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88051.

A polymer of prop-2-ene nitrile is called:

Answer»

Saran
Orlon
Dacron
Terylene

Answer :B
Discussion
88052.

A polymeris formedwhen simple chemical units

Answer»

combineto FORM LONG CHAINS
combine to form helicalchains
BREAK up
become ROUND

Answer :A
Discussion
88053.

A polymer contains 10 molecules with molecularmass 10,000 and 10 molecules with molecular mass 1,00,000.Calculate number-average molecular mass.

Answer»

SOLUTION :`BAR(M_(N))=(sumN_(i)M_(i))/(sumN_(i))=(10xx10000+10xx100000)/(10+10)=55,000`
Discussion
88054.

A polymer is formed between ethylene dichloride and sodium disulphide. Discuss.

Answer»

Solution :Ethylene DICHLORIDE REACTS with sodium disulphide to GIVE a polymer called thiocol. This is a condensation copolymer obtained by the elimination of sodium chloride.
`nCl-CH_(2)-CH_(2)-Cl+nNa-S-S-Na rarr (--CH_(2)-CH_(2)-S-S--)_(n)+2n NACL`
Discussion
88055.

A polymer commonly used for making nonstick cookware is

Answer»

SBR
Teflon
PVC
POLY ethyl acrylate

Answer :B
Discussion
88056.

A polar covalent bond with positive and negative charge centres at its ends is called a dipole. The polarity of a dipole is measured by its dipole moment. Mathematically it is expressed as dipole moment, mu=q xx d where q and d are the net charge and the distance between the two charges respectively. Dipole moment is a vector quantity. The net dipole moment of a polyatomic molecule is the resultant of the various bond moments present in the molecule. The values of dipole moment are expressed in Debye (D) or in SI units in terms of coulomb- metre (Cm). One of the most important applications of dipole moment is in the determination of geometry and shape of molecules besides prediction of a number of properties of the molecules. A diatomic molecule has a dipole of 1.2 D, if the bond distance is 1 Ã…, what percentage of electronic charge exists on each atom.

Answer»

25% of E
29% of e
19% of e
20% of e.

Solution :`mu=e/d=(1.2D)/(1.0A)=(1.2xx10^(-18))/(1XX10^(-18)"CM")=1.2xx10^(-10)` esu
`therefore` % of e=`(1.2xx10^(-10))/(4.8xx10^(-10))xx100`=25%
Discussion
88057.

A polar covalent bond with positive and negative charge centres at its ends is called a dipole. The polarity of a dipole is measured by its dipole moment. Mathematically it is expressed as dipole moment, mu=q xx d where q and d are the net charge and the distance between the two charges respectively. Dipole moment is a vector quantity. The net dipole moment of a polyatomic molecule is the resultant of the various bond moments present in the molecule. The values of dipole moment are expressed in Debye (D) or in SI units in terms of coulomb- metre (Cm). One of the most important applications of dipole moment is in the determination of geometry and shape of molecules besides prediction of a number of properties of the molecules. Debye is equivalent to

Answer»

`3.33 xx 10^(-30)` e.s.u. cm
`1.662xx10^(-27)` e.s.u.cm
`1 xx 10^(-18)` e.s.u. cm
`3.33 xx 10^(-12)` e.s.u.cm.

Answer :C
Discussion
88058.

A polyhydroxy alcohol having molecular weight =168. On acetylation the molecular weight increases to 294. Determine the number of -OH groups present in the alcohol.

Answer»

SOLUTION :
So when one -OH group is replaced by acetyl group MOLECULAR weight increases by `59-17=42`.
Here TOTAL increase in molecular weight `294-168=126`
So number of `-OH` groups `=(126)/(42)=3`.
Discussion
88059.

