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50401.

A compound (P) , obtained as an ozonolysis product of (Q) gives brisk effervescence with Na, violet coloration with netutral FeCl_(3) and silver mirrorwith Tollen's reagent . (Q ) may be

Answer»




All of these

Solution :
Discussion
50402.

A compound on decomposition in the laboratory produces 24.5 g of nitrogen and 70 g of oxygen. Calculate the empirical formula of the compound.

Answer»

SOLUTION :
Discussion
50403.

A compound on analysis was found to contain 53.33 % C , 15.5 % H , and the rest nitrogen . The formula of the compound is

Answer»

`C_(3) H_(7) N`
`C_(2) H_(7) N`
`C_(4) H_(10) N`
`C_(2) H_(8) N`

Answer :B
Discussion
50404.

A compound on analysis gave the following percentage composition: C - 54.54%, H = 9.09%, 0 = 36.36%

Answer»

SOLUTION :`C_2H_4O`
Discussion
50405.

A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09% 0 = 36.36. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.

Answer»

SOLUTION :`C_2H_4O`
Discussion
50406.

A compound on analysis gave the following percentage composition: C = 24.47%, H= 4.07 %, Cl = 71.65%. Find out its empirical formula.

Answer»

SOLUTION :
`:.` The empirical FORMULA - `CH_(2)Cl`
Discussion
50407.

A compound on analysis gave the following percentage composition: Na=14.31% S = 9.97%, H = 6.22%, O = 69.5%, calcualte the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32, H = 1, 0 = 16].

Answer»

Solution :
All H combines with 10 oxygen ATOMS to form as `10H_(2)O`
So the empirical formula is `Na_(2)SO_(4).10H_(2)O`
Empirical formula mass = `(23xx2) + (32 XX 1) + (16xx4) + (10xx 18)`
= 46 + 32 + 64 + 180 = 322
`n=("Molecular mass")/("Emperical formula mass")=322/322=1`
Molecular formula=`Na_(2)SO_(4).10H_(2)O`
Discussion
50408.

The compound of vanadium has magnetic moment of 1.73 BM. The vanadium chloride has the formula:

Answer»


SOLUTION :Magnetic moment `=sqrt(n(n+2))= 1.73 , n=1`
Electron configuration of V(23) = `1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 4s^2` .
As one unpaired electron is PRESENT in vanadium `V^(n+)` , so the CORRECT OXIDATION state is +4.
Discussion
50409.

A compound of the formula C_(4)H_(10)O reacts with sodium and undergoes oxidation to give a carbonyl compound which does not reduce Tollen's reagent, the original compound is

Answer»

Diethyl ether
n-Butyl alcohol
Isobutyl alcohol
sec-Butyl alcohol

Solution :Because the compound react with Na, it may be alcohol, phenol or a carboxylic acid. As the compound on oxidation gives a carbonyl compounds which does not reduce Tollen's reagent, the compound FORMED on oxidation is a reagent, the compound is a ketone. Because only SECONDARY `(2^(@))` alcohols on oxidation yield KETONES, thethe original compound is a secondary alcohol.
`underset(2^(@)-"Butyl alcohol (D)")(CH_(3)-overset(OH)overset(|)(CH)-CH_(2)-CH_(3))overset([O])rarrunderset("Butane-2-one")(CH_(3)-overset(O)overset(||)(CH)-CH_(2)-CH_(3))`
Discussion
50410.

A compound of sodium does not give CO_(2) when heated but it gives CO_(2) when treated with dilute acids. A crystalline compound is found to have 37.1 percent Na and 1.6 percent Ho_(2). Hence the compound is

Answer»

`NaHCO_(3), 10H_(2)O`
`NaHCO_(3).5H_(2)O`
`Na_(2)CO_(3).10H_(2)O`
`Na_(2)CO_(3). H_(2)O`

Solution :`Na_(2)CO_(3)xxH_(2)O`
`4600=3932.6 +667.8x`
`667.4=667.8x,x=1`
Discussion
50411.

