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52151.

0.01 mol TNT was completely decomposed as, {:(C_(7)H_(5)N_(3)O_(6)(s), rarr, CO(g),+,H_(2)(g),+, N_(2)(g),+,C(s)),(""TNT,,,,,,,,):} The gases evolved occupied 2.24 L at constantpressure and 273 K in the eudiometer . Select the correct option on the basis of above information.

Answer»

Partial pressure of CO is 0.6 atm in the evolved gas
If just sufficient `O_(2)`si introduced in the CONTAINER to combust CO and `H_(2)`completely, then final volume FO gases would be 0.68 L at 1 atm and 273 K
If just sufficient `O_(2)` is introduced in the container to combust CO and `H_(2)`completely, then fianl volumeof gases would be 0.896 L at 1 atm and 273 K
If after combustion the mixture of gases at 273 K is passes through KOH(aq) , contraction of 1.344 L would take place

Answer :A::B::D
Discussion
52152.

0.01 M solution of an organic acid is found to have a pH of 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_(a).

Answer»


Solution :PH = 4.15 means `- log [H^(+)] = 4.15 or log [H^(+)] = - 4.15 = bar(5) . 85 or [H^(+)] = 7.08 xx 10^(-5)M`
`HA hArr H^(+) + A^(-)`
HENCE, at equilibrium `[H^(+)]=[A^(-)] = 7.08 xx 10^(-5)M ~= 7.1 xx 10^(-5)M`
`[HA] = (0.01-7.1xx10^(-5))=(0.01-0.000071)M = 0.009929 M`
`K_(a) = ([H^(+)][A^(-)])/([HA])=((7.1xx10^(-5))^(2))/(9.929xx10^(-3))=5.08 xx 10^(-7)`
`pK_(a) = - log K_(a) = - log (5.08 xx 10^(-7))= 6.29`.
Discussion
52153.

0.01 M solution is given it's pH is .......... (K_a=6.6xx10^(-4))

Answer»

7.6
8
2.6
`5.0`

Solution :`alpha=SQRT(K_a/C)=sqrt((6.6xx10^(-4))/0.01)`
`=sqrt(6.6xx10^(-2))`=0.257
`[H^+]=C-alpha`
=0.01 x 0.257
`=2.57xx10^(-3)`
`[H^+]=sqrt(K_a.C_o)`
`=sqrt(6.6xx10^(-4)xx10^(-2))`
`[H^+]= 2.57xx10^(-3)`
`therefore` pH=-log `(2.57xx10^(-3))`
=3-log 2.5
=(3-0.4)=2.60
Discussion
52154.

0.01 M NH_3 and (0.01 M of 50 mL NH_3) + (0.01 M, 25 mL H) from these two solution which has pH more ?

Answer»

Solution :In the MIXTURE of (HCl + `NH_3`) after neutralization more `NH_3` left without neutralization so PH decrease.
Initial pH of only BASE is more pH of only base `gt`(more base + LESS acid) pH
Discussion
52155.

0.0093 g of Na_(2)H_(2)EDTA.2H_(2)O is dissolved in 250 mL of aqueous solution. A sample of hard water containing Ca^(2+) and Mg^(2+) ions is titrated with the above EDTA solution using a buffer of NH_(4)OH+NH_(4)Cl using eriochrome balck-T as indicator. 10 mL of the above EDTA solution requires 10 mL of hard water at equivalent point. another sample of hard water is titrated with 10 mL of above EDTA solution using KOH solution (pH=12). using murexide indicator, it requires 40 mL of hard water at equivalence point. a. Calculate the ammount of Ca^(2+) and Mg^(2+) present in 1L of hard water. b. Calculate the hardness due to Ca^(2+), mG^(2+) ions and the total hardness of water in p p m of CaCO_(3). (Given MW(EDTA sal t)=372 g mol^(-1),MW(CaCO_(3))=100 gmol^(-1))

Answer»

