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86851.

A teacherr gave a student two salts (A) and (B) told him to identiify these salts. The student heated salt (A) strongly and observed two oxides to sulphur. He added NaOH solution to aqueous solution of (A) and observed a green precipitate, which turned brown on exposure to air. When he tok salt (B) to flame test, green colour was observed. On heating salt (B) with a solid compound (X) and concentrated sulphuric acid, orange red vapours are evolved. when this gas is passed through an aqueous solution of a base, the solution turns yellow. Q. Compound (X) is:

Answer»

`K_(2)Cr_(2)O_(7)`
`NH_(4)Cl` BOILED with NaOH
`CaF_(2)`
`Na_(2)B_(4)O_(7)*10H_(2)O`

Answer :A
Discussion
86852.

A teacherr gave a student two salts (A) and (B) told him to identiify these salts. The student heated salt (A) strongly and observed two oxides to sulphur. He added NaOH solution to aqueous solution of (A) and observed a green precipitate, which turned brown on exposure to air. When he tok salt (B) to flame test, green colour was observed. On heating salt (B) with a solid compound (X) and concentrated sulphuric acid, orange red vapours are evolved. when this gas is passed through an aqueous solution of a base, the solution turns yellow. Q. The salt (A) can be:

Answer»

`FE(SO_(4))_(3)`
`FeSO_(4)`
`FeSO_(4)*7H_(2)O`
both (b) and (c)

Answer :D
Discussion
86853.

A teacher one day pointed out to his students the peculiar fact that water is unique liquid which freezes exactly at 0^@C and boils exactly at 100^@C. He asked the students to find the correct statement based on this fact :

Answer»

Water dissolves ANYTHING HOWEVER sparingly the DISSOLUTION may be
Water is a polar molecule
Boiling and freezing temperature of water were used to DEFINE a temperature scale
Liquid water is denser than ice

Answer :C
Discussion
86854.

A tangent drawn on the curve obtained by plotting concentration of product ("mole"L^(-1)) of a first order reaction vs. time (min) at the point corresponding to time 20 minute makes an angle to 30^(@) with concentration axis. Hence, the rate of formations of product after 20 minutes will be

Answer»

`0.580 "mole" L^(-1)min^(-1)`
`1.723 "mole" L^(-1) min^(-1)`
`0.290 "mole" L^(-1) min^(-1)`
`0.866 "mole" L^(-1) min^(-1)`

ANSWER :B
Discussion
86855.

A : Tailing of Hg caused by ozone is due to formation of HgO . R : In the presence of O_(3) , Hg does not loses its meniscus .

Answer»

If both Assertion & Reason are true and the reason is the correct EXPLANATION of the assertion , then mark (1)
If both Assertion & Reason are true but the reason is not the correct explanation of the assertion , then mark (2)
If Assertion is true statement but Reason is false , then mark (3)
If both Assertion and Reason are false STATEMENTS , then mark (4)

Answer :D
Discussion
86856.

(A): Sythetic detergents do not contain any soap but exhibit all the properties of soaps. (R): Synthetic detergents give foam even in hard water.

Answer»

Both (A) and (R) are true and (R) is the correct EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are false

Answer :C
Discussion
86857.

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system.A liquid vapourises to form a more disordered gas.When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness.As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendency to freeze.In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution.Elevation of B.Pt.(DeltaT_b) and depression of F.Pt.(DeltaT_f) of a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identify.For dilute solutions. DeltaT_b and DeltaT_f are proportional to the molality of the solute in the solution. DeltaT_b=K_bm , K_b=Ebullioscopic constant=(RT_(b)^(@^(2))M)/(1000 DeltaH_(vap)) And DeltaT_f=K_fm , K_f=Cryoscopic constant=(RT_(f)^(@^(2))M)/(1000 DeltaH_(fus)) (M=molecular mass of the sovent) The values of K_b and K_f do depend on the properties of the solvent.For liquids, (DeltaH_(vap))/T_b^@ is almost constant. [Troutan's Rule, this constant for most of the unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state ) is equal to 90 J//"mol".] For solutes undergoing changes of molecular state is solution (ionization or association), the observed DeltaT values differ from the calculated ones using the above relations In such situations, the relationships are modified as DeltaT_b=i K_bm, DeltaT_f=i K_fm where i=Van't Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules. A mixture of two Immiscible liquids at a constant pressure of 1 atm boils at a temperature

Answer»

equal to the NORMAL boiling point of more volatile liquid
equal to the mean of the normal boiling point of the TWO liquids
greater than the normal boiling point of either of the liquid
smaller than the normal boiling point of either of the liquid

Solution :The vapour PRESSURE of a mixture of two IMMISCIBLE liquids is the sum of their vapour pressures in the pure states, independent of their relative amounts.Hence, B.Pt. of the mixture will be less than that of either of the liquids, remaining CONSTANT throughout.
Discussion
86858.

