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28501.

The molecular formulae of methanoic acid and propanoic acid differ by:

Answer»

`C_2H_4`
`CH_4`
`CH_2`
`CH_2CH_2CH_3`

ANSWER :B
Discussion
28502.

The major product of the reaction is

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ANSWER :(C )
Discussion
28503.

The major product of the reaction is…….

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ANSWER :A
Discussion
28504.

The molecular electronic configuration of B_2 is

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`KK(SIGMA2S)^2(sigma^ast2s)^2(pi2p_x^1=pi2p_y^1)`
`KK(sigma2s)^2(sigma^ast2s)^2(pi2p_x)^2`
`KK(sigma2s)^2(sigma^ast2s)^2(sigma2p)^2`
`KK(sigma2s)^2(sigma^ast2s)^2(sigma2p)^1(PI2P)^1`

Answer :A
Discussion
28505.

The molecular conductivity of strong electrolyte

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increases lineraly with concentration
increases LINEARLY with concentration in a linear fashion
decreases lineraly with concentration
decrases with SQUARE ROOT of concentration in a linearfashion.

Solution :`^^_(m)^(@)= ^^_(m)^(@)-Asqrt(C)`
Discussion
28506.

The major product of the reaction is…….

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Solution :`C_(2)H_(5)O^(Θ)Na^(+)` being a strong BASE FAVOURS ELIMINATION.
Discussion
28507.

The molarity of NaOH solution obtained by dissolving 4g of it in 250 mL solution is :

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0.4 m
0.8m
0.2m
0.1 M Glucose.

Solution :Molality (m)
`=("Mass of mass//Molar mass")/("Mass of solvent in kg")`
`=((4G)//(40g mol^(-1)))/(250//(1000KG))=0.4 m`
It is ASSUMED that DENSITY of WATER is `1g//mL`.
Discussion
28508.

The major product of the reaction is :

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ANSWER :C
Discussion
28509.

The molecular conductance and equivalent conductances are same for the solution of

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1 M th `(NO_(3))_(4)`
1 M `BA(NO_(3))_(2)`
1 M `LA(NO_(3))_(3)`
1 M NaCl

Answer :D
Discussion
28510.

The major product of the reaction between m-dinitrobenzene and NH_(4)HS is

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ANSWER :B
Discussion
28511.

The mole .weight of NaCl determined by studying freezing point depression of its 0.5% aqueous solution is 30. The apparent degree dissociation of NaCl is

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`0.95`
`0.5`
`0.6`
`0.3`

ANSWER :A
Discussion
28512.

The major product of the reaction between CH_(3)CH_(2)ONa and (CH_(3))_(3)C Cl in ethanol is

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`CH_(3)CH_(2)OC(CH_(3))_(3)`
`CH_(2)=C(CH_(3))_(2)`
`CH_(3)CH_(2)C(CH_(3))_(3)`
`CH_(3)CH=CHCH_(3)`

Solution :`CH_(3)-underset(CH_(3))underset(|)OVERSET(CH_(3))overset(|)C-Clundersetunderset(("base"))("Alkoxide")overset(CH_(3)-CH_(2)-overset(Theta)Ooverset(o+)Na)toCH_(3)-overset(CH_(3))overset(|)C=CH_(2)`
`3^(@)` (halide)
Alkoxide ion is strong nucleophile and strong base & with `3^(@)` Alkyl halide Alkenes is the MAJOR PRODUCT `[E_(2)"ELIMINATION"]`
Discussion
28513.

The mole percentage of oxygen in a mixture of 7.0 g of nitrogen and 8.0 g of oxygen is

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8
16
21
50

Answer :D
Discussion
28514.

The major product of the reaction :

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ANSWER :A
Discussion
28515.

The mole percentage of hydrogen in a mixture of 6 g of hydrogen and 28 g of nitrogen is :

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25
50
75
100

Answer :C
Discussion
28516.

