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.

Out of AlCl_(3) and NaCl which is more effective in causing coagulation of a negative sol and why ?

Answer»
Discussion
2.

Out of AgF, CaF_(2), BeF_(2), MgF_(2)

Answer»

AgF and `BeF_(2)` are soluble, and `CaF_(2) and MgF_(2)` are INSOLUBLE in water
AgF is soluble, others are insoluble
`CaF_(2), BeF_(2), MgF_(2)` are soluble, AgF is insoluble
all of the above

Answer :A
Discussion
3.

Out of alanine and phenyalanine, ..........is an essential amino acid.

Answer»

SOLUTION :PHENYLALANIE
Discussion
4.

Out of acetophenone and benzophenone, which one will give iodoform test? Write the reaction involved. (The compound should have CH_3CO group to show the iodoform test).

Answer»

SOLUTION :ACETOPHENONE `(C_6H_5 COCH_3)`CONTAINS the group `(CH_3CO)`attached to the carbon atom and hence give iodoform TEST while benzophenone does not CONTAIN this group and hence does not give iodoform test.
Discussion
5.

Out of acetophenone and benzophenone, which gives iodoform test? Write the reaction involved.

Answer»

Solution :Acetophenone `(C_(6)H_(5)COCH_(3))` contains the GROUP `CH_(3)CO` attached to carbon and hence gives iodoform test while benzophenone does not contain this group and hence does not give idoform test.
`underset("Acetopheone")(C_(6)H_(5)COCH_(3))+3I_(2)+4NaOH underset("Iodoform")(CHI_(3)) + underset("Sod. benzoate")(C_(6)H_(5)COONa)+3NaI+3H_(2)O`
`underset("Benzophenone")(C_(6)H_(5)COC_(6)H_(5)) overset(I_(2)//NAOH)to ` No reaction.
Discussion
6.

Out og 300 g substance [decomposes as per 1st order], hou much will remains after 18 hr? (t_(0.5)=3hr)

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4.6 gm
5.6 gm
9.2 gm
6.4 gm

Answer :A
Discussion
7.

Out of 2- chloroethanol and ethanol, which is more acidic and why?

Answer»

Solution :2 - chloroethanol, DUE to `-I` effect of CHLORINE ATOMS.
Discussion
8.

Out of 2-chloroethanol and ethanol which is more acidic and why?

Answer»

Solution :Due to -I-effect of the CL atom, ELECTRON density in the O-H bond decreases. As a result, O-H bond in 2-chloroethanol BECOMES WEAKER than in ethanol and hence 2-chloroethanol is a stronger ACID than ethanol.
Discussion
9.

Out of 1*0 g dioxygen, 1*0 g (atomic) oxygen and 1*0 g ozone, the maximum number of oxygen atoms are contained in

Answer»

`1*0` g of atomic oxygen
`1*0` g of ozone
`1*0` g of oxygen gas
All the CONTAIN same number of atoms.

Solution :Since mass of oxygen is same. Therefore, the number of gm-atom the number of atoms will be same in all the samples
i.e., Atoms of oxygen `= (1xx2)/(32)` in `O_(2)`
`=(1)/(16)` in atomic oxygen `=(1xx3)/(48)` in `O_(3)`
Discussion
10.

Out of 2-chloroethanol and ethanol, which is more acidic and why ?

Answer»

Solution :`to` 2-chloroethanol is more ACIDIC than ETHANOL. This is because of the presence of chlorine atom that exerts the STRONG -I effect that facilitates the release the PROTON making the ethoxide ion more STABLE.
Discussion
11.

Out of 1M glucose solution and 2M glucose solution which has higher boiling point and why ?

Answer»

SOLUTION :2M GLUCOSE solution
Discussion
12.

Out of 1M BaCl_2 and 1M Al_2(SO_4)_3, which will have more osmotic pressure ?

Answer»

Solution :` IM Al_2(SO_4)_3` will have more OSMOTIC pressure but it will produce more IONS in solution per MOLE as compared to 1M `BaCl_2` solution.
Discussion
13.

Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?

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Solution :The ELEVATION in BOILING of a solution is a colligative property. Hence, it depends upon the number of moles of the solute present in the solution. HIGHER the concentration, greater will be elevation in boiling point. Therefore, 2 M GLUCOSE will have higher boiling point than 1 M glucose solution.
Discussion
14.

