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.

Benzene can be obtained by heating either benzoic acid with Xor phenol with Y. X and Yare respectively

Answer»

 ZINC DUST and SODA LIME
soda lime and zinc dust
 zinc dust and SODIUM hydroxide
soda lime and copper.

Solution :
Discussion
2.

Benzene can be converted to isopropyl benzene (Cumene) by the following reagent.

Answer»

`CH_(3) CH_(2) CH_(2)` Cl + ANHY. `AlCl_(3)`
`CH_(3)–CH = CH_(2)` + Anhy.` AlCl_(3)`
Both of the above
none of these

Answer :C
Discussion
3.

Benzene can be converted into toluene by

Answer»

KOLBE's REACTION
Sabatier and Sendrens reaction
Reimer-Tiemann reaction
Friedel-Craft's reaction

Answer :D
Discussion
4.

Benzene can be converted into toluene by :

Answer»

WURTZ REACTION
Wurtz FITTING reaction
Friedel Crafts reaction
Kolbe's reaction

Answer :A
Discussion
5.

Benzene (C_(6)H_(6)) and toluene (C_(7)H_(8)) from a nearly ideal solution at 313 K. The vapour pressure of pure benzene and toluene are 160 mm of Hg and 60 mm of Hg respectively. Calculate the partial pressure of benzene and toluene and the total pressure over the following solutions : (iii) containing equal molecules of benznen and toluene.

Answer»


SOLUTION :(III) `n_(B)=N//T_(0), n_(T)=N//N_(0)." Hence, "n_(B)=n_(T)=0.5.""("N = No. of molecules, "N_(0)=" AVOGADRO's No.")`
Discussion
6.

Benzene (C_(6)H_(6)) and toluene (C_(7)H_(8)) from a nearly ideal solution at 313 K. The vapour pressure of pure benzene and toluene are 160 mm of Hg and 60 mm of Hg respectively. Calculate the partial pressure of benzene and toluene and the total pressure over the following solutions : (ii) containing 1 mole of benzene and 4 moles of toluene.

Answer»


SOLUTION :(II) `x_(B)=1//(1+4)=1//5=0.2, x_(T)=0.8.`
Discussion
7.

Benzene can be converted into acetophenone by treating it with :

Answer»

acetone in the presence of HCl
acetyl chloride in the presence of `AlCl_(3)`
METHYL chloride in the presence of `AlCl_(3)`
ACETALDEHYDE in the presence of FE

Answer :C
Discussion
8.

Benzene (C_(6)H_(6)) and toluene (C_(7)H_(8)) from a nearly ideal solution at 313 K. The vapour pressure of pure benzene and toluene are 160 mm of Hg and 60 mm of Hg respectively. Calculate the partial pressure of benzene and toluene and the total pressure over the following solutions : (i) containing equal weights of benzene and toluene.

Answer»


Solution :Mass of benzene = mass of toluene = w g
Molar mass of benzene, `C_(6)H_(6)="78 g mol"^(-1)," Molar mass of toluene, "C_(7)H_(8)="92 g mol"^(-1)`
`therefore"Moles of benzene "=(w)/(78)" and Moles of toluene "=(w)/(92)`
`"Total no. of moles in the solution "=(w)/(78)+(w)/(92)=(92w+78w)/(78xx92)=(170w)/(78xx92)`
`therefore"Moles FRACTION of benzene,"x_(B)=(w//78)/(170w//(78xx92))=(92)/(170)=0.541`
`"and MOLE fraction of toluene, "x_(T)=1-0.541 = 0.459`
Discussion
9.

Benzene (C_6 H_6) and toluene (C_7 H_8) form a nearly ideal solution. At 313 K, the vapour pressure of pure benzene is 150 mm Hg and of pure toluene is 50 mm Hg. Calculate the vapour pressure of a mixture of these two containing their equal masses at 313 K.