The mole of fraction of nitrogen, in a mixture of 7 g of N_(2) and 16 g of O_(2) is

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0.5
0.75
0.66
0.33

Solution :MOLES of `N_(2) = 7/28=0.25` and
Moles of `O_(2) = 16/32 = 0.5`
MOLE FRACTION of `N_(2) = 0.25/(0.25 + 0.5)=0.33`
Discussion
28517.

The major product of thereaction :

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`CH_(3) CH = CH_(2) + CI// H_(2) O `
`CH_(3) UNDERSET(OH) underset(|) (C HC H_(2)) CI`
`CH_(3) CH_(2) CH_(2) OCI`
`CH_(3) CH(C )CH_(2) OH`

Answer :B
Discussion
28518.

The mole fractionof Xin the vapours in equilibrium withhomogenous mixtuerof liquids X andY is 0.42. Thevapour pressure of liquids X and Y at the sametemperatureare 406.5 and 140 torr respectively. Calculate the mole fractionof Xin the solution.

Answer»

Solution :According the Raoult.s law,
`p_(X) = p_(X)^(@) xx x_(X)""…..(i)`
`p_(Y) = p_(Y)^(@) xx x_(Y) = p_(Y)^(@) (1-x)""…….(ii)`
In VAPOUR STATE , MOLEFRACTIONS of X and Y are:
`y_(X) = (p_(X))/(p_("(total)"))`
0.42 `= (p_(x))/(p_("total")) orp_(X) = 0.42 p_("total")""....(III)`
Similarly , `y_(Y) = (p_(y))/(p_("total"))`
`0.58 = (p_(y))/(p_("total")) orp_(Y)= 0.58 p_("total") "".......(iv)`
Dividing eq.(iii) by eq,(iv).
or ` ""(p_(y))/(p_("total")) = (0.42)/(0.58) = 0.724`
From eq.(i) ,(ii) and eq.(v) ,
`(p_(x))/(p_(y)) = (p_(x)^(@) xx x_(x))/(p_(y)^(@) xx (1-x_(x))) = 0.724`
` or (406.5x)/(140(1-x)) = 0.724`
or `"" 406.5 x_(x) = 101.36 - 101.36 x_(X)`
`507.86 x_(X) = 101.36`
`therefore"" x _(X) = 0.20`
Discussion
28519.

The major product of the given reaction is: .

Answer»




SOLUTION :The PRODUCT is formed as a RESULT of electrophilic addition.
.
Discussion
28520.

Themole fraction of water in a solution containing 50g of water and 50g of ethyl alcohol is :

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`(C_(6)H_(5)CH_(3))`
`18/(18+46)`
`1.09/(1.09+2.78)`
`2.78/(1.09+2.78)`

Solution :`"MOLE fraction of water"=("Mass of water"/"Molar Mass of water")/("Mass of water"/("Molar mass of water"))+("Mass of ETHYL ALCOHOL")/("Molar mass of ethyl alcohol")`
`(50/18)/(50/18+50/46)=2.78/(2.78+1.09)`
Discussion
28521.

The major product of the following reactions sequence is

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ANSWER :A
Discussion
28522.

The major product of the following reactions is :

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ANSWER :A
Discussion
28523.

The major product of the following reaction sequence is

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ANSWER :B
Discussion
28524.

The mole fraction of water in a solution containing 50 g of water and 50 g of ethyl alcohol is:

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50/(50+50)
18/(18+46)
1.09/(1.09+2.78)
2.78/(1.09+2.78)

ANSWER :D
Discussion
28525.

The mole fraction of water in 20% aqueous solution of H_(2)O_(2) is

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`(77)/(68)`
`(68)/(77)`
`(20)/(80)`
`(204)/(231)`

Answer :B::D
Discussion
28526.

Themajor product of the following reaction sequence is :

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ANSWER :A
Discussion
28527.

The mole fraction of water in a mixture containing equal number of moles of water and ethanol is

Answer»

1
0.5
2
0.25

Answer :B
Discussion
28528.

