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90451.

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 and B atoms are at the body centres. The simplest formulta of the compound is :

Answer»

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

Solution :No. of A atoms `= 8 XX ( 1)/( 8 ) = 1 `
No. of B atoms `=1 ` ( at body centre )
FORMULA `: AB`
Discussion
90452.

A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is :

Answer»

`C_4A_3`
`C_2A_3`
`C_3A_2`
`C_3A_4`

Solution :ANION A is in hcp ion.
The NUMBER of A ion in unit cell = 6
The number of octahedral VOIDS = 6
Cation C ion occupies 75% of space.
`THEREFORE`Number of `C = (6 xx 75)/(100) = 9/2`
`therefore ` Molecular formula = `C_(9/2)`
`therefore ` Molecular formula `= C_9A_12`
Molecular formula `= C_3A_4` (General ratio) .
Discussion
90453.

A compound is formed by elements A and B . Thiscrystallises in the cubic structure where the A atoms are the corners of the cube and B atoms are at the body centres . The simplest formula ofthe compound is

Answer»

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

Solution :AB compound crystallises in the cubic structure. A ATOMS are at the corners of the cube. So , No. of a atoms per unit CELL `=8cc1//8=1`.
No . Of B atoms per unit cell =1
`therefore` Simplest FORMULA of the compound = AB.
Discussion
90454.

A compound is chiral even if

Answer»

a MIRROR plane is present
a CENTRE of INVERSION exists
a ROTATION axis axists
an IMPROPER rotation axis is present

Answer :C
Discussion
90455.

A compound in the hydrated state gives an ionic oxide when heated strongly. Its aqueous solution does not give any precipitate with H_(2)S, however, it does give white ppt. with AgNO_(3) solution which is soluble in aqueous ammonia. The given compound contains a metal which resembles Li and an integral part of Grignard reagent. Identify the compound withh the equation of its oxide formation.

Answer»

Solution :The metal which is diagonally RELATED to LI and forms an integral PART of Grignard reagent is Mg. Besides Mg, the hydrated salt ALSO contain `Cl^(-)` as it gives white ppt. of AgCl with `AgNO_(3)`. AgCl is soluble in `NH_(4)OH`. Thus, the given salt is MgO as FOLLOWS:
`MgCl_(2).6H_(2)O overset("heat")to MgO + 2HCl + 5H_(2)O`
`MgCl_(2) + 2AgNO_(3) to 2AgC l + Mg(NO_(3))_(2)`
`AgCl + 2NH_(4)OH to [Ag(NH_(3))_(2)Cl^(-) + 2H_(2)O]`
`Mg + RX overset("ether") to underset("Gingard reagent")(RMgX)`
Ans: The compound is `MgCl_(2).6H_(2)O`
Discussion
90456.

A compound having n independent asymmetric centres will have

Answer»

N recemates
2n RACEMATES
`2^(n+1)` racemates
`2^(n-1)` racemates

Solution :`2^(n-1)` racemates.
Discussion
90457.

A compound having n dissimilar asymmetric centres will have :

Answer»

`2^n` OPTICAL ISOMERS
Zero
`2^(n+1)` optical isomers
`2^(n-1)` optical isomers

Answer :D
Discussion
90458.

A compound having n dissimilar asymmetric carbon atoms can exist in stereoisomers equal to :

Answer»

`2^(N)`
`2N^(2)`
`2sqrt(n)`
`2^(n+1)`

ANSWER :A
Discussion
90459.

A compound having molecular formula C_(5)H_(10)O_(5) was reacted with excess of CH_(3)COCl to give a compound having molecular weight 318 gm. Predict the number of hydroxy groups present into the unknown compound.

Answer»


Solution :In REACTION IST 6 moles are used.2 moles are used for acid-base reaction.
So,x=6
In 2nd reaction 3 moles are usedy=3
In 3rd reaction 3 moles are usedz=3
x+y+z=6+3+3
=12
Discussion
90460.

A compound having elements X and Y crystallises in a cubic structure, where X is at the corneer position and Y is at the center of the cube. The correct formula of the compound is

Answer»

XY
`x_(3)y`
`xy_(2)`
`xy_(3)`

ANSWER :A
Discussion
90461.

A compound hasmoleculemass42. it should be :

Answer»

PROPANE
cycloprpane
Propene
butane

Answer :B,C
Discussion
90462.

A compoundhas an empirical formula C_(2)H_(4)O. An independent analysisgave a value of 132.16 for its molecular mass.What is the correct molecular formula

Answer»

`C_(4)H_(4)O_(5)`
`C_(10)H_(12)`
`C_(7)O_(3)`
`C_(6)H_(12)O_(3)`

Solution :Empirical formula WIEGHT `C_(2)H_(4)O`
`= (12xx 2+ 4+16) = 44`
Molecular formula` = (" mol. Wt")/(" eq. formula wt.") XX " Emp. Formula"`
`= (132.1)/(44) xx" Empiriacal formula " = 3 xx C_(2)H_(4)O = C_(6)H_(12)O_(3)`.
Discussion
90463.

