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30901.

The DeltaH_(t)^(@) for CO_(2)(g), CO(g) and H_(2)O (g) are -393.5, -110.5 and -241.8 kJ mol^(-1) respectively. The standard enthalpy change (in kJ) for the reaction. CO_(2)(g) + H_(2)(g) to CO(g) + H_(2)O(g) is

Answer»

524.1
41.2
`-262.5`
`-41.2`

ANSWER :B
30902.

the DeltaH_(t)^(@)for CO_(2)(g) , CO(g) and H_(2)O(g) are -393.5 , -110.5 and -241.8 kJ mol ^(-1) respectively, the standed enthalpy change ( in kJ) for the reaction CO_(2)(g)+ H_(2)(g)to CO(g) + H_(2)O(g) is

Answer»

524. 1
41.2
`-262.5`
`-412`

SOLUTION :`C+O_(2)toCO_(2)(G),DeltaH_(t)^(@)=-393.5 kj mol^(-1)`
`C=1/2O_(2)toCO(g),DeltaH_(f)^(@)=-110.5 kJ mol^(-1)`
` H_(2)+1/2O_(2)to H_(2)O(g), DeltaH_(f)^(@)=-241.8 kJ mol^(-1)`
Eq.(ii) + Eq.(III) -Eq.(i)
` Co_(2)(g)+H_(2)Oto CO(g)+ H_(2)O(g), DeltaH_(r)=412.2 kJ`
30903.

The DeltaH_(f)^(@) of O_(3),CO_(3),NH_(3) and HI are 142.2, -393.3, -46.2 and +25.9 kJ per mol, respectively. The order of their increasing stabilities will be

Answer»

`O_(3),CO_(2),NH_(3),HI`
`CO_(2),NH_(3),HI,O_(3)`
`O_(3),HI,NH_(3),O_(2)`
`NH_(3),HI,CO_(2),O_(3)`

Solution :Energy absorbed `PROP` STABILITY of COMPOUND
Energy released `prop` stability of compound
`142.2 gt 25.9 gt -46.2 gt -393.2`
i.e. `O_(3) gt HI gt NH_(3) gt CO_(2)`
30904.

The DeltaH_(f)^(@) for CO_(2)(g), CO(g) and H_(2)O(g) are - 393.5, - 110.5 and - 241.8 kJ mol^(-1) respectively. The standard enthalpy change (in kJ) for the reaction CO_(2)(g)+H_(2)(g)rarrCO(g)+H_(2)O(g) is

Answer»

524.1
41.2
`- 262.5`
`- 41.2`

SOLUTION :`DeltaH=SigmaDeltaH, "(Product)"-SigmaDeltaH`,
`"(REACTANT)"=(-110.5-241.08)-(-393.5)=41.2 KJ mol^(-1)`.
30905.

The DeltaH^(0), DeltaG^(0) and DeltaS^(0) values for the reaction 2H_(2)O_(2)(l) to 2H_(2)O(l)+O_(2)(g) at 25^(@)C are - 196.0 KJ//"mole"-233.6 kJ//"mole" and +125.6 J//"mole". K respectively . Is there any temperature at which H_(2)O_(2)(l) is stable at 1 atm ? Assume that DeltaH and DeltaS values do not change with temperature.

Answer»


ANSWER :(THEORETICALLY at -1586 K)
30906.

The DeltaH values for reactions of ,C(s) + 1/2 O_2 (g) to CO(g) DeltaH =-100 KJCO(g) + 1/2 O_2 (g) to CO_2 (g) DeltaH = 200 KJThe heat of reaction for C(s) + O_2 (g) to CO_2 (g)is

Answer»

`-300 KJ`
`-100 KJ`
`-150 KJ`
`-50 KJ`

Answer :A
30907.

The DeltaH value of the reaction H_(2) + Cl_(2) hArr 2 HCl is -44.12 kcal . If E_(1) is the activation energy of the products , then for the above reaction

Answer»

`E_(1) gt E_(2)`
`E_(1) LT E_(2)`
`E_(1) = E_(2)`
`DeltaH` is not related to `E_(1)` and `E_(2)`

Solution :Because REACTION is exothermic .
30908.