Answer»

Solution : Let the mass of both benzene and toluene=Wg
` therefore `Moles of benzene`= (W )/(78) ,`Moles of toluene `=(W)/(92)`
Mole fraction of benzene` =((w)/(78))/((w/(78)+W/(92)))=0.541`
Mole fraction of toluene = 1-0-541 = 0.459
partialvapourpressureof benxzene`(P_A ) =P_(A)^(@)X_A`
`=150 XX 0.541 = 81.15m m`
Particalvapourpressureof toluene `(P_B) =P_(B)^(@)X_B`
`=50 xx 0.459 =22.95 m m`
totalvapourpressureof MIXTURE=`81.15 +22.95= 104.1 m m`
Discussion
10.

Benzene burns according to the following equation 2C_(6)H_(6)(l)+15 O_(2)(g) rarr 12 CO_(2) (g)+6H_(2)O(l)""DeltaH^(@)= -6542 KJ What is the DeltaE^(@) for the combustion of 1.5 mol of benzene

Answer»

`-3271 KJ`
`-9813 kJ`
`-4906.5 kJ`
None of these

Solution :`DELTAN=-3`
`DeltaH^(@)=DeltaE^(@)+(Deltan_(G))RT`
`rArr ""-6542=DeltaE^(@)+(-3)(8.31)(298)xx10^(-3)""rArr""DeltaE^(@)=-6534 kJ`
Discussion
11.

Benzene and toluene from nearly ideal solutions. If at 300k, p_("toluene")^(0)=32.06 mm and p_("benzene")^(0)=103.01mm (a) Calculate the vapour pressure of a solution containing 0.6 molefraction of toluene (b) Calculate the mole fraction of toluene in the vapour for this composition of liquid

Answer»

SOLUTION :(a) We have,
`p=x_(A)*x_(A)^(0)+x_(B)*p_(B)^(0)`……………..(Eqn.3)
Given that `x_("toluene")=0.6`, `x_("BENZENE")=1-0.6=0.4`
`:.p=0.6xx32.06+0.4xx103.01`
`=60.44mm`
Again we have,
`(1)/(p)=(x_(A).)/(p_(A)^(0))+(x_(B)^(.))/(p_(B)^(0))`...........(Eqn.5)
`(1)/(60.44)=(x_("toluene")^(.))(32.06)+(x_("benzene").)/(103.01)`
But `x_("toluene").+x_("benzene").=1`
`:. (1)/(60.44)=(x_("toluene").)/(32.06)+(1-x_("toluene").)/(103.01)`
`:.x_("toluene").=0.3182`
Discussion
12.

Benzene and tolune from nearly ideal solution. At a certain temperature, the vapour pressure os pure benzene is 150 mm Hg and of pure toluene is 50 mm Hg. Calculate the vapour pressure of the solution containing equal weight of benzene andtolune.

Answer»


Solution :Let the mass of benzene and toluene in the solution = W g
`"Molar mass of benzene" = 778 " g MOL"^(-1)`
`"Molar mass of toluene"= 92" g mol"^(-1)`
`"No. of MOLES of benzene"=((Wg))/((78" g mol"^(-1)))=W/78 mol`
`"No. of moles of toluene"=((Wg))/((92" g mol"^(-1)))=W/92mol`
`"MOLE fraction of benzene"=(W/78)/(W/78+W/92)=0.541`
Mole fraction of toluene= 1-0.541=0.459
`"Pratial vapour pressure of benzene"=0.541xx150mm=81.15mm Hg`
`"Partial vapour pressure of toluene"0.459xx50mm=22.95mmHg`
`"Total vapour pressure of solution"=81.15+22.95=104.1 mm Hg`.
Discussion
13.

Benzene and toluene have equal mole fractions in their mutual solution. What do you expect about their mole fraction n the vapour phase at the same temperature? Explain. (Given : p_("Benzene")^(@)=3xxp_("Toluene")^(@))

Answer»

<P>

Solution :`p_(B)//p_(T)=p_(B)^(@)//p_(T)^(@)(because x_(B)=x_(T)).` But `p_(B)^(@)//p_(T)^(@)=3` (GIVEN). Hence, `p_(B)//p_(T)=3.` But `p_(B)//p_(T)=3` implies that `y_(B)//y_(T))=3` in the VAPOUR phase because vapour pressure in the vapour phase are in the ratio of their no. of moles of mole fractions.
Discussion
14.

