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25751.

The plots of (1)/(x_(A)) (on y-axis) (1)/(y_(A)) (on x-axis) is linear with slope and intercept respectively. p^(@)A = vapour pressure of pure liquid A p^(@)B = vapour pressure of pure liquid B

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<P>`(p_(A)^(@))/(p_(B)^(@))` and `((p_(A)^(@)-p_(B)^(@)))/(p_(B)^(@))`
`(P_(a)^(@))/(P_(b)^(@))` and `((P_(B)^(@)-P_(A)^(@)))/(p_(B)^(@))`
`(p_(B)^(@))/(p_(A)^(@))` and `((p_(A)^(@)-p_(B)^(@)))/(p_(B)^(@))`
`(p_(B)^(@))/(p_(A)^(@))` and `((p_(B)^(@)-p_(A)^(@)))/(p_(B)^(@))`

ANSWER :B
25752.

The plot shows the variation of -InK_(p) versus temperature for the two reactions. M(s)+(1)/(2)O_(2)(g)toMO(s)andC(s)+(1)/(2)O_(2)(g)toCO(s) Identify the correct statement :

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At `Tlt1200K`, oxidation of CARBON is unfavourable.
Oxidation of carbon is favourable at all temperatures.
At `Tgt1200K`, the REACTION `MO(s)+C(s)toM(s)+CO(g)` is spontaneous.
At `Tgt1200K`, carbon will REDUCE MO(s) to M(s).

Solution :
At `Tlt1200K`, carbon will reduce MO(s) to M(s) HENCE, chemical reaction
`MO(s)+C(s)toM(s)+CO(g)` is spontaneous.
25753.

The plot of x/m versus temperature at constant pressure is called

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ADSORPTION ISOTHERM
Adsorptionisobar
Adsorption isochore
Freundlich isotherm

Answer :B
25754.

The plot of partial vapour vapour pressure of solvent verses its mole fraction in the solution of a constant temp.is

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<P>a STRAIGHT LINE
a straight line parallel to one axis
a straight line passing through origin
none of the above

Solution :`P_("solvent") prop X_("solvent")`
25755.

The plot of t(1)/(2) V/s [R]0 for a reaction is a straight-line parallel to x-axis.The unit for the rate constant of this reaction is

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MOL `L^(-1)` s
mol `L^(-1) s^(-1)`
L `mol^(-1) s^(-1)`
`s^(-1)`

Solution :(d): Plot of `t_(1//2)` vs `[R]_(0)` for a REACTION is a STRAIGHT line parallel to x-axis, means `t_(1//2)` is independent of `[R]_(0)` hence, the reaction is first order.
`:.` Unit of RATE CONSTANT `= s^(-1)`
25756.

The plot of 1/X_(A) vs 1/Y_(A) (where X_(A) and Y_(B)) are the mole fractions oif A in liquid and vapour phases respectively) is linear whose slope and intercept respectively are given as

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<P>`P_(A)^(@)/P_(B)^(@).(P_(A)^(@) - P_(B)^(@))/(P_(B)^(@))`
`P_(A)^(@)/P_(B)^(@).(P_(B)^(@) - P_(A)^(@))/(P_(B)^(@))`
`P_(B)^(@)/P_(A)^(@).(P_(A)^(@)-P_(B)^(@))/(P_(B)^(@))`
`P_(B)^(@)/P_(A)^(@).(P_(B)^(@)-P_(A)^(@))/(P_(B)^(@))`

ANSWER :B
25757.

The plot of (1)/(Y_(A))" Vs "(1)/(x_(A))((1)/(Y_(A))" on y - axis") where A and B form a ideal solution. Y is mole fraction in vapour phase and X is mole fraction in liquid phase, is linear with slope and inercept respectively

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<P>`(P_(A)^(0))/(P_(B)^(0)) and (P_(A)^(0)-P_(B)^(0))/(P_(B)^(0))`
`(P_(A)^(0))/(P_(B)^(0)) and (P_(B)^(0)-P_(A)^(0))/(P_(B)^(0))`
`(P_(B)^(0))/(P_(A)^(0)) and (P_(A)^(0)-P_(B)^(0))/(P_(A)^(0))`
`(P_(B)^(0))/(P_(A)^(0)) and (P_(B)^(0)-P_(A)^(0))/(P_(B)^(0))`

ANSWER :C
25758.

