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26101.

The partial pressure of PCL_3,CL_2 and PCL_5are 0.1,0.2 and 0.008 atmosphere respectively for reaction , PCL_5 hArr PCL_3+CL_2. The value of K_p is :

Answer»

2.5
5
0.25
25

Answer :A
Discussion
26102.

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

Answer»

<P>

Solution :Molar mass of ethane `(C_(2)H_(6))`
`= 2xx12+6xx1`
= 30 G `mol^(-1)`
`therefore` Number of molar present in `6.56xx10^(-3)` g of ethane `= (6.56xx10^(-2))/(1)`
`= 2.187xx10^(-4)`
Let the number of MOLES of the solvent be 55.55 assuming solvent is water.
According to Henry.s low,
`p = K_(H)X`
1 bar `= KH(2.187xx10^(-4)//55.455)`
[Assuming DILUTION condition i.e., moles of solvent `gt gt` moles of solute]
`K_(H)=1` bar / `(0.039xx10^(-4))`
If mass of ethane = 0.05 g then moles of ethane
= 0.05/30 mol
= 0.00166 mol
So `p=K_(H)X`
`p=1//0.039xx10^(-4)xx(0.00166//55.55)`
= 7.66 bar
Discussion
26103.

The partial pressure of ethane over a solution containing 6.56 xx 10^(-2) g of ethane is 1 bar. If the solution contains 5.0 xx 10^(-2) g of ethane, then what will be the partial pressure of the gas?

Answer»

SOLUTION :According to Henry.s law,
m = `K_(H)`P
`K_(H) = (m)/(p) = (6.56 xx 10^(-2) (g))/(1 ("bar"))`
= `6.56 xx 10^(-2) g "bar"^(-1)`
In the second case,
P `= (m)/(K_(H))`
`= (5.00 xx 10^(-2)(g))/(6.56 xx 10^(-2) (g " bar"^(-1))) `
= 0.762 bar
Discussion
26104.

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

Answer»

<P>

Solution :Applying the RELATIONSHIP `m=K_(H)xxp`
`"In the FIRST CASE, "6.56xx10^(-2)g=K_(H)xx"1 bar or "K_(H)=6.56xx10^(-2)"g bar"^(-1)`
`"In the second case, "5.00xx10^(-2)g=(6.56 xx10^(-2)" g bar"^(-1))xxp or p=(5.00xx10^(-2)g)/(6.56xx10^(-2)" g bar"^(-1))="0.762 bar."`
Discussion
26105.

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

Answer»


ANSWER :7.62 BAR
Discussion
26106.

The partial pressure of CH_3OH(g), CO(g) and H_2(g) in equilibrium mixture for the reaction, CO(g) + 2H_2(g) hArr CH_3OH(g) are 2.0,1.0 and 0.1 atm respectively at 427^Oc. The value of K_p for the decomposition of CH_3OH to CO and H_2 is :

Answer»

`10^2` ATM
`2xx10^2` `atm^-1`
50 `atm^2`
`5XX10(-3)` `atm^2`

ANSWER :D
Discussion
26107.

The partial hyddrolysis of XeF_(4) at low temperature gives :

Answer»

`XeO_(3)`
`XeOF_(2)`
`XeOF_(4)`
`XeF_(2)`.

SOLUTION :`XeF_(4)+H_(2)Ooverset(-80^(@)C)RARR XeOF_(2)+2HF`
Discussion
26108.

The paramagneticoxides of nitrogen are

Answer»

dinitrogen monoxide and nitrogen monoxide
nitrogen monoxide and nitrogen dioxide.
nitrogen dioxide and dinitrogen trioxide.
dinitrogen trioxide and dinitrogen tetraoxide.

Solution :`NO` has 11 valence electrons and HENCE it contains one unpaired electron :
`:N::O:`
`NO_(2)` MOLECULES is ALSO an ODD electron molecule.
`:O:N::O:or`
Discussion
26109.

The paramagnetic nature of oxygen molecules is best explained on the basis of

Answer»

Valence bond THOERY
Resonance
Molecular ORBITAL theory
HYBRIDIZATION

Answer :C
Discussion
26110.

The paramagnetic character in 3d transition series elements increases upto Mn and then decreases . Explain why.

Answer»

Solution :As we move from `. _(21)SC` to `._(25)Mn`, the numberof UNPAIRED electrons increases and HENCE paramagneticcharacter increases. After Mn, pairing of ELECTRON in the d-subshell starts and the number of unparied electrons decreases and hence paramagneticcharacter decreases.
Discussion
26111.