A compound has 62 % carbon, 10.4% hydrogen and 27.5% oxygen. If molar mass of compound is 58, find numbe of H-atom per molecule of the compound.

Answer»


ANSWER :6
Discussion
90464.

A compound has a vapour density of 29. on warming an aqueous solution of alkali, it gives a yellow precipitate. The compound is

Answer»

`CH_(3)CH_(2)CHO`
`CH_(3)CHOHCH_(3)`
`CH_(3)COCH_(3)`
`CH_(3)CH_(2)COOH`

Solution :Molecular WEIGHT of the COMPOUND `= 2xx` Vapour density `= 2 xx 29 = 58`
Molecular weight of `CH_(3)CH_(2)CHO, CH_(3)CHOHCH_(3), CH_(3)COCH_(3)` and `CH_(3)CH_(2)COOH` are 58,60,58 and 74 RESPECTIVELY. Both `CH_(3)CH_(2)CHO` and `CH_(3)COCH_(3)` have molecular weight 58 but only aldehyde i.e., `CH_(3)CH_(2)CHO` on warming with AQUEOUS alkali gives yellow precipitate.
Discussion
90465.

A compound has haemoglobin-like structure. Ithas one Fe atom. It contains 4.6% of Fe. The approximate molecular mass is

Answer»

`100 g "mol"^(-1)`
`1200 g "mol"^(-1)`
`1400 g "mol"^(-1)`
`1600 g "mol"^(-1)`

Solution :1g of atom of Fe (56gFe) is present in 1 mol of COMPOUND . As 4.6 g of Fe are present in 100 g of the compound,
56 g of Fe will be present in `(100)/(4.6) XX 56 = 1217 g ` of compound .
` THEREFORE ` Approximate molecular mass = 1200
Discussion
90466.

A compound has 3 chiral carbon atoms. The number of possible optical isomers it can have

Answer»

3
2
8
4

Solution :The number of possible optical ISOMERS `= 2^(N)`
(where 'n' is the number of CHIRAL C-atoms) `= 2^(3) = 8`
Discussion
90467.

A compound H_(2)X with molar weight of 80 g is dissolved in a solvent having density of 0.4 g ml^(-1). Assumingno change in volume upon dissolution, the molality of a 3.2 molar solution is

Answer»


Solution :Given 3.2 M solution
`THEREFORE` moles of solute = 3.2 MOL
CONSIDER 1 L Solution.
`therefore` volume of SOLVENT = 1 L
`P_("solvent")=0.4 g.mL^(-1) therefore m_("solvent")=P xx V=400 g`
`therefore` molality `=(3.2 mol)/(0.4 kg)= 8` molal.
Discussion
90468.

A compound giving NH_(3) on boiling with sodium hydroxide is:

Answer»

`CH_(3)CONH_(2)`
`CH_(3)NH_(2)`
`CH_(3)CN`
`C_(6)H_(5)CHO.`

ANSWER :A
Discussion
90469.

A compound H_(2)X with molar weight of 80 g is dissolved in a solvent having density of "0.4 g mL"^(-1). Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

Answer»

Solution :3.2 molar solution contains 3.2 moles of `H_(2)X` in 1000 mL of the solution. As there is no change in volume on dissolution, volume of solvent= 1000 mL.
`"Mass of solvent= 1000 mL "xx0.4"g mL"^(-1)=400g`
= 0.4 kg
`"MOLALITY"=("3.2 moles")/("0.4 kg")="8 mol kg"^(-1)=8M`
Discussion
90470.

A compound give violet flamerestand givesa white ppt with AgNO_(3) .Thecompound is

Answer»

`NACI`
`KCI`
`BaCI_(2)`
`CaCI_(2)`

Answer :b
Discussion
90471.

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 :NUMBER of ATOMS in 0.5 mol = `0.5 xx 6.022 xx 10^(23)= 3.011 xx 10^(23)`.
Number of octahedral voids = Number of atoms = `3.011 xx 10^(23)`
Number of tetrahedral voids =`2 xx "Number of atoms" = 2 xx 3.011 xx 10^(23)= 6.022 xx 10^(23)`.
It may be remembered that the number of octahedral voids is EQUAL to the number of atoms in close-packing while the number of tetrahedral voids is TWICE the number of atoms in close- packing.
Hence, total number of voids = `3.011 xx 10^(23) + 6.022 xx 10^(23) = 9.033 xx 10^(23)`
Discussion
90472.

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 :Total number of atoms in 0.5 MOL of HCP lattice `= 0.5 xx NA`
`therefore ` Total number of TETRAHEDRAL voids are
`(2 xx 0.5 NA) = 6.022 xx 10^23`
`therefore` Total number of voids = No. of Octahedral voids + No. of Tetrahedral voids.
`= 3.011 xx 10^23 + 6.022 xx 10^23 `
` = 9.033 xx 10^23`
Discussion
90473.

A compound forms hcp structure, what is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?