The plot of log k vs 1/T helps to calculate

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ENERGY of activation
RATE constant of the REACTION
ORDER of the reaction
Energy of activation as well as the frequency factor

Answer :D
25759.

The plot of In k versus 1/T is liner with slope of

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`E_a/R`
`E_a/2.303R`
In A
`-E_a/R`

ANSWER :D
25760.

The plotof 1//x_(A) versus 1//y_(A) (where X_(A) and Y_(A) are the mole fraction of A in liquid and vapour phases, respectively ) is linear with slope and intercept respectively are given as (y-axis=1//y_(A), x-axis =1//x_(A))

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<P>`P_(A)^(0)//P_(B)^(0), (P_(A)^(A)-P_(B)^(0))//P_(B)^(0)`
`P_(A)^(0)//P_(B)^(0), (P_(B)^(0)-P_(A)^(0))//P_(B)^(0)`
`P_(B)^(0)//P_(A)^(0), (P_(A)^(0)-P_(B)^(0))//P_(A)^(0)`
`P_(B)^(0)//P_(A)^(0),(P_(B)^(0)-P_(A)^(0))//P_(B)^(0)`

ANSWER :C
25761.

The plot between concentration versus time for a zero order reaction is represented by

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SOLUTION :For zero ORDER reaction, `k = (1)/(t){[A]_(0) - [A]}`
or `[A] = - kt + [A]_(0)`. Thus, Plot of [A] vs t is LINEAR with `-ve` SLOPE `( = -k)`
25762.

The plots of (1)/(X_(A))vs.(1)/(Y_(A)) (where X_(A) and Y_(A) are the mole fraction of liquid A in liquid and vapour phase respectively) is linear with slope and intercept respctively are given as:

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`P_(A)^(@)/P_(B)^(@),((P_(A)^(@)-P_(B)^(@)))/(P_(B)^(@))`
`P_(A)^(@)/P_(B)^(@),((P_(B)^(@)-P_(A)^(@)))/(P_(B)^(@))`
`P_(B)^(@)/P_(A)^(@),((P_(A)^(@)-P_(B)^(@)))/(P_(B)^(@))`
`P_(B)^(@)/P_(A)^(@),((P_(B)^(@)-P_(A)^(@)))/(P_(B)^(@))`

SOLUTION :ACCORDING to Raoult's LAW
`P_(A)=X_(A)P_(A)^(@),P_(B)=X_(B)P_(S)^(@)`
`Y_(A)P_(A)/(P_(A)+P_(B))=(x_(B)P_(S)^(@))/(X_(A)P_(A)^(@)+X_(B)P_(B)^(@))`
`or 1/Y_(A)=(P_(A)^(@)-P_(B)^(@))/P_(A)^(@)+P_(B)^(@)/P_(A)^(@)xx1/X_(A)`
`or P_(B)^(@)/(X_(A)P_(A)^(@))=1/Y_(A)-[(P_(A)^(@)-P_(B)^(@)]/P_(A)]`
`1/X_(A)=P_(A)^(@)/P_(B)^(@)[1/Y_(A)+((P_(B)^(@)-P_(B)^(@)))/P_(A)^(@)]`
`=P_(A)^(@)/P_(B)^(@)xx1/Y_(A)+((P_(B)^(@)-P_(A)^(@)))/P_(B)^(@)`
Hence PLOT of `1/X_(A)vs1/Y_(A)` will be liner with slope =`P_(A)^(@)/P_(B)^(@)"and intercept"=((P_(B)^(@)-P_(A)^(@)))/P_(B)^(@)`
25763.

The platinum catalyst used in the oxidation of SO_2by contact process is poisoned by_______.

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`As_2O_3`
`V_2O_5`
`Fe_2O_3`
`CuCl_2`

ANSWER :A
25764.

The platinised plate in SHE is used as __________.