The paramagnetic lanthanoid ion among the following is

Answer»

`CE^(4+)`
`YB^(2+)`
`Lu^(3+)`
`Sm^(2+)`

ANSWER :D
Discussion
26112.

The paramagnetic complexes is (are)

Answer»

`[CR (CN)_(6)]^(4-)`
`[Co (NH_(3))_(6)]^(3+)`
`[Fe (CN)_(6)]^(3-)`
`[CoCl_(6)]^(3-)`

ANSWER :A::C::D
Discussion
26113.

The paramagnetic behaviour of B_(2) is due to the presence of

Answer»

2 unpaired electrons in `pi_(b)MO`
2 unpaired electrons in `pi^(***)MO`
2 unpaired electrons is `sigma^(***)MO`
2 unpaired electron in `sigma_(b)MO`

SOLUTION :`B_(2)` (electrons=10)
`sigma(1s)^(2),sigma*(1s)^(2),sigma(2s)^(2),sigma*(2s)^(2),pi(2p_(X))^(1),pi(2p_(y))^(1)`
THUS, paramagnetic behaviour of `B_(2)` is due to the presence of two unpaired electrons in `pi`- BONDING molecular orbital.
Discussion
26114.

The paramagnetic behaviour of O_2 molecules is best explained by :

Answer»

MOLECULAR ORBITAL theory
Resonance theory
VSEPR theory
Valence BOND theory

Answer :A
Discussion
26115.

The pale yellow coloured gas is:

Answer»

`Cl_2`
`F_2`
`Br_2`
`I_2`

ANSWER :B
Discussion
26116.

The pair(s) of reagents that yeild paramagnetic species is/are

Answer»

Na and excess of `NH_(3)`
K and excess of `O_(2)`
CU and dilute `HNO_(3)`
`O_(2)` and 2-ethylanthraquinol.

Solution :`Na+UNDERSET(("excess"))((x+y)NH_(3))to[Na(NH_(3))_(x)]^(+)+underset(("Paramagnetic"))underset("SOLVATED" e^(-))(e^(-)(NH_(3))_(y))`
`underset(("excess"))(K+O_(2))tounderset(("Paramagnetic"))underset("Potassium superoxide")(KO_(2))`
`3Cu+8HNO_(3(DIL"))to3Cu(NO_(3))_(2)+underset(("Paramagnetic"))(2NO+4H_(2)O)`
Discussion
26117.

The pairs of compounds which cannot exist together in aquous solution are: (I)NaH_(2)PO_(4) and Na_(2)HCO_(3)(II)Na_(2)CO_(3) and NaOH (III)NaOH and NaH_(2)PO_(4)(IV)NaHCO_(3) and NaOH

Answer»
Discussion
26118.

The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)

Answer»

`[Cr(NH_(3))_(5)Cl]Cl_(2)` and `[Cr(NH_(3))_(4)Cl_(2)]Cl`
`[Co(NH_(3))_(4)Cl_(2)]^(+) and [Pt(NH_(3))_(2)(H_(2)O)Cl]^(-)`
`[CoBr_(2)Cl_(2)]^(2-) and [PtBr_(2)Cl_(2)]^(2-)`
`[Pt(NH_(3))_(3)(NO_(3))]Cl and [Pt(NH_(3))_(3)Cl]Br`

SOLUTION :Each complex ion in the pair `[Co(NH_(3))_(4)Cl_(2)]^(+),[Pt(NH_(3))_(2)(H_(2)O)Cl]^(-)` shows geometrical isomerism. Each complex in the pair `[Pt(NH_(3))_(3)(NO_(3))]Cl,[Pt(NH_(3))_(3)Cl]Br` shows ionization isomerism. In other pairs, the two complexes/ions do not show the same type of isomerism.
Discussion
26119.

The pair(s) of ions where both the ions are precipitated upon passing H_(2)S gas I presence of dilute HCl, is (are)

Answer»

`Ba^(2+),Zn^(2+)`
`Bi^(3+),Fe^(3+)`
`Cu^(2+),Pb^(2+)`
`HG^(2+),Bi^(3+)`

Solution :Precipitate is formed on pasing `H_(2)S` in ACIDIC medium i.e., ION must be of group II
`Cu^(2+),Pb^(2+),Hg^(2+),Bi^(3+)`
Discussion
26120.

The pairs of compounds having metals in their highest oxidation state is

Answer»

`MnO_(2),FeCl_(3)`
`MnO_(4)^(-),CrO_(2)Cl_(2)`
`[Fe(CN)_(6)]^(3-),CO(CN)_(3)`
`[NiCl_(4)]^(2-),[CoCl_(4)]^(-)`

SOLUTION :`MnO_(4)^(-)(+7)andCrO_(2)Cl(+6)`have highestoxidation states.
Discussion
26121.