Solution :Case I: Using erichrome black-`T` indicator
`M` of `EDTA` solution `=(0.093xx1000)/(372xx250)=0.001 M`
Volume of `EDTA` used `=10 mL`
Volume of water sample=`40 mL`
`M_(1)V_(1)(EDTA)=M_(2)V_(2)(Ca^(2+)` and `Mg^(2+)` in hard water)
`0.001xx10=M_(2)xx10`
`M_(2)=0.001`
`:.` Molarities of `(Ca^(2+)+Mg^(2+))`ions `=0.001 M`
`=0.1` mmoles `L^(-1)`
Case II: Using murexide indicator
`M_(1)V_(1)(EDTA)=M_(2)V_(2)` (Hard water)
`0.001xx10=M_(2)xx40`
`M_(2)=0.25xx10^(-3)=0.25 mmol L^(-1)`
`:.` Total `mmol L^(-1)`of `Ca^(2+)` and `Mg^(2+)=1.0`
`mmolL^(-1)` of `Ca^(2+)` and `Mg^(2+)=1.0`
`mmol L^(-1)` of `Mg^(2+)=1.0-0.75 mmol L^(-1)`
mmoles `L^(-1)` of `Ca^(2+)=0.25 mmol L^(-1)`
`:.` Amount of `Ca^(2+) L^(-1)implies0.25xx40xx10^(-3)`
`=0.01gL^(-1)`
Amount of `Mg^(2+) L^(-1)implies0.75xx24xx10^(-3)=0.018 gL^(-1)`
(b) `[MW (CaCO_(3))=100 g mol^(-1)]`
Total `mmol of Ca^(2+)` and `Mg^(2+)` ions `L^(-1)`
`=1.0=0.001 molL^(-1)`
`=0.001 M`
`0.1 mol of Ca^(2+)` and `Mg^(2+)-=0.1 mol CaCO_(3) L^(-1)`
`-=(0.001xx100xx10^(6))/(10^(3))`
`-=100 p p m`
`:.` total hardness due to `Ca^(2+)` and `Mg^(2+)` ions
of the sample in grams of `CaCO_(3)` in `10^(6) mL of H_(2)O`
`=("Total MxMW"+N19(CaCO_(3))xx10^(6))/(10^(3))`
Hardness due to `Ca^(2+)` ions of the sample in gram of `CaCO_(3)` in `10^(6) mL H_(2)O=(0.25xx10^(-3)xx100xx10^(6))/(10^(3))=25 p p m`
Hardness due to `Mg^(2+)` ions of the sample in grams of `CaCO_(3)` in `10^(6)mL of H_(2)O=100-25 =75 p p m`
Discussion
52156.

0.004 gof NaOHis present in 1Lof a solution. The pH of the solution is

Answer»

`14.0`
` 12.0`
` 10.0`
` 8.0 `

Solution :`N= (0.004) /(40 ) = 10 ^(-4),V =1L `
` therefore [OH^(-) ] =(10 ^(-4))/( 1) = 10 ^(-4)`
` thereforePOH = 4 rArr pH = 14- 4 = 10 `
Discussion
52157.

0.0026 M I_(2) solution having unknown volume is reacted with excess of ferrous thiocynate solution to form Fe_(2)O_(3),SO_(4)^(2-),CN^(-) along with I^(-) . If all the sulphate ions formed are precipitated using BaCl_(2) such that 16776 gm of BaSO_(4) is obtained, calculate volume of I_(2) consumed in litre. (At. mass : Ba=137)

Answer»


ANSWER :9
Discussion
52158.

0.002 M potassium thiocyanate added to 1 mL 0.2 M Fe (III) nitrate than what happen ?

Answer»

Solution :The iron (III) Nitrate `Fe^(3+)` is react with potassium thiocyanate `SCN^-` and FORM `[Fe(SCN)]_((aq))^(2+)` and FOLLOWING equilibrium is established. The solution is red because the ferric thiocyanate ION is red.
`Fe_((aq))^(3+) + SCN_((aq))^(-) hArr UNDERSET"dark red"(`[Fe(SCN)]_((aq))^(2+))`
Discussion
52159.

0.00025 has how many significant figures?

Answer»

5
3
`-4`
2

Answer :D
Discussion
52160.

0.00050 mole of NaHCO_(3) is added to a large volume of a solution buffer at pH = 8.00. How much material will exist in each of the three forms, H_(2)CO_(3), HCO_(3)^(-) and CO_(3)^(2-) ? K_(1) and K_(2) for H_(2)CO_(3) are 4.5xx10^(-7) and 4.5xx10^(-11) respectively.