Benzene and toluene from nearly ideal solutions. If at 300k, p_("toluene")^(0)=32.06 mm and p_("benzene")^(0)=103.01mm (a) A liquid mixture is composed of 3 moles of toluene and 2 moles of benzene. If the pressure over the mixture at 300K is reduced, at what pressure does the first vapour form ? (b) What is the composition of the first trace of vapour formed ? (c ) If the pressure is reduced further, at what pressure does the last trace of liquid disappear? (d) What is the composition of the last trace of liquid ?

Answer»

<P>

SOLUTION :(a) `p=(3)/(5)xx32.06+(2)/(5)xx103.01`…………..(Eqn.3)
`=60.44mm`
(B) `(1)/(60.44)=(x_(t))/(32.06)+((1-x_(1))/(103.01))`…………(Eqn.5)
`x_(t)=0.3181`
(c ) `(1)/(p)=((3)/(5))/(32.06)+((2)/(5))/(103.01)`………….(Eqn5)
`p=44.25mm`
(d) `44.25=x_(t)xx32.06+(1-x_(t))103.01` ...........(Eqn 3)
`x_(t)=0.8281`
Discussion
15.

Benzene and toluene forms nearly an ideal solution. At 300 K, P_("toluene")^(@)=32.06" mm" and P_("benzene")^(@)=103.01" mm" (of Hg) (i) A liquid mixture is conposed of 3 mole of tolune and 2 mole of benzene. If the pressure over the mixture at 300 K is reduced, at what pressure does the first vapour form ? (ii) What is the composition of the first trace of vapour formed ? (iii) If the pressure is reduced futher,at what pressure does the last trace of liquid dissappear ? (iv) What is the composition of last trace of liquid ?

Answer»

(i)50.44mmg
(II)x_("toluene")=0.3181
(iii) P=46.28mmhg
(IV) x_("toluene")=0.852
(i)P=42.44mmhg
(ii)x_("toluene")=0.786
(iii) P=38.23mmhg
(iv) x_("toluene")=0.879
(i) 60.44 mm Hg
(ii) `x_("toluene")=0.3181`
(iii) P=44.25 MMHG
(iv)`x_("toluene")=0.8281`
(i) 74.44 mm Hg
(ii) `x_("toluene")=0.218`
(iii) P=32.33 mmHg
(iv)`x_("toluene")=0.9281`

Solution :(i) 60.44 mm Hg
(ii) `x_("toluene")=0.3181`
(iii) P=44.25 mmHg
(iv)`x_("toluene")=0.8281`
Discussion
16.

Benzene and toluene form nearly ideal solutions . If at 27^(@)C the vapour pressures of pure toluene and pure benzene are 32.06mm and 103.01mm respectively (a) Calculate the vapour pressure of a solution containing 0.60 mole fraction of toluene (b) Calculate the mole of fraction of toluene in vapour for this composition of the liquid

Answer»


ANSWER :(a) `60.44mm` (B) `0.318`
Discussion
17.

Benzene and toluene form nearly ideal solution. At a certain temperature, the vapour pressure of the pure benzene is 150 mm Hg and of pure toluene is 50 mm Hg. For this temperature, calculate the vapour pressure of solution containing equal weights of two substances. Also calculate their composition in the vapour phase.

Answer»


SOLUTION :PROCEED as in Hint to Practice PROBLEM 2(i)
Discussion
18.

Benzene and toluene form nearly ideal solutions. At 20^(@)C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20^(@)C for a solution containing 78 g of benzene and 46 g of toluene in torr is

Answer»

<P>50
25
37.5
53.5

Solution :`X` (benzene)
`=((78)/(78))/((78)/(78)+(46)/(92))=(1)/(1+0.5)=0.667` `p"(benzene)"=p^(@)"(benzene)"XX x"(benzene)"`
`=75xx0.667=50` TORR
Discussion
19.

