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26051.

The percentage of Fe​^( +3)​ ion present in Fe0.93O1.00 is :

Answer»

`15% `
`5.5% `
`10.0% `
`11.5% `

SOLUTION :
`X(+2)+3(0.93 -x )-2 =0`
`X=0.79`
`Fe ^(+2)= 0.79 Fe^(+3)= 0.14`
`% Fe ^(+3)=(0.14)/(0.93 )xx 100 =15 %`
Discussion
26052.

The percentage of ethyl alchol by weight is 46% in a mixture of ethanol and water. The mole fraction of alcohol in this solution is

Answer»

0.25
0.75
0.46
0.54

Answer :A
Discussion
26053.

The percentage of empty space in a body centred cubic arrangement is

Answer»

74
68
32
26

Solution :In BCC lattice, SPACE occupied = 68% `""THEREFORE`Empty space = 100 – 68 = 32%
Discussion
26054.

The percentage of copper (II) salt can be determined by using a thiosulphate titration. 0.305g of a copper (II) salt was dissolved in water and added to an excess of potassium iodide solution, liberating iodine according to the following equation 2Cu^(2+)(aq) + 4I^(-)(aq) rarr 2CuI(s)+ I_(2) ( aq) The iodine liberated required 24.5cm^(3) of a 0.100 mol dm^(-3) solution of sodium thiosulphate. 2S_(2)O_(3)^(2-) (aq) + I_(2) (aq)rarr 2I^(-) ( aq) + S_(4) O_(6)^(2-) (aq) The percentage of copper, by mass, in the copper ( II) salt is

Answer»

64.2
`51.0`
48.4
25.5

Answer :B
Discussion
26055.

The percentage of copper in a copper ( II) salt can be determined by using a thiosulphate titration. 0.305g of a copper (II) salt was dissolved in water and added to an excess of potassium iodide solution, liberating iodine according to the following equation 2Cu^(2+) ( aq) + 4I^(-)( aq) rarr 2CuI(s) + I_(2) ( aq) The iodine liberated required 24.5cm^(3) of a 0.100 mol dm^(-3) solution of sodium thiosulphate. 2S_(2) O_(3)^(-) (aq) + 4I^(-)( aq) rarr 2I^(-) ( aq) + S_(4) O_(6)^(2-)(aq) The percentage of copper , by mass, in the copper ( II) salt is

Answer»

64.2
`51.0`
48.4
25.5

Answer :B
Discussion
26056.

The percentage of empty space in a body centred cubic arrangement is .........

Answer»

 74 
68 
32 
26 

Solution :PACKING EFFICIENCY in bcc = 68 % . EMPTY space = 32 %
Discussion
26057.

The percentage of copper, tin and zinc metals present in 'Gun metal' respectively are

Answer»

88,2,10
88,10,2
80,20,Zero
80,zero,20

Solution :88,10,2
Discussion
26058.

The percentage of carbon is same in :

Answer»

CAST IRON and pig iron
Cast iron and steel
Pig iron and steel
Pig iron and WROUGHT iron

ANSWER :A
Discussion
26059.

The percentage of carbon in steel is approx

Answer»

0.01
0.03
0.02
0.1

Answer :C
Discussion
26060.

The percentage of carbon in pig iron, wrougth iron and cast iron is in the order of ________.

Answer»

pig IRON `LT ` CAST iron `gt ` wrought iron
pig iron `gt` cast iron `gt ` wrought iron
pig iron = cast iron `lt` wrought iron
pig iron `lt ` cast iron `lt ` wrought iron

Answer :B
Discussion
26061.

The percentage of carbon in high carbon steel is ________.

Answer»

`0.5 - 1%`
`0.15 - 1.5%`
`0.15 - 1.5%`
`0.15 - 0.3%`

ANSWER :B
Discussion
26062.

The percentage of carbon in cast iron is

Answer»

`5-10`
`0.250-2.5`
`2.5-4.5`
`0.12-0.2`

SOLUTION :CAST IRON has 2.5- 4.5 percentof CARBON.
Discussion
26063.

The percentage of C_(2)H_(5)OH in wash is (approximatly)

Answer»

0.95
0.1
0.5
0.75

Answer :B
Discussion
26064.

The percentage of C_(2)H_(5)OH in wash is

Answer»

0.95
0.1
0.5
0.75

Answer :B
Discussion
26065.

The percentage of available chlorine in a sample of bleaching powder, CaOCl_(2),2H_(2)O, is:

Answer»

30
50
43.5
59.9

Answer :C
Discussion
26066.

