94420 + Interview Questions in Chemistry in Class 12 Page 1769 InterviewSolution

88401.	A molten ionic hydride on electrolysis gives:
Answer» `H^+` IONS moving towards the cathode `H^+` ions moving towards the ANODE `H_2` is LIBERATED at anode `H_2` is liberated at cathode Answer :C

Discussion

88402.	a' moles ofPCl_(5) are heated in a closed container to equilibriate PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) at pressure of p atm . Ifx moles of PCl_(5) dissociate at equilibrium , then
Answer» <P>`(X)/(a)=((K_(P))/(P))^(1//2)` `(x)/(a)=(K_(P))/(K_(P)+p)` `(x)/(a)=((K_(P))/(K_(P)+p))^(1//2)` `(x)/(a)=((K_(P)+p)/(K_(P)))^(1//2)` ANSWER :C

Discussion

88403.	A molecule XY_(2) contains two sigma, two pibond and one lone pair of electron in the valence shell of X. The arrangement of lone pair as well as bond pair is :
Answer» SQUARE pyramidal Linear Trigonal planar Unpredictable Solution :`[(sigma=2),(Ip=1)] HYB : SP^(2),` so trigonal planar

Discussion

88404.	A molecule with zero dipole moment among the following is :
Answer» `CHCl_(3)` `CH_(3)CL` `CH_(2)Cl_(2)` `C Cl_(4)` Answer :D

Discussion

88405.	A molecule which can not exist theoretically is:
Answer» `SF_4` `OF_2` `OF_4` `O_2F_2` ANSWER :C

Discussion

88406.	A molecule with highest bond energy is
Answer» `Br_(4)` `F_(2)` `cl_(2)` `l_(2)` Solution :the BOND energy of `Cl_(2)`is the highest among the GIVEN.

Discussion

88407.	A molecule of sulphur may be represented as
Answer» S `S_2` `S_4` `S_8` SOLUTION :`S_8`

Discussion

88408.	A molecule of Stachyose contains how many carbon atoms?
Answer» 6 12 18 24 Solution :The molecular formula of stchyose is `C_(24)H_(42)O_(21)`.

Discussion

88409.	A molecule of stachyose containshow many carbonatoms ?
Answer» 6 12 18 24 Solution :Dependingupon the number ofmonosaccharidemoleculescondensedto FORM oligasaccharidessuch asdisaccharides (e.g.,sucrose)trisaccharides(e.g., raffinose), tetrassaccharides) (e.g, STACHYOSE) atc. Stachyose contains24 CARBONATOMS INITS STRUCTURE.

Discussion

88410.	A molecule of nitrogen occupies 1.62 xx 10^(-19) m^(2). The volume of nitrogen at 0^@C and 1.013 atm required to cover a sample of silica gel with unimolecular layer is 129 cm^(3) g^(-1) of gel.Calculate the surface area per gram of gel?
Answer» ANSWER :`5.7 XX 10^(2) m^(2)`

Discussion

88411.	A molecule MX_(3) has no dipole moment. The sigma bonding orbital used by M (atomic no < 21) is :
Answer» `sp^(3)` hybridised p unhybridised sp hybridised `sp^(2)` hybridised. Solution :Since the MOLECULE has ZERO dipole moment. It MUST be linar and involves sp hybridisation.

Discussion

88412.	A molecule may be represented by three structures having energies E_(1), E_(2) and E_(3), respetively. The energies of these structures follow the order E_(2)ltE_(2)ltE_(1), respectively. If the experimantal bond energy of the molecule is E_(0), the resonance energy is :
Answer» `(E_(1)+E_(2)+E_(3))-E_(0)` `E_(0)-E_(3)` `E_(0)-E_(1)` `E_(0)-E_(2)` ANSWER :B

Discussion

88413.	A molecule is said to be chiral if it
Answer» contains a plane of SYMMETRY contains a CENTRE of symmetry cannot be SUPERIMPOSED on its mirror IMAGE EXISTS as cis-trans-isomers Answer :C