Out of (1) butane(2) butan -1- ol (3) butanal(4) butanone . The decreasing order of their B.P is

Answer»

`1 GT 2 gt 3 gt 4`
` 2 gt 4 gt 4 gt 1`
` 2 gt 3 gt 4 gt 1`
` 4 gt 2 gt 3 gt 1`

ANSWER :C
Discussion
15.

Out of 1-bromobutane, 2-bromobutane and 2-bromo-2 methylpropane, which has the lowest boiling point ?

Answer»

SOLUTION :2-bromo -2-methylpropane
Discussion
16.

Our of C,N,O and S, the smallest is size of :

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O
N
C
S

Solution :Size DECREASES from C,N andO (same period) and increases from O to S(same GROUP).
Discussion
17.

Other than the X-ray diffactions, how could be the followingpair of isomers be distinguised from one another by ? [Cr(NH_(3))_(6)][Cr(NO_(2))_(6)] and[Cr(NH_(3))_(4)(NO_(2))_(2)] [Cr(NH_(3))_(2)(NO_(2))_(4)]

Answer»

cryoscopic method
measurement of molar CONDUCTANCE
measuring magnetic MOMENTS
OBSERVING their colours

ANSWER :B
Discussion
18.

Other things being equal , the EMF of a Daniel cell may be increased by

Answer»

keeping low temperature
using large COPPER ELECTRODES
using large zinc electrodes
decreasing concentration of `Cu^(2+)` ions .

Solution :For Daniel cell ,
`ZN + Cu^(2+) to Zn^(2+) + Cu` ,
`E_("cell") = E_("cell")^(@) - (RT)/(nF) "ln" ([Zn^(2+)])/([Cu^(2+)])` .
Decrease of temperature and increase of `[Cu^(2+)]` will increase the EMF .
Discussion
19.

Other theories explaining the bonding in coordination compounds are

Answer»

Ligand field theory
Molecular orbital theory
VSEPR theory
Atomic orbital theory

ANSWER :A::B
Discussion
20.

Other forms of iron can be produced from :

Answer»

COST IRON
WROUGHT iron
PIG iron
steel

Answer :C
Discussion
21.

otal number of pi-electrons in benzene is

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2
3
4
6

Answer :D
Discussion
22.

Ostwald's process for the manufacture of HNO_3 involves the

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OXIDATION of `N_(2)" to "NO`
oxidation of `NH_(3)" to "NO` in presence Pt/Rh catalyst
COMBINATION of `N_2 and O_2`
combination of `H_O and N_O_5`

ANSWER :B
Discussion
23.

Ostwald's dilution law is applicable in the case of the solution of:

Answer»

`CH_3COOH`
NaCl
NaOH
`H_2SO_4`

ANSWER :A
Discussion
24.

Ostwald's dilution law is applicable in the case of the solution of

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`CH_3 COOH`
`NA CI`
`NAOH`
`H_2SO_4`

ANSWER :A
Discussion
25.

Ostwald dilution law is expressed as:

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`K_a = (C CDOT ALPHA^2)/(1- alpha)`
`K_a = (C cdot alpha)/(1- alpha)`
`K_a = (1- alpha)/(C cdot alpha^2)`
`K_a=(C(1-alpha))/(alpha^2)`

ANSWER :A
Discussion
26.

Ostwald process is used to prepare -

Answer»

`NH_(3)`
`N_2`
`HNO_2`
NO

Answer :C
Discussion
27.

Ostwald dilution law is applicable to the dissociation of

Answer»

AMMONIUM HYDROXIDE
SULPHURIC acid
hydrochloric acid
sodium chloride

Solution :Thislaw is applicableto thedissociationof weakelectrilyteand ammoniumhydroxideis aweakelectrolyte
Discussion
28.

"Ostwald dilution law" constituents one of the postulates of the "Arrhenius theory of electrolytic dissociation". It is valid for

Answer»

STRONG ELECTROLYTES
WEAK electrolytes
both strong and weak electrolytes
non-electrolytes.

Solution :weak electrolytes
Discussion
29.

Osteomalacia is caused due to deficiency of ………………….