A compound of sodium does not give CO_(2) when heated but it gives CO_(2) when treated with dilute acids. A crystalline compound is found to have 37.1% Na and 14.52% H_(2)O. Hence, compound is

Answer»

`NaHCO_(3)*10H_(2)O`
`NaHCO_(3)*5H_(2)O`
`Na_(2)CO_(3)*10H_(2)O`
`Na_(2)CO_(3)*H_(2)O`

Solution :`CO_(2)` is obatined only when the compound is decomposed by acid thus, it is `Na_(2)CO_(3)`. It cannot be `NaHCO_(3)` since, it given `CO_(2)` on heating.
Probable compound is `Na_(2)CO_(3).xH_(2)O`
`{:(,,"MOLAR RATIO","Ratio"),(Na_(2)CO_(3),85.48,0.8064,1),(H_(2)O,14.52,0.8064,1):}`
Thus, compound is `Na_(2)CO_(3)*H_(2)O`
MOLECULAR WEIGHT `=124" g "mol^(-1)`
percentage of sodium `=(2xx23)/(124)xx100=37.1%`
Discussion
50412.

A compound of formula C_3H_8 does not react with bormine in "CC"I_4 in the dark. The compound could be

Answer»

ALKANE
Cycloalkane
Alkene
Cycloalkene

Answer :A
Discussion
50413.

A compound of C, H and N contains the three element in the ratio of 9 : 1 : 3.5. Calculate the molecular formula given that the molecular mass of the compound is 108 amu Hint : Percentgae of C=(9)/(13.5)xx100=66.7,H=(1)/(13.5)xx100=7.4,N=(3.5)/(13.5)xx100=25.9.

Answer»


Solution :Step I. Empirical formula of the compound
`{:("Element","PERCENTAGE","Atomic mass","Gram ATOMS (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",66.7,12,(66.7)/(12)=5.56,(5.6)/(1.85)=3.0,3),("H",7.4,1,(7.4)/(1)=7.40,(7.40)/(1.85)=4.0,4),("N",25.9,14,(25.9)/(14)=1.85,(1.85)/(1.85)=1.0,1):}`
Empirical formula of compound `= C_(3)H_(4)N`
Step II. Molecular formula of the compound
Empirical formula mass `= 3 xx 12 + 4 xx 1 + 14 = 54 U`
Molecular mass = 108 AMU (GIVEN)
`nn=("Molecular mass")/("Empirical formula mass")=((108u))/((54u))=2`
`:.` Molecular formula `= 2 xx C_(3)H_(4)N=C_(6)H_(8)N_(2)`.
Discussion
50414.

A compound M_(p)X_(q) has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown on the side. The empirical formula of the compound is

Answer»

MX
`MX_(2)`
`M_(2)X`
`M_(5)X_(14)`

Solution :`"No. of M atoms" = underset(("Edge centre"))(4times1/4) + underset(("Body centre"))1=2`
`"No. of atom" = underset("corners")(8times1/8)+underset("face centre")(6times1/2)=4`
`therefore "UNIT FORMULA" = M_(2)X_(4)`
`therefore "empirical formula" = MX_(2)`
Discussion
50415.

A compound M_p X_q has cubic close packing (ccp) arrangement of X. Its unit cell structure . The empirical formula of the compound is

Answer»

MX
`MX_2`
`M_2X`
`M_5X_14`

SOLUTION :No. of M ATOMS `{:(=4xx1/4,"" 1,=2),("(EDGE centres)","(body-centred)",):}`
No. of X atoms `{:(=8xx1/8,6xx1/2,=4),("(CORNERS)","(face-centres)",):}`
`THEREFORE` Unit formula =`M_2X_4` and empirical formula =`MX_2`
Discussion
50416.

A compound made up of elements A and B crystallises in the cubic sttruture. Atoms A are present on the corners as well as face centres wherease atoms B are present on the edge centres as will as body centre. What is the formula of the compound ? Draw the struture of its unit cell.

Answer»


Solution :`{:("ATOMS B/unit cell" , =8xx1/8 , +, 6 XX 1/2=4),(, ("CORNERS"), , ("face-centres")),("Atoms B/unit cell" , = 12xx1/4 , +, 1=4),( , ("edge-centres") , , ("body-centre")):}`
A:B = 4 :4 = 1:1 , so the formula is AB.
Discussion
50417.

A compound made up of elements A and B crystallises in the cubic structure. Atoms A are present on the corners as well as face centres whereas atoms B are present on the edge centres as well as body centre. What is the formula of the compound ? Draw the structure of its unit cell.

