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93301.

1 g of a mixture of NaHCO_(3) and Na_(2)CO_(3) is heated to 150^(@)C. The volume of the CO_(2) produced at STP is 112.0 mL. Calculate the percentage of Na_(2)CO_(3) in the mixture (Na=23, C=12, O=16)

Answer»

20
46
84
16

Solution :`2NaHCO_(3) overset(150^(@)C)(rarr) Na_(2)CO_(3)+CO_(2)+H_(2)O`
`n_(NaHCO_(3))/n_(CO_(2))=2/1`
`n_(NaHCO_(3))=2n_(CO_(2))`
`=2xx112/22400=0.01` mole
`W_(NaHCO_(3))=0.01xx84=0.84 g`
`W_(Na_(2)CO_(3))=1.00-0.84=0.16 g`
`% Na_(2)CO_(3)=16`.
Discussion
93302.

1 g of a monobasic acid when dissolved in 100 g of water lowers the freezing point by 0.168^(@)C.0.2g of the same acid when dissolved and titrated required 15.1 mL of N/10 alkali solution. Calculate the degree of dissociation of the acid.

Answer»

Solution :Observed MOLECULAR mass of the acid.
`M_(2)=(1000K_(f)w_(2))/(w_(1)DeltaT_(f))=(1000xx1.86xx1)/(100xx0.168)=110.71`
`"15.1 mL of "(N)/(10)" alkali neutrilize acid = 0.2 g"`
`therefore " 1000 mL of 1 N alkali will neutralize acid "=(0.2)/(15.1)xx1000xx10=132.45g`
This is the normal (calculated) molecular mass of the acid.
`therefore""i=("Normal molecular mass")/("Observed molecular mass")=(132.45)/(110.71)=1.196`
`{:(,HA,hArr,H^(+),+,A^(-),),("Initial",1,,0,,0,),("After disso.",""1-ALPHA,,alpha,,alpha",","Total "=1+alpha):}`
`i=1+alpha"or"alpha=i-1=1.196-1=0.196=19.6%`
Discussion
93303.

1 g of a carbonate (M_2CO_3) on treatment with excess HCl produces 0.01186 mole of CO_2. The molar mass of M_2O_3 is g "mol"^(-1) is:

Answer»

1186
84.3
118.6
11.86

Answer :B
Discussion
93304.

1 g of ""_79AU^(198(t_(1//2)=65 h) gives stable mercury by beta-emission. What amount of mercury will left after 260 h?

Answer»

0.9374 g
0.3758 g
0.7586 g
0.9000 g

Solution :`_(79)AU^(198)OVERSET(-BETA)to _(80)HG^(198)`
`k=(0.0693)/(t_(1//2))=(0.0693)/(65)`
After 260 h, `" "k=(2.303)/(260)"log"(a)/(a-x)`
`(0.693)/(65)=(2.303)/(2.60)"log"(a)/(a-x)`
`x=(1498)/(1598)"g"=0.9374"g"`
Since,all radioactive reactions follow first order kinetics.
Discussion
93305.

1 g mixture of glucose and urea present in 259 mL aqueous solution shows the osmotic pressure of 0.74 atm at 27^(@)C. Assuming solution to be dilute, select the correct statements:

Answer»

<P> Percentage of urea in mixture is `17.6`
Relative lowering in vapour pressureof this solution is `5.41 xx 10^(-4)`
(c) The solution will boil at `100.015^(@)C`, if `K_(b)` of water is `0.5 K "molality"^(-1)`
If glucose is replaced by same amount of sucrose, the solution will show higher osmotic pressure at `27^(@)C`