Answer»

VITAMIN D
Vitamin B_(2)`
Vitamin E
Vitamin `B_(6)`.

SOLUTION :Vitamin D
Discussion
30.

OsO_(4) exists but OsF_(8) does not exist. Why?

Answer»

Solution :In both CASES O.S. of OS in +8 but in `OsO_(4)`. Os is bonded to OXYGEN by 4 double bonds whereas in `OsF_(8)`, will have to form 8 SINGLE bonds which require much more space around central atom.
Discussion
31.

OsO_4 when treated with R-CH=CH-R gives

Answer»




NONE of the above.

Solution :
Discussion
32.

Osmotic pressure of the solution can be increased by,

Answer»

INCREASING the temperature of the solution
decreasing the temperature of the solution
increasing the volume of the VESSEL
diluting the solution

Solution : `:pi=CRT= (N)/(V)RT`
`pialphaT`
Discussion
33.

Osmotic pressure of a solution containing 0.1 mole of solute per litre at 273 K is:

Answer»

`(0.1)/1 xx0.08205xx273 ` ATM
`0.1 xx2 xx0.08205xx273 ` atm
`(1)/0.1 xx0.08205xx273 ` atm
`(0.1)/1 XX(273)/0.08205xx273`atm

Answer :A
Discussion
34.

osmotic presure of a sugar solution at 27^(@)C is 2.46 atmosphere. The concentration of the solution in mol per litre is :

Answer»

1 M
0.01 M
0.0125 M
0.1 M

Solution :`PI V = NRT, pi = C R T `
`:. C= pi/(RT)= 2.46/(0.082 xx (27 + 273))=0.1 M`
Discussion
35.

Osmotic pressure of solution (density is 1 g/ml) containing 3g of glucose(molecular weight=180 g) in 60 g of water at 15^(@)C is

Answer»

5.57 ATM
0.65 atm
6.25 atm
0.34 atm

ANSWER :C
Discussion
36.

Osmotic pressure of naphthalene solution at 27^@C is 500 mm Hg. When thesolution is diluted and temperature is raised to 47^@C, osmotic pressure of the solutionbecomes 105 mm Hg. The extent of dilution approximately is :-

Answer»

3 TIMES
5 times
7 times
4 times

SOLUTION :`pi_1/pi_2 =(C_1T_1)/(C_2T_2)`
`500/105 = (n_1 /V_1 xx300)/(n_1/V_2 xx 320)`
`500/105=V_2/V_1 xx 300/320`
`500/105xx320/300 =V_2/V_1`
`5 = 320/63=V_2/V_1`
Discussion
37.

Osmotic pressure of insulin solution at 298K is found to be 0.0072 atm. Hence height of water column due to this osmotic pressure is (density of Hg = 13.6 g cm^(-3))

Answer»

`0.76mm`
`70.28mm`
`74.42mm`
`760MM`

SOLUTION :Pressure = hdg
`P = hdg` (for Hg)
1 atm = 76 MM of Hg
`:. 0.0072` atm `-= 0.0072 xx 760mm` of Hg
`P = 0.0072 xx 760 xx 13.6 xx 9.8` (for Hg)
`P = h xx 1 xx 9.8` (for water) (density of water `= 1g//cm^(3))`
`:. 9.8 xx h = 0.0072 xx 60 xx 13.6 xx 9.8`
`h = 0.0072 xx 760 xx 13.6mm`
`= 74.42 mm`
Discussion
38.

Osmotic pressure of non-aqueous solution is measured by:

Answer»

BERKELEY and HARTLEY METHOD
Pfeffer's method
MORSE and Frazer method
Townend's method

Answer :D
Discussion
39.

Osmotic pressure of insulin solution at 298 K is found to be 0.0072 atm. Hence, height of the water column due to this pressure will be (Given density of Hg = "13.6 g mL"^(-1))

Answer»

7.4 mm
7.4 CM
74 cm
760 mm

Solution :p=hdg
But P = 0.0072 ATM. Hence, `h=0.0072xx76cm` of Hg column, d = density of `Hg="13.6 g cm"^(-3), g ="981 cm s"^(_2)`
Hence, `P=0.0072xx76xx13.6xx981"(for Hg column)"`
For WATER, `d="1 g cm"^(-3)`. Hence, for water column
`P=hxx1xx981`
Thus, `hxx981=0.0072xx75xx13.6xx981`
`"or"h=7.4 cm`
Discussion
40.

Osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose that will be isotonic with blood is …….wt./vol.

Answer»

`5.41%`
`3.54%`
`4.53%`
`53.4%`

ANSWER :A
Discussion
41.

Osmotic pressure of aqueous solution of urea is 500 mm at 283 K. If its temperature is maintained at 298 K, then on how much dilution its osmotic pressure would be 105.3 mm ?

Answer»

10 times
5 times
2.5 times
4 times

Solution :`(pi_(1))/(pi_(2))=(T_(1))/(V_(1))xx(V_(2))/(T_(2))`
`THEREFORE (500//760)/(105.3//760)=(283xxV_(2))/(V_(1)xx298) therefore (V_(2))/(V_(1))=5`
Discussion
42.

Osmotic pressure of a urea solution at 10^(@)C is 500mm. Osmotic pressure of the solution become 105.3 mm. When it is diluted and temperature raised to 25^(@)C. The extent of dilution is

Answer»

6 Times
5 Times
7 Times
4 Times

SOLUTION :`pi V = nRT`
`(500 V_(1))/(105.3V_(2))=(nRxx283)/(nR xx298), (V_(1))/(V_(2))=(1)/(5)` so `V_(2)=5V_(1)`
Discussion
43.

Osmotic pressure of a sugar solution at 24^(@)C is 2.5 atm. The concentration of the solution in gm mole per litre is :

Answer»

10.25
1.025
102.5
0.1025

Solution :`pi=CRT`
or `C= pi//RT=(2.5)/(0.082xx 297)=0.1025` M
Discussion
44.

Osmotic pressure of a sugar solution at 24^@C is 2.5 atmosphere. The concentration of the solution in mol per litre is:

Answer»

10.25
1.025
1025
0.1025

Answer :D
Discussion
45.

Osmotic pressure of a solution is

Answer»

directly PROPORTIONAL to MOLAR concentration of the solution
inversely proportional to the molecular WEIGHT of the solute
inversely proportional to the temperature
directly proportional to the volume of the solution

Answer :A::B
Discussion
46.

Osmotic pressure of a solution is 0.0821 atm at temperature of 300 K. The concentration of solution in mol//litre will be

Answer»

0.33
0.066
`0.3times10^(-2)`
3

Answer :C
Discussion
47.

Osmotic pressure of a solution containing 7 g of a protein per 100 mL of solution is 25 mm Hg at 310 K. Calculate the mass of the protein. (R=0.0821L atm K^(-1)mol^(-1))

Answer»


SOLUTION :`W_(B)=7g, V=100mL=0.1 L, pi=25" mm of Hg"=25/760=0.032895 atm`
`T=310 K, R=0.0821" L atm K"^(-1)mol^(-1), M_(B)=?`
` M_(B)=(W_(B)xxRxxT)/(pixxV)=((7g)xx(0.021"L atm K"^(-1)mol^(-1))xx(310 K))/((0.032895" atm")xx(0.1L))=54159.29" g mol"^(-1)`
Discussion
48.

Osmotic pressure of a solution containing 7 g of a protein present in deciliter of a solutionis 3.3xx10^(-2) bar at 37^(@)C. Calculate the molar mass of protein.

Answer»

SOLUTION :`54579" G "MOL^(-1)`
Discussion
49.

Osmoticpressureof a solutioncontaining6.8 xx 10^(-3)kg of aproteinper 1xx 10^(-4)m^(3) of solutionis 3.02 xx 10^(3)Pa at37^(@) C Calculatethe molarmassof protein (R= 8.314J K^(-1) mol^(-1))

Answer»

Solution :58. 06 kg `mol^(-1)`
Discussion
50.

Osmotic pressure of a solutioncontaining3.5g of dissolved protein per50 cc ofa solution is 25 mm (Hg) at 37^(@)C. Calculate the molar mass of protein.

Answer»


SOLUTION :`M_(B) = (w_(B) RT)/(piV) =(3.5 xx 0.0821 xx 310)/(25 //760xx 50 // 100)`
` = 54159.7`
Discussion