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system.A liquid vapourises to form a more disordered gas.When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness.As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendency to freeze.In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution.Elevation of B.Pt.(DeltaT_b) and depression of F.Pt.(DeltaT_f) of a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identify.For dilute solutions. DeltaT_b and DeltaT_f are proportional to the molality of the solute in the solution. DeltaT_b=K_bm , K_b=Ebullioscopic constant=(RT_(b)^(@^(2))M)/(1000 DeltaH_(vap)) And DeltaT_f=K_fm , K_f=Cryoscopic constant=(RT_(f)^(@^(2))M)/(1000 DeltaH_(fus)) (M=molecular mass of the sovent) The values of K_b and K_f do depend on the properties of the solvent.For liquids, (DeltaH_(vap))/T_b^@ is almost constant. [Troutan's Rule, this constant for most of the unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state ) is equal to 90 J//"mol".] For solutes undergoing changes of molecular state is solution (ionization or association), the observed DeltaT values differ from the calculated ones using the above relations In such situations, the relationships are modified as DeltaT_b=i K_bm, DeltaT_f=i K_fm where i=Van't Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules. A liquid possessing which of the following characteristics will be most suitable for determining the molecular mass of a compound by cryoscopic measurements ?

Answer»

That having low freezing point and small ENTHALPY of freezing
That low high freezing point and small enthalpy of freezing
That having high freezing point and small enthalpy of vapourisation
That having large SURFACE tension

Solution :`K_f=(RT_f^2M)/(1000DeltaH_("fus"))` would be larger for larger value of `T_f` and SMALLER value of enthalpy of fusion of the SOLID solvent.
Discussion
86859.

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system.A liquid vapourises to form a more disordered gas.When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness.As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendency to freeze.In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution.Elevation of B.Pt.(DeltaT_b) and depression of F.Pt.(DeltaT_f) of a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identify.For dilute solutions. DeltaT_b and DeltaT_f are proportional to the molality of the solute in the solution. DeltaT_b=K_bm , K_b=Ebullioscopic constant=(RT_(b)^(@^(2))M)/(1000 DeltaH_(vap)) And DeltaT_f=K_fm , K_f=Cryoscopic constant=(RT_(f)^(@^(2))M)/(1000 DeltaH_(fus)) (M=molecular mass of the sovent) The values of K_b and K_f do depend on the properties of the solvent.For liquids, (DeltaH_(vap))/T_b^@ is almost constant. [Troutan's Rule, this constant for most of the unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state ) is equal to 90 J//"mol".] For solutes undergoing changes of molecular state is solution (ionization or association), the observed DeltaT values differ from the calculated ones using the above relations In such situations, the relationships are modified as DeltaT_b=i K_bm, DeltaT_f=i K_fm where i=Van't Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules. Depression of freezing point of which of the following solutions does represent the cryoscopic constant of water ?

Answer»

6% by MASS of urea in aqueous solution
100g of aqueous solution CONTAINING 18 G of glucose
59 g of aqueous solution containing 9 g of glucose
1 M KCL solution in water.

Solution :Cryoscopic constant `K_f=DeltaT_f` of solution having unit molality of normal solutes
Molality of glucose solution in `( C)=(9xx1000)/((59-9)xx180)=1`
Discussion
86860.