Discussion

88414.	A molecule is said to be chiral if it :
Answer» EXISTS as cis and trans isomers contains a plane of SYMMETRY cannot be SUPERIMPOSED on its MIRROR image contains a centre of symmetry. Answer :C

Discussion

88415.	A molecule in which sp^(2) hybrid orbitals are used by the central atom in forming covalent bonds is :
Answer» `PCl_(5)` `NH_(3)` `SO_(2)` `XeF_(2)` ANSWER :C

Discussion

88416.	A molecule has two lone pairs and two bond parts around the central atom. The molecular shape is expected to be :
Answer» 1. V - shaped 2. TRIANGULAR 3. Linear 4. Tetrahedral Answer :A

Discussion

88417.	A molecule H - X will be 50% ionic if electronegativity difference of H and X is
Answer» `1.2 EV ` `1.4 eV` `1.5eV` `1.7 eV` ANSWER :D

Discussion

88418.	A molecule A_(x) dissovle in water and is non volatile. A solution of certain molality showed a depression of 0.93K in freezing point. The same solution boiled at 100.26^(@)C. When 7.87g of A_(x) was dissovled in 100g of water, the solution boiled ate 100.44^(@)C. Given K_(f) for water =1.86K kg "mol"^(-1) atomic mas of A=31u. Assume no association or dissociatioin of solute. Calculate the value of x........
Answer» Solution :`DeltaT_(f)=K_(f)m` `m=0.93/1.86=0.5` `DeltaT_(B)=K_(b)m` `K_(b)=0.26/0.5=0.52K KG "mol"^(-1)` Similarly `DeltaT_(b)^(')=K_(b)m^(')` `DeltaT_(b)^(')=(K_(b)xxwxx1000)/(Wxxm.wt)` `:.` m.wt of solute `=(K_(b)xxwxx1000)/(wxxDelta_(b)^('))` `=(0.52xx7.87xx1000)/(100xx0.44)` `31x~~93` `x=3`

Discussion

88419.	(A) Molecularity has no meaning for a complex reaction. (R ) The overall molecularity of a complex reaction is equal to the molecularity of the slow step.
Answer» Both A & R are TRUE, R is the correct EXPLANATION of A Both A & R are true, R is not correct explanation of A A is true, R is FALSE A is false, R is true Answer :C

Discussion

88420.	(A) : Molecular mass of polymers is always expressed as an average (R) : Generally, a polymer sample contains chains of verying lengths
Answer» Both A & R are TRUE, R is the CORRECT EXPLANATION of A Both A & R are true, Ris not correct explanation of A A is true, R is false A is false, R is true Answer :A

Discussion

88421.	(A) Molecular solids are characterized by low melting point. (R) Molecular solids are made up of covalent molecules.
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 :B

Discussion

88422.	(A): Molecular mass of polymers is always expressed as an average. (R) : Generally, a polymer sample contains chains of varying lengths.
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

88423.	(A) Molecuels that are not superimposable on their mirror images are chiral. (R) All chiral molecules have chiral centre.
Answer» ANSWER :C

Discussion

88424.	A mole ofN_2 H_4 loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation number of hydrogen.)
Answer» `-1` `-3` `+3` `+5` Solution :`N_2 H_4- 10 e^(-)toY ` ( withtwoN atoms ) totalO.Nof TWON atoms in `N_2 H_4` is `2x +4 =0or2x=- 4` sinceitloses10molesof ELECTRONS therefore , totalO.Nof twoN atomsincreasesby10i.e..,the totalO.NoftwoN atoms in ` y=- 4 +10= + 6` ` thereforeO.N ` of eachN atominY=+3