The DeltaH^(@), of thermal decomposition of BeCO_(3), MgCO_(3), SrCO_(3) and BaCO_(3) are in the order:

Answer»

`BeCO_(3) LT MgCO_(3) lt CaCO_(3) lt SrCO_(3) lt BaCO_(3)`
`BaCO_(3) lt SrCO_(3) lt CaCO_(3) lt MgCO_(3) lt BeCO_(3)`
`BeCO_(3) lt BaSO_(3) lt CaCO_(3) lt SrCO_(3) lt MgCO_(3)`
can't be predicted.

Answer :A
30909.

The DeltaH and DeltaS for a reaction at one atmospheric pressure are +30.558 kJ and 0.066 kJk^(-1) respectively. The temperature at which the free energy change will be zero and below of this temperature the nature of reaction would be

Answer»

483 K, spontaneous
443 K, non-spontaneous
443 K, spontaneous
463 K, spontaneous

SOLUTION :`DeltaG=DeltaH-TDeltaS`
`0=+30.558-Txx0.066`
`or T=(30.558)/(0.066)=463 K`
If `(dG)_(T,P)=0` sign '=' mean. It is REVERSIBLE PROCESS.
30910.

The Delta_(f) H^(@) (N_(2)O_(5), g) in kJ/mol on the basic of the following data is: {:(2NO(g)+O_(2)(g) rarr 2NO_(2)(g),,Delta_(r)H^(@)= -114" kJ/mol"),(4NO_(2)(g)+O_(2)(g) rarr 2N_(2)O_(5)(g),,Delta_(r)H^(@)= -102.6" kJ/mol"),(Delta_(f)H^(@) (NO, g)=90.2" kJ/mol",,):}

Answer»

`15.1`
`30.2`
`-36.2`
NONE of these

Solution :`{:(1/2 N_(2)(g)+1/2 O_(2)(g) rarr NO(g),Delta_(f)H^(@)=90.2,),(N_(2)(g)+O_(2) rarr 2NO(g),Delta_(R)H^(@)=90.2xx2,...(1)),(2NO(g)+O_(2)(g) rarr 2NO_(2)(g),Delta_(r)H^(@)=-114,...(2)),(2NO_(2)(g)+1/2 O_(2) (g)rarr N_(2)O_(5) (g),Delta_(r)H^(@)=(-102.6)/(2)=-51.3,...(3)):}`
Addition of (1), (2) and (3) EQUATION
`N_(2)(g)+5/2O_(2)(g) rarr N_(2)O(g)""Delta_(f)H^(@) (N_(2)O_(5), g)=15.1 kJ//mol`
30911.

The Delta H^(theta) for the reaction 4S(s) + 6O_(2)(g) rarr 4 SO_(3)(g) is -1583.2kJ. Standard enthalpy of formation of sulphur trioxide is:

Answer»

`-3166.4kJ`
`3166.4kJ`
`-395.8kJ`
395.8kJ

Solution :`Delta_(F)H = (-1583.2)/(4) = -395.8 KJ`
30912.

The dehydromination of 2-bromobutane gives CH_(3)CH=CHCH_(3). The product is

Answer»

HOFMANN product
SAYTZEFF product
Hofmann-Saytzeff product
Markownikoff product

Solution :`CH_(3)-UNDERSET(Br)underset(|)(C)H-CH_(2)-CH_(3)+underset((alc))(KOH)overset("Saytzeff's rule")toCH_(3)-CH=CH-CH_(3)+KBR +H_(2)O`
30913.

The dehydro halogenation of neopentyle bromide with alco.KOH mainly given:

Answer»

2-methy1-1butane
2-methyl-2-butene
2,2-dimethy1-1-butene
2-butene

Answer :B
30914.