The pair with more ionic nature among lithium halides :-

Answer»

LiF
LICL
LiBr
LII

Solution :`UNDERSET({:("SIZE of halogen"(uarr)),("POLARISATION"(uarr)),(C.C.(uarr)M.P.(uarr)):})overset("LiF LiCl LiBr LiI")to`
Discussion
26122.

The pairs of bases in DNA are hled together by :

Answer»

IONIC BONDS
covalent bonds
phosphate group
HYDROGEN bonds.

Answer :D
Discussion
26123.

The pair whose both species are used in anti-acid medicinal preparation is

Answer»

`NaHCO_3` and `MG(OH)_2`
`Na_2CO_3` and `CA(HCO_3)_2`
`Ca(HCO_3)_2` and `Mg(OH)_2`
`Ca(OH)_2` and `NaHCO_3`

SOLUTION :`NaHCO_3` and `Mg(OH)_2`
Discussion
26124.

The pair that yields the same gaseous product on reaction with water :

Answer»

K and `KO_2`
CA and `CaH_2`
NA and `Na_2O_2`
Ba and `BaO_2`

ANSWER :B
Discussion
26125.

The pair that requires calcination is

Answer»

`ZnCO_3 and CAO`
`Fe_2O_3 .xH_2Oand CaCO_3.MgCO_3`
`ZNO and Fe_2O_3.xH_2O`
All of these

ANSWER :D
Discussion
26126.

The pair that contains two P-H bonds in each of the oxoacids is ..........

Answer»

`H_(3)PO_(2)` and `H_(4)P_(2)O_(5)`
`H_(4)P_(2)O_(5)` and `H_(4)P_(2)O_(6)`
`H_(3)PO_(3)` and `H_(3)PO_(2)`
`H_(4)P_(2)O_(5)` and `H_(3)PO_(3)`

SOLUTION :
Discussion
26127.

The pair that contains no P-H bonds in each of the oxoacids is :

Answer»

`H_4P_2O_5 and H_3PO_3`
` H_3PO_3and H_3PO_2`
` H_3PO_4and H_4P_2O_6`
` H_3PO_2 and H_4P_2O_5`

ANSWER :C
Discussion
26128.

The pair that act as both oxidising and as well as reducing agent is

Answer»

`NO,SO_3`
`NO_(2),H_(2)O_2`
`CO_(2),SO_(2)`
`N_(2)O_(5),O_3`

Solution :In order to act as REDUCING agent the compound should be in its MINIMUM OXIDATION STATE (or) should contains valence electrons `NO_2,H_2O_2` - can ACTS reducing as well as oxidising agents .
Discussion
26129.

The pair of coordination complex exhibiting the same kind of isomerism is .

Answer»

`[Cr(NH_(3))_(5)Cl]Cl_(2)` and `[Cr(NH_(3))_(4)Cl_(2)]Cl`<BR>`[Co(NH_(3))_(4)Cl_(2)]^(+)` and `[Pt(NH_(3))_(2) (H_(2)O)Cl]^(+)`
`[CoBr_(2)Cl_(2)]^(2-) ` and `[PtBr_(2)Cl_(2)]^(2-)`
`[Pt(NH_(3))_(2)(NO_(2))]Cl` and `[Pt(NH_(3))_(2)Cl]Br`

Answer :B::D
Discussion
26130.

The pair of the equimolar compound that would be give a single condensation product when treated with an alkali is

Answer»

`CH_(3)-CHO+CH_(3)CH_(2)CHO`


`CH_(3)CHO+HCHO`

ANSWER :D
Discussion
26131.

The pair of the compounds in which both the metals are in the highest possible oxidation state is,

Answer»

`[FE(CN)_6]^(3-),[CO(CN)_6]^(3-)`
`CrO_2Cl_2,MnO_4^-`
`TiO_2,MnO_2`
`[Co(CN)_6]^(3-),MnO_2`

ANSWER :B
Discussion
26132.

The pair of structures given below represents :

Answer»

Enatiomers
DIASTEREOMERS
Conformations
STEREOISOMERS

Answer :C
Discussion
26133.

The pair of species having square planar shapes for both is :

Answer»

`[Fe(CO)_(4)]^(-), XeF_(4)`
`[Ni(CN)_(4)]^(2-), CCl_(4)`
`[Cu(NH_(3))_(4)]^(2+),[PtCl_(4)]^(2-)`
`SF_(4), PCI_(4)^(+)`

ANSWER :C
Discussion
26134.