A polar covalent bond with positive and negative charge centres at its ends is called a dipole. The polarity of a dipole is measured by its dipole moment. Mathematically it is expressed as dipole moment, mu=q xx d where q and d are the net charge and the distance between the two charges respectively. Dipole moment is a vector quantity. The net dipole moment of a polyatomic molecule is the resultant of the various bond moments present in the molecule. The values of dipole moment are expressed in Debye (D) or in SI units in terms of coulomb- metre (Cm). One of the most important applications of dipole moment is in the determination of geometry and shape of molecules besides prediction of a number of properties of the molecules. The dipole moment of HBr is 0.78 xx 10^(-18) esu cm. The bond length of HBr is 1.41 Ã…. The percentage ionic character of HBr is

Answer»

7.54
11.52
15.7
27.3

Solution :% IONIC character
`=(0.78xx10^(-18)XX100)/(1.41xx10^(-8)xx4.8xx10^(-10))=11.52%`
Discussion
88060.

point that is located at the corner of a unit cell is shared by how many unit cells ?

Answer»

2
4
6
8

Answer :D
Discussion
88061.

A poisonous gas is formed by the reaction of R-NH_(2) with CHCl_(3), KOH.What is this test known as ?

Answer»


ANSWER :CARBYLAMINE TEST
Discussion
88062.

A plot of the number of neutrons (n) against the number of protons (p) of stable nuclei exhibits upward deviation from linearily for atomic number, Z gt 20. For an unstable nucleus having n/p ratio less than 1, the possible mode(s) of decay is(are)

Answer»

`beta^(-)` -DECAY (`beta`-EMISSION)
orbital and K-electron capture
neutron emission
`beta^(+)` -decay (POSITRON emission)

ANSWER :B::D
Discussion
88063.

A plot of the number of neutrons (n) against the number of protons (p) of stable nuclei exhibits upward deviation from linearity for atomic number, Z > 20. For an unstable nucleus having n, p ratio less than 1, the possible mode(s) of decay is(are)

Answer»

<P>`beta^(-)`-decay `(beta`emission)
orbital or K- ELECTRON capture
neutron emission
`beta^(+)`-decay (POSITRON emission).

Solution :NUCLIDES with Z > 20 lying below the STABILITY belt decay by `beta+` (positorn) emission or K-electron capture so, that n,p ratio increases to `(N + 1)//(Z - 1)`.
Discussion
88064.

A plot of the number of neutrons (n) against the number of protons (p) of stable nuclei exhibits upward deviation from linearity for atomic number, Z gt 20. For an unstable nucleus having n//p ratio less than 1, the possible mode (s) of decay is (are)

Answer»

`beta^(-) -` decay (`beta` EMISSION)
Orbital or K-electron capture
Neutron emission
`beta^(+) -` decay (positron emission)

Solution :`(n)/(p) lt 1`
Positron emission `OVERSET(.^(a)X^(b))rarr + ._(1)^(0)e + ._(a -1)Y^(b)`
`rArr K -` electron capture
`._(a)X^(b) + ._(-1)e^(0) rarr ._(a -1)Y^(b)`
both process cause an increase in n/p ratio towards 1 THUS STABILISING the nucleus
Discussion
88065.

A plot of mass of the gas adsorbed per gram of the adsorbent varsus pressure at constant temperature is called…………

Answer»

SOLUTION :ADSORPTION ISOTHERM
Discussion
88066.

A plot of In K against 1/T (abscissa) is expected to be a straight line with intercept on ordinate axis equal to

Answer»

`(Delta S^@)/(2.303 R)`
`(Delta S^@)/(R )`
`-(Delta S^@)/(R )`
`R xx Delta S^@`

SOLUTION :
`Delta G^@ = -RT lnK`
`Delta H^@ - T Delta S^@= -RT ln K`
`ln K = (-Delta H)/(RT) + (Delta S^@)/(R )`
comparing with
`y = mx+ c `
` THEREFORE ` y INTERCEPT is `(Delta S^@)/(R )`
Discussion
88067.

A plot of k vs 1//T for a reaction gives the slope -1 xx 10^(4) K. The energy of activation for the reaction is:

Answer»

8314 J `mol^(-1)`
`1.202 kJmol^(-1)`
`12.02 J mol^(-1)`
`83.14 kJ mol^(-1)`

Solution :d) Plot of In k vs `1//T` GIVES
Slope = `-E_(a)/R`
`-1xx10^(4) = -(EA)/R`
`Ea=Rxx(1xx10^(4))`
`=8.314 XX 10^(4)Jmol^(-1)=83.14 xx 10^(3)J mol^(-1)`
`=83.14kJ mol^(-1)`
Discussion
88068.