Discussion

88425.	A mole of a monoatomic ideal gas at 1 atm and 273 K is allowed to expand adiabatically against a constant pressure of 0.395 bar will until equilibrium is reached. (a) What is the final temperature ? (b) What is the final volume ? (c ) How much work is done by the gas ? (d) What is the change in internal energy ?
Answer» Solution :LET the initial and FINAL volumes of the gas be `V_(1)` and `V_(2)m^(3)` respectively. Given that the initialtemperature is 273K. Let the final temperature be `T_(2)` We have, `p_(1)V_(1)=n_(1)RT_(1)` `V_(1)=(1xx8.314xx273)/(1xx10^(5))=0.022697m^(3)` For an ADIABATIC expansion of 1 mole of monoatomic ideal gas against a constant external pressure `(p_(2))`, work done is given as `W=-p_(2)(V_(2)-V_(1))=C_(V)(T_(2)-T_(1))=(3R)/(2)(T_(2)-T_(1))` or `-0.395xx10^(5)(V_(2)-0.022697)=(3xx8.314)/(2)(T_(2)-273)`............(1) Again, `p_(2)V_(2)=nRT_(2)` `0.395xx10^(5)xxV_(2)=1xx8.314xxT_(2)` ..............(2) Solving eqns. (1) and (2) , we get (a) the final temperature, `T_(2)=207K` (b) the final VOLUME `V_(2)=0.043578m^(3)` (c ) the work done by the gas `W=-p_(ext)(V_(2)-V_(1))` `=-0.395xx10^(5)(0.043578-0.022697)` `=-825J//mole` (d) as `q=0`, and `q=DeltalI-W` `DeltalI=W=-825J//mole`

Discussion

88426.	Assertion : Mole fraction has no units Reason:Mole fraction is a ratio of number of moles of solute to number of moles of solvent
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 (A) is false but (R) is true Answer :C

Discussion

88427.	Assertion : Molarity of a solution decreases with an increase of temperature Reason :As temperature increases volume of the solution increases.
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 (A) is false but (R) is true ANSWER :A

Discussion

88428.	Molarity of 3N H_3PO_4 solution is
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 (A) is false but (R) is true ANSWER :C

Discussion

88429.	Assertion : Molality is independent of temperature Reason : There is no volume factor in the expression of molality
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 (A) is false but (R) is true ANSWER :A

Discussion

88430.	A molal solution is one that contanis one mole of a solute in
Answer» 1000 G of the solvent one LITRE of solvent one litre ofsolution `22.4` LITRES of solution Answer :A

Discussion

88431.	A molal solution is one that contains one mole of a solute in
Answer» 1000 gm of the solvent One litre of the solvent One litre of the solution 22.4 LITRES of the solution Solution :The number of moles of solute DISSOLVED in 1000 gm of the solvent is called MOLAL soltuion.

Discussion

88432.	(A) MnO_(4)^(-1) is strong oxidizing agent. (R) Mn^(+2) is stable with fully filled d sublevel.
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 Solution : (A) is true but (R) is false

Discussion

88433.	(A) Mn atom loses ns electrons first during ionization as compared to (n-l)d electrons . (R) The effective nuclear charge experienced by (n-l)d electrons is greater than that by ns electrons.
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 Solution :Both (A) and (R) are true and (R) is the correct explanation of (A)

Discussion

88434.	(A) Mn _(2) O _(7) and Cr O _(3) are acidic oxides. (R ) In Mn _(2) O _(7)and CrO _(3) the Mn and Cr are in their highest oxidation states.
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

88435.	(A) Mn & Hg exist in simple crystal structures. (R) Both Mn & Hg are solids.
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 Solution :Both (A) and (R) are false