The percentage of an element M is 53 in its oxide of molecular formula M_(2)O_(3).Its atomic mass is about

Answer»

45
9
18
27

Solution :If m is the atomic mass of the element M, then
`2m+48G" of "M_(2)O_(3)" contain O"=48g.`
`therefore %" of O in "M_(2)O_(3)=(48)/(2m+48)xx100=53"(Given)"`
`therefore""2m+48=(4800)/(53)=90.56`
`"or"2m=42.56"or"m=26.28~~27`
Discussion
26067.

The percentage of an element M is 53 in its oxide of molecular formula M_(2)O_(3). Its atomic mass is about:

Answer»

45
9
18
27

Solution :Equivalent MASS of element `=("Mass of element")/("Mass of oxygen")XX8`
`=(53)/(47)xx8=9`
ATOMIC mass=Equivalent mass `xx` Valency
`=9xx3=27` AMU.
Discussion
26068.

The percentage of available chlorine in a commercial sample of bleaching powder is:

Answer»

`12%`
`35%`
`58%`
`85%`

ANSWER :B
Discussion
26069.

The percentage ionization of a weak base is given by

Answer»

`(sqrt((K_(a))/C))xx 100`
`((1)/(1+10^(pK_(b)-pOH))) xx 100`
`(sqrt((K_(b))/c))xx 100`
`(sqrt((K_(w))/(c xx K_(a) "of conjugate acid")))100`

SOLUTION :Let weak base be `= BOH HARR B^(+) + OH^(-)`
`{:("Initial conc.",c,0,0,),("Final conc.",c-c alpha,c alpha,c alpha,):}`
Let bolume of container be 1 LITRE
`rArr` % dissociation `= (c alpha)/(c) xx 100 = 100 alpha`
Also `(c alpha^(2))/(1-alpha) = K_(b)`
As `alpha` is negligible `rArr K_(b) = c alpha^(2) , alpha = sqrt((K_(b))/(c))`
`rArr` % ionization `= 100 alpha = (sqrt((K_(b))/(c))) 100`
`rArr 100alpha = (sqrt((K_(b))/(c))) 100 = (sqrt((K_(w))/(K_(a)c))) 100`
`[ because K_(b) = (K_(w))/(K_(a))]` (where `c alpha = [OH^(-)]`)
`= log_(10) c alpha = pOH = -log_(10) (K_(b)c)^(1//2) rArr pOH = -(1)/(2)log K_(b)c`
`pOH = -(1)/(2)log K_(b)c = -log c alpha`
`rArrpOH = (1)/(2)[pK_(b) - log c] = -log c alpha`
`2pOH = pK_(b) - logc rArr = pK_(b) - 2pOH`
`rArr c = 10^(pK_(b) - 2pOH)`
Hence choices (c) and are correct while (a) and (b) are not correct.
Discussion
26070.

The percentage composition of ferrous ammonium sulphate is 14.32%Re^(2+), 9.20% NH_(4)^(+), 49.0% SO_(4)^(2-) and 27.57% H_(2)O. What is the empirical formula of the compound?

Answer»

Solution :Calculation of EMPIRICAL formula

Hence, the empirical formula of the COMPOUNDS is `(Fe^(2+))(NH_(4)^(+))(SO_(4)^(2-))_(2)(H_(2)O)_(6) or Fe(NH_(4))_(2)(SO_(4))_(2).6H_(2)O`.
Discussion
26071.

The percentage composition of carbon by mole in methane is :

Answer»

`80%`
`25%`
`75%`
`20%`

ANSWER :D
Discussion
26072.

The percentage composition by weight of an aqueous solution of a solute (molecule mass 150) which boils at 373.26 is

Answer»

5
15
7
10

Answer :C
Discussion
26073.

The percentage composition (by weight) of a solution is 45% X, 15% Y 40% Z. Calculate the mole fraction of each component of the solution. (Mol. Wt. of X = 18, Y = 60, Z = 60 gms/mole)

Answer»

X = 0.2 , Y = 0.61, Z = 0.194
X = 0.73, Y = 0.073, Z = 0.194
X = 0.73, Y = 0.25, Z = 0.194
X = 0.3, Y = 0.2, Z = 0.5

SOLUTION :
Discussion
26074.

The percentage composition by mass of an aqueous solution of a solute (molecular mass 150 which boils at 373.26 is (K_(b) = 0.52)

Answer»

5
15
7
none of the above.