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OXIDISING agent
adsorbent
REDUCING agent
adsorbate

Answer :B
25765.

The plastics if are hard, become soft and readily workable by addition of certain compound called

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Catalyst
Telomers
Pasticisers
VULCANISER

SOLUTION :PVC is EXTREMELY stiff and hard but the addition of di-n-butyl phthalate plasticisers MAKES it soft and rubber like.
25766.

What is plaster of paris ?

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`CaSO_4.2H_2O`
`CaSO_4`
`2CaSO_4.2H_2O`
`2CaSO_4.H_2O`

ANSWER :D
25767.

The plant cell will shrink when placed in:

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WATER
A HYPOTONIC solution
a HYPERTONIC solution
an siotonic solution

ANSWER :C
25768.

The plant cells will shrink when placed in

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Water
A HYPOTONIC solution
A hypertonic solution
An ISOTONIC solution

Answer :C
25769.

The planar complex [M_(ABCD)] gives:

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TWO OPTICAL isomers
Two GEOMETRICAL isomers
Three optical isomers
Three geometrical isomers

Answer :D
25770.

The pK_(SP) of Agl is 16.07. If the E^(@) value for Ag^(+)//Ag is0.7991 V, find the E^(@) for hlaf cell reaction : AgI_((s)) + e rarr Ag + I^(-)

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ANSWER :`-0.1490 V ;`
25771.

The pK_c of equimolecular sodium acetate and acetic acid mixtuare is 4.74. If pH is :

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1.4
4.74
9.2
7

Answer :B
25772.

The pkb values of Ammonia, methanamine and Benzenamic (aniline) are 4.75, 3.38 and 9.38 respectively. Arrange them in the increasing order of their basic strength.

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SOLUTION :`UNDERSET(3.38)("Methanomine")GT underset(4.75)("AMMONIA")gt underset(9.38)("bensenomine")`
25773.

The pK_b value of ammonium hydroxide is 4.75. An aqueous solution of ammonium hydroxide is titrated with HCl. The pH of the ammonium hydroxide has been neutralized will be

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9.25
8.25
`7.50`
4.75

Answer :A
25774.

The pK_(a_1), pK_(a_2) and pK_(a_3) value for the amino acid cysteine (HS-CH_2-undersetunderset(NH_2)(|)CH-COOH) are respectively 1.8, 8.3, 10.8. What is isoelectric point of cysteine amino acid ?

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<P>

SOLUTION :`p^(|)=(pK_(a_2))+pk_(a_3))/2=(8.3+10.8)/2=8.85`
25775.

The pK_(b) value is highest for :

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<P>p - METHOXY aniline
p - chloroaniline
p - nitroaniline
p - methylaniline

Answer :A
25776.

The pK_(a0 of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is :

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`2.5`
`9.5`
`7.0`
`4.5`

SOLUTION :If the given WEAK acid is 50% ionised then its conjugate BASE will have the CONCENTRATION same as that of weak acid.
`pH=pK_(a)+"log"(["conjugate base"])/(["Acid"])`
`=4.5+"log"(0.5c)/(0.5c)=4.5`
`therefore POH = 14-4.5=9.5`
25777.

The pka values for the three ionizable groups x, y & z of glutamic acid are 4.3, 9.7 & 2.2 respectively underset((x))(HOOC)-CH_(2)-underset((y))underset(+NH_3)underset(|)(CH)-underset((z))(COOH) The isoelectric point for the amino acid is :

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ANSWER :3.25
25778.

The Pk_(a) value of the C-H acids HC -= CH (I), MeCOCH_(2)COOEt (II), MeCOOEt (III) and PhMe (IV) decreases as

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`I gt III gt IV gt II`
`II gt IV gt III gt I`
`IV gt III gt I gt II`
`I gt IV gt III gt II`

Solution :The carbanion left after removal of a proton from MeCOOEt is more STABLE than that obtained from PhMe since in the latter case, the aromatic character of the benzene ring is DESTROYED. THUS, MeCOOEt is a stronger C-H acid than PhMe.
`CH_(3)-overset(O)overset(||)C_OEt overset("base")to overset(-)CH_(2)-overset(O)overset(||)C-OEt harr CH_(2)=overset(O^(-))overset(|)C-OEt harr "etc."`

