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92651.

2 moles of an ideal gas initially present in a piston fitted cylinder at 300 K, and 10 atm are allowed to expand against 1 atm but the piston was stopped before it stablished the mechanical equilibrium. If temperature were maintained constantthrough out the change and system delivers 748.26 J of work, determine the final gas pressure and describe the process on PV diagram.

Answer»

SOLUTION :`W_("irrv")=-748.26`
`W_("irr")=-P_("EXT")[1//P_(2)-1//P_(1)]nRT`
`P_(2)` = 4 ATM
92652.

Two moles of ammonia were found to occupy a volume of 5 L at 27^(@)C. Calculate the pressure using van der Waals equation (a="4.17 bar L"^(2)"mol"^(-2), b="0.0371 L mol"^(-1)).

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Solution :n= 2
V = 5 LITRES
`T = 273 + 27 = 300 K`
`a = 41.7 "atm lit"^(2) "mole"^(-2)`
`b = 0.0371 "lit mole"^(-1)`
`R = 0.082 "lit atm deg"^(-1) "mole"^(-1)`
Applying van der Waals equation for n moles
`(p + (a n^(2))/(V^(2))) (V-n b) = nRT`
`(p+(4.17 xx 2^(2))/(5^(2))) (5-2 xx 0.0371) = 2 xx 0.082 xx 300`
p = 9.33 atm.
92653.

2 moles of acetaldehyde are warmed in the presence of 10% Ba(OH)_2 to give a product which dehydrates to a compound X. The compound on reduction with NaBH_4 will give

Answer»

`CH_(3)CH=CHCHO`
`CH_(3)CH=CHCH_(2)OH`
`CH_(3)CH_(2)CH_(2)CH_(2)OH`
`CH_(3)underset(OH)underset(|)(C)HCH_(2)CH_(2)OH`

SOLUTION :Aldol condensation
`CH_(3)-overset(O)overset(||)underset(H)underset(|)(C)+overset(H)overset(|)(C)H_(2)-overset(O)overset(||)(C)-H overset(BA(OH)_(2))toCH_(3)-overset(OH)overset(|)underset(H)underset(|)(C)-overset(H)overset(|)(C)H-overset(O)overset(||)(C)-H overset(DELTA)underset(-H_(2)O)to CH_(3)-CH=CH-overset(O)overset(||)(C)-H overset(NaBH)underset({:("Selective"),("reduction"):})toCH_(3)-CH=CH-CH_(2)OH`
92654.

2 moles monoatomic ideal gas is expanded isothermally and reversibly from 5 L to 20 L at 27°C. deltaH, deltaU, q and w for the process respectively are (log 2 = 0.3)

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ZERO, Zero, Zero and Zero
Zero, Zero, Zero and -6.9 KJ
Zero, +6.9 kJ, +6.9 kJ and -6.9 kJ
Zero, Zero, +6.9 kJ and -6.9 kJ

Answer :3
92655.

2 moles carbon and 1.5 moles of oxygen gas are reacted in a container to produced CO and CO_(2) or both such that none of reactant is left. Find moles of CO, CO_(2) produced.

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ANSWER : `(##ALN_NC_CHM_MC_E01_030_A01##)`
`CO=1"MOLE",CO^(2)=1"mole"`
92656.

2 MOLE of PCl_5 were heated in a closed vessel of 2 litre capacity. AT equilibrium 40% Of PCl_5 dissociated into PCl_3and Cl_2.Find the value of equilibrium constant.

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0.267
0.53
2.63
5.3

Answer :A
92657.

2 mole of ideal momoatmic gas was subjected toreversibleadiabaticcompression from initianl state of P= 1 andT = 300 Ktill the pressrue is 4sqrt(2) atm and temperature is TK . The gas is then subjected to reversible isothermal expansion atT K tilltheinternalpressureis one atm . The gas is nowisobarically cooled to attaininitialstate. Find W_(Net) ( in calorie) forwhole process . [Use In2=0.7]

Answer»


Solution :`T_(2)= T_(1)(P_(1)/(P_(2)))^(((1-gamma)/(gamma))`
` = 600 K`
`W_(I)= W_("rev")= nC_(V)[T_(2)- T_(1)] = 2 xx (3)/(2) R xx 300`
`= 900 R`
`W_(II)= - 2.303 nRT LOG .(P_(1))/(P_(2)) = - 2xx R xx 600 In ((4sqrt(2))/(1))`
`W_(III) = - PDeltaV = - nRT = - 1200R xx (5)/(2) xx 0.7 = - 2100 R`
`W_("Total") = 900 R - 2100 R + 600 R`
` = - 600 R = - 1200 CAL`
92658.