Benzene and toluene form nearly ideal solution. At 20^(@)C, the vapour pressure of benzene is 75 torr and that of toluene ios 22 torr. The partial vapour pressure of benzene at 20^(@)C for a salution containing 78 g of benzene and 46 g of toluene in torr is :

Answer»

<P>50
25
37.5
53.5

Solution :`P_(B)^(@)=75"TORR"`
`X_(B)=(((78g))/((78gmol^(-1))))/(((78g mol^(-1)))/((78g mol^(-1)))+((46G))/((92gmol^(-1))))`
`P_(B)^(@)=P_(B)^(@)X_(B)=75"torr"xx1/1.5=50 "torr"`
Discussion
20.

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures of pure benzeneand toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Answer»

SOLUTION :Molar mass of benzene `(C_(6)H_(6))="78 g mol"^(-1),"Molar mass of toluene "(C_(6)H_(5)CH_(3))="92 g mol"^(-1)`
`therefore"No. of moles in 80 g of benzene "=("80 g")/("78 g mol"^(-1))="1.026 mole"`
`"No. of moles in 100 g of toluene "=("100 g")/("92 g mol"^(-1))="1.087 mole"`
`therefore"In the solution, mole fraction of benzene "=(1.026)/(1.-26+1.087)=(1.026)/(2.113)=0.486`
mole fraction of toluene `=1-0.486=0.514`
`p_("Benzene")^(@)="50.71mm, "p_("Toluene")^(@)="32.06 mm"`
`"Applying RAOULT's law,"p_("Benzene")=x_("Benzene")xxp_("Benzene")^(@)=0.486xx"50.71 mm = 24.65 mm"`
`""p_("Total")=x_("Toluene")=0.514xx"32.06 mm = 16.48 mm"`
`therefore"Mole fraction of benzene in the vapour phase"=(p_("Benzene"))/(p_("Benzene")+p_("Toluene")) =(24.65)/(24.65+16.48) =(24.65)/(41.13) = 0.60`
Discussion
21.

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Answer»

Solution :MOLAR mass of `(C_(6)H_(5))`
`= 6xx12+6xx1`
`= 78 g mol^(-1)`
Molar mass of toluene `(C_(6)H_(5)CH_(3))`
`= 7xx12+8xx1`
`= 92 g mol^(-1)`
Now, no. of moles present in 80 g of benzene
`= (80)/(78)=1.026` mol
And, no. of moles present in 100 g of toluene
`= (100)/(92)=0.087` mol
Therefore, Mole FRACTION of benzene,
`x_(b)=(1.026)/(1.026+1.087)`
= 0.486
And, mole fraction of toluene,
`x_(t)=1-0.486`
= 0.514
It is given that vapour presure of PURE benzene,
`p_(b)^(0)=50.71` mm Hg
Therefore, partial vapour presure of benzene,
`p_(b)=x_(b)xx p_(b)`
= 24.645 mm Hg
And, partial vapour pressure of toluene,
`p_(1)=x_(t)xx p_(t)`
`= 0.514xx32.06`
= 16.479 mm Hg
Hence, mole fraction of benzene in vapour PHASE is given by :
`=(p_(b))/(p_(b)+p_(t))`
`= (24.645)/(24.645+16.479)`
`= (24.645)/(41.124)`
= 0.599
=0.6
Discussion
22.

Benzene and toluene form an ideal solution. The vapour pressure of benzene and toluene are respectively 75mm and 22mm at 20^(@)C. If the mole fractions of benzene and toluene in vapour are 0.63 and 0.37 respectively, calculate the vapour pressure of the ideal mixture.

Answer»

<P>

Solution :SINCE the mole fractions of the components GIVEN in the problem are of the vapour which is in equilibrium with the ideal mixture, we have,
`(1)/(p)=(x_(A).)/(p_(A)^(0))+(x_(B).)/(p_(B)^(0))` …………..(Eqn.5)
`(1)/(p)=(0.63)/(75)+(0.37)/(22)`
`p=39.65mm`
Discussion
23.