Answer»


Answer :`n_(H_(2)CO_(3))= 1.069xx10^(-5), n_(HCO_(3)^(-))= 4.86xx10^(-4)`,
`n_(CO_(3)^(2-))= 2.28xx10^(-6)`;
Discussion
52161.

, X is :

Answer»




SOLUTION :
Discussion
52162.

........ was used as solvent for dry cleaning of clothes.

Answer»

TETRACHLOROETHANE
DICHLORO ethene
Both (A) and (B)
NONE of these

Solution :Tetrachloroethane
Discussion
52163.

............ used in refrigerator is hazardous to ozone layer.

Answer»

CFC
CMC
CNC
FCC

Answer :A::C
Discussion
52164.

-underset(|)overset(|)CHgroupin alkanesis called_____. Group .

Answer»

carbene
methylene
isomethine
methine

Answer :D
Discussion
52165.

………. undergoes oxidation in the following reaction. Zn_((s)) + 2NaOH_((aq)) to Na_2ZnO_(2(aq)) + H_(2(g))

Answer»

Zn
`NA^+`
`OH^-`
`H_2`

Solution :The OXIDATION number of Zn CHANGES from 0 to +2
Discussion
52166.

, the product can be :

Answer»




SOLUTION :It is BIRCH REDUCTION.
Discussion
52167.

, the product can be :

Answer»




ANSWER :B
Discussion
52168.

. The product A will be-

Answer»




ANSWER :A
Discussion
52169.

. The molecular weight difference between X and Y is

Answer»


SOLUTION :BENZENE PROPERTIES
Discussion
52170.

. The possible product (s) of the above reaction

Answer»




both (1) & (2)

SOLUTION :
Discussion
52171.

…….. statement is true for Al_2O_3 and Al(OH)_3

Answer»

Both are ACIDIC.
Both are basic.
Both are amphoteric.
`Al_2O_3` is acidic while `Al(OH)_3` is basic.

Answer :C
Discussion
52172.

---------- solids do not possess sharp melting points and can be considered as ------ liquids.

Answer»

SOLUTION :AMORPHOUS , SUPERCOOLED
Discussion
52173.

+Rpower is not possessed by given groups

Answer»

`-O^(-)`
`-NH_2`
`-OH`
`-COOH`

ANSWER :D
Discussion
52174.

. Q. Which combination is correct for Ni.

Answer»

(II) (i) (R)
(II) (ii) (Q)
(II) (iii) (S)
(II) (IV) (P)

ANSWER :B
Discussion
52175.

. Q. Which combination is correct:

Answer»

<P>(I) (i) (P)
(II) (ii) (Q)
(III) (iii) (S)
(IV) (iv) (R)

ANSWER :B
Discussion
52176.

. Q. Which combination is/are not correct

Answer»

(I) (ii) (Q)
(III) (i) (R)
(I) (iii) (S)
(IV) (iv) (P)

Answer :B
Discussion
52177.

. Product P, Q ,and R are :

Answer»

`P implies I, Q and R implies II`
`P and Q` both I and II, `R implies II`
P is both I and II, Q and `RimpliesII`
P, Q and R `implies ` all II

Solution :Unsymmetrical alkyne gives two PRODUCTS with catalytic dehydration and with HBO reaction, where as with `Sia_2BH + H_2O_2+O^(-) H`, it gives only one product. more BULKY `Sia_2BH` attacks on the less hindered triple bond to give only one product.
Discussion
52178.

, Product B and C are respectively :

Answer»


`CH_(3)-CH_(2)-UNDERSET(O)underset(||)(C)-H` and `CH_(3)-underset(O)underset(||)(C)-H`
`H-underset(O)underset(||)(C)-H` and `CH_(3)-CH_(2)-underset(O)underset(||)(C)-CH_(3)`
`CH_(3)-underset(O)underset(||)(C)-CH_(3)` and `H-underset(O)underset(||)(C)-H`

Solution :
Discussion
52179.

........ present in polluted water are also the food per bacteria.