2mole of Fe_(2)C_(2)O_(4) is oxidized by x mole of Cr_(2)O_(7)^(2-)in acidic medium , x is

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SOLUTION :N/A
92659.

2 mole of ethanol is dissolved in 8 mole of water. The mole fraction of water in the solution is

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0.2
0.8
0.4
0.1

SOLUTION :Total NUMBER of MOLES in the solution =10
`:. "MOLE fraction of water"=("Moles of water")/("Total moles")`
`=8/10=0.8`
92660.

2 mole of ethanol are burnt. The amount of CO_2 obtained will be:

Answer»

132 g
44 g
176 g
88 g

Answer :C
92661.

2 mole of CO_(2) is required to preapre:

Answer»


ANSWER :B,d
92662.

2 mol of N_(2) is mixed with 6 mol of H_(2) in a closed vessel of one litre capacity. If 50% of N_(2) is converted into NH_(3) at equilibrium, the value of K_(c) for the reaction N_(2(g))+3H_(2(g))iff2NH_(3(g)) is

Answer»

`4//27`
`27//4`
`1//27`
27

Solution :`{:(,N_(2),+,3H_(2),iff,2NH_(3)),("INITIAL","2 mol",,"6 mol",,),("At. EQM.","1 mol",,"3 mol",,),("Molar conc.",1,,3,,"2 mol ""^(-1)):}`
`K_(C)=((2)^(2))/(1xx(3)^(3))=(4)/(27)`
92663.

2 mol of N_(2) is mixed with 6 mol of H_(2) in a closed vessel of one litre capacity. If 50% of N_(2) is converted into NH_(3) at equlibrium, the value of K_(c) for the reaction N_(2(g))+3H_(2(g))hArr2NH_(3(g)) is

Answer»

`4//27`
`27//4`
`1//27`
`24`

Solution :`underset((2-x))(N_(2))+underset((6-3x))(3H_(2))hArrunderset((2x))(2NH_(3))`
`50%` Dissociation of `N_(2)` TAKE PLACE so,
At equlibrium `(2xx10)/(100)=1,` value of x=1
`K_(C)=([2]^(2))/([1][3]^(3))=4/27so, K_(c)=4/27`
92664.

2 mol of an ideal gas expands isothermally and reversibly from 1 litre to 10 litres at 300 K. What is the enthalpy change?

Answer»

4.98 KJ
11.47 kJ
`-11.47` kJ
0 kJ

Answer :D
92665.

2-Methylpropene upon hydroboration-oxidation gives

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2-Methyl-1 propanol
2-Methyl-2-propanol
2-Propanol
1-Propanol

Answer :A
92666.

2-Methylpropane on monochlorination under photochemical conditions give

Answer»

2-chloro-2-methylpropane as major PRODUCT
(1:1) mixture of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane
1-chloro-2-methylpropane as a major product
(1:9) mixture of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane

Solution :`CH-UNDERSET(CH_(3))underset(|)overset(H)overset(|)(C)-CH_(3)-underset((A))(CH_(3)-underset(CH_(3))underset(|)overset(Cl)overset(|)(C)-CH_(3))+underset((B))(CH_(3)-underset(CH_(3))underset(|)(C)H-CH_(2)-Cl`
RATIO of (A): (B) is 5:9
92667.

2-Methylpropane onmonochlorination under photochemical condition give

Answer»

2-chloro-2-methylpropane as MAJOR PRODUCT
(1:1) MIXTURE of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane
1-chloro-2-methylpropane as a major product
(1:9) mixture of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane

Answer :C
92668.

2-Methylpropan-2-ol on oxidation underdrastic condition gives _________ as the final product.

Answer»

methanoic acid
ETHANOIC acid
PROPANOIC acid
butanoic acid

ANSWER :B
92669.

2-methylpropan-2-ene overset(H_2SO_4//H_2O)

Answer»

Solution :`UNDERSET(("2 - methyl propan - 2- ENE"))(CH_3-underset(CH_3)underset(|)C=CH_2)OVERSET(H_2SO_4//H_2O)to underset(("2-methyl PROPANE - 2-ol"))(CH_3 -underset(CH_3)underset(|)overset(OH)overset(|)C-CH_3)`
92670.

2-methylpentane is :

Answer»




ANSWER :B
92671.

2-methylpropan -1-ene overset(H_(2)SO_(4)//H_(2)O) to ?

Answer»

Solution :`underset("2-methyl propan-1-ene") (CH_(3)-underset(CH_(3)) underset(|)C= CH_(2)+H_(2)O underset(H_(2)SO_(4)) to CH_(3) -underset(CH_(3)) underset(|)OVERSET(OH) overset(|)C-CH_(3)`
It follows Markownikoff's addition.
92672.