Themajorproduct of thefollowingreactionis (##MOD_SPJ_CHE_XII_P2_C12_E07_108_Q01.png" width="80%">

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SOLUTION :
`NaBH_4 `can REDUCE ` - UNDERSET(I) C=O`group-OHbutcannotreduceC=C
Discussion
28529.

The mole fraction of the solvent in the solution of a non-volatile solute is 0.980. The relative lowering of vapour pressure is :

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0.01
0.98
0.02
0.49

Answer :C
Discussion
28530.

The major product of the following reaction is CH_(3)-overset(" "CH_(3))overset(|)underset(" "OH)underset(|)(C)-CH_(2)-OHoverset(H_(2)SO_(4))rarr

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`(CH_(3))_(2)C =CH_(2)`
Butan-2-one
`(CH_(3))_(2)-overset(OH)overset(|)(C )-CHO`
Isobutyraldehyde

Solution :`CH_(3)-overset(CH_(3))overset(|)underset(OH)underset(|)(C)-CH_(2)-OHoverset(H^(+))RARR`
`CH_(3)-underset(o+)underset(OH_(2))underset(|)overset(CH_(3))overset(|)(C)-CH_(2)-OH underset(-H_(2)O)rarr CH_(3)-underset(DARR)underset(o+)overset(CH_(3))overset(|)(C)underset("HYDROGEN shift")(-CH_(2)OH)`
`CH_(3)-overset(CH_(3))overset(|)(CH)-CHO underset(-H^(+))larr CH_(3)-overset(" "CH_(3))overset(|)(C)-overset(o+)(C)HOH`
Discussion
28531.

The mole fraction of solute is 0.2 at that time the decrease in vapoure pressure is 10 mm. If the decrease in vapour pressure is 20 mm, then what will be the mole fraction of solute ?

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<P>`0.2`
`0.4`
`0.6`
`0.8`

Solution :`(p^(@)-p)/(p^(@))=(N)/(n+N)`
`(DELTA p)/(p^(@))=0.2` (given)
`(10)/(p^(@))=0.2 "" THEREFORE p^(@)=50`
Now `(p^(@)-p)/(p^(@))=(n)/(n+N)`
`therefore (20)/(50) = x "" x =0.4`
Discussion
28532.

The mole fraction of solvent in 0.1 molal aqueous solution is

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`0.9982`
`0.0017`
`0.017`
`0.17`

ANSWER :A
Discussion
28533.

The major product of the following reaction is: C_(6)H_(5)CH_(2)-underset(Br)underset(|)overset(CH_(3))overset(|)(C)-CH_(2)-CH_(3) underset(C_(2)H_(5)OH)overset(C_(2)H_(5)Na)to

Answer»

`C_(6)H_(5)CH_(2)-UNDERSET(OC_(2)H_(5))underset(|)OVERSET(CH_(3))overset(|)(C)-CH_(2)-CH_(3)`
`C_(6)H_(5)CH=underset(CH_(3))underset(|)(C)-CH_(2)-CH_(3)`
`C_(6)H_(5)CH_(2)-underset(CH_(3))underset(|)(C)=CHCH_(3)`
`C_(6)H_(5)CH_(2)-underset(CH_(2)CH_(3))underset(|)(C)=CH_(2)`

Solution :
Discussion
28534.

The molefractionofnitrogen, in amixtureof asolutionhavea sumof ______.

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`0.5`
`0.75`
`0.66`
`0.33`

ANSWER :D
Discussion
28535.

The major product of the following reaction is

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a hemiacetal
an acetal
an ether
an ester

Answer :B
Discussion
28536.

The major product of the following reaction is

Answer»




ANSWER :A
Discussion
28537.