The dehydrohalogenation of neopentyl bromide with alcoholic KOH mainly gives

Answer»

2-METHYL-1-butene
2-methyl-2-butene
2,2-dimethyl-1-butene
2-butene

Solution :`CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-CH_2-Br+underset"(ALC)""KOH"to CH_3-overset(CH_3)overset|C=CH-CH_3+KBr +H_2O`
In this reaction `1^@` carbonium ion is FORMED which rearranges to form `3^@` carbonium ion from which base obstruct proton Hence 2-methyl-2-butene is formed as a main product. `underset(1^@ "carbonium less soluble")(CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-overset+(CH_2) )overset"Methyl SHIFT"to CH_3-underset+oversetoverset(CH_3)(|)C-CH_2-CH_3 overset("Elimination of proton from" beta "carbon which is less hydrogenated")to underset"2-Methyl-2-Butene"(CH_3-oversetoverset(CH_3)(|)C=CH-CH_3)`
30915.

The dehydration yield of cyclohexanol of cyclohexene is 75%. The yeild obtained when 100 g of cyclohexanol is dehydrated will be

Answer»

82.35 g
61.76 g
38.34 g
17.65 g

Solution :
102 g of cyclohexanol GIVES cyclohexene = 84 g
`THEREFORE " 100 g of cyclohexanol will give cyclohexene"`
`=(84)/(102)xx100g=82.35g`
This is the theoretical yield. As ACTUAL yield is 75%`,
therefore, actual yield `=(75)/(100)xx82.35g=61.76g`
30916.

The dehydration reaction is represented by :

Answer»

<P>(II)(III)(S)
(III)(i)(P)
(IV)(i)(Q)
(I)(iii)(R)

SOLUTION :`Ph-undersetunderset(OH)|oversetoverset(Ph)|C-undersetunderset(OH)|oversetoverset(Ph)|C-PhundersetDeltaoverset((iii)Conc.H_2SO_4)toPh-oversetoverset(O)(||)C-undersetunderset(Ph)|oversetoverset(Ph)|C-Ph`
30917.

The dehydration of neo=pentanol gives mainly:

Answer»




None

Answer :C
30918.

The dehydration of neopentanol gives mainly:

Answer»




None

Answer :C
30919.

The dehydration of 2-methyl butanol with conc. H_(2)SO_(4) gives

Answer»

2-METHYL butene as major product
Pentene
2-methyl but-2-ene as major product
2-methyl pent-2-ene

Solution :`underset("2 Methyl BUTANOL")(H_(3)C-CH_(2)-underset(CH_(3))underset(|)CH-CH_(2)-OH )underset("dehydration")overset("CONC." H_(2)SO_(4))to underset({:(2-"Methyl butene"),("Major product"):})(H_(3)C-H_(2)C-underset(CH_(3))underset(|)C=CH_(2))`
30920.

The dehalogenation of vicinal dihalides with zinc dust gives :

Answer»

ALKENES
ALCOHOLS
ALKANES
ALKYNES

ANSWER :A
30921.

The dehydration of alcohol is an example of …………………….. .

Answer»

Bimolecular ELIMINATION REACTION
Nucleophilic SUBSTITUTION reaction
Uimolecular elimination reaction
Internal substitution reaction

SOLUTION :Uimolecular elimination reaction
30922.

The dehydration of alcohols to form ethers in presence of concentrated acid follows:

Answer»

UNIMOLECULAR NUCLEOPHILIC SUBSTITUTION REACTION.
Biomolecular nucleophilic substitution reaction
Unimolecular ELIMINATION reaction
Bimolecular elimination reaction

Solution :Biomolecular nucleophilic substitution reaction
30923.

The degreen of crystallinity of which of the following is highest

Answer»

Atactic polyvinylchloride
Isotactic polyvinylchloride
Syndiotactic polyvinylchloride
All of these

Solution :Syndiotactic polyvinylchloride
`(-CH_(2)-UNDERSET(Cl)underset(|)(Cl)-CH_(2)-overset(Cl)overset(|)(C)H-CH_(2)-underset(Cl)underset(|)(C)H-CH_2-overset(Cl)overset(|)(C)H-CH_(2)-underset(Cl)underset(|)(CH)-)_(n)`
In this arrangementthe chlorine ATOMS are alternately arranged. the polymer is stereoregular and has high crystallinity.
30924.