Solution :`DELTA T_(b) = K_(b) xx m, i.e, 0.26 = 0.52m "or" M = 0.5`
`m = (W_(2) xx 1000)/(M_(2) xx W_(1)(g)) :. 0.5 = (W_(2) xx 100)/(150 xx 1000)`
`W_(2) = 150 xx 0.5 = 75 g `
%age by mass `=75/1075 xx 100 = 7% "APPROX".`
Discussion
26075.

The percentage by mole of NO_(2) in a mixture of NO_(2)(g) and NO(g) having average molecular mass 34 is

Answer»

0.25
0.2
0.4
0.75

Answer :A
Discussion
26076.

The percentage by weightof hydrogen in H_2O_2 is :

Answer»

50
25
6.25
5.88

Answer :D
Discussion
26077.

The percentage by volume of Argon in atmosphere

Answer»

0.01
0.02
0.1
0.002

Answer :A
Discussion
26078.

The percentage abundance of Neon gas in air by volumeis 1.8 xx 10^(-x) and by weight is 1.0 xx 10^(-y). Than x - y = ________

Answer»


Solution :`X - Y = 0Ne to Y to 1.5 XX 10^(-3), W to 110^(-3)`
Discussion
26079.

The percent of void space in a body - centred cubic lattice is :

Answer»

`32%`
`48%`
`52%`
`68%`

ANSWER :A
Discussion
26080.

The percent of enol content will be highest in

Answer»

`CH_3 CHO`
`CH_3COCH_3`
`CH_3 COCH_2 COCH_3`

Solution :`CH_3- overset(O ) overset(||)C-CH_2 - overset(O) overset(||)C- CH_3`PERCENTAGEOF ENDIS more
Discussion
26081.

The percent loss in weight after heating a pure sample of pottasium chlorate (mol.wt. =122.5 ) will be :

Answer»

12.25
24.5
39.18
49

Answer :C
Discussion
26082.

The penultimate shell of f-block elements contains how many electrons ?

Answer»

19 to 32
8 to 9
8 to 14
19 to 36

Answer :B
Discussion
26083.

The peptide bond joining amino acid into proteins is a specific example of

Answer»

ESTER
CARBONYL
GLYCOSIDIC
AMIDE

ANSWER :D
Discussion
26084.

The peptide linkage is

Answer»

`-OVERSET(|)UNDERSET(|)CH-COO-NH`
`-overset(|)underset(|)CH-CO-NH-`
`-overset(|)underset(|)CH-CH_(2)-CO-NH_(2)`
`-overset(|)CH-NH-NH-CO-`

ANSWER :B
Discussion
26085.

The pentose sugars in DNA and RNA have the structure of

Answer»

FURANOSE
OPEN CHAIN
Pyranose
Four membered ring

Answer :A
Discussion
26086.

The pentapeptide composed of Val, Ala, Phe and 2 Gly, gives no N_2", with "HNO_2. Among its hydrolysis products are Ala, Gly and Gly, Ala. Identify the structure

Answer»

Gly-Ala-Gly-Phe-Val
Val-Gly-Ala-Gly-Phe

Answer :C
Discussion
26087.

The pentose sugar in DNA and RNA has

Answer»

OPEN CHAIN structure
Pyranose structure
Furanose structure
None of the above 

ANSWER :C
Discussion
26088.

The PDI(polydispersity index) is the ratio of weight to number-average molecular masses(Mbar(w))//(M(bar(n)) .In natural polymers,which are generally monodispersed,PDIis………and in synthetic polymers which are always polydispersed, PDI is …….. because Mbar(w) is always .......than Mbar(n) .

Answer»

Greater then`1,1`,HIGHER
`1`,greater then`1`,higher
LESS than `1,1`,LOWER`1,`less than `1`,lower

Answer :B
Discussion
26089.

The passageof electricity through dil H_(2)SO_(4)for 16 minutes liberates a total of 224 ml of H_(2).The strength of the current in ampres will be

Answer»

5A
3A
4A
2A

Answer :D
Discussion
26090.

The passivity of iron results due to theformation of a thin protective layer of

Answer»

iron oxide
ferric HYDROXIDE
`Fe(NO_(3))_(3)`
`Fe_(2)O_(4)`

ANSWER :A
Discussion
26091.

The passivity of iron in concentrated nitric acid is due to

Answer»

Ferric NITRATE coating on the METAL
Ammonium coating on the metal
A thin oxide layer coating on the metal
A hydride coating on the metal

Solution :A thin layer of `Fe_(3)O_(4)` is formed on the FE metal.
Discussion
26092.