Discussion

88436.	A mixture which contains 0.550g of camphor and 0.045g of an organic solute freezes at 157^(@)C . The solute contains 93.46% of C and 6.54% of H by weight . What is the molecular formula of the compound ? (Freezing point of camphor =178.4^(@)C and K_(f)=37.70)
Answer» Solution :Molality `=(DeltaT_(f))/(K_(f))=(178.4-157)/(37.70)=(21.4)/(37.70)` If the molecular wt. of SOLUTE is M then molality `=("moles of solute")/("wt.of solvent in grams")xx1000` `=(0.045//M)/(0.550)xx1000=(4500)/(55M)` Thus, `(4500)/(55M)=(21.4)/(37.70)`, `M=144.14` Now, from the given weight `%` of C and H , we get ` {:("moles of " C=(93.46)/(12)=7.79),("moles of "H=(6.54)/(1)=6.54):}}`(supposing 100g of the solute) `:.` molar ratio of C and `H=7.79 : 6.54`, i.e. `1.2 : 1`, i.e. `12 : 10` `:.` EMPIRICAL FORMULA is `C_(12)H_(10)` Empirical-formula weight `=12xx12+1xx10=154` The empirical-formula weight is slightly greater than the molecular weight which MIGHT be due to some molecular change of the solute in the solvent. However, the molecular formula will be `C_(12)H_(10)`.

Discussion

88437.	A mixture when heated with dil. H_2SO_4 does not evolve brown vapours but when heated with conc. H_2SO_4brown vapours are obtained. The vapours when brought in contact with aqueous AgNO_3 solution do not give any precipitate. The mixture contains :
Answer» `NO_2^(-)`<BR>`CL^(-)` `Br^(-) ` `NO_3^(-)` ANSWER :D

Discussion

88438.	A mixture when heated with dil H_(2)SO_(4) does not coolve brownvapours butwith cone H_(2)SO_(4)brown withAgNO_(3) soin do notgiveanyprecipitate .The mixturecontain
Answer» `NO_(2)^(THETA)` `NO_(3)^(Theta)` `I^(Theta)` `Br^(Theta)` ANSWER :b

Discussion

88439.	A mixture of ZnCl_(2) and PbCl_(2) can be sepqrated by
Answer» Distillation Crystallization Sublimation Adding aceitic acid Answer :D

Discussion

88440.	A mixture on heatinggave a gas used as ananaesthetic solublein waterforming cis , and trans dibasic acid 1.1 g of gasoccupes 0.56 at STP mixturecontain
Answer» `NaNO_(3) + NH_(4)CI` `NaNO_(2) + NH_(4)CI` `CaCO_(3) + MgCO_(3)` `NH_(4)CI+ NaSO_(4)` ANSWER :a

Discussion

88441.	A mixture of X and Y was loaded in the column of silica. It was eluted by alcohol water mixture. Compound Y eluted in preference to compound X. Compare the extent of adsorption of X and Y on column.
Answer» SOLUTION :COMPOUND X is more STRONGLY adsorbed than the compound Y. That is why compound Y is eluted in PREFERENCE to X.

Discussion

88442.	A mixture of two white substances is solubel in water. This solution gives brown colour gas on passing chlorine gas. Another sample of solution gives white precipitate with BaCl_(2) which is soluble in concentrated HCl. The original solution of the mixture gives white precipitate with large excess of NaOH solution whose suspension is used as an antacid. After filtering off this precipitate, the filtrate was boiled with excess NaOH. This solution gave a yellowish precipitate on addingNaClO_(4). Identify the mixture.
Answer» Solution :`2Br^(-)+Cl_(2)rarr 2CL^(-) +underset("brown")(Br_(2))` `CO_(3)^(2-)+BaCl_(2)rarr underset("white")(BaCo_(3)darr)+2Cl^(-)` `MG^(2+)+2NaOH rarr Mg(OH)_(2)+2NA^(+)` Therefore mixture consists `K_(2)CO_(3)` and `MgBr_(2)`.