Solution :Let `a` glucose, `b g` urea be present in `1 g`
`:. a+b = 1` ….(1)
Thus, `0.74 xx (250)/(1000) = ((1)/(180)+(b)/(60)) xx 0.0821 xx 300` ….(2)
By eqs. (1) and (2),
`b = 0.176, a = 0.824`
`(P^(@)-P_(S))/(P^(@)) = (w xx M xx 1000)/(m xx W xx 1000)`
`:.` Molality `= (P^(@)-P_(S))/(P^(@)) xx (1000)/(M)`
or `(PI)/(ST) = (P^(@)-P_(S))/(P^(@)) xx (1000)/(M) ,(because M = m)`
`:. (P^(@) -P_(S))/(P^(@)) = (0.74 xx 18)/(0.0821 xx 300 xx 1000)`
`= 5.41 xx 10^(-4)`
`Delta T_(b) = "molality" xx K_(b) = (pi)/(ST) xx K_(b)`
`= (0.74)/(0.0821 xx 300) xx 0.5 = 0.015`
`b.pt. = 100.015^(@)C`
On replacing glucose `(mol. WT. 180)` by sucrose (mol wt. 342), `pi` will DECREASES
as `pi prop (1)/(mol.wt)`
On replacing glucose `(mol. wt. 180)` by `NaCl (mol. wt. 58.5), pi` will increases as
`pi prop (1)/(mol. wt.) xx 2` for `NaCl`
Discussion
93306.

1 g mixture of cuprous oxide and cupric oxide was quantitatively reduced to 0.839 g of metallic copper. What was the weight of cupric oxide in the original sample ? (Cu = 63. 5, O = 16).

Answer»


SOLUTION :N/A
Discussion
93307.

1 g Mg was burnt in a closed vessel containing 2 g oxygen. Which of the following are not correct?

Answer»

0.25 G of Mg will be left unburnt
1.33 of `O_(2)` will be left unreacted
2.5 g of MgO will be formed
The mixture at the end will weigh 3 g.

Solution :`{:(2Mg,+,O_(2),rarr,2MgO),(2xx24g,,32g,,2xx40),(=48G,,,,=80g):}`
`"48 g Mg react with "O_(2)=32 g`
`therefore"1 g Mg will react with "O_(2)=(32)/(48)g=0.67g.`
Thus, whole of Mg will react
`O_(2)" left unreacted "=2-0.67=1.33g`
`MgO" formed"=(80)/(48)g=1.67g.`
Total `=1.67+1.33=3.0g`
Also, by law of conservation of mass, mixture at the end should weigh `1g+2g=3g`
Discussion
93308.

1 g equivalent of Na metal is formed from electrolysis of fused NaCl. Number of mol of Al from fused Na _(3)AlF_(6) will be :

Answer»

1
3
`1/3`
2

Answer :C
Discussion
93309.

1 g equiv of a substance is the weight of that amount of a substance which is equivalent to:

Answer»

0.25 MOL of `O_(2)`
0.50 mol of `O_(2)`
1 mol of `O_(2)`
8 mol of `O_(2)`

ANSWER :A
Discussion
93310.

1 g atom of nitrogen represents:

Answer»

14 G nitrogen
11.2 LITRE of `N_(2)` at NTP
22.4 litre of `N_(2)` at NTP
`6.023xx10^(23)` molecules of `N_(2)`

Answer :A
Discussion
93311.

1 g equivalent of Na metal is formed from electrolysis of fused NaCl. Number of mol of Al from fused Na_(3)AlF_(6) Will be :

Answer»

1
3
`1//3`
2

Answer :C
Discussion
93312.

1 g-atom of Ca was burnt in excess of oxygen and the oxidewas dissolved in water to make up a one - liter solution. Calculate the normality of thealkaline solution.

Answer»

Solution : FirstMethod
`Ca + O_(2) CO overset(H_(2))to Ca (OH)_(2)`
Ca atoms are CONSERVED `:.`applying POACfor Ca atoms,
molesof Ca ATOMSIN reactant = moles of Ca in Ca `(OH)_(2)`
`1 = 1 xx ` moles of `Ca(OH)_(2)`
(`:.` MOLE of `Ca(OH)_(2)` contains1 mole of Ca)
`:.`moles of `CaO=(OH)_(2)= 1 `
Sincethe solutionis madeup to 1 liter ,
molarityof `Ca(OH)_(20`= 1 M
andnormalityof `Ca(OH)_(2) = 2 N`
[ Note : In the above problems,therules applied have been referred to wherevernecessary just to MAKE everstep easy to follow. Itmay, thus, be expected that thestudentsmust now have beenwell acquainted with therulesandtheirapplication, and so in the following problemsrules have been rarely mentioned except where theyare at all required.]
Discussion
93313.