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system.A liquid vapourises to form a more disordered gas.When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness.As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendency to freeze.In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution.Elevation of B.Pt.(DeltaT_b) and depression of F.Pt.(DeltaT_f) of a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identify.For dilute solutions. DeltaT_b and DeltaT_f are proportional to the molality of the solute in the solution. DeltaT_b=K_bm , K_b=Ebullioscopic constant=(RT_(b)^(@^(2))M)/(1000 DeltaH_(vap)) And DeltaT_f=K_fm , K_f=Cryoscopic constant=(RT_(f)^(@^(2))M)/(1000 DeltaH_(fus)) (M=molecular mass of the sovent) The values of K_b and K_f do depend on the properties of the solvent.For liquids, (DeltaH_(vap))/T_b^@ is almost constant. [Troutan's Rule, this constant for most of the unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state ) is equal to 90 J//"mol".] For solutes undergoing changes of molecular state is solution (ionization or association), the observed DeltaT values differ from the calculated ones using the above relations In such situations, the relationships are modified as DeltaT_b=i K_bm, DeltaT_f=i K_fm where i=Van't Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules. Dissolution of a non-volatile solute into a liquid leads to the-

Answer»

decrease of entropy
increase in tendency of the LIQUID to FREEZE
increases in tendency to pass into the VAPOUR phase
decrease in tendency of the liquid to freeze

Solution :SINCE the solution has greater entropy (disorder) then the pure liquid, so former has lesser tendency to freeze i.e., the temperature has to be lowered to freeze the solution.
Discussion
86861.

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system.A liquid vapourises to form a more disordered gas.When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness.As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendency to freeze.In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution.Elevation of B.Pt.(DeltaT_b) and depression of F.Pt.(DeltaT_f) of a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identify.For dilute solutions. DeltaT_b and DeltaT_f are proportional to the molality of the solute in the solution. DeltaT_b=K_bm , K_b=Ebullioscopic constant=(RT_(b)^(@^(2))M)/(1000 DeltaH_(vap)) And DeltaT_f=K_fm , K_f=Cryoscopic constant=(RT_(f)^(@^(2))M)/(1000 DeltaH_(fus)) (M=molecular mass of the sovent) The values of K_b and K_f do depend on the properties of the solvent.For liquids, (DeltaH_(vap))/T_b^@ is almost constant. [Troutan's Rule, this constant for most of the unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state ) is equal to 90 J//"mol".] For solutes undergoing changes of molecular state is solution (ionization or association), the observed DeltaT values differ from the calculated ones using the above relations In such situations, the relationships are modified as DeltaT_b=i K_bm, DeltaT_f=i K_fm where i=Van't Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules. To aqueous solution of Nal increasing amounts of solid HgI_2 is added.The vapour pressure of the solution

Answer»

decrease to a CONSTANT value
increase to a constant value
increases first and then decreases
remains constant because `HgI_2` is sparingly soluble in water

Solution :`2Na^+(aq)+2I^(-)(aq)+HgI_2(s)to2Na^+(aq)+HgI_4^(2-)(aq)`
The number of mole particle decreases from `4(2Na^+ + 2I^(-)) "to" 3(2Na^+ + HgI_4^(2-))` Hence, the COLLIGATIVE property will decrease or, the VAPOUR pressure will increase to a constant value until NaI is completely consumed.
Discussion
86862.

A system is changed from state A to state B by one path and from B to A by another path. If U and U_2 are the corresponding change in internal energy, then:

Answer»

`U_1+U_2=+ve`
`U_1+U_2=-ve`
`U_1+U_2=0`
None

Answer :C
Discussion
86863.

A system is provided 50 joule of heat and work done on the system is 10 J. The change in internal energy during the process is:

Answer»

40 J
60 J
80 J
50 J

Answer :B
Discussion
86864.

A system is changed from state A to state B by one path and from B to A another path. If E_(1) and E_(2) are the correspondingchanges in internal energy, then

Answer»

`E_(1)+E_(2)=-ve`
`E_(1)+E_(2)=+ve`
`E_(1)+E_(2)=0`
None of these

Solution :`DeltaE=0` for a cyclic process.
Discussion
86865.

A system is provided 50 joule of heat and work done on the system is 10 J. The change in internal energy during the process is

Answer»

40 J
60 J
80 J
50 J

ANSWER :B
Discussion
86866.

A system is changed from an initial state to a final state to a final state by a maaner such that DeltaH=q . If the same change from the initial state to the final state were made by a different path, would DeltaH be the same as that to the first path ? Would q ?

Answer»


ANSWER :(same, most PROBABLY DIFFERENT )
Discussion
86867.

A system changes its state irreversibly at 300 K in which it absorbs 300 cals of heat. When the same changes is carried out reversibly the amount of heat absorbed is 900 cals. The change in entropy of the system is equal to

Answer»

`1 cal K^(-1)`
`3 CALS K^(-1)`
`2 cal K^(-1)`
`1.5 cals K^(-1)`

Answer :B
Discussion
86868.