Discussion

88443.	A mixture of two volatile liquids having little difference in their boiling points can be purified by
Answer» Distillation Crystalization Column chromatography Fractional distillation Answer :D

Discussion

88444.	A mixture of two volatile liquids A and B for 1 and 3 moles respectivley has a V.P. of 300 mm at 27^@C. If one mole of A is further added to this solution, the vapour pressure becomes 290 mm at 27^@C. The vapour pressure of pure A is -
Answer» 250 mm 316 mm 220 mm 270 mm Answer :A

Discussion

88445.	A mixture of two solid elements with a mass of 1.52g was treated with an excess of hydrochloric acid. A volume of 0.896 dm^(3) of a gas was liberated in this process and 0.56 g of a residue remained which was undissolved in the excess of the acid. In another experiment, 1.52 g of the same mixture were allowed to react with an excess of a 10% sodium hydroxide solution. In this case 0.896 dm^(3) of a gas were also evolved but 0.96 g of an undissolved residue remained. In the third experiment, 1.52 g of the initial mixture were heated to high temperature without acess of the air. In this way a compound was formed which was totally soluble in hydrochloric acid and 0.448 dm^(3) of an unknown gas were released. All the gas obtained was introduced into a one litre closed vessel filled with oxygen. After the reaction of the unknown gas with oxygen the pressure in the vessel decreased by approximately ten times (T=const). Write chemical equations for the above reactions and prove their correctness by calculations. In solving the problem consider that the volumes of gases were measured at STP and round up the relative atomic masses to whole numbers.
Answer» ANSWER :A

Discussion

88446.	A mixture of two salts were treated as follows (i) The mixture was heated with MnO_(2) & conc H_(2)SO_(4). A yellowish green gas was liberated. (ii) The mixture on heating with NaOH solution gave a gas, which turned red litmus blue. (iii) The solution in water gas brown colouration with Kj_(3)Fe(CN)_(6) and red colouration with NH_(4)CNS. (iv) The mixture was boiled with KOH and the liberated gas was bubbled through an alkaline solution of K_(2HgIj_(4) to give brown precipitate. Identify the two salts. Give ionic equations for reactions involved in the test (i), (ii) & (iii).
Answer» Solution :`FeCl_(3) & NH_(4)Cl`

Discussion

88447.	A mixture of two salts was treated as follows. (i)The mixture was heated with precipitated MnO_(2) and concentrated H_(2)SO_(4) when a yellowish green gas was liberated. (ii)The mixture on heating with NaOH solution gave a gas which turned red litmus blue. (iii)Its solution in water gave red colouration with dimethylglyoxime in alkaline solution and white precipitate with K_(4)[Fe(CN)_(6)] in absence of air. (iv)The mixture was boiled with KOH and the liberated gas was bubbled through an alkaline solution of K_(2)HgI_(4) to give a brown precipitate. Identify the ions present in the mixture.
Answer» SOLUTION :N//A