1 fermi is equal to:

Answer»

`10^(-13)` cm
`10^(-10)` cm
`10^(-4)` cm
`10^(-8)` cm

Answer :A
Discussion
93314.

1dm^(3) of2MCH_(3)COOH is mixed with 1dm^(3) of 3M ethanol to form ester. The decrease in the initial rate if each solution is diluted with an equal volume of water would be

Answer»

2 TIMES
4 times
0.25 times
0.5 times

ANSWER :C
Discussion
93315.

1 coulomb of electricity produces m kg of a substance 'X'. The electrochemical equivalent of 'X' is

Answer»

m
`m XX 10^(3)`
`m xx 10^(-3) `
0.1 m

Answer :B
Discussion
93316.

1 coulomb of electricity is passed through a solution of AlCl_(3), 13.5 g of Al is deposited. The number of faradays must be

Answer»

m
`m XX 10^(3)`
`m xx 10^(-3)`
all wrong

Answer :B
Discussion
93317.

1 coulomb of charge passes through solution of AgNO_(3) and CuSO_(4) connected in series and the concentration of two solution being in the ratio 1 : 2 . The ratio of amount of Ag and Cu deposited on Pt electrode is

Answer»

`107.9 : 63.54`
`54 : 31.77`
`107.9 : 31.77`
`54 : 63 .54`

SOLUTION :The ratio of amount deposited during the same CHARGE take place in the ratio of their equivalent weights and is INDEPENDENT of solution concentration .
`therefore` 1g Eq. of Ag = 1 g Eq. of CU
`therefore 107.9 -= 31.77`
Discussion
93318.

1-chloromethyl-4-methyl-1, 3-cyclopentadiene is solvolysed in aqueous acetone based on the above solve the following questions. The number of enantiomeric pair formed during the reaction is

Answer»

1
2
3
4

Answer :B
Discussion
93319.

1-chloromethyl-4-methyl-1, 3-cyclopentadiene is solvolysed in aqueous acetone based on the above solve the following questions. Which of the intermediates is not possible?

Answer»




ANSWER :D
Discussion
93320.

1-chloromethyl-4-methyl-1, 3-cyclopentadiene is solvolysed in aqueous acetone based on the above solve the following questions. which of the following product is not formed?

Answer»




ANSWER :D
Discussion
93321.

1-Chlorobutane on reaction with alcoholic potash gives :

Answer»

1-Butene
1-Butanol
2-Butene
2-butanol

Solution :`CH_(3)CH_(2)CH_(2)CH_(2)Cloverset("ALC. KOH")rarrunderset("1-Butene")(CH_(3)CH_(2)CH=CH_(2))`
Discussion
93322.

1- chlorobutane on reaction with alcohlic KOH gives:

Answer»

1-butene
butanol
2-butene
2-butanol

Answer :A
Discussion
93323.

1-chlorobutane on elimination gives,

Answer»

1-butene
2-butene
1-butanol
2-butanol

Answer :A
Discussion
93324.