A plot of (1)/(T) Vsk for a reaction gives the slope - 1 xx 10^(4)K. The energy of activation for the reaction is (Given R = 8.314 JK^(-1) " mol"^(-1))

Answer»

`8314 "J MOL"^(-1)`
`1.202 "kJ mol"^(-1)`
`12.02 "J mol"^(-1)`
`83.14 "kJ mol"^(-1)`

Solution :Correction in the quenstion: It should be ln k vs 1/T
PLOT of ln k vs 1/T GIVES
Slope `= - (E_(a))/(R)`
`:.-1xx10^(4)= - (E_(a))/(8.314)`
`E_(a)= 8.314xx10^(4)"J mol"^(-1)= 83.14xx10^(3)"J mol"^(-1)`
or `E_(a)= 83.14 "kJ mol"^(-1)`
Discussion
88069.

A pleasant smelling optically active ester (A) has M.W. = 186. It does not react with Br_(2) in "CC"l_(4) . Hydrolysis of (A) gives two optically active compounds (B) soluble in NaOH and (C). Compound (C) gives a positive iodoform test and on warming with conc. H_(2)SO_(4) gives (D) (Saytzeff-product) with no geometrical isomers. (C) on treatment with benzene sulphonyl chloride gives (E) which on treatment with NaBr gives optically active (F). When Ag^(+) salt of (B) is treated with Br^(2) ,racemic (F) is formed. Give the structures of (A) to (F).

Answer»

SOLUTION :
Discussion
88070.

A plot between time and the amount of reactant consumed is found to be a straight line passing through origin for a reaction. The order of the reaction is

Answer»

0
1
2
3

Answer :A
Discussion
88071.

A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called .....

Answer»

SOLUTION :ADSORPTION ISOTHERM
Discussion
88072.

(A): Platinum and gold occur in native state in nature. (R): Platinum and gold are noble metals

Answer»

Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A)
Both (A) and (R) are true and (R) is not the correct EXPLANATIONOF (A)
(A) is true but (R) is false
Both (A) and (R) are false

Answer :A
Discussion
88073.

A plastic bakelite is a compound of HCHO with :

Answer»

BENZENE
Phenol
Ammonia
Hydrocarbon

Answer :B
Discussion
88074.

A plant virus is found to consist of uniform cylindrical particles of 150 Å indiameter and 5000 Å long. The specific volume of the virus is 0.75 cm^3/g. If the virus is considered to be a single particle, find its molecular weight.

Answer»

SOLUTION :MOL wt. = Mass of 1 MOLECULE x AV. constant.
`7.0939 xx 10^7`
Discussion
88075.

A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The specific volume of the virus is 0.75cm^(3)//g. If the virus is considered to be a simple particle, find the molecular weight.

Answer»

SOLUTION :`"Volume of one virus particle"=pir^(2)h=(22)/(7)XX((150)/(2)xx10^(-8)cm)^(2)xx(5000xx10^(-8)cm)=8.839xx10^(-17)cm^(3)`
`"Mass of one virus particle "=8.839xx10^(-17)cm^(3)xx(1g)/(0.75cm^(3))=(11.785xx10^(-17))g`
`THEREFORE" Molar mass (i.e. mass of Avogadro's no. of particles)"=(11.785xx10^(-17))xx(6.022xx10^(23))`
`=7.098xx10^(7)" g mol"^(-1)`.
Discussion
88076.