Answer»


Solution :`{:("ATOMS A/unit cell",=8xx1/8"" +,6xx1/2=4),(,"(CORNERS)","(face-centres)"),("Atoms B/unit cell",=12xx1/4" " +,1=4),(,"(edge-corners)","(body-centres)"):}`
`THEREFORE` A:B=4:4=1:1 . So the formula is AB.
Discussion
50418.

A compound is soluble is conc. H_(2)SO_(4). It does not decolourise bromine in C Cl_(4) but oxidised by chromic anhydride in aqueous sulphuric acid within two second, turning orange solution to blue green , then opaque. The original solution cotnains

Answer»

A SECONDARY alcohol
An alkene
An ether
A primary alcohol

SOLUTION :Since oxidation is EASIER and occuts very QUICKLY (just 2 seconds), hence it must be primary alcohol. The DICHROMATE solution changes from organe to blue green.
Discussion
50419.

A compound is soluble in water. If ammonia is added to aqueous solution of compound, a reddish brown precipitate appears which is soluble in dill. HCl. The compound is a salt of :

Answer»

aluminium
zinc
iron
cadmium

Solution :It is the CORRECT ANSWER
Discussion
50420.

A compound is soluble in conc. H_(2)SO_(4), it does not decolourise bromine in carbon tetrachloride but is oxidised by chromic anhydride in aqueous solution to blue, green and then opaque. The original compound is

Answer»

primary alcohol
tertiary alcohol
ALKENE
ETHER

Solution :As the compound is soluble in conc. `H_(2)SO_(4)`, it may be as alkene, alcohol or ether.
The compound does not decolourise `KMnO_(4)` solution, it cannot be ETHENE.
Further, it is oxidised by `CrO_(3)` in `H_(2)SO_(4)` it is not an ether. It is easily oxidised within two seconds it is primary alcohol.
Discussion
50421.

A compound is soluble in conc. H_(2)SO_(4). It doesdecolourise bromine in carbon tetrachloride but oxidised by chromic anhydride in aqueous sulphuric acid within two seconds, turning organge solution to blue, green and then opaque. The original compound is :

Answer»

an alkane
a TERTIARY alcohol
a PRIMARY alcohol
an ether

Answer :C
Discussion
50422.

A compound is known to be hydrated doubel salt of potassium oxalate and oxalic acid of thetuype aK_(2)C_(2)O_(4), bH_(2)C_(2)O_(4), cH_(2)O, where a, b and c are unknown. When 1.613g of this compound is dissolved in water and solution is neutralised by 19.05mL of 0.1Nalkali and reduces25.40mL of this solution. What is the formula of the salt?

Answer»


Answer :`K_(2)C_()O_(4) 3H_(2)C_(2)O_(4) .4H_(2)O`
Discussion
50423.

A compound is heated with zinc dust and ammonium chloride followed by addition of t Tollen's reagent. Formation of silver mirror indicates the presence of following group

Answer»

`-CHO`
`-overset(O)overset(||)(C)-`
`-NO_(2)`
`-NH_(2)`

ANSWER :C
Discussion
50424.

A compound is forned by two elemets X and Y , Atoms of the element Y (asanions)make ccp and those of the element X (as cations)occupy all the octahbedral voids. What is the formula of the compound ?

Answer»

SOLUTION :SUPPOSE number of atoms Y in ccp =n
As number of OCTAHEDRAL voids = No. of atoms in ccp : No of cotahedral voids = n
As alll the octahedral voids AR occupied by atoms X , therefore, number of atoms X = n .
Ratio of X :Y = n: n = 1:1.
Hence the formula of the compound is XY.
Discussion
50425.

A compound is formed by two elements X and Y. Atoms of the element Y (as anions) make ccp and those of the element X (as cation) occupy all the octahedral voids. What is the formula of the compound ?

Answer»

Solution :Suppose number of ATOMS Y in ccp=N
As number of octahedral VOIDS = No. of atoms in ccp `therefore` No. of octahedral voids=n
As all the octahedral voids are occupied by atoms X, therefore , number of atoms X=n
`therefore` Ratio of X :Y =n:n=1:1
Hence the formula of the compound is XY
Discussion
50426.

A compound is formed by two elements M and N. The element N forms cep and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound ?