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg, respectively, Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Answer»

<P>

Solution :Molar mass of benzene `(C_6H_6) = 78 g "mol"^(-1)`
Molar mass of toluene `(C_6H_3CH_3) = 92 g "mol"^(-1)`
` therefore ` Number of moles in 80 g of benzene`= (80g)/(78 g"mol"^(-1)) = 1.026` mole
Number of moles in 100 g toluence ` = (100g)/(92 g "mol"^(-1)) = 1.087` mole
Mole FRACTION of benzene ` = (1.026)/(1.026 + 1.087) = (1.026)/(2.113) = 0.486`
Mole fraction of toluene = 1 - 0.486 = 0.514
`p_("benzene")^0 = 50.71 mm, p_("toluence")^0 = 32.06 mm`
`p_("benzene") = x_("benzene") XX p_("benzne")^0 = 0.486 xx 50.71 mm= 24.65 mm`
`p_("toluence")= x_("toluence") xx p_("toulence")^0 = 0.514 xx 32.06mm = 16.48 mm`
Mole fraction of benzene in the vapour phase
` = (p_("benzene"))/(p_("benzene") + p_("toluence")) = (24.65)/(24.65+ 16.48) = (24.65)/(41.13) = 0.60`
Discussion
24.

Benzene and toluene form an ideal solution at room temperature. Which Of the following is not true for this process?

Answer»

`DeltaV_(MIX)=0`
`DeltaH_(mix)=0`
`DeltaS_(mix)for system=0`
`DeltaG_(mix)lt 0`

Answer :C
Discussion
25.

Benzene and naphthalene from an ideal solution at room temperature. For this process, the true statement(s) is(are)

Answer»

`DeltaG` is posiotive
`DeltaS_("SYSTEM")` is poitive
`DeltaS_("SURROUNDINGS") = 0`
`DeltaH=0`

SOLUTION :(a,c,d) are CORRECT options.
Discussion
26.

Benzene and naphthalene form an ideal solution at room temeprature. For this processthe true statement(s) is (are)

Answer»

`DELTAG` is positive
`DeltaS_("SYSTEM")` is positive
`DeltaS_("SURROUNDING")` = 0
`DeltaH=0`

Answer :B::C::D
Discussion
27.

Benzene and naphthalein form an ideal solution at room temperature. For this process, the true statement (s) is (are)

Answer»

`DeltaG` is positive
`DeltaS_(SYS)` is positive
`DeltaS_("SUR")=0`
`DeltaH=0`

ANSWER :B::C::D
Discussion
28.

Benzenamine reacts with benzoyl chloride in presence of aqueous sodium hydroxide to form_____and the reaction is called____reaction.

Answer»

SOLUTION :N-phenylbenzamide or BENZANILIDE, Schotten-Baumann.
Discussion
29.

Benzamide to toluene

Answer»

SOLUTION :
Discussion
30.

Benzamide to benzoic acid

Answer»

SOLUTION :BENZAMIDE to BENZOIC ACID :
Discussion
31.

Benzamide on reaction with POCl_(3) gives

Answer»

ANILINE
chlorobenzene
benzyl amine
benzonitrile.

Solution :`C_(6)H_(5)CONH_(2) underset(-H_(2)O)overset(POCl_(3))(RARR) underset("(Cenzonitrile)")(C_(6)H_(5)CNO)`
Discussion
32.

Benzamide reacts with nitrous acid with the evolution of :

Answer»

`N_(2)`
`CO_(2)`
`NH_(3)`
`O_(2)`.

Solution :
Discussion
33.

Benzamide on reduction with Na and C_(2)H_(5)OH gives:

Answer»

BENZYL alcohol
Benzoic acid
Benzylamine
Benzene.

Solution :
Discussion
34.

Benzamide on reaction with POCl_(3) gives.

Answer»

aniline
chlorobenzene
benzyl amine
benzonitrile

Answer :D
Discussion
35.

Benzamide on reaction with POCl_3 gives :

Answer»

ANILINE
CHLOROBENZENE
BENZYLAMINE
BENZONITRILE

ANSWER :D
Discussion
36.

Benzamide is reacted with sodium metal in ethanol gives

Answer»

ANILINE
BENZYL amine
benzoic acid
Benzaldehyde

Answer :B
Discussion
37.