The degree of polarity of a covalent bond is given by the dipole moment (mu). Identify the incorrect order of dipole moment in the following.

Answer»

`NH_(3)gtNF_(3)`
`CH_(3)FltCH_(3)CL`
`HFgtHCl`

SOLUTION :
30925.

The degree of unsaturation of following compound C_(8)H_(12)O, C_(3)H_(5)N, C_(4)H_(8)O are respectively :

Answer»

`4,3,2`
`3,2,1`
`2,1,3`
`2,2,3`

Solution :DEGREE of unsaturation `= ("No. of CARBON atom +1")-("No. of H-atom + No. of HALOGEN atom-No. of NITROGEN atom")//2`.
30926.

The degree of ionization of HF in 1.00 m aqueous solution is (freezing = 1.86^(@)C) is :

Answer»

0.06
0.12
0.03
0.09

Solution :`DeltaT_(F)=ixxK_(f)xxm`
`i=(DeltaT_(f))/(K_(f)xxm)=(0.197)/(1.86xx0.1)=1.059`
For dissociation
`ALPHA=((i-1))/((n-1))=((1.059-1))/((2-1))-0.059`
`=5.9%~~6%`
30927.

The degree of ionization of 0.2 m weak acid HX is 0.3. If the K_(f) for water = 1.85. What will be the freezing point of solution ?

Answer»

`-0.360^(@)C`
`-0.206^(@)C`
`+0.480^(@)C`
`-0.480^(@)C`

Solution :`{:(,HX,hArr,H^(+),+,X^(-)),("INITIAL mole","1mole",,0,,0),("Final mole",(1-0.3),,0.3,,0.3):}`
Total mole `= (1-0.3)+0.3+0.3=1.3`
`i=(1.3)/(1)=1.3`
`Delta T_(f)=ixxmxxK_(f)`
`= 1.3xx1.85xx0.2`
`= 0.480^(@)C`
`THEREFORE` Freezing point `= 0^(@)C-0.480^(@)C`
`= -0.480^(@)C`
30928.

The degree of ionization increases

Answer»

with increase in concentration of the SOLUTION
on ADDITION of excess of water to the solution
on DECREASING the temperature of the solution
on STIRRING the solution vigorousoly.

Solution :DEGREE of ionization increases with dilution .
30929.

The degree of hydrolysis of a salt of weak acid and weak base in its 0.1 M solutionis 0.2 M, the precentage hydrolysis of the salt should be

Answer»

`50%`
`35%`
`75%`
`100%`

Solution :The degree of HYDROLYSIS of a salt of weak ACID and weak BASE is independent of CONCENTRATION of salt.
30930.

The degree of hydrolysis of a salt of weak acid and strong base is = 0.5. The equation to be used to calculate the accurate value of the degree of hydrolysis (h) is

Answer»

`h=sqrt(K_(W)/(K_(a)*K_(b)))`
`h=sqrt(K_(w)/(K_(a)*C))`
`h=sqrt(K_(w)/(K_(b)*C))`
NONE of these

Answer :D
30931.

The degree of hydrolysis in hydrolic equilibrium A^(-) + H_(2)O hArr HA + OH^(-) at salt concentration of 0.001 M is (K_(1) = 1 xx 10^(-5))

Answer»

`1 xx 10^(-3)`
`1 xx 10^(-4)`
`5 xx 10^(-4)`
`1 xx 10^(-6)`

SOLUTION :`K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(1 xx 10^(-5)) = 10^(-9)`
`K_(h) = alpha^(2)C, alpha = sqrt((K_(h))/(C)) = sqrt((1 xx 10^(-9))/(0.001)) = 1 xx 10^(-3)`.
30932.

The degree of hardness of water is usually expressedin terms of

Answer»

ppm by WEIGHT of `MgSO_4`
g/L of `CaCO_3` and `MgCO_3` PRESENT
ppm by weight of `CaCO_3` IRRESPECTIVE of whetherit is actually present.
ppm of `MgCO_3` actually present in water.