A system absorbs 640 J heat and does work of 260 J, the change in internal energy of the system will be

Answer»

`+380` J
`-380` J
`+900` j
`-900` J

ANSWER :C
Discussion
86869.

A system absorbs 10 kJ of heat at contant volume and its temperature riess from 27^@C to 37^@C. The triangleU of reaction is:

Answer»

100 kJ
10 kJ
Zero
1 kJ

Answer :B
Discussion
86870.

A synthetic polyamide prepared by prolonged heating of caprolactum is

Answer»

nylon-6, 6
nylon-6
nylon-6,10
glyptal

Solution :
Discussion
86871.

A system absorb 600 J of heat and work equivalent to 300 J on its surroundings. The change in internal energy is

Answer»

300 J
400 J
500 J
600 J

Solution :We KNOW that `DeltaE=Q+W=600+(-300)=300 J`
W=300, because the WORK done by the system.
Discussion
86872.

A system absorbs 10 kJ of heat at constant volume and its temperature rises from 27^(@)C to 37^(@)C. The value of Delta U is :

Answer»

<P>100 kJ
10 kJ
0
1 kJ

Solution :`DELTA U = q + p Delta V = 10 + 0 = 10 kJ`
Discussion
86873.

A synthetic polyamide prepared by prolongedheating of caporlactam is

Answer»

NYLON 6,6
nylon 6
nylon 6,10
glyptal.

ANSWER :B
Discussion
86874.

A symmetricla bridged complex cation made of Co(III), NH_(3) molecules and oxygen (in the proper ligands form) is found to contain the following mass composition [Atomic mass of Co=59] Co=36.875%, NH_(3)=53.125%, O=10% The complex cation exists in three ionic forms with cationic charges A : (n+), B(n-1)^(+)" and "C: (n-2)^(+) such that O-O bond length in all of them is found to be more than that in O_(2)[Ptf_(6)]. Calculate the value of n,(n-1)" and "(n-2).

Answer»


SOLUTION :`{:(CO,:,NH_(3),:,O),((36.875)/(59),:,(53.125)/(17),:,(10)/(16)),(0.625,:,3.125,:,0.625),(1,:,5,:,1):}`
`Co_(2)(NH_(3))_(10)O_(2)`
0.5 of `O_(2)` can be 0,-1,-2
x can be 6,5,4
Discussion
86875.

A symmetrical organic compound of C_(4)H_(11)N give yellow oily layer on treatment with HNO_(2) then find the structure of the compound.

Answer»


SOLUTION :N//A
Discussion
86876.

A symbol not only represents the name of the element but also its:

Answer»

ATOMIC MASS
atomic NUMBER
atomicity
atomic volume

ANSWER :A::B::C
Discussion
86877.

A sweety smelling organic compound 'A' is slowly oxidised by air in the presence of light to a highly poisonous gas. On warming with silver powder, it forms a gaseous substance 'B' whish is also formed by the action of calcium carbide on water. Identify 'A' and 'B'. writ ethe chemical equations involved.

Answer»

SOLUTION :`A(CHCl_(3)),B(CH-=CH)`
Discussion
86878.

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. calculte the internal energy of vaporisation at 100^(@), [Delta_("vap")H^(@) for water at 373K = 40. 66 kJ mol^(-1)]

Answer»

`35.67 kJmol ^(-1)`
`35.67 kJmol ^(-1)`
`36.57kJmol ^(-1)`
`38.75kJmol ^(-1)`

SOLUTION :we can REPESENT the process of evaporation as`18gH_(2)O(l)OVERSET("veporistion")to 18 g H_(2)O(g)`
number of moles in ` 18 g H_(2)O(l)`
`(18g)/(18g mol^(-1))=1 mol`
`Delta_(VAP)U=Delta_(vap)H^(-)-pDeltaV`
`Delta_(vap)H^(-)-Deltan_(g)RT`
assume steam behave as an ideal ga`Delta_(vap)U=(40.66)-(1)(8.314xx10^(-3))(373) `
40.66 - 3.10
` 37.56 KJ mol^(-1)`
Discussion
86879.

A sulphuricacid solutions has PH =2its molarity is :

Answer»

1/100
1/50
1/2
1/200

Answer :D
Discussion
86880.