(1) Chlorine forms number of oxides such as Cl_(2)O , ClO_(2), Cl_(2) O_(7) chlorine and oxygen donot combine together directly. So, these are prepared by indirect method. HgO. Hgo + Cl_(2) rarr Cl_(2)O + Hg, (2) Some of these oxides undergo disproportionation reactions. For Ex : Cl_(2) O_(6) + 2KOH rarr KClO_(3) + KClO_(4) + H_(2)O ( 3) The oxides of chlrine are acidic in nature. Cl_(2)O_(7) is the anhydride of perchloric acid Cl_(2) O_(7) + H_(2) O rarr 2HClO_(4) Which of the given statement is incorrect

Answer»

`ClO_(3)` exists in equilibrium with `Cl_(2) O_(6)` the dimer diamagnetic while monomer is paramagnetic
`Cl_(2)O` is an ANGULAR molecule
`Cl_(2) O_(6)` is mixed anhydride of chloric acid & perchloric acid
`ClO_(2)` reacts with KOH to give a mixture of `Cl_(2)` & `KClO_(2)`

Solution :`2ClO_(2) + 2KOH rarr KClO_(2) + KClO_(3) + H_(2)O`
Discussion
93325.

(1) Chlorine forms number of oxides such as Cl_(2)O , ClO_(2), Cl_(2) O_(7) chlorine and oxygen donot combine together directly. So, these are prepared by indirect method. HgO. Hgo + Cl_(2) rarr Cl_(2)O + Hg, (2) Some of these oxides undergo disproportionation reactions. For Ex : Cl_(2) O_(6) + 2KOH rarr KClO_(3) + KClO_(4) + H_(2)O ( 3) The oxides of chlrine are acidic in nature. Cl_(2)O_(7) is the anhydride of perchloric acid Cl_(2) O_(7) + H_(2) O rarr 2HClO_(4) Euochlorine is a mixture prepared by heating KClO_(3 with con HCl euchlorine is a mixture of

Answer»

`Cl_(2)` and `Cl_(2)O_(7)`
`Cl_(2) O` and `ClO_(2)`
`Cl_(2)` and `ClO_(2)`
`Cl_(2)O_(5)` and `Cl_(2)`

Solution :MIXTURE of `Cl_(2)` and `ClO_(2)` is Euochlorine
`2KClO_(3) + 4HCL RARR 2KCl + 2ClO_(2) + Cl_(2) + 2H_(2)O`
Discussion
93326.

1-chloro-2, 4-dinitrobenzene to 2, 4-dinitrophenol

Answer»

SOLUTION :1-chloro-2, 4-dinitrobenzene to 2, 4-dinitrophenol :
Discussion
93327.

1-chloributane is treated with alcoholic potash the major product formed is

Answer»

but-1-ene
but-2-ene
butane-1-ol
butane-2-ol

Solution :`CH_(3)CH_(2)CH_(2)CH_(2)Cl+KOH_((ALC)) to CH_(3)CH_(2)CH=CH_(2)+KCl+H_(2)O`
Discussion
93328.

1 cf at liquid of molar volume 100 mL is kept in an adiabatic container under a pressure of 1 bar.The pressure is steeply increased to 100 bar.Under this constant pressure of 100 bar, the volume of the liquid decreases by 1 mL.Calculate DeltaU of the process.

Answer»


SOLUTION :`DeltaU=10J`
`DeltaU=DeltaQ+DeltaW`
for adiabatic process `DeltaQ=O`
`:. DeltaU=DeltaW=-PDeltaV= -100xx10^5(-10^(-6))=10 J`
Discussion
93329.

1 c.c. of 0.01 M HCl is added to 99.9 cc of NaCl solution . PH of resulting solution will be :

Answer»

7
4
2
1

Answer :B
Discussion
93330.

1 C electricity deposits:

Answer»

`10.8` G of Ag
96500 g of Ag
ELECTROCHEMICAL EQUIVALENT of Ag
half of electrochemical equivalent of Ag

Answer :C
Discussion
93331.

1-butyne reacts with cold alkaline KMnO_4 to produce

Answer»

`CH_3CH_2COOH`
`CH_3CH_2CH_2COOH`
`CH_3CH_2COOH+CO_2`
`CH_3CH_2COOH+HCOOH`

SOLUTION :`CH_3-CH_2-C-=CHunderset(ALC. KMnO_4)OVERSET"COLD"to CH_3CH_2COOH+CO_2`
Discussion
93332.