A plant virus consists of uniform cylindrical particles of 150 overset(@)A in diamter and 5000overset(@)A long. The specific volume of virus is 0.75 cm^(3)/g. if the virus is considered to be a single particle, its molecular mass is:

Answer»

`7.09xx10^(7)g "mol"^(-1)`
`7.90xx10^(7)g "mol"^(-12)`
`9.07xx10^(7)g "mol"^(-1)`
`9.70xx10^(7)g "mol"^(-1)`

Solution :Volume of SINGLE virus `=pir^(2)h`
`=3.14xx(75xx10^(-8))^(2)xx(5000xx10^(-8))`
`=8.836xx10^(-17)cm^(3)`
MASS of single virus `=("Volume")/("Specific volume")=(8.836xx10^(-17)cm^(3))/(0.75cm^(3)//g)`
`=1.178xx10^(-16)g`
Molar mass of virus `=1.178xx10^(-16)xx6.023xx10^(23)`
`=7.09xx10^(7)g "mol"^(-1)`
Discussion
88077.

A piston filled with 0.4 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0^(@)C. As it does so, absorb 208 J of heat. The values of q and w for the process will be : (R = 8.314 J//mol K) (ln 7.5 = 2.01)

Answer»

`q = + 208 J, w = - 208 J`
`q = - 208 J, w = - 208 J`
`q = - 208 J, w = + 208 J`
`q = + 208 J, w = + 208 J`

SOLUTION :q = + 208 J
Since the process is ISOTHERMAL, `Delta U = 0`
:. From FIRST law of thermodynamics,
`Delta U = q + w`
`0 = q + w` or `w = - q`
`= - 208 J`
Discussion
88078.

(A) Pka value of HF is positive (R ) HF is strongest acid among HX

Answer»

Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A)
Both (A) and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is false
Both (A) and (R ) are false

ANSWER :C
Discussion
88079.

A piston filled with 0.04 mole of an ideals gas expands revesrsiby from 50.0 mL to 375 ml at a canstant heat. The values of q and w for the process will be (R= 8.314 J/ mol. K in 7.5 = 2.01)

Answer»

`q=+208j , W= -208`
`q=-208j , W= -208`
`q=-208j , W= +208`
`q=+208j , W= +208J`

ANSWER :a
Discussion
88080.

A piston filled with 0.04 "mole" of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37^(@)C. As it does so it absorbs 208 J of heat. The value of q and W for the process will be (R=8.314 J//"mol" K, In 7.5=2.01)

Answer»

Q=+208 J, W=-208 J
q=-208 J, W=-208 J
q=-208 J, W=+208J
q=+208 J, W=+208 J

Solution :SEE SIGN convebsion for q and W
Discussion
88081.

A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0^(@)C. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be: (R=3.314J//molK)(Ln7.5=2.01)

Answer»

q = + 208 J, W = + 208 J
q = + 208 J, w = – 208 J
q = – 208 J, w = – 208 J
q = – 208 J, w = + 208 J

Answer :B
Discussion
88082.

A piston filled with 0.04 mol of an ideal gas expands eversibly from 50.0 mL to 375 mL at a constant temperature of 37.0^(@)C. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be (R=8.314 J//mol K) (ln 7.5 = 2.01)

Answer»

Q=+208 J, w=-208 J
q=-208 J, w=-208 J
q=-208 J, w=+208 J
q=+208 J, w=+208 J

Solution :The PROCESS is isothermal expansion HENCE, q =-w
`Deltau=0`
q=+208 J
w=-208 J (expansion work).
Discussion
88083.

A pink coloured salt turns blue on heating. The presence of which cation is most likely?

Answer»

`CU^(2+)`
`FE^(2+)`
`Zn^(2+)`
`Co^(2+)`

ANSWER :D
Discussion
88084.

(A)Pin oil is fused as a foaming agent in frothfloatation process (R) Adsorption principle is involved in froth floatation process

Answer»

Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct EXPLANATIONOF (A)
(A) is true but (R) is FALSE
Both (A) and (R) are false

Answer :A
Discussion
88085.

A piece of zinc metal is dipped in a 0.1 M solution of zinc salt. The salt is dissociated to the extent of 20%. Calcualte the electrode potential of Zn^(2+)//Zn (Given E_(Zn^(2+)//Zn)^(@) = -0.77 volt

Answer»


Solution :`[Zn^(2+)]=0.1 xx 0.2 =0.02 M`
`Zn^(2+)+2E^(-) RARR Zn"(REDUCTION)"`
`E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)+0.0591/2 log_(10) [Zn^(2+)]`
Discussion
88086.