Answer :C
30933.

The degreeof hardness of water is usually expressed in terms of

Answer»

ppm by weightof `mgSO_(4)`
g/L of `CaCO_(3)` and `MgCO_(3)` present
ppm by weight of `CaCO_(3)` IRRESPECTIVE ofwhether it is actually present
ppm of `CaCO_(3)`presentin water

Answer :A
30934.

The degree of halode in isobutyl chloride is :

Answer»

`4^(@)`
`2^(@)`
`3^(@)`
`1^(@)`

ANSWER :D
30935.

The degree of dissociationof water is 1.8 xx10^(-9) at 298 K . Calculatetheionization constant and Ionic product of water at 298 K .

Answer»

SOLUTION :`1XX10^(-14)`
30936.

The degree of dissociation ofPCl_5 (alpha) obeying the equiliberium, PCl_5 ⇌ PCl_3 + Cl_2is approximately realted to the pressure at equilibrium by :

Answer»

<P>``alpha PROP P`
`alpha prop 1/sqrtP`
`alpha prop 1/P^2`
`alpha prop 1 / P^4`

Answer :B
30937.

The degree of dissociation of water at 25^(@)C is 1.9 xx 10^(-7)% and density of 1.0 gm^(-3). The ionic constant for water is

Answer»

`1.0 xx 10^(-14)`
`2.0 xx 10^(-16)`
`1.0 xx 10^(-16)`
`1.0 xx 10^(-8)`

Solution :For water `K_(w) = ([H^(+)][OH^(-)])/([H_(2)O]), K = 1.9 xx 10^(-9), C = (100)/(18)`
`K = 1.9 xx 10^(-9) xx 1.9 xx10^(-9) xx 1000/18`
`= 2.0 xx 10^(-16)`
30938.

The degree of dissociation of water in a 0.1 M aqueous solution of HCl ata certain temperature t^(@)C is 3.6 xx 10^(-15). The temperature t must be : [density of water at t^(@)C = 1 gm//mL.]

Answer»

`lt 25^(@)C`
`= 25^(@)C`
`GT 25^(@)C`
INSUFFICIENT data to predict

Solution :`K_(W) = 55.5 x 3.6 XX 10^(-15) xx 0.1 = 2 xx 10^(-14)`
Hence temperature MUST be `gt 25^(@)C`.
30939.

The degree of dissociation of Ca(NO_(3))_(2) in a dilute aqueous solution containing 7g of the salt per 100g of water at 100^(@)C is 70 per cent. If the vapour pressure of water at 100^(@)C is 760mm, calculate the vapour pressure of the solution.

Answer»

Solution :`{:(1,,,,,,,,),(Ca(NO_(3))_(2),,to,,Ca^(2+),,+,,2NO_(3)^(-)),(1-x,,,,x,,,,2x):}`
van.t Hoff factor `i=(1-x+x+2x)/(1)=(1+2x)/(1)=1+2xx0.7=2.4`
Now, `i=("observed lowering of VAPOUR PRESSURE")/("normal lowering of vapour pressure")`
`:.` obs. lowering of vap. PRESS. `=ixx` nor.lowering of vap.press.
`p^(0)-p_(obs.)=ixx{(n)/(n+N).p^(0)}`
`760-p_(obs)=2.4xx(7//164)/((7)/(164)+(100)/(18))xx760` `{{:(Ca(NO_(3))_(2)=164),(H_(2)O=18):}}`
`=13.90`
`:.p_(obs.)=760-13.90=746.10mm`
30940.

The degree of dissociation of N_2O_4 according to the equation N_2O_4 iff 2NO_2 at 70^@C and 1 atm is 65.6%. Calculate the apparent mol. wt. of N_2O_4under the given conditions.

Answer»


ANSWER :55.56
30941.

The degree of dissociation of dinitrogen tetroxide N_(2)O_(4) (g) to 2NO_(2)(g) atature T and total pressure P is alpha. Which one of the following is the correct expression for the equilibrium constant (K_(p)) at this temperature ?