2-Methylpenta-2,3-diene is achiral because it has:

Answer»

a PLANE of SYMMETRY
a centre of symmetry
` a c_(2)` AXIS of symmetry.
both a plane and a centre of symmentry.

Answer :C
92673.

2-methylpent-3-enoic acid shows:

Answer»

OPTICAL isomerism
Geometrical isomerism
BOTH (A) AND (B)
NONE of the above

Answer :C
92674.

2- methylbutaneon reactionwithbrominein thepresenceofsunlightgivesmainly :

Answer»

1- bromo-2-METHYLBUTANE
2- bromo-2-methylbutane
2-bromo-3- methylbutane
1-bromo-3- methylbutane

ANSWER :B
92675.

2-Methylbutane on reacting with bromine in the presene of AlBr_(3) gies mainly

Answer»

2-bromo 2-methylbutane
1-bromo 2-methylbutane
1-bromo 3-methylbutane
1-bromo 3-methylbutane

Answer :A
92676.

2-methyl propene is isomeric with butene-1 They can be distinguished by:

Answer»

BAEYER's reagent
Ammoniacal `AgNO_3`
`Br_2` SOLUTION
`o_3 Zn//H_2O`

ANSWER :D
92677.

2-methyl propanoic acid is obtained from dry ice and what

Answer»




ANSWER :C
92678.

2-methyl propan-1-ol is obtained from 2-methyl prop-1-ene by using

Answer»

DIL. `H_2SO_4`
`B_2H_6`/THF, `H_2O_2` and NAOH
`(CH_3COO)_2Hg + NaBH_4`
`H_2`/Ni

Answer :B
92679.

2- Methyl propaneonmonochlorination underphtochemical condition give :

Answer»

2-choro-2-Methylpropaneas MAJORPRODUCT
(1:) mixtureof 1- chloro-2-methypropane2-chloro-2-mehtylprorpane
1-chloro-2-methylpropane
(1:9)mixturefo 1- chloro-2- methylpropane

ANSWER :C
92680.

2-methyl propanal is formed from isopropyl megnesium halide and what ?

Answer»

`CH_(3)CN`
HCHO
HCN
`CH_(3)COCH_(3)`

SOLUTION :`HCN +(CH_(3))_(2) CHMgX underset(H_(3)O^(+)) overset("DRY ETHER") to (CH_(3))_(2) CH - CHO + MgXON`
92681.

2- methyl propane nitrile on acid hydrolysis give

Answer»

BUTYRIC ACID
isobutyric acid
propionic acid
pentanoic acid

ANSWER :B
92682.

2 - Methyl phenol is

Answer»

m - cresol
o- cresol
m - xylene
o- xylene

Answer :B
92683.

2-methyl phenol is

Answer»

m-cresol
Catechol
p-cresol
o-cresol

Answer :D
92684.

2-methyl pentene-2 on ozonolysis will gives :

Answer»

Only propanal
propanal and ethanal
propan-2-one and ethanal
propan-2-one and propanal

Answer :D
92685.

2- methyl butane on reacting with bromine in the presence of sunlight gives mainly:

Answer»

1- BROMO-3- methylbutane
2- bromo -3- methyl BUTANE
2 -bromo- 2 methyl butane
1- bromo -2- methylbutane

ANSWER :C
92686.

2-methyl but-3-en-2-ol belongs to which type of alcohol?

Answer»

`3^@` ALCOHOL
`2^@` alcohol
`1^@`alcohol
Aromatic alcohol

SOLUTION :`3^@` alcohol
92687.

2-methyl 2-propanol with Fenton's reagent gives-

Answer»

1,2-methyl propene-1
2-methyl propene-2
2,5-dimethyl hexanediol-2,5
2,2,3,3-tetramethyl butane

Answer :C
92688.

2-Methyl-2-propanol may be prepared by reacting methylmagnesium iodide with

Answer»

PROPANONE
ETHYL ethanoate
ethanal
ethylene oxide

Solution :Both propanone and ethyl ethanoate on TREATMENT with MEMGI GIVE 2-methyl-2-propanol.
92689.

2-methyl 2-pentanol is prepared from acetone and what?

Answer»

`C_2H_5MgI`
`(CH_3)_3CMgI`
2-molecules of `CH_3MgI`
`CH_3CH_2CH_2MgI`

Solution : 2-methyl 2-pentanol is `3^@` alcohol PREPARED form acetone and n-propyl magnesium iodide
`H_3COCH_3+CH_3CH_2CH_2MgI underset(H_3O^+)OVERSET"DRY ether "to (CH_3)_2C(OH)CH_2CH_2CH_3+MgIOH`
92690.