Answer»

Organic COMPOUNDS
Inorganic compounds
SOLUBLE salts
Metals

SOLUTION :Inorganic compounds
Discussion
52180.

_____playsan importantroleinWurtz synthesis .

Answer»

LI
NA
K
Cs

Answer :B
Discussion
52181.

................. plantation can metabolise nitrogen oxide and control photochemical smog.

Answer»

SOLUTION :PINUS TREE
Discussion
52182.

______ on reductive ozonolysis gives ethanal, methanal and carbon dioxide.

Answer»


ANSWER :1,2-Butadiene (`CH_3CH=C=CH_2`)
Discussion
52183.

% of H inH_(2)O is .........

Answer»

11.11
88.89
2
20

Solution :MOLECULAR MASS of `H_(2)O = 18 gm//"mole"^(-1)`
Discussion
52184.

% of H in CH_(3)COOH= ..........

Answer»

40
6.66
10
15

Answer :B
Discussion
52185.

....... obtained on passing of dry ammonia over sodium metal at high temperature.

Answer»

`Na_(2)NH_(2)`
`NaNH_(2)`
`NaNH_(3)`
`Na_(2)NH_(3)`

Solution :`2Na+2NH_(3) OVERSET(573-673K)to 2NaNH_(2)+H_(2)`
Discussion
52186.

........ monument (wonder of world) is getting demage due to acid rain.

Answer»

SOLUTION :TAJMAHAL
Discussion
52187.

"_____________" modified Linde process such that the cooling becomes more efficient.

Answer»

CLAUDE
JOULE
Thomson
Planck

Answer :A
Discussion
52188.

........... method is used to prepare sodium carbonate.

Answer»

SOLUTION :SOLVAY PROCESS
Discussion
52189.

.... metals having very high hydration fraction.

Answer»

LI
NA
K
RB

ANSWER :A
Discussion
52190.

……. method is used to obtain H_2O_2 from 2 ethylanthraquinol ?

Answer»

SOLUTION :OXIDATION
Discussion
52191.

....... metals are used to form windows of X-ray tubes.

Answer»

MG
BA
CU
Be

ANSWER :D
Discussion
52192.

...... metal is used to form Grignard reagent.

Answer»

MG
LI
CU
Be

ANSWER :A
Discussion
52193.

________ metal act as co-factor in phosphate transfer of ATP by enzymes.

Answer»

SOLUTION :MAGNESIUM
Discussion
52194.

Mention the uses of deuterium.

Answer»

SOLUTION :(1) Deuterium is USED as a tracer element.
(II) Deuterium is used to study the movement of ground water by isotopic EFFECT.
Discussion
52195.

(+)- mandelic acid has a specific rotation of 158^(@) What would be the observed specific rotation of a mixture of 25% (-)mandelic acid and 75% (+)- mandlic acid?

Answer»

`+118.5^(@)`
`-118.5^(@)`
`-79^(@)`
`+79^(@)`

SOLUTION :RACEMIC = 50%, `(+)to 50%`
Discussion
52196.

{: ( "List -I ", "List - II"),( "(A)Additional reaction", "1. Nitration") , ("(B)Elimination reaction" , "2. Hydration"),("(C)Nucleophilic substitution" ,"3.Dehydration"),("(C)Electrophilic substitution" ,"4.Hydrolysis of alkyl halides") :}

Answer»

1,2,3,4
4,3,2,1
2,3,4,1
3,4,2,1

Answer :C
Discussion
52197.

…………… law is used to explain gas-solution equilibrium processes.

Answer»

SOLUTION :Henry.s LAW
Discussion
52198.

(+-) Lactic acid is optically inactive due to

Answer»

INTERNAL COMPENSATION
EXTERNAL compensation
Presence of PLANE of symmetry
Absence of ASYMMETRIC carbon

Solution :External compensation
Discussion
52199.

Soda lime test is used to detect one of the following element of organic compound

Answer»

PHOSPHORS
chlorine
nitrogen
Oxygen

Answer :D
Discussion
52200.

________ isusefulin purificationof bauxite .

Answer»

Sodiumcarbonate
SODIUM HYDROXIDE
SODIUMCHLORIDE
Soidumhydrogencarbonate

Answer :B
Discussion