Among `HC -= CH` and `MeCOCH_(2)COOEt` is a stronger acid than `HC -= CH` since the carbanion from `MeCOCH_(2)COOEt` is resonance stabilized by two carbonyl groups.
`CH_(3)-overset(O)overset(||)C-CH_(2)-overset(O)overset(||)C-OEt overset("base")to CH_(3)-overset(O)overset(||)C-CH-overset(O)overset(||)C-OEt harr CH_(3)-overset(O^(-))overset(|)C=CH-overset(O)overset(||)C_OEt harr CH_(3)-overset(O)overset(||)C_CH=overset(O)overset(|)C-OEt`
while the carbanion, i.e., `CH -= C^(-)` from `CH -= CH` is stabilized only by the greater electronegativity of the sp-carbon. Thus the overall acidity decreases in the order, `MeCOCH_(2)COOEt gt HC -= CH gt MeCOOEt gt PhMe`. Since a stronger acid has a lower `pK_(a)` value, `pK_(a)` VALUES decrease in the opposite order, i.e., `PhMe gt MeCOOEt gt HC -= CH gt MeCOCH_(2) COOEt`. In other words, `pK_(a)` values decrease in the order :
`IV gt III gt I gt II`
25779.

The pKa values for the three acidic group P,Q,R are 4.3, 9.7 and 2.2 respectively, Calculate the isoelectric point of the amino acid?

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`7.00`
`3.25`
`4.95`
`5.95`

ANSWER :B
25780.

The pKa ofHCN is 9.30. The pH of a solution prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making up the total volume of 500 ml is :

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`9.30`
`7.30`
`10.30`
`8.30`

SOLUTION :`pH=pK_(a)+"LOG"(["SALT"])/(["ACID"])`
`pH=9.30+"log"((2.50))/((2.50))`
`= 9.30`.
25781.

The pK_a of HCN is 9.30. The pH of a solution prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making up the total volume of 500 mL is :

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`9.30`
`7.30`
`10.30`
`8.30`

ANSWER :a
25782.

The pK_a of an indicator is 4. Its working range lies in beetween pH:

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1-5
3-5
5-8
8-12

Answer :B
25783.

The pK_(a) of an indicator is 4. Its pH range is

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`1-3`
`3-5`
`5-8`
`8-12`

ANSWER :B
25784.

The pk_a of acetylsalicylic acid (aspirin) is 3.5. The PH of gastric juice in human stomach is about 2-3 and the pH in the small intestine is about 9. Aspirin wil be

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UNIONIZED in the SMALL intestine and in the stomach
Completely IONIZED in the small intestine and in the stomach
Ionized in the stomach and almost unionized in the small intestine
ionized in the small intestine and almost unionized in the stomach

Answer :D
25785.

The pK_(a) of acetylsalicylic acid (aspirin) is 3.5. The pH of gastric juice in human stomach is about 2-3 and the pH in the small intestine is about 8. Aspirine will be

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Unionised in the small intestineand the STOMACH
Completely ionised in the small INTESTINE and in the stomach
Ionised in the stomach and ALMOST unionised in the small intestine
Ionised in the small intestine and almost unionised in the stomach

Solution :Under strong acidic conditions present in stomach (pH 2-3), weak acids such as acetylsalicylic ACID with `pK_(a) = 3.5` remains unionized but under basic conditions (pH 8) present in small intestine, it is ionized. Thus option (d) is correct.
25786.

ThepK_(a)of acetylsaliclyicacid (aspirin) is 3.5.The pH ofgastric juicein human stomachisabout2 - 3and pH in thesmall intestine is about 8. Aspirin will be .

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UNIONIZED in the SMALLINTESTINE and in the STOMACH
completelyionizedin the stomach and almostunionizedin the SMALL intestine.

Answer :D
25787.

