductivity of KCl, NaCl and KNO_(3) are 150, 126 and 109 SC m^(2)mol^(-1) respectively, then what is the molar conductivity of NaNO_(3) ?

Answer»

385 S `cm^(2)mol^(-1)`
133 S `cm^(2)mol^(-1)`
167 S `cm^(2)mol^(-1)`
85 S `cm^(2)mol^(-1)`

Solution :`Lamda_(NaNO_(3))^(@)=Lamda_(NaCl)^(@)+Lamda_(KNO_(3))^(@)-Lamda_(KCL)^(@)`
`=126-109-150`
=85 S `cm^(2)mol^(-1)`
Discussion
32.

Molar conductivity of ionic solution depends on_____.

Answer»

temperature
distance between electrodes
concentration of ELECTROLYSIS in solution
surface AREA of electrodes.

Answer :A::C::D
Discussion
33.

Molar conductivity of ionic solution depends on………. 1) Temperature 2) Distance between electrodes. 3) Concentration of electrolytes in solution. 4) Surface area of electrodes.

Answer»

1 & 2
2 & 3
3 & 4
1 & 4

Answer :C
Discussion
34.

Molar conductivity of ionic solution depends on ______

Answer»

TEMPERATURE
Distance between electrodes
Concentration of ELECTROLYTES in solution
Surface AREA of electrodes.

Solution :On INCREASING the temperature the temperature molar conductivity increases whereas molar conductivity decreases on increasing the concentration.
(A) As temperature increases, molar conductivity increases.
(B) Molar conductivity increases with decrease in concentration.
Discussion
35.

Molar conductivity of an electrolyte is the conductance of all the ions produced by one gram mole of the electrolyte in solution and is denoted as Lambda_(m). Lambda_(m)=(kxx1000)/(c ) Here k is the specific conductance while c is the molar concentration of the electrolyte. The molar conductance of the strong electrolytes at infinite dilution (Lambda_(m)^(oo)) can be obtained graphically by extrapolation while the same for weak electrolytes cannot be obtained graphically. It can be calculated theoretically with the help of Kohrausch's. Law. Lambda_(m(AxBy))^(oo)=xlambda_(m (A^(y+)))^(oo)+ylambda_(m(B^(x)))^(oo) Which of the following solution of KCl will have the maximum value of specific conductance ?

Answer»

1.0 N
0.1 N
`1.0xx10^(-2)" N"`
0.5 N.

Solution :(a) SPECIFIC CONDUCTANCE (k) is directly proportional to the number of ions per unit volume. 1.0 NKCl will GIVE maximum number of ions in solution.
Discussion
36.

Molar conductivity of Ba^(2+) and Cl^(-1) ions present in aqueous solution of BaCl_(2) are 127.32"S"cm^(2)mol^(-1) and 76.34" S "cm^(2)mol^(-1) respectively, then what is Lamda_(m) of BaCl_(2) solutioin ?

Answer»

`280" S "CM^(2)mol^(-1)`
`330.98" S "cm^(2)mol^(-1)`
`90.98" S "cm^(2)mol^(-1)`
`203.6" S "cm^(2)mol^(-1)`

Solution :`Lamda_(m)(BaCl_(2))=lamda(BA^(2+))+2lamda(Cl^(-))`
`=127.32+2xx76.34`
`=280" S "cm^(2)mol^(-1)`.
Discussion
37.

Molar conductivity of an electrolyte is the conductance of all the ions produced by one gram mole of the electrolyte in solution and is denoted as Lambda_(m). Lambda_(m)=(kxx1000)/(c ) Here k is the specific conductance while c is the molar concentration of the electrolyte. The molar conductance of the strong electrolytes at infinite dilution (Lambda_(m)^(oo)) can be obtained graphically by extrapolation while the same for weak electrolytes cannot be obtained graphically. It can be calculated theoretically with the help of Kohrausch's. Law. Lambda_(m(AxBy))^(oo)=xlambda_(m (A^(y+)))^(oo)+ylambda_(m(B^(x)))^(oo) Equivalent conductance of 1 M methanoic acid solution is 10" ohm"^(-1)cm^(2)" equiv"^(-1) and at infinite dilution it is 200" ohym"^(-1)cm^(2)" equiv"^(-1). The pH of methanoic acid solution is :

Answer»

7
3.3
1.3
6.8.