A pigment protein in animals is

Answer»

Chlorophyll
Insulin
Keratin
Haemoglobin

Answer :D
Discussion
88087.

A piece of wood was found to have .^(14)C//^(12)C ratio 0.6 times that in a living plant. Calculate the period when the plant died (Half-life of .^(14)C=5760 years)

Answer»

SOLUTION :`t=(2.303xxt_(1//2))/(0.693)xxlog((N_(0))/(N))`
`=(2.303xx5760)/(0.693)xxlog((1)/(0.6))`
`=(2.303xx5700)/(0.693)xx0.2201=4213` YEARS
Discussion
88088.

A piece of wood was found to have ""^(14)C//^(12)C ratio 0.7 times that in a living plant. Calculate the period (in years) when the plant died. (t_((1)/(2)) " for" C^(14)= 5760yr)

Answer»


ANSWER :2964 YR
Discussion
88089.

A piece of wood was found to have C^(14)//C^(12) ratio 0.7 times that in a living plant. The time period when the plant died is (Half-life of C^(14) = 5760 yr)

Answer»

2770 YR
2966 yr
2980 yr
3070 yr

Solution :GIVEN, `N_(0) = 1N, N_(t) = 0.70 and t_(1//2) = 57600 yr`
`k = (0.693)/(t_(1//2)) = (0.693)(5760)`
We also know, `k = (2.303)(t) "LOG" (N_(0))/(N_(t)) = (0.693)/(5760)`
or `t = (2.303 xx 5760 xx 0.155)/(0.693) = 2966yr`
Discussion
88090.

A piece of wood from an aechaeological source shows a C - 14 activity which is 60% of the activity found today. Calculate the age of the sample t_(1//2)for""_(6)^(14)C = 5770years).

Answer»

3515 years
300 YEAR
4253 year
4000 years

Solution :`LAMBDA = (0.693)/(5770) YR^(-1), t = (2.303)/(0.693//5770) "log" (100)/(60)`
`= 4253` years.
Discussion
88091.

A piece of wood from an aechaeological sample has 5.0 counts "min"^(-1) per gram of C - 14, while a fresh sample of wood has a count of 15.0 "min"^(-1). If half-life of C-14 is 5770 years, the age of the archaeological sample is

Answer»

8,500 years
9,200 years
10,000 years
11,000 years

SOLUTION :Radioactive decay FOLLOWS first order kinetics.
Now, `t_(1//2) = 5770` years
`lambda = (0.693)/(t_(1//2)) = (0.693)/(5770) "years"^(-1)`
where , `lambda` = decay constant
If age is t years, then
`t = (2.303)/(lambda) log_(10) ("activity of the FRESH SAMPLE")/("activity of archaecological sample")`
`implies t = 2.303 xx (5770)/(0.693) xx log_(10) 15/5`
`implies t = 9149` years `~~ 9200` years (approx).
Discussion
88092.

A piece of silica gel is placed in the water vapours. Will the gel adsorb or absorb water vapours?

Answer»

SOLUTION :It will ADSORB the WATER VAPOURS.
Discussion
88093.

A piece of metal is 3 inch (represented by in ) long. What is its length in cm?

Answer»


SOLUTION :`"3 in"="3 in"XX(2.54cm)/("1 in")=7.62cm`
Discussion
88094.

A piece of magnesium ribbon was heated to redness in an atmosphere of nitrogen and then cooled with water. The gas evolved is

Answer»

ammonia
hydrogen
nitrogen
oxygen

Answer :A
Discussion
88095.

A piece of magnesium ribbon was heated to redness in an atmosphere of nitrogenand then treated with water, the gas evoloved is:

Answer»

Ammonia
Hydrogen
NITROGEN
Oxygen

Answer :A
Discussion
88096.