The pK_(a) of a weak acid is 4.8. What should be the ratio of [Acid]/[Salt] of a buffer if pH = 5.8 is required

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10
0.1
1
2

Solution :`PH = pK_(a) + log.(["SALT"])/(["ACID"])`
`5.8 = 4.8 + log.(["Salt"])/(["Acid"])` or `log.(["Salt"])/(["Acid"]) = 1.0`
`(["Salt"])/(["Acid"]) = "antilog 1.0" = 10`
`:. (["Salt"])/(["Acid"]) = (1)/(10) = 0.1`.
25788.

The pK_(a) of a weak acid is 4.8. What should be the ratio of [Acid]/[Salt] of a buffer if pH = 5.8is required ?

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`0.1`
`4.0`
`4.3`
`3.3`

ANSWER :A
25789.

The pk_(a) of a weak acid, HA is 4.8. The pK_(b) of a weak base, BOH is 4.78. The pH of an aqueous solution of the corresponding salt, BA will be

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4.79
7.01
9.22
9.58

Solution :`pK_(a) = 4.8 , pK_(B) = 4.78`
`PH = 7+(1)/(2)(pK_(a) - pK_(b)) = 7 + (1)/(2)(4.8 - 4.78) = 7.01`.
25790.

The pK_(a) of a weak acid (HA) is 4.5 the pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is

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4.5
2.5
9.5
`7.0`

Solution :For buffer solution
`PH = pK_(a) + log.(["SALT"])/(["Acid"]) = 4.5 + log.(["Salt"])/(["Acid"])`
as HA is 50% ionized so [Salt] = [Acid]
pH = 4.5
pH + pOH = 14
`rArr pOH = 14 - 4.5 = 9.5`.
25791.

The pK_(a) for acetyl salicylic acid (aspirin) is 3.5. The pH of gastric juice in the human stomach is about 2-3 and the pH in the small intestine is about 8. Aspirin will be

Answer»

un-ionised in the SMALL INTESTINE and in the STOMACH
completely ionised in the small intestine and in the stomach
ionised in the stomach and ALMOST un-ionised in the small intestine
ionised in the small intestine and almost un-ionised in the stomach

Answer :D
25792.

The pink colour of phenolphthalein in alkaline medium is due to :

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NEGATIVE ION
Positive ion
`OH^-` ION
neutral ion

Answer :A
25793.

The pink colour of phenolphthalein in alkaline medium is

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DUE to negative form
Due to positive form
Due to `OH^(-)` ions
Due to NEUTRAL form

Answer :A
25794.

The pi electron cloud of C_(1)-C_(2) is present in the plane of paper than which of the following is incorrect?

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`Z` is perpendicular to the PLANE of PAPER
`X` is present in the plane of paper
`sigma` bond of `C_(2)-C_(3)_` is perpendicular to the plane of paper.
`PI` ELECTRON cloud of `C_(2)-C_(3)` bond and `X` is present in same plane

Solution :Only (D) option is CORRECT `pi` electron cloud of `C_(2)-C_(3)` is present in same plane.
25795.

The pigment responsible for photosynthesis is a coordination compound of magnesium.

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ANSWER :1
25796.

The pi-acid ligand which uses its d-orbital during synergic bonding in its complex compound:

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`NO^(+)`
`PR_(3)`
`C_(6)H_(6)`
`CO`

Solution :In `PR_(3)` there is VACANT atomic d-orbital on P-atom, which can be involved in synergic BONDING.
25797.

The pi acid ligand which uses it d-orbital during synergic bonding in its complex compound.

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`CN^(-)`
`PR_(3)`
`NO`
`N_(2)`

Solution :Vaccant non-axial d-orbital participiate in `M OVERSET(PI)toPR_(3)` BACK bonding in its complexes.
25798.

The pi bonds in SO_2 molecule arises due to _________ overlap.

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<P>`SIGMA`
`p PI - p pi and sigma`
`p pi -d pi and p pi - p pi `
only `p pi- d pi`

ANSWER :C
25799.

The pi- bond in carbonyl group is formed by

Answer»

s-s-overlapping
p-p- overlapping
s-p- overlapping
p-d- overlapping

Answer :B
25800.

The physical properties of a material show different values when measured in different directions. What is this property known as ?

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SOLUTION :ANISOTROPY