The passage of current liberates H_2 at cathode and Cl_2 at anode. The solution is

Answer»

copper chloride in water
NACL in water
`H_2SO_4`
Water

SOLUTION :Since DISCHARGE potential of watier is greater than that of sodium so water is reduced at cathode instead of `Na^(+)`.
Cathode: `H_(2)O+e^(-)to(1)/(2)H_(2)+OH^(-)`
Anode: `CL^(-)to(1)/(2)Cl_(2)+e^(-)`.
Discussion
26093.

The passage of current through the solution of a dcertain electrolyte results in the liberation of H_(2) at the cathode and chlorine at the anode. The solution in the container could be:

Answer»

`NACL(aq.)`
`CuCl_(2)(aq.)`
`Kcl(aq.)`
`MgCl_(2)(aq.)`

ANSWER :A::C::D
Discussion
26094.

The passage of a constant current through a solution of dilute H_(2)SO_(4) with 'Pt' electrodes liberated 340.5 cm^(3) of a mixture of H_(2) and O_(2) at S.T.P. The quantity of electricity that was passed is:

Answer»

96500 C
965 C
1930 C
(1/100) Faraday

Answer :C
Discussion
26095.

The passage of electricity in the Daniell cell when Zn and Cu electrodes are connected :

Answer»

From CU TOZN inside the cell
From Cu to Zn OUTSIDE the cell
From Zn to Cu outside the cell
None

Answer :B
Discussion
26096.

The particles of true solution can pass through a semi permeable membrane, but those of a colloidal solution cannot. Explain why?

Answer»

SOLUTION :The SIZE of colloid particles are LARGER than the size of the pores PRESENT in a semipermeable membrane. Hence particles of colloidal solution cannot PASS through a semipermeable membrane. On the other hand, particles of true solution can pass through a semipermeable membrane because size of such particles are smaller than the size of the pores present in a semipermeable membrane.
Discussion
26097.

The particle size of suspension should be greater than 10 xx A^@, x is………

Answer»


SOLUTION :`> 1000 A^@`
Discussion
26098.

The partial pressures of CH_(3)OH,CO and H_(2) in the equlibrium mixture for the reaction CO+2H_(2)hArrCH_(3)OH at 427^(@)C are 2.0, 1.0 and 0.1 atm respectively. The value of k_(p) for the decommposition of CH_(3)OH to CO and H_(2) is

Answer»

`1XX10^(2)atm`
`2xx10^(2)atm^(-1)`
`50atm^(2)`
`5xx10^(-3)atm^(2)`

Solution :`CH_(3)OH toCO+2H_(2)`
`([H_(2)]^(2)[CO])/([CH_(3)OH])=(0.1xx0.1xx1)/(2)=(0.01)/(2)=(10xx10^(-3))/(2)`
`=5xx10^(-3).`
Discussion
26099.

The partial pressure ratio P_(A)^(@):P_(B)^(@) for the two volatile liquid A and B and P_(A)^(@): P_(B)^(@)=1:2 and mole ratio is X_(A): X_(B)=1:2. What is the mole fraction of A ?

Answer»

<P>`0.33`
`0.25`
`0.20`
`0.52`

Solution :`p_(A)=X.p , p_(B)=x.p`
`p_(A)=x_(p),p_(B)=2x.2p`
Total pressure `p =p_(A)+p_(B)`
`p=xp+4xp=5xp`
`THEREFORE` Mole fraction of,
`A(X_(A))=(p_(A))/(p)=(xp)/(5xp)=(1)/(5)=0.2`
Discussion
26100.

The partial pressure of ethane over a solution containing 6.56 xx10^(-3) g of ethane is 1 bar. If thesolution contains 5.00 xx 10^(-2)g of ethane, then what shall be the partial pressure of the gas ?

Answer»

<P>

SOLUTION : Applying the relationship `m = K_Hxx p` (Henry.s law)
In the first case, `6.56 xx 10^(-2) g = K_H X `
1 bar or `K_H = 6.56 xx 10^(-2)g "bar"^(-1)`
Let the PARTIAL pressure in the second case be p.
In the second case, `5.00 xx 10^(-2) g = (6.56 xx 10^(-2) g "bar"^(-1)) xxp`
`p = (5.00 xx 10^(-2) g)/(6.56 xx 10^(-2) g "bar"^(-1)) = 0.762` bar
Discussion