1-Butyne on treatment with hot alkaline KMnO_(4) gives

Answer»

`CH_(3)CH_(2)CH_(2)COOH`
`CH_(3)CH_(2)COOH`
`CH_(3)CH_(2)COOH+CO_(2)`
`(CH_(3)COOH+HCOOH)`

Answer :C
Discussion
93333.

1-Butyne can be prepared by treating sodium acetylide with :

Answer»

ethyl bromide
METHYL bromide
ethyl ALCOHOL
acetic acid

ANSWER :A
Discussion
93334.

1-butyne on reaction with hot alkaline KMnO_4 gives:

Answer»

`CH_3CH_2CH_2COOH`
`CH_3CH_2COOH + CO_2`
`CH_3CH_2COOH`
`CH_3CH_2COOH + HCOOH`

ANSWER :D
Discussion
93335.

1-butyneand 2-butyne are allowed to react separately with the reagents given below: Na, liquid NH_3

Answer»

SOLUTION :Both 1-butyne & 2-butyne REACT separately with reagents 1 and 2 to produce 1 - butene and 2-butene respectively.
However reagents 3 &4 react with 1-butyne & 2-butyne separately to YIELD same products (2-butanone &butane ).
Discussion
93336.

1-Butene may be converted to butane by reaction with

Answer»

Zn-HCl
Sn-HCl
Zn-Hg
`Pb//H_2`

SOLUTION :PD ACTS as CATALYST
Discussion
93337.

1-butanol is treated with PCC gives,

Answer»

`CH_3CH_2CH_2COOH`
`CH_3CH_2CH_2CHO`
`CH_3COCH_2CH_3`
`CH_3COOH`

SOLUTION : 1-butanol is treated with ALKALINE `KMnO_4` undergoes CONTROLLED oxidation gives butanal and when it is treated with strong oxidising AGENT it give DIRECTLY butanoic acid.
Discussion
93338.

1-butanol is reacted with diazomethane to give,

Answer»

1-methoxy butane
2-methoxy butane
1-ethoxy butane
2-ethoxy butane

Answer :A
Discussion
93339.

1-butanol is oxidised by acidified K_2Cr_2O_7 give,

Answer»

butanal
BUTANOIC ACID
butene
butane

Solution :1-butanol is treated with acidified `K_2Cr_2O_7` oxidation GIVES butanoic acid and. when treated with mild OXIDISING AGENT itgive butanal.
Discussion
93340.

1-butane reacts with HBr in presence of peroxide given

Answer»

`CH_(3)-CHBrC_(2)H_(5)`
`CH_(3)-CH_(2)-CH_(2)-CH_(2)-Br`
`(CH_(3))_(2)CH-CH_(2)-Br`
`(CH_(3))_(3)C-Br`

ANSWER :B
Discussion
93341.

1-Bromopropane on reaction with LiAlH_4 yields

Answer»

PROPANE
HEXANE
Propane
PROPYNE

ANSWER :A
Discussion
93342.

1-Bromo-3-chlorocyclobutane when treated with two equivalents of Na, in the presence of ether, which of the following will be formed?

Answer»




SOLUTION :Since ALKYL bromide is more REACTIVE than alkyl chloride, therefore, intramolecular Wurtz reaction occurs on the side of Br atom.
Discussion
93343.

1 - Bromo -2, 2 -dimethylcyclohexane on treatment with methanol gives

Answer»




all of these

Answer :D
Discussion
93344.

1 atm is

Answer»

`236.7 KJ KG^(-1)`
`333.4 kJ kg^(-1)`
`-333.4 kJ kg^(-1)`
`-236.7 kJ kg^(-1)`

ANSWER :C
Discussion
93345.