Discussion

88448.	A mixture of two salts was treated as follows: (i) The mixture was heated with manganese dioxide and concentrated sulphuric acid when yellowish green gas was liberated. (ii) The mixture on heating with sodium hydroxide solution gave a gas which turned red litmus blue. (iii) Its solution in water gave blue precipitate with potassium ferricyanide and red colouration with ammonium thiocyanate. (iv) The mixture was boiled with potassium hydroxide and the liberated gas was bubbled through an alkaline solution of K_(2)Hgl_(2) to give brown precipitate. Identity the two salts. Give ionic equations for reactions involved in the tests (). (i) and (int).
Answer» Solution :(i) `2Cl^(-)+MnO_(2)+H_(2)SO_(4)(conc.)+2h^(+)toMnSO_(4)+UNDERSET(groon)underset(YELLOW wish)(Cl_(2)uarr)+2H_(2)O` (ii) `NH_(4)^(+)+2NaOHtoNH_(3)+H_(2)O+Na^(+)` (iii) `3Fe^(2+)+2[Fe(CN)_(6)]^(3-)toFe_(3)underset((blue ppt.))underset(Prussian blue)([Fe(CN_(6))]_(2))` (i) `Mixture+MnO_(2)+H_(2)SO_(4)(conc.)to` yellowish green gas M The reaction suggests thal the mixture contains `Cl^(-)` ions. `therefore 2Cl^(-)+MnO_(2)+H_(2)SO_(4)(conc.)+2H^(+)MnSO_(4)+underset(Yellowwish green)(CI_(2)uarr)+2H_(2)O` (ii) `Mixture+NaOHto` Gas turningred litmus blue The reaction suggests that the gas is ammonia (ie, basic in character turning red litmus blue). Ammonia will be evolved from an ammonium SALT. Thus, `NH_(4)^(+)+NaOHtoNH_(3)+H_(2)O+Na^(+)` (iii) Solution of mixture `+K_(2)[Fe(CN)_(6)] toblue ppt`. The reaction suggests that the mixture contains a Fe (II) of prussian blue complex. `3Fe^(2+)+2[Fe(CN)_(6)]^(3+)Fe_(3)underset((Blue ppt.))underset(Prussian blue) ([Fe(CN)_(6)]_(2))` Red colourless with ammonium thioyanate suggests that some Fe (III) is also present. It is likely that a part of Fe (II) is oxidized (III) by AIR. `Fe^(3+)+3NH_(4)SCN toFeunderset(“red colourless”)((SCN))+3NH_(4)^(+)` `2Fe^(2+)+[O]+2H^(+)to 2Fe^(3+)+H _(2)O` (iv) `Mixture+KOHtoGasoverset(K_(2)Hgl_(4))tobrown ppt`. the above reaction can be explained, when ammonia is liberated on boiling the mixture with potassium hydroxide. The liberated ammonia reacts with an alkaline solutions of `K_(2)Hgl_(4)` Hgle to giv a brown ppt. Hence, the mixture contains, `Fe^(2+)`,`NH_(4)^(+)` and cr ions (with an impurity of `Fe^(3+)` ions). The two salts are `FeCl_(2)` and `NH_(4)CI`.

Discussion

88449.	A mixture of two salts was treated as follows : (i) The mixture was heated with maganese dioxide and concentrated sulphuric acid, when yellowish green gas was liberated. (ii) The mixture on heating with sodium hydroxide solution gave a gas which turned red litmus blue. (iii) Its solution in water gave blue precipitate with potassium ferricyanide and red colouration with ammonium thiocyanate. (iv) The mixture was boiled with potassium hydroxide and the liberated gas was bubbled through an alkaline solution of K_(2)HgI_(4) to give brown precipitate. Identify the two salts, Give ionic equations for reactions involved in the tests (i), (ii) and (iii).
Answer»

Discussion

88450.	A mixture of two salts is not water soluble but dissolves completely in dilute hydrochloric acid to form a colourless solution. The mixture could be
Answer» `AgNO_(3) and KBr` `BaCO_(3) and ZNS` `FeCl_(3) and CaCO_(3)` `Mn(NO_(3))_(2) and MgSO_(4)` SOLUTION :`BaCO_(3)+2HCl to BaCl_(2)+H_(2)O+CO_(2)` `ZnS+underset((dil.))(2HCl)toZnCl_(2)+H_(2)S`

Discussion

Explore topic-wise InterviewSolutions in .

A molten ionic hydride on electrolysis gives:

a' moles ofPCl_(5) are heated in a closed container to equilibriate PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) at pressure of p atm . Ifx moles of PCl_(5) dissociate at equilibrium , then

A molecule XY_(2) contains two sigma, two pibond and one lone pair of electron in the valence shell of X. The arrangement of lone pair as well as bond pair is :

A molecule with zero dipole moment among the following is :

A molecule which can not exist theoretically is:

A molecule with highest bond energy is

A molecule of sulphur may be represented as

A molecule of Stachyose contains how many carbon atoms?

A molecule of stachyose containshow many carbonatoms ?

A molecule of nitrogen occupies 1.62 xx 10^(-19) m^(2). The volume of nitrogen at 0^@C and 1.013 atm required to cover a sample of silica gel with unimolecular layer is 129 cm^(3) g^(-1) of gel.Calculate the surface area per gram of gel?