Benzamide cam be converted into aniline by the action of

Answer»

<P>`Br_2//C Cl_4`
` Br_2//"WATER"`
` Br_2//KOH`
`Br_2//"RED P"`

ANSWER :C
Discussion
38.

Benzamide (A) is prepared by heating benzoic acid with ammonia. It undergoes the following reactions: A underset("reaction 1")overset(Br_(2),KOH, Delta)(to) B underset(273-278K "reaction 2")overset(NaNO_(2),HCl)(to) C underset(Delta "reaction 3")overset(H_(3)PO_(2),H_(2)O)(to) D underset(anhyd AlCl_(3) "reaction 4")overset(CH_(3)Cl)(to) E E on heating with acidic KMnO_(4) gives back benzoic acid. 3. Write the chemical reaction for reaction 2 if temperature is 298K.

Answer»

SOLUTION :
Discussion
39.

Benzalkonium chloride is a:

Answer»

cationic surfactant and antiseptic
ANIONIC surfactant and soluble in most of ORGANIC SOLVENTS
cationic surfactant and INSOLUBLE in most of organic solvents
cationic surfactant and antimalarial

Answer :A
Discussion
40.

Benzalkonium chloride is a .

Answer»

CATIONIC SURFACTANT and antiseptic
anionic surfactant and soluble in most of the organic solvents
cationic surfactant and insoluble is most of the organic solvents
cationic surfactant and ANTIMALARIAL

Answer :A
Discussion
41.

Benzalehyde can be obtained from benzal chloride. Write reactions for obtaining benzal chloride and then benzaldehyde from it.

Answer»

SOLUTION :
Discussion
42.

Benzaldoxime exists in how many forms ?

Answer»

4
2
3
1

Answer :B
Discussion
43.

Benzaldoxime exists in how many forms :

Answer»

1
2
3
4

Answer :B
Discussion
44.

Benzaldelhyde can be converted to styrene by the following reaction.

Answer»

PERKINS REACTION
BENZOIN CONDENSATION
Wittig reaction
Reformatsky reaction

SOLUTION :
Discussion
45.

Benzaldehyde undergoes oxidation and reduction in the presence of

Answer»

`NaHCO_3`
CONCENTRATED NAOH
`Na_2CO_3`
HCL

ANSWER :B
Discussion
46.

Benzaldehyde Undergoes Claisen condensation with another aldehyde to give cinnamaldehyde.The aldehyde is:

Answer»

Formaldehyde
Acetaldehyde
Crotonaldehyde
Propanaldehyde

Answer :B
Discussion
47.

Benzaldehyde undergoesauto-oxidation and reduction in presence of

Answer»

CONC. NAOH
`Na_(2)CO_(3)`
`NaHCO_(3)`
dil. NaOH

ANSWER :A
Discussion
48.

Benzaldehyde to prop-hydroxyphenyl acetic acid

Answer»

SOLUTION :
Discussion
49.

Benzaldehyde to benzophenone

Answer»

SOLUTION :
Discussion
50.

Benzaldehyde reduces Tollens' reagent but not the Fehling's or the Benedict's solution. Explain.

Answer»

Solution :The electron-donating resonance effect (+R-effect) of the benzene ring increases the electron density in the carbonly group of benzaldehyde. This, in turn, increases the electron density in the C-H bond of the aldehyde group. As a result, the C-H bond becomes STRONGER and hence only stronger oxidising agents like Tollens' reagent, `Ag(NH_(3))_(2)^(+)(E_(Ag^(+)//Ag)^(@)=+0.80V)` can oxidise C-H to C-OH toform carboxylic Acids but weaker oxidisinig agents like FEHLING's solution (`Cu^(2+)+` tartrate ion +base) and Benedict (`Cu^(2+)` + CITRATE ion+Base) solution `(E_(Cu^(2+)//Cu^(+))^(@)=0.18V)` fail to oxidise benzaldehyde to benzoic acid . Thus, in general, all these three reagents oxidise aliphatic aldehydes but only tollens' reagent oxidise aromatic aldehydes.
Discussion