Answer»

SOLUTION :SUPPOSE the atoms N in the ccp= n `therefore`No. of TETRAHEDRAL voids = 2n
As `1/3`rd of the tetrahedral voids are occupied by atoms M, therefore, no. of atoms `M="2n"/3`
`therefore `RATIO of `M:N="2n"/3:n=2:3` . Hence, the formula is `M_2N_3`
Discussion
50427.

A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. What is the number of structural isomers possible ?

Answer»

SOLUTION :Four : 1, 1-dichloropropane `(CH_(3)CH_(2)CHCl_(2))`, 1, 2-dichloropropane `(CH_(3)CHClCH_(2)Cl)` , 2,2-dichloropropane `(CH_(3)C Cl_(2)CH_(3))` and 1, 3-dichloropropane `(ClCH_(2)CH_(2)CH_(2)Cl)`.
Discussion
50428.

A compound is formed by elements A and B . This crystallizes in the cubic structure where the A atoms are at the corners of the cube can B atoms are at the body centers. The simplest formula of the compound is

Answer»

`A_(8)B_(4)`
`AB_(6)`
AB
`A_(6)B`

Solution :ATOMS of A perunit of the cell =`8times1/8=1` Atoms of B per unit cell=1
`THEREFORE` The SIMPLEST formula of the compound is AB
Discussion
50429.

A compound having the empirical formula C_(6)H_(6)O has the vapour density 47. find its Molecular formula (ii) The Simple A romatic Hydrocarbon compound (A) reacts with Bromine

Answer»

Solution :Empirical formula ` = C_(6)H_(6)O`
Empirical formula mass `=(6 xx 12)+(6 xx 1)+(16xx1)`
=72 + 6+16
= 94
VAPOUR density =47
MOLECULAR mass `=2 xx ` Vapour density
`= 2 xx 47`
=94
Molecular formula `=n xx` Empirical formula
`n = ("Molecular mass")/("Empirical formula mass ")`
`n = 94 //94 rArr n = 1 `
Molecular formula`= 1xx C_(6)H_(6)O`
`=C _(6)H_(6)O`
Discussion
50430.

A Compound having the empirical formula C_(6) H_(6) O has the vapour density 47 . Find its Molecular formula.

Answer»

Solution :EMPIRICAL Formula ` = C_(6) H_(6) O`
`n = ("MOLAR MASS")/("Calculated empirical formula mass")`
`= (2 xx "vapour DENSITY")/(94) = (2 xx 47)/(94) = 1 `
molecular formula `(C_(6)H_(6)O) xx 1 = C_(6) H_(6) O`
Discussion
50431.

A compound having bcc geometry has atomic mass 50. Calculate the density of the unit cell, if the edge length is 290pm.

Answer»

`6.81gcm^(-3)`
`3.40gcm^(-3)`
`13.62gcm^(-3)`
NONE of these

Solution :(a) density of unit CELL, `d=(ZxxM)/(N_(A)xxa^(3))`
`d=(2xx50)/((290xx10^(10))^(3)xx6.023xx10^(23))=6.81g//cm^(3)`
Discussion
50432.

A compound has molar of 98.98g. The analysis gave 24.27%C, 4.07%H and 71.6% CI. Determine the molecular formula of alkane from which the chlorinated compound is obtained.

Answer»

Solution :`C : H : CI = (24.27)/(12) : (4.07)/(1) : (71.6)/(35.5) = 1 : 2 : 1 `
Empirical FORMULA is ` CH_(2)CI`
NUMBER of repeating empirical units in the MOLECULAR formula `= ("Molecular formula mass" )/("Empirical formula mass")=(98.98)/(49.5) = 2`
Molecular formula is `C_(2)H_(4)CI_(2)`
The formula of alkane from which the halogenated compound is obtained `= C_(2)H_(6)`.
Discussion
50433.

A compound has the following composition Mg = 9.76%,S = 13.01%, 0 = 26.01, H_2O = 51.22, what is its empirical formula?[Mg = 24, S = 32, O = 16, H = 1]

Answer»

SOLUTION :
HENCE the empirical formula is Mg `SO_4 .7 H_2O`.
Discussion
50434.

A compound has C, H,N&S. The weight percentages of C&H are 41.37 and 5.75 respectively. On Kjeldahlising, the ammonia obtained from 1.01 g of the substance was neutralised by 11.6 ml of 1N HCI. In Carius method 0.2066 g of the substance gave 0.5544 g of BaSO_4.The molecular formula of the compound is

Answer»

`C_3H_4N`
`C_3H_5NS`
`C_2H_4NS`
`C_6H_5NS`

ANSWER :B
Discussion
50435.