A molecule MX_(3) has no dipole moment. The sigma bonding orbital used by M (atomic no < 21) is :

A molecule may be represented by three structures having energies E_(1), E_(2) and E_(3), respetively. The energies of these structures follow the order E_(2)ltE_(2)ltE_(1), respectively. If the experimantal bond energy of the molecule is E_(0), the resonance energy is :

A molecule is said to be chiral if it

A molecule is said to be chiral if it :

A molecule in which sp^(2) hybrid orbitals are used by the central atom in forming covalent bonds is :

A molecule has two lone pairs and two bond parts around the central atom. The molecular shape is expected to be :

A molecule H - X will be 50% ionic if electronegativity difference of H and X is

(A) Molecularity has no meaning for a complex reaction. (R ) The overall molecularity of a complex reaction is equal to the molecularity of the slow step.

(A) : Molecular mass of polymers is always expressed as an average (R) : Generally, a polymer sample contains chains of verying lengths

(A) Molecular solids are characterized by low melting point. (R) Molecular solids are made up of covalent molecules.

(A): Molecular mass of polymers is always expressed as an average. (R) : Generally, a polymer sample contains chains of varying lengths.

(A) Molecuels that are not superimposable on their mirror images are chiral. (R) All chiral molecules have chiral centre.

A mole ofN_2 H_4 loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation number of hydrogen.)

Assertion : Mole fraction has no units Reason:Mole fraction is a ratio of number of moles of solute to number of moles of solvent

Assertion : Molarity of a solution decreases with an increase of temperature Reason :As temperature increases volume of the solution increases.

Molarity of 3N H_3PO_4 solution is

Assertion : Molality is independent of temperature Reason : There is no volume factor in the expression of molality

A molal solution is one that contanis one mole of a solute in

A molal solution is one that contains one mole of a solute in

(A) MnO_(4)^(-1) is strong oxidizing agent. (R) Mn^(+2) is stable with fully filled d sublevel.

(A) Mn atom loses ns electrons first during ionization as compared to (n-l)d electrons . (R) The effective nuclear charge experienced by (n-l)d electrons is greater than that by ns electrons.

(A) Mn _(2) O _(7) and Cr O _(3) are acidic oxides. (R ) In Mn _(2) O _(7)and CrO _(3) the Mn and Cr are in their highest oxidation states.

(A) Mn & Hg exist in simple crystal structures. (R) Both Mn & Hg are solids.

A mixture which contains 0.550g of camphor and 0.045g of an organic solute freezes at 157^(@)C . The solute contains 93.46% of C and 6.54% of H by weight . What is the molecular formula of the compound ? (Freezing point of camphor =178.4^(@)C and K_(f)=37.70)

A mixture when heated with dil. H_2SO_4 does not evolve brown vapours but when heated with conc. H_2SO_4brown vapours are obtained. The vapours when brought in contact with aqueous AgNO_3 solution do not give any precipitate. The mixture contains :

A mixture when heated with dil H_(2)SO_(4) does not coolve brownvapours butwith cone H_(2)SO_(4)brown withAgNO_(3) soin do notgiveanyprecipitate .The mixturecontain

A mixture of ZnCl_(2) and PbCl_(2) can be sepqrated by

A mixture on heatinggave a gas used as ananaesthetic solublein waterforming cis , and trans dibasic acid 1.1 g of gasoccupes 0.56 at STP mixturecontain

A mixture of X and Y was loaded in the column of silica. It was eluted by alcohol water mixture. Compound Y eluted in preference to compound X. Compare the extent of adsorption of X and Y on column.

A mixture of two volatile liquids having little difference in their boiling points can be purified by

A mixture of two volatile liquids A and B for 1 and 3 moles respectivley has a V.P. of 300 mm at 27^@C. If one mole of A is further added to this solution, the vapour pressure becomes 290 mm at 27^@C. The vapour pressure of pure A is -

A mixture of two salts is not water soluble but dissolves completely in dilute hydrochloric acid to form a colourless solution. The mixture could be