A piece of aluminium weighing 2.7 g is heated with 75.0 mL of sulphuric acid ( sp . Gr . 1.18 containing 24.7 % H_(2)SO_(4) be weight ) .Afterthe metal is carefully dissolved the solutions is dilutedto 400 mL . Calculate the molarity of the free H_(2)SO_(4)in the resulting solution .

Answer»

Solution :Normality of the given `H_(2)SO_(4) = 5.95 N `
` :. ` m.e of 75 mL of `H_(2)SO_(4) = 5.95 xx75`
` = 446.25 `
Equivalent of Al ` = (2.7)/9 = 0.3 "" …(Eqn .4)`
`{ " EQ .wt. of Al " = ("al .wt")/("valency") = 27/3 = 9 }`
` :." m.e of Al " = 0.3 xx 1000 = 300 "" ...(Eqn .3)`
Since 300 m.e of Al will react with 300 m.eof `H_(2)SO_(4)`
m.e of FREE `H_(2)SO_(4)`= TOTAL m.e of `H_(2)SO_(4)-300`
`= 446.25 - 300`
`= 146.25`
Now the free `H_(2)SO_(4)` is DILUTED to 400 mL and we knw that the m.e of `H_(2)SO_(4)` does not change on dilution .
` :. ` normality of the diluted free `H_(2)SO_(4)`in the resulting solution
` = ( " m.e of free" H_(2)SO_(4))/(" volume (mL)")`
` = (146.25)/400 = 0.366` N .
` :. "molarity " = (0.366)/2 = 0.183 M"" ...(Eqn . 6i)`( basicity of `H_(2)SO_(4) = 2`)
Discussion
88097.

A piece of Cu is added to an aqueous solution of FeClg

Answer»

No iron will PRECIPITATED from the solution
COPPER will not-DISSOLVE in the solution
Copper will appear to dissolve in the solution
Iron will be precipitated from the solution

Answer :A::C
Discussion
88098.

A piece of copper weighs 0.634g. How many atoms of copper does it contain?

Answer»

Solution :Gram -ATOMIC MASS of COPER =63.5g
Number of moles in 0.635 g of copper `=(0.635)/(63.5)=0.01`
Number of copper atoms in one MOLE `=6.02xx10^(23)`
Number of copper atoms in 0.01 moles `=0.01xx6.02xx10^(23)`
`=6.02xx10^(21)`
Discussion
88099.

A piece of aluminium weighing 2.7 g is heated with 75 mL of H_(2)SO_(4) which has a density of 1.18"g mL"^(-1) and contains 24.7% by mass. When whole of the metal had dissolved, the solution was diluted to 400 mL. Calculate the molarity of free H_(2)SO_(4) in the resulting solution. Hence, molarity =(7.17)/(98)xx(1)/(400)xx1000=0.183M

Answer»


Solution :`24.7%" of "H_(2)SO_(4)" by mass means 24.7 g of "H_(2)SO_(4)" are PRESENT in 100 g of the solution."`
`"VOLUME of 100 g of "H_(2)SO_(4)" solution "=("Mass")/("Density")=(100g)/(1.18gmL^(-1))=84.7mL`
Thus, `"84.7 mL of "H_(2)SO_(4)" soluiton contain "H_(2)SO_(4)=24.7g`
`therefore "75 mL of "H_(2)SO_(4) " solution will contain "H_(2)SO_(4)=(24.7)/(84.7)xx75g=21.87g`
The equaiton for the reaction is
`underset(=54g)underset(2xx27g)(2Al)+underset(=294g)underset(3xx98g)(3H_(2)SO_(4))rarrAl_(2)(SO_(4))_(3)+3H_(2)`
`"54 g Al react with "H_(2)SO_(4)=294g`
`therefore "2.7 g Al will react with "H_(2)SO_(4)=(294)/(54)xx2.7g=14.7g`
`thereforeH_(2)SO_(4)" left unreacted "=21.87-14.7=7.17g=(7.17)/(98)" moles"`
This is present in 400 mLof the solution (after dilution).
Discussion
88100.

A pi- bond is formed by sideways overlapping of:

Answer»

s-s orbitals
p-p orbitals
s-p orbitals
s-p-s orbitals

Answer :B
Discussion