(1) Arrange sp, sp^(2) & sp^(3) - orbitals in increasing order of : (a) bond length (b) bond angle ( c ) bond energy (d) size of orbitals and ( e ) s - character. (2) Organic compounds are usually water insoluble. Why? (3) Write the structure of the smallest hydrocarbon having empirical formula C_(2)H. What is the shape of the molecule? (4) Draw the p - orbitals involed in forming pi - bonds in the molecule, CH_(2) = C = CH_(2) and predict whether the molecule is planar or not.

Answer»

Solution :(1) (a) `SP-spltsp^(2)-sp^(2)ltsp^(3)-sp^(3)`
(b) `sp^(3)ltsp^(2)ltsp
( c ) )sp^(3)-sp^(3)ltsp^(2)-sp^(2)ltsp-sp`
(d) `spltsp^(2)ltsp^(3)`
( E ) `sp^(3)ltsp^(2)ltsp`
(2) Organic COMPOUNDS are covalent in nature. Thus they do not get ionised. Moreover, they are usually less polar ot non - polar compounds and hence do not dissolve in highly polar solvent, water.
(3) The compound is `(C_(2)H)_(2) or C_(4)H_(2)` and its STRUTURE is `HC-=C-C-=CH`. The shape of the molecule is linear because all the C - atoms are sp - hybridised.
(4) The molecule is non- planar because the two planes containing one C - atom and two H - atoms are perpendicular to each other.
Discussion
93346.

1% aqueoussolutionofCa(NO_(3))_(2)has freezing point

Answer»

`0^(@)C`
LESS than ` 0^(@)C`
`1^(@)C`
`2^(@)C`

ANSWER :B
Discussion
93347.

1 amu is

Answer»

`1.66X10^-24` KG
`1.66X10^-27` kg
`frac{1}{NA}`
Both (2) & (3)

ANSWER :D
Discussion
93348.

1//[A] vs time is a straight line. The order of the reaction is

Answer»

1
2
3
0

Solution :(B) : LOG [A] vs time is LINEAR for 1ST order reactions, `1//[A]` vs time is linear for 2ND order reactions , `1//[A]^(2)` vs time is linear for 3rd order reactions.
Discussion
93349.

1. 84 g of a mixture of CaCO_(3) nad Mg CO_(3) was heated to a constant weight. The constant weightof theresiduewas found to be 0.96g. Calculate thepercentage compositionofthemixture. ( Ca = 40 , Mg = 24 , c = 1 , o = 16)

Answer»

Solution :On heating `caCO_(3) and MgCO_(3)` , one of the products , `CO_(2)` , escapes out .
We have,
`{:(CaCO_(3)+MgCO_(3)""to ""CaO+MgO+CO_(2)uarr),("x g(-84 - x) gy g (0.96 - y ) g "),("(say)(say)"):}`
APPLYING POAC for Ca atoms,
moles of Ca atoms in `CaCO_(3)` = moles of Ca atomsin CaO
`1 xx ` moles of `CaCO_(3) = 1 xx ` molesof CaO
`(x)/(100) = (y)/( 56)""[{:(CaCO_(3)=100),(" "CaO = 56):}]. . . (i)`
Again applying POAC for Mg atoms,
moles of `MgCO_(3) ` = molesof Mg in MgO
`1 xx` molesof `MgCO_(3)= 1 xx ` moles of MgO
`(1 . 84 - x)/( 84) = (0.96 - y)/( 40 ) [{:(MgCO_(3)=84),(" MgO = 40 "):}]"". . . (ii)`
From eqns . (i) and (ii), we GET x = 1 g, y = 0. 84 g
`% of Ca CO_(3)= (1)/( 1 . 84) xx 100 = 54.34%`
and `% of MgCO_(3) = 45.66%`
Second Method Applying POAC for C atoms.
Discussion
93350.

(1)/(4)[Li_(CH_(3))_(4)]+[Mo(CO)_(6)] to [X] Here X is Lithium acetylpentacarbonyllmolybdate(1-), then find the total number of Mo-C linkages in the complex X

Answer»


SOLUTION :`X=Li[Mo(CH_(3)CO)(CO)_(5)]`
Discussion