(A) Sulphuric acid forms two series of salts with alkalies (R ) Sulphuric acis is a diabasic acid

Answer»

Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A)
Both (A) and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is false
Both (A) and (R )are false

ANSWER :A
Discussion
86881.

A super saturated solution is a metastable state of solution in which solute concentration:

Answer»

Is EQUAL to the SOLUBILITY of that SUBSTANCE in water
Exceeds than its solubility
Less than its solubility
Continuously change

Answer :B
Discussion
86882.

(A): Sulphur is hexavalent in the ground state (R): Sulphur can form a minimum of six bonds

Answer»

Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A)
Both (A) and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is false
Both (A) and (R )are false

Answer :D
Discussion
86883.

(a) Sulphur dioxide gas turns a dichromate solution green. Write the reaction. (b) Also write the reaction when nickel salts in basic medium react with dimethyl glyoxime.

Answer»

SOLUTION :`({:(Cr_(2)O_(7)^(" "2-)+14H^(+)+6bar(e)to2Cr^(3+)+7H_(2)O),([SO_(2)+2H_(2)OtoSO_(4)^(" "2-)+4H^(+)+2e^(-)]xx3),(3SO_(2)+6H_(2)Oto3SO_(4)^(""2-)+12H^(+)+6bar(c)):})/(Cr_(2)O_(7)^(" "2-)+3SO_(2)+2H^(+)tounderset(("green"))(2Cr^(3+)+3SO_(4)^(" "2-)+H_(2)o)`
(B) `Ni^(2+)` salts, in a basic medium, on reaction with dimethylglyoxime, give a cherry red precipitate of nickel dimethyl glyoximate.
Discussion
86884.

(A) Sulphur dioxide acts as antichlor (R ) In presence of charcoal SO_(2) combine with Cl _(2) to form sulphuryl chloride

Answer»

Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A)
Both (A) and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is FALSE
Both (A) and (R )are false

Answer :A
Discussion
86885.

(a) Sulphur has greater tendency for catenation than oxygen. Give reasons. or Oxygen shows catenation behaviour less than sulphur. (b) Sulphur exhibits greater tendency for calenation than selenium. Give reasons.

Answer»

Solution :(a) Due to SMALLER SIZE of oxygen than sulphur, the lone pairs of electrons on the oxygen atoms repel the bond pair of the O-O bond to a greater EXTENT than the lone pairs of electrons on the sulphur atoms repel the bond pair of S-S bond. As a result, S-S bond is much stronger (213 kJ `mol^(-1)`) than O-O bond (138 kJ `mol^(-1)`) and hence sulphur has a much greater tendency for catenation than oxygen.
(b) As we move from S to Se, the atomic size increases and hence the strength of Element-Element bond DECREASES. THUS, S-S bond is much stronger than Se-Se bond and consequently, S shows greater tendency for catenation than selenium.
Discussion
86886.

A sulphur containing species that connot be an oxidising agent is:

Answer»

`H_2SO_4`
`H_2S`
`SO_2`
`H_2SO_3`

ANSWER :B
Discussion
86887.

A sulphur containing amino acid is

Answer»

Glycine
Cysteine
ALANINE
Leucine

Answer :B
Discussion
86888.

(A) : Sulphide ores are concentrated by froth floatation process.(R) : Pine oil acts as a frothing agent in froth floatation process.

Answer»

Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanationof (A)
(A) is true but (R) is FALSE
Both (A) and (R) are false

Answer :B
Discussion
86889.

A sulphide ore (A) on roasting leaves a residue (B). (B) on heating with chlorine gives (C), soluble in water, addition of excess potassium iodide to a solution of (C) gives a solution (D). A brown precipitate (E) is formed when a solution of ammonium sulphate is added to an alkaline solution of (D). Identify (A) to (E).

Answer»

Solution :`{:(underset((A))(HgS),overset("roasting")rarr,underset((B))(HgO),overset(Cl_(2))rarr,underset((C))(HgCl_(2)),overset("excess of KI"-)rarr,""underset((D))(K_(2)HgI_(4))),(,,,,,,"alkaline" DARR (NH_(4))SO_(4)),(,,,,,,underset((E))underset("BROWN ppt.")(HgO.Hg(NH_(2))I)):}`
Discussion
86890.

A sulphide of Fe contains 46.5% of Fe by weight. Find the empirical formula of the sulphide.

Answer»


ANSWER :`(feS_(2))`
Discussion
86891.