Answer»

<P>`(2 ALPHA)/((1-alpha^(2)))`
`(alpha^(2)P)/(1-alpha)`
`(4alpha^(2))/((1-alpha^(2)))`
`(4alpha^(2)P)/((1-alpha^(2)))`

Solution :`underset(1-alpha)(N_(2)O_(4)) hArr underset(2ALPHA)(2NO_(2))`
Total moles at equ `=1-alpha+2alpha=1+alpha`
`K_(e )=(P_(NO_(2))^(2))/(P_(N_(2)O_(4)))=(((2alpha)/(1+alpha)P)^(2))/((1-alpha)/(1+alpha)P)=(4alpha^(2)P)/(1-alpha^(2))`
30942.

The degree of dissociation of dinitrogen tetroxide N_(2)O_(4)(g)rarr 2NO_(2)(g)at temperature T and total pressure P is alpha. Which one of the following expressions is correct for the equilibrium constant at this temperature ?

Answer»

<P>`(2ALPHA)/(1-ALPHA^(2))`
`(alpha^(2)P)/(1-alpha)`
`(4alpha^(2))/(1-alpha^(2))`
`(4alpha^(2)P)/(1-alpha^(2))`

ANSWER :D
30943.

The degree of dissociation of acetic acid in 0.1 M solution is 0.04 . Calculate K_a for acetic acid. Where alpha is the degree of dissociation,C is the concentrationof the acid in moles/ lit.

Answer»

SOLUTION :`K_a=(alpha^2 C)/(1-alpha)=(02xx.02xx0.1)/(1-0.02)`
`K_a=(0.4xx10^(-4))/(0.98) = 4.08xx10^(-5)`
30944.

The degree of dissociation of an electrolyte does not depend on

Answer»

NATURE of electrolyte
catalyst
dilution
temperature

ANSWER :B
30945.

The degree of dissociation of an electrolyte in aqueous solution depends on A) Temperature B) Concentration of the electrolyte C) Nature of the electrolyt

Answer»

Only A
Only A,B
Only B,C
A,B,C

30946.

The degree of dissociation of 100 mL of pure water at 25^(@)C is

Answer»

`1.8xx10^(-16)`
`1XX10^(-14)`
`1.8xx10^(-9)`
`1.0`

ANSWER :C
30947.

The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 mL of 1.0MHA solution is diluted to 32 mL the degree of dissociation of acid and H_(3)O^(+)ion concentration in the resulting solution will be respectively

Answer»

`0.02and3.125xx10^(-4)`
`1.25xx10^(-3)and0.02`
`6.02and1.25xx10^(-3)`
`0.02and8.0xx10^(-12)`

SOLUTION :`alpha_(1)=0.005=sqrt(K_(a)xxC_(1))`
MOLARITY of diluted solution `2xx1=32xxM,M=1//16(C_(2))`
`alpha_(2)=sqrt((K_(a))/(C_(2)))=0.005sqrt(16)=0.02`
`H_(3)O^(+)=C_(2)alpha_(2)=(0.02)/(16)=1.25xx10^(-3)M`
30948.

The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be

Answer»

`10^(-3)`
`10^(-5)`
`10^(-7)`
`10^(-9)`

SOLUTION :`HA hArr H^(+) + A^(-)`
`K = C ALPHA^(2) = 0.1 xx (10^(-4))^(2) = 10^(-9)`.
30949.

The degree of dissociation is 0.4 at 400 K and 1 atm for the gaseous reactionPCl_5 iff PCl_5 + Cl_2Assuming ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400 K and 1 atm.

Answer»

<P>

SOLUTION :APPLY `p = (DRT)/(M_(MIX)`
4.535g/L
30950.

The degree of dissociation in a weak electrolytic increases

Answer»

On INCREASING
On increasing pressure
On DECREASING
NONE of these

Solution :In weak electrolyte the degree of DISSOCIATION is very small. So it increases with increasing dilution. (OSTWALD's law).