A compound gives 'X' gas on addition of dilute HCI. The gas 'X', when passed through lime water or baryta water, gives white turbidity. Identify the compound.

Answer»

`Na_(2)CO_(3)`
`Na_(2)SO_(3)`
`Na_(2)S_(2)O_(3)`
`BaCO_(3)`

Answer :A::B::C::D
Discussion
50436.

A compound forms hexagonl close-packed structure. What is the total number of voids in 0.5 mol of its? How many of these are tetrahedral voids ?

Answer»

Solution :No. of ATOMS in the olose packing = 0.5 mol = ` 0.5 xx 6.022 xx 10^(23) = 3.011 xx 10^(23)`
No. of OCTAHEDRAL VOIDS =No, of atoms in the packing = ` 3.011 xx 10^(23)`
No. of TETRAHEDRAL voids = ` 2xx` No, of atoms in the packing = ` 2xx 3.011 xx 10^(23)= 6.022 xx 10^(23)`
Total no. of voidsl `= 3.011 xx 10 + 6.022 xx 10^(23) = 9.033 xx 10^(23)`
Discussion
50437.

A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids ?

Answer»

Solution :No. of ATOMS in the CLOSE packing =0.5 mol =`0.5 xx 6.022xx10^23 =3.011xx10^23`
No. of octahedral VOIDS= No. of atoms in the packing =`3.011xx10^23`
No. of tetrahedral voids= 2 x No. of atoms in the packing =`2xx3.011xx10^23 =6.022xx10^23`
Total no. of voids =`3.011xx10+6.022xx10^23=9.033xx10^23`
Discussion
50438.

A compound formed by elements X and Y has a cubic structure in which X atoms are at the corner of the cube and also at alternate face centers. Y atoms are present at the body center and also at the alternate edge center of the cube. If all the atoms are removed from one of the axis passing through the middle of the cube not containing face center atoms, formula of the compounds 64. If all the

Answer»

XY
`X_(2)Y`
`XY_(2)`
`XY_(3)`

SOLUTION :
Formula `Z_(EFF(X))=2,Z_(eff(Y))=1`
`therefore` Formula `=X_(2)Y`
Discussion
50439.

A compound formed by elements X and Y has a cubic structure in which X atoms are at the corner of the cube and also at alternate face centers. Y atoms are present at the body center and also at the alternate edge center of the cube. If all the atoms are removed from one of the plane passing through the middle of the cube which contains atoms both on the edge center as well as on the face center, calculate

Answer»

1
2
3
4

Solution :
`XrArr =2X` atoms from face center are removed
`YrArr=4Y` atoms from edge center are removed `YrArr=1Y` atom from BODY center is removed
`Z_(EFF(X))=(n_(c))/(8)+("Alternate"n_(f))/(2)=(8)/(8)+((2-2))/(2)=1+0=1`
`Z_(eff(Y))=(r_(B))/(1)+("Altranate"n_(c))/(4)=((1-1))/(1)+((4-4))/(4)=0+0=`zero
`Z_(eff(X+Y))=1+0=1`
Discussion
50440.

A compound formed by elements X and Y has a cubic structure in which X atoms are at the corner of the cube and also at alternate face centers. Y atoms are present at the body center and also at the alternate edge center of the cube. If all the atoms are removed from one of the plane passing through the middle of the cube which neither contains atoms on the edge center nor on the face center, total number of atoms in a cube.

Answer»

8
12
14
18

Solution :
`YrArr=1Y` atom from body center is removed Total number of atoms in a CUBE =8(corner)+2 (ALTERNATE FACE center)+4 (alternate edge center) +ZERO (body center)=14 atoms/unit cube.
Discussion
50441.

A compound formed by elements X and Y has a cubic structure in whch X atoms are at the corner of the cube and also at the face centers. Y atoms are present at the body center and at the edge center of the cube. If all the atoms are removed from one of the axis passing through one of the face centers of the cube, then formula of the compound is

Answer»

XY
`XY_(2)`
`X_(2)Y`
`XY_(3)`

Solution :
`YrArr1` body CENTER atom is removed
`XrArr2` FACE center atoms are removed
Formula : `Z_(eff(X))=3,Z_(eff(Y))=3`
Formula is `X_(3)Y_(3)` or 3XY.
Discussion
50442.