Solution :(c ) `alpha=(Lambda_(m)^(c ))/(Lambda_(m)^(prop))=((10" OHM"^(-1) cm^(2) " equiv"^(-1)))/((200" ohm"^(-1) cm^(2)" equiv"^(-1)))=0.05`
`[H^(+)]= C alpha=1Mxx0.05=0.05 M`
`pH=-LOG [H^(+)]=-log(0.05)`
`=-log(5xx10^(-2))=(2-log 5)`
=2-0.6990=1.3
Discussion
38.

Molar conductivity of an electrolyte is the conductance of all the ions produced by one gram mole of the electrolyte in solution and is denoted as Lambda_(m). Lambda_(m)=(kxx1000)/(c ) Here k is the specific conductance while c is the molar concentration of the electrolyte. The molar conductance of the strong electrolytes at infinite dilution (Lambda_(m)^(oo)) can be obtained graphically by extrapolation while the same for weak electrolytes cannot be obtained graphically. It can be calculated theoretically with the help of Kohrausch's. Law. Lambda_(m(AxBy))^(oo)=xlambda_(m (A^(y+)))^(oo)+ylambda_(m(B^(x)))^(oo) 48250 C of electricity was required to deposit all the copper present in 0.5 L of CuSO_(4) solution using inert electrodes. The molarity of solution was (Assume volume constant).

Answer»

0.50 M
2.50 M
0.25 M
1.0 M

Solution :(a) `Cu^(2+)(aq)+UNDERSET(2F)2e^(-) to underset(63 g)(Cu(s))`
`2xx96500" C"` of charge deposit Cu=63 g
48250 C of charge deposit Cu
`=((63 g)xx(48250" C"))/((2xx96500" C"))=15.75" g"`b
Molarity (M)`=(((15.75 g))/((63.0 g" MOL"^(-1))))/((0.5 L))`
=0.5 mol `L^(-1)`=0.5 M
Discussion
39.

Molar conductivity of a solution is 1.26 xx 10^(2)Omega^(-1) cm^(2) "mol""^(-1). Its molarity is 0.01M. Its specific conductivity will be

Answer»

`1.26xx10^(-5)`
`1.26xx10^(-3)`
`1.26xx10^(-4)`
0.0063

Answer :B
Discussion
40.

Molar conductivity (^^_m) is defined as conducting power of the ions produced by 1 mole of an electrolyte in a solution. ^^_m =(K)/(C) where K is conductivity (in S-cm^(-1)), ^^_m is molar conductivity (in Scm^(2) mol^(-1)) and C is molar concontration (in "mole"//cm^(3)) The molar conductivity of 0.04 M solution of MgCl_2 is 200Scm^2 mol^(-1) at 298 K. A cell with electrodesthat are 2.0cm^(2) in surface area and 0.50 cm apart is filled with MgCl_2 solution. The cell constant is

Answer»

`8 xx 10^(-3) S cm^(-1)`
`32S cm^(-1)`
`0.032 S cm^(-1) `
None of these

SOLUTION :`^^_M = (K xx 1000)/(M) , K= M xx (^^_M)/(1000) = (0.04 xx 200)/(1000) = 8 xx 10^(-3)`
Discussion
41.

Molar conductivity(L) mof aqueous solution of sodium stearate, which behaves as a strong electrolyte, is recorded at varying concentrations (c) of sodium stearate. Which one of the following plots provides the correct representation of micelle formation in the solution? (Critical micelle concentration (CMC) is marked with an arrow in the figures)

Answer»





SOLUTION :Molar conductivity DECREASES with CONCENTRATION till CMC & further decreases rapidly beyond CMC.
Discussion
42.

Molar conductivity increases with dilution. Is the above statement true? Justify your answer

Answer»

Solution :Yes, the above given statement is TRUE.
When the DILUTION increases, the ions are far APART and the attractive forces decrease. At infinite dilution the ions are so far apart, the interaction between them becomes insignigicant and HENCE, the molar conductivity increases and REACHES a maximum value at infinite dilution.
Discussion
43.

Molar conductivities of Li^(-),Na^(+),K^(+) and Rb^(+) ions in aqueous solutions are in the following order.