The pair of species having identical shaped for molecules of both species is

Answer»

`XeF_2,CO_2`
`BF_3,PCl_3`
`PF_5, IF_5`
`CF_4, SF_4`

ANSWER :A
Discussion
26135.

The pair of species having identical shape is :

Answer»

`CF_4,SF_4`
`PCl_3,BF_3`
`XeF_2, CO_2`
`PF_5`, `IF_5`

ANSWER :C
Discussion
26136.

The pair of isochoric processes among the transformatin of states is

Answer»

K to L and L to M
L to M and N to K
L to M and M to N
M to N and N to K

Answer :B
Discussion
26137.

The pair of lanthanoide ions used as raducing agents is/are

Answer»

`Eu ^(3+) and Ce ^(4+)`
`Ce ^(4+) and Tb ^(4+)`
`Eu ^(2+) and DY^(4+)`
`ND ^(2+) and SM ^(2+)`

Answer :D
Discussion
26138.

The pair of metals which dissolves in NaOH(aq) is :

Answer»

AL , CU
Zn , Cd
PB , SN
Zn , Al

Answer :D
Discussion
26139.

The pair of ions which do not have same number of unpaired electrons is

Answer»

`Mn^(2+) and FE^(3+)`
`Ti^(2+) and Ni^(2+)`
`Cu^(2+) and Ti^(3+)`
`Fe^(2+) and Ni^(2+)`

SOLUTION :`Fe^(2+) and Ni^(2+)`
Discussion
26140.

The pair of ions which do not have diamag netic nature

Answer»

`Cu^(1+) and ZN^(2+)`
`Sc^(3+) and Ti^(4+)`
`Ca^(2+) and Zn^(2+)`
`V^(2+) and Fe^(2+)`

ANSWER :D
Discussion
26141.

The pair of ions having same electronic configuration is

Answer»

`Cr^(3+),FE^(3+)`
`Fe^(3+),MN^(2+)`
`Fe^(3+),Co^(3+)`
`Sc^(3+),Cr^(3+)`

Answer :B
Discussion
26142.

The pair of exothermic hydrides of VI A group are

Answer»

`H _(2) O , H_(2)S`
`H_(2)S,H_(2)SE`
`H_(2)Se, H_(2)TE`
`H_(2)O, H_(2)Te`

ANSWER :A
Discussion
26143.

The pair of elements which can show +1 odidation state.

Answer»

`Cr, Zn`
`FE, Zn `
`Cr, Cu`
`Cu, Zn`

ANSWER :C
Discussion
26144.

The pair of elements which are not having similarly in atomic and ionic radii are

Answer»

Zr and HF
Mo and Ta
Nb and Ta
Mo and W

ANSWER :B
Discussion
26145.

The pair of elements like Zr - Hf, Nb - Ta and Mo -W. Due to lanthanoide contraction

Answer»

having SIMILARLY in atomic and IONIC size
these are called chemical TWINS
these ELEMENTS have similar properties
all of these

Answer :D
Discussion
26146.

The pair of elements having approximately equal ionisation potential is

Answer»

Al,Ga
Al,Si
Al,Mg
Al,B

Solution :In case of Ga there are `10`d - electrons in the penultimate energy SHELL which shield the nuclear charge less effectively, the outer electron is held firmly by nucleus. As a result, the IONISATION energy remains nearly the same as that of aluminium inspite of the fact that atomic size INCREASES.
Discussion
26147.

The pair of compounds which cannot exist together in solution is:

Answer»

`NaHCO_3` and NAOH
`Na_2CO_3`and `NaHCO_3`
`NaH_2PO_2` and NaOH
`NaHCO_3` and NaCl

Answer :A
Discussion
26148.

The pair of compounds which can exist together in aqueous solution is:

Answer»

`Na_(2)CO_(3) and NaHCO_(3)`
`NaHCO_(3) and NaOH`
`NaOH and NaH_(2)PO_(4)`
`NaOH+NaHPO_(3)`

ANSWER :B::C
Discussion
26149.

The pair of compounds which can form a co-ordinate bond is :-

Answer»

`(C_(2)H_(5))_(3)` B and simplest `3^(@)` amine
HCl and HBr
`BF_(3)` and Diamond
(1) and (3) both

Answer :A
Discussion
26150.

The pair of molecules having strongest intermolecular hydrogen bonds is :

Answer»

`H_(2)`and `H_(2)O`
`CH_(3)COOH`and HCOOH
`CH_(3)COOH`and`CH_(3)COCH_(3)`
`SiH_(4)`and`SiCl_(4)`

Answer :B
Discussion