(A) : Sulphate is estimated as BaSO_(4) and not as MgSO_(4) (R) : Ionic radius of Mg^(2+) is smaller than that of Ba^(2+).

Answer»

Both A and R are correct, and R is the correct EXPLANATION of the A
Both A and R are correct, and R is not the correct explanation of the A
A is correct but R is incorrect
A is incorrect but R is correct

Answer :B
Discussion
86892.

A suitable combination of reagents leading to the preparation of tert butyl ether in Williamson's synthesis is :

Answer»




NONE of the above

Answer :A
Discussion
86893.

A sugar syrup of weight 214.2 g contains 34.2 g of sugar ( molar mass = 342). The molality of the solution is :

Answer»

0.0099
0.56
0.28
0.34

Solution :MOLALITY `=("MOLES of SUGAR")/("WT. of sugar")xx1000`
Wt. of WATER `=214.2-34.2=180g`
`=(34.2//342)/(180)xx1000`
`=0.56m`
Discussion
86894.

(a) Suggest a quantitative method for estimation of a gas which protects us from U.V. rays of the sun. (b) Nitrogen oxides emitted from the exhaust system of supersoni jet aeroplanes slowly deplete the concentration of ozone layer in upper atmosphere. Comment.

Answer»

Solution :(a) Ozone `(O_(3))` absorbs ultraviolet radiations (220 - 290 NM) and thus protects us from the harmful radiations of the sun.
(b) NO present in the exhaust of supersonic AEROPLANES combines rapidly with `O_(3)` FORMING `O_(2)` thereby DEPLETING the concentration of `O_(3)` in the UPPER atmosphere.
`NO(g) + O_(3)(g) rarr NO_(2)(g) + O_(2)(g)`
Discussion
86895.

A sugar syrup of weight 214.2g contains 34.2 g of sugar (C_(12)H_(22)O_(11)). Calculate (i) molal concentration and (ii) mole fraction of sugar in the syrup

Answer»


ANSWER :`0.556m,0.0099`
Discussion
86896.

A sugar syrup of weight 214.2 grams contains34.2 grams of sugar.The molal concentration is:

Answer»

0.55
5.5
55
0.1

Solution : Molecular WT. of sugar is 342
0.1 mole of sugar PRESENT in 214.2 gms of solution.
`( 214.2 - 34.2 = 180`gms of SOLVENT )
Molality `= ( "Number of moles ")/("1000 gms of solvent")`
Discussion
86897.

A sugar that is not a disaccharide among the following is

Answer»

GALACTOSE
lactose
maltose
sucrose

Solution :Among the GIVEN carbohydrates sucrose, lactose and maltose are disaccharides. Lactose is a condensation product of GLUCOSE and galactose. Galactose is a monosaccharide.
Discussion
86898.

A sugar is classified as a D-isomer if the hydroxyl group

Answer»

On the CHIRAL carbon nearest carbon nearest to the CARBONYL points to the left
On the chiral carbon nearest to the carbonyl POINT to the right
On the chiral carbon FARTHEST from the carbonyl point to the left
On the chiral carbon farthest from the carbonyl points to the right

ANSWER :D
Discussion
86899.

A sugar syrup of weight 183.42 g contains 3.42 g of sugar (C_(12)H_(22)O_(11)). Calulate the mole fraction of sugar.

Answer»


Solution :`"MASS of sugar" = 3.42 g: "Moles of sugar" =((3.42g))/((342 g MOL^(-1)))=0.01 mol`
`"Mass of water"=183.42-3.42=180g : "Moles of water"=(180g)/((342 "g mol"^(-1)))=10 mol`
`"Mole fration of sugar" = ("No. of moles of sugar")/("No. of moles of sugar + No. of moles of water")`
`=((0.01mol))/((0.01mol)+(10 mol))=(0.01)/(10.01)=0.00099`.
Discussion
86900.

A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C_(12)H_(22)O_(11)). Calculate (i) molal concentration (ii) mole fraction of sugar in the syrup.

Answer»


Solution :`"Solute "(C_(12)H_(22)O_(11))=34.2g,"Solution = 214.2 G,Solvent = " 214.2-34.2=180 g = 0.180 KG.`
`"Molal conc. "=(34.2//342mol)/(0.180kg)="0.556 mol kg"^(-1),"Mole FRACTION "=(34.2//342)/(180//18+34.2//342)=0.0099`
Discussion