Answer»

`Li^(+) GT Na^(+) = K^(+) lt Rb^(+)`
`Li^(+) gt Na^(+) gt K^(+) = Rb^(+)`
`Rb^(+) gt K^(+) gt Na^(+) gt Li^(+)`
`Li^(+) gt Rb^(+) gt K^(+) gt Na^(+)`

Solution :The smaller the size of the ion, the greater is the degree of hydration. Therefore, the extent of hydration decreases from `Li^(+) to Rb^(+)`So, hydrated ionic radii decreases in the order: `Li^(+) gt Na^(+) gt K^(+) gt Rb^(+)`
As a result, hydrated Li ion being largest in ionic size, has lowest mobility in AQUEOUS solution. On the other hand, hydrated `Rb^(+)` ion being smallest in size has the HIGHEST mobility in aqueous solution. Higher the ionic mobility, greater is the molar conductivity. Thus, the DECREASING order of molar conductivity is,`Rb^(+) gt K^(+) gt Na^(+) gt Li^(+)`
Discussion
44.

Molar conductivities (wedge_(m)^(infty)) at infinite dilution of NaCl, HCl and CH_(3)COONa are 126.4, 425.9 and 91.0 S cm^(2) mol^(–1) respectively. wedge_(m)^(infty) for CH_(3)COOH will be:

Answer»

`290.8 S CM^(2) MOL^(–1) `
`390.5 S cm^(2) mol^(–1) `
`425.5 S cm^(2) mol^(–1) `
`180.5 S cm^(2) mol^(–1) `

ANSWER :C
Discussion
45.

Molar conductivity decreases with decrease in concentration

Answer»

For STRONG electrolytes
For WEAK electrolytes
Both for strong and weak electrolytes
For NON electrolytes

Answer :D
Discussion
46.

Molarconductivities atinfinitedilution ofMg^(2+) and Br^(-) are105.8 Omega^(-1) cm^(-2) mol^(-1)and 78.2 Omega^(-1) cm^(-2)mol^(-1)respectively . Calculatemolarconductivityat zeroconcentrationof MgBr_(2)

Answer»

<BR>

SOLUTION :`gamma_(Mg^(2+))^(o)= 105.8 Omega^(-1) cm^(2) mol^(-1)`
`gamma_(Br^(-))^(0) = 78.2 Omega^(-1) cm^(2) mol^(-1)`
`A_(o(MgBr_(2))) = gamma_(Mg^(2+))^(0) = 2gamma_(Br^(-))^(0)`
`=105.8 + 2 xx78 .2= 105.8 + 156 .4`
` =262 .2 Omega^(-1) cm^(-2) mol^(-1)`
Discussion
47.

Molar conductivity.

Answer»

Solution :It is defined as a conductance of a VOLUME of the solution containing IONS from one MOLE of an electrolyte when placed between two PARALLEL plate electrodes 1 cm apart and LF large area, sufficient to accommodate the whole solution between them, at constant temperature. It is denoted by `^^_(m)`.
Discussion
48.

Molar conductivities at infinitedilution of KCl, HCl and CH_3COOK are 0.013o and 0.038 and 0.09 S m^(2) mol^(-1) respectively at 291 K . If conductivity of 0.001 M CH_3COOH is2.72 xx 10^(-3) Sm^(-1) then find % degree of ionization of CH_3COOH.

Answer»


Solution :`^^_(M)^(0)[{:(^^_M^(0)+ , ^^_M^(0)- , ^^_M^(0)),(CH_3COOK, HCl , KCl):}]=(0.009 + 0.038 - 0.013)`
`^^_M^(0)(CH_3COH)=0.034 xx 10^(4) = 340S -cm^(2) -"mole"^(-1)`
`^^_M^(0) = (K xx 10^(3))/(M)=(2.72 xx 10^(-3)xx 10^(-2))/(0.001) = 2.72 xx 10`
`^^_M = 2.72 xx 10^(32) = 27.2 , ALPHA = (^^_M)/(^^_M^(0))=(27.2)/(340)=0.08, alpha % = 8%`
Discussion
49.

Molar conductance of electrolytic solution wedge_(m) is

Answer»

`PROP1`
`prop(1//A)`
`prop(1//C)`
`prop(SQRT(C))`

Solution :Molar conductivity or molar CONDUCTANCE
`(wedge_(m))=kappaxxV`
`wedge_(m)=kappa=(1000)/(C_(m))`
where `C_(m)` is molar concentration (mol `l^(-1)`)
`therefore`Molar conductance `(wedge_(m))prop(1)/(C)`.
Discussion
50.

Molar conductance of Al_(2)(SO_(4))_(3) equals

Answer»

EQUIVALENT conductance `XX 1/6`
equivalent conductance
equivalent conductance `xx6`
none of the above

Answer :C
Discussion