Write the relavant chemical equations : 2-Methyl--1-propene to 2-chloro-2-methyl propane

Answer»

Solution :`CH_(3)- overset(overset(CH_(3))(|))(C )=CH_(2) overset(HCl)rarr CH_(3)- UNDERSET(underset(Cl)(|))overset(overset(CH_(3))(|))(C )-CH_(3)`
92691.

2-methyl-1- butene on ozonolysis forms

Answer»

Butanone
Methanal
Both (a) and ( C )
None

Answer :C
92692.

2 M of 100 mL Na_2 SO_4 is mixed with 3 M of 100 mL NaCl solution and 1 M of 200 mL CaCl_2 solution . Then the ratio of the concentration of cation and anion is

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`1:1`
`2:1`
`2:3`
`1:2`

ANSWER :A
92693.

2-Methyl-1- bromopropane reacts with alcoholic KCN and the product formed is treated with alkaline H_(2)O_(2) to give A. The structure of A is :

Answer»

`CH_(3)-UNDERSET(CH_(3))underset(|)(CH)-overset(O)overset(||)(C)-NH_(2)`
`CH_(3)-underset(CH_(3))underset("|")(CH)-CH_(2)-CH_(2)COOH`
`CH_(3)-underset(CH_(3))underset("|")(CH)-CH_(2)CONH_(2)`
`CH_(3)-underset(CH_(3))underset("|")(CH)-CH_(2)-CH_(2)-NH_(2)`

SOLUTION :`CH_(3)-overset(CH_(3))overset("|")(CH)-CH_(2)Brunderset(alc.)overset(KCN)rarrCH_(3)-overset(CH_(3))overset("|")(CH)-CH_(2)CNoverset(alk.H_(2)O_(2))rarrCH_(3)-overset(CH_(3))overset("|")(CH)-CH_(2)CONH_(2)`
92694.

2-Methychlcyclohexanone is allowed to react with matachloroperbenzoic acid. The major product in the reaction is

Answer»




SOLUTION :
92695.

2 litres of ammonia at 13^(@) and 0.90atmospheric pressure is neutralised by 134 mL of H_(2)SO_(4) solution .Find the normality of the acid .

Answer»

SOLUTION :`0.57` N
92696.

2-Hexyne gives trans-2- Hexene on treatment with

Answer»

`Pt//H_2`
`Li//NH_3`
`Pd//BaSO_4`
`LiAlH_4`

SOLUTION :
92697.

1-Hexanol or Hexane which has has higher bolling points ?

Answer»

SOLUTION :2-Hexanol
92698.

2^@ gt 3^@ gt 1^@

Answer»

Fescher-Speier esterification
Clemmensen condensation
Claisen condensation
None

Answer :A
92699.

2 g of brass containing Cu and Zn only reacts with 3M HNO_(3) solution. Following are the reactions taking place Cu(s) + HNO_(3) (aq) to Cu^(2+) (aq) +NO_(2)(g) + H_(2)O(I)Zn(s) + H^(+)(aq) + NO_(3)^(-)(aq) to NH_(4)^(+) + Zn^(2+)(aq) + H_(2)O(l)The liberated NO_(2)(g) was found to be 1.04 L at 25^(@)C and 1 atm [Cu=63.5 , Zn=65.4]The volume of HNO_(3) consumed during the reaction with brass is

Answer»

9.52 ml
14.5 ml
16.25 ml
10.5 ml

Answer :B
92700.

2 g of benzoic acid (C_6H_5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg "mol"^(-1). What is the percentage association of acid if it forms dimer in solution ?

Answer»

Solution :The data provided is listed below :
`W_B = 2g, K_f = 4.9 K kg "mol"^(-1) , W_A = 25 g, DELTA T_f = 1.62 K`
Applying the following relation :
` Delta T_f = m xx K_f = (W_B)/(M_B) xx (1000)/(W_A) xx K_f`
Substituting the values, we get
`1.62 = (2)/(M_B) xx 1000/25 xx 4.9`
`M_B = (2000 xx 4.9)/(25 xx 1.62) = 241.98 g"mol"^(-1)`
`underset(1 -x)underset(1)(2C_6H_5COOH) iff underset(x//2)underset(0)((C_6H_5COOH)_2)`
Total number of particles at equilibrium = 1 - x + x/2 = 1 - x/2
`i = (1 - x//2)/(1) = 1 - x//2`
i = Calculated molecular MASS / Observed molecular mass
` =(122)/(241.98)`
Thus `(122)/(241.98) = 1 - x/2`
`x/2 = 1 - (122)/(241.98) = (241.98 - 122)/(241.98) = (119.98)/(241.98) = 0.4958`
` x = 0.9916 = 99.16%`
The degree of association of BENZOIC ACID in benzene is 99.16%.