The mole fraction of saturated solution is 1.2 xx 10^(-6) . The pressure of the above the solution is -(K_(H)= 1.44.97 bar )

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0.174 bar
17.4 bar
27.4 bar
0.274 bar

Solution :`P_(He)=K_(H) XX X_(He)`
`P_(He)=(144.97 xx 10^(3) "bar") xx (1.2 xx 10^(-6))=0.174 "bar"`
Discussion
28538.

The major product of the following reaction is:

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SOLUTION :
Discussion
28539.

The major product of the following reaction is:

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ANSWER :C
Discussion
28540.

The major product of the following reaction is

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Solution :`CH_(3)O^(-)` is both a strong base as WELL as a strong nucleophile, thereofre both `S_(N_(2))` and `E_(2)` reactions are favourable. The given alkyl HALIDE is secondary `(2^(@))` in nature and `beta`-position is quite hindered. Therefore `E_(2)` is more favourable than `S_(N_(1))` pathway. since the polarity of the solvent `CH_(3)OH` is not quite high, `E_(1)` is also not favourable. under the conditions, `E_(2) `pathway is favourable.
.
Discussion
28541.

The major product of the following reaction is:

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Solution :It is easier to PERFORM nucleophilic SUBSTITUTION in alkyl halides as COMPARED to aryl halides. Therefore the nucleophilic substitution `(S_(N^(2)))` takes place inside chain where the attacking nucleophilic `PhS^(-)` ION replaces `Br^(-)` ion accompanied by inversion of configuration.
.
Discussion
28542.

The mole fraction of oxalic acid (Molar mass 63) required to prepare 0.10 m solution in water is

Answer»

1
0.0018
6.3
0.0992

Solution :`X_(2)= (mM_(1))/(1000+mM_(1))`
`=(0.1 XX 18)/(1000 + 0.1 xx 18)`
=0.0018
Discussion
28543.

The major product of the following reaction is

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Solution :Aralkyl halids are more REACTIVE than aryl HALIDES towards nucleophilic occurs at the more reactive `CH_(2)Cl` INSTEAD of at `Br` as shown below.
Discussion
28544.

The mole fraction of helium in a saturated solution at 20^(@)C is 1.2xx10^(-6). Find the presssure of helium above the solution. Given Henry's constant at 20^(@)C=144.97" bar".

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<P>

Solution :`p_(He)=K_(H)xx x_(He)=(144.97xx10^(3)" bar")(1.2xx10^(-6))=0.174" bar."`
Discussion
28545.

The major product of the following reaction is

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ANSWER :(a)
Discussion
28546.

The major product of the following reaction is

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a HEMIACETAL
an ACETAL
an ETHER
as ESTER.

SOLUTION :
Discussion
28547.

The major product of the following reaction is:

Answer»




SOLUTION :
Discussion
28548.

The mole fraction of methanol in its 4.5 molal aqueous solution is

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`0.250`
`0.125`
`0.100`
`0.075`

Solution :Let a mole of `CH_(3)OH` and b mole of WATER be PRESENT in the solution.
Mass of water `=b xx 18`
Molality of solution
`=(a)/(b xx 18)xx1000=4.5`
or `a/b=(18xx4.5)/(1000)=0.081`
Mole FRACTION of `CH_(3)OH`
`=(a)/(a+b)` or `(1)/(1+b//a)`
`=(1)/(1+(1)/(0.081))=(1)/(1+12.34)=0.075`
Discussion
28549.

The major product of the following reaction is .

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.
.
.
.

Solution :The product (a) will be formed
Nucleophilic substitution of an alkyl halide is easier as compared to that of an ARYL halide
`ph-S` is a strong nucleophile and dimethyl FORMAMIDE
`(overset(O)overset(||)HCNMe_(2))` is a highly polar aprotic solvent These
REAGENTS favour `S_(N)2` reaction the major product formed is INVERSION product
.
Discussion
28550.

The major product of the following reaction after acidification is Me_(2) "CHCHO" + H_(2) C = CH - "COCH"_(3) overset(OH^(-))(rarr)

Answer»




ANSWER :C
Discussion