A compound formed by elements X and Y has a cubic structure in whch X atoms are at the corner of the cube and also at the face centers. Y atoms are present at the body center and at the edge center of the cube. If all the atoms from the diagonals of one of the face of the cube are removed, then Z_(eff) is

Answer»

3
4
7
8

Solution :
`XrArr4` corner atoms are removed
`XrArr1` FACE center atom is removed
`Z_(eff(X))=(n_(C))/(8)+(n_(f))/(2)=((8-4))/(8)+((6-1))/(2)=(1)/(2)+(5)/(2)=3`
`Z_(eff(Y))=(n_(b))/(1)+(n_(f))/(4)=(1)/(1)+(12)/(4)=1+3=4`
`Z_(eff(X))=3+4=7`
Discussion
50443.

A compound formed by elements X and Y has a cubic structure in whch X atoms are at the corner of the cube and also at the face centers. Y atoms are present at the body center and at the edge center of the cube. If all the atoms are removed from one of the plane passing through the middle of the cube, calculate total number of atoms in the cube,

Answer»

8
9
12
18

Solution :PLANE PASSING through the middle of cube (other atoms are not SHOWN)

`YrArr1` body center atom is removed
`YrArr4` edge center atoms are removed
`XrArr4` FACE center atoms are removed
Total number of atom in the cube =8 (corner)+2 (face centeres) +zero (body center)+8 (edge centers).
Discussion
50444.

A compound formed by elements X and Y crystallizes in the cubic structure where Y atoms are at the corners of the cube and X atoms are at the alternate faces. What is the formula of the compound ?

Answer»


Solution :As there are 8 Y ATOMS at the corners and CONTRIBUTION of each =`1/8`, therefore , no. of Y atoms/unit CELL `=8xx1/8=1` . There can be only two X atoms on ALTERNATE faces. As contribution of each of them =`1/2`, therefore, no. of X-atoms/unit cell =`2xx1/2=1`. Hence, formula is XY
Discussion
50445.

A compound formed by elements X and Y crystallises in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the face-centres. The formula of the compound is

Answer»

`XY_(2)`
`XY_(3)`
`X_(3)Y`
XY

Solution :No. of A STOMS present per UNIT cell =`8times1/8=1`
No. of B atoms present per unit cell =`6times1/2=3` ltbegt `therefore` Correct composition of `A_(X)B_(y)"is"AB_(3)`
Discussion
50446.

A compound formed by elements A and B has a cubic structure in which A atoms are at the corners of the cube and B atoms are at the face centres. Derive the formula of the compound.

Answer»

Solution :As A atoms are PRESENT at the 8 CORNERS of the cube, therefore , number of atoms of A in the unit CELL `=1/8xx8=1`
As B atoms are present in the face centres of the 6 FACES of the cube, therefore, number of atoms of B in the unit cell`=1/2xx6=3`
`therefore` Ratio of atoms A:B =1:3
Hence, the formula of the compound is `AB_3`
Discussion
50447.

A compound exists in the gaseous phase both as monomer (A) and dimer (A_(2)) The mo1 wt of A is 48 In an experiment 96g of compound was confined in a container of volume 33.6 litre and heated to 273^(@)C Calulate the pressure developed if the compound exists as dimer to the extent of 50% by weight under these conditions .

Answer»


ANSWER :`2ATM`
Discussion
50448.

A compound decolourises Baeyer's reagent and gives a mixture of propanoic acid and ethanoic acid when treated with a hot and conc. solution of KMnO_4 . The compound is

Answer»

pent-1-ene
pent-2-ene
2-methylbut-1-ene
2-methylbut-2-ene

Answer :B
Discussion
50449.

A compound contains two types of atoms X and Y. It crystallizes in a cubic lattice with atom X at the corners of the unit cell and atom Y at the body centres. The simplest possible formula of this compound is

Answer»

`X_()3)Y`
`X_(2)Y`
XY
`XY_(6)`

Solution :X atoms PER unit CELL =`1//8times8=1`
Y atoms per unit cell =1
`THEREFORE` formula of compound :XY
Discussion
50450.

A compound contains two dissimilar asymmetric carbon atoms. The number of optical isomers is

Answer»

2
3
4
5

Solution :4
Discussion