Answer»

SOLUTION :Number of particles in the close PACKED structure = `0.5xx6.02xx10^(23)=3.01xx10^(23)`
Number of TETRAHEDRAL voids = ` 2xx3.01xx10^(23)=6.02xx10^(23)`
Number of ctahedral voids `= 3.01xx10^(23)`
`THEREFORE` total number of voids `=6.02xx10^(23)+3.01xx10^(23)=9.03xx10^(23)`
Discussion
90474.

A compound formed by elements X, Y and Z has a cubic structure in which X atoms are at the corner of the cube and also at alternate face centres. Y atoms are present at the body centre and Z atoms are present at the alternate edge centre. Then the molecular formula of the compound is

Answer»

`XYZ`
`XY_2 Z`
`XYZ_3`
`X_2 YZ`

ANSWER :D
Discussion
90475.

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

Answer»

Solution :The Total no. of voids = The no. of octahedral voids + the no. of tetrahedral voids present in the compound.
Given : no of moles = 0.5 MOL
`because` 1 mol contains `6.022 xx 10^(23)` ATOMS
`therefore` 0.5 mol contains `6.022 xx 10^(23) xx 0.5`
`= 3.011 xx 10^(23)` atoms
`because` No. of octahedral voids present in the
Compound `= 3.011 xx 10^(23)` voids
`therefore` No. of tetrahedral voids
`= 2 xx` no. of octahedral voids
`= 2 xx 3.011 xx 10^(23)`
`= 6.022 xx 10^(23)`
`therefore` Total no. of voids
`= 3.011 xx 10^(23) + 6.022 xx 10^(23)`
`= 9.033 xx 10^(23)` voids
Discussion
90476.

A compound formed by elements X & Y. Crystallizes in the cubic structure, where X is at the corners of the cube and Y is at the six face centres. What is the formula of the compound? If side length is 5A^(@), estimate the density of the solid assuming atomic weight of X and Y as 60 and 90 respectively.

Answer»

Solution :From eight corner atoms one atoms (X) contribute to one unit cell.
From six face centres, three atoms (Y) contribute to one unit cell.
So, the formula of the COMPOUND is `XY_(3)`.
As we know that,
`rho =(n xx M_(m))/(N_(A) xx a^(3))`, here n=1
MOLAR mass of `XY_(3)`
`M = 60 + 3 xx 90 = 330` gm
`rho =(1 xx 300)/(6.023 xx 10^(23) xx (5 xx 10^(-8))^(3)) gm//cm^(3)`
`a = 5A = 5 xx 10^(-6)` cm
`=(330)/(6.023 xx 10^(23) xx 125 xx 10^(-34)) gm//cm^(3) = 4.38 gm //cm^(3)`
Discussion
90477.

A compound formed by elements X and y crystallises in the cubic structure , where X atoms at the corners of a cube and Y atoms are at the centres of the body . The formula of the compound is

Answer»

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

SOLUTION :`X=(8xx1/8)=1,Y=1 :.` FORMULA `=XY`
Discussion
90478.

A compound formed by elements X & Y, Crystallizes in a cubic structure, where X is at the corners of the cube and Y is at six face centers. What is the formula of the compound ? If side length is 5Å, estimate the density of the solid assuming atomic weight of X and Y as 60 and 90 respectively.

Answer»


ANSWER :`XY_(3)`, 4.38 G/`CM^(3)`
Discussion
90479.

A compound formed by elements X and Y crystallizes 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»

`x_(3)y`
XY
`xy_(2)`
`xy_(3)`

ANSWER :D
Discussion
90480.

A compound formed by elements X and Y crystallizes in cubic arrangement in which X atoms are at the corners of a cube and Y atoms are at the face centres. What is the formula of the compound ?

Answer»

SOLUTION :In one unit cell:
`"No. of ATOMS of X" = 8 xx (1)/(8)=1`
`"No. of atoms of Y" = 6 xx (1)/(2)=3`
`"Formula of compound" =X_(1) Y_(3) = XY_(3)`
Discussion
90481.

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

Answer»


ANSWER :`AB_3`
Discussion
90482.

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

Answer»

`X_2Y` 
`XY `
`XY_2` 
`XY_3`

SOLUTION :Number of X-atom = `(8 XX 1/8) = 1`
Number of Y-atoms = `(6 xx 1/2) = 3`
`THEREFORE X : Y = 1 : 3 IMPLIES XY_3`
Discussion
90483.

A compound formed by elements A and B crystallizes in the cubic structure where A atoms are at the corners of a cube and B atoms are at the face centre. Determine the simple formula of the compound.

Answer»


ANSWER :`AB_(3)`
Discussion
90484.

A compound formed by elements A and B has a cubic structure in which A atoms are at thecorners of the cube and B atoms are at the face centres. The formula of the compound is :

Answer»

`AB_(3)`
`A_(2) B_(3)`
`A_(2) B`
`A_(4)B_(3)`

Solution :The number of A ATOMS in a cube =` 8 xx ( 1)/(8) =1 `
Number of B atoms at faces `= 6 xx ( 1)/( 2) = 3`FORMULA `: AB_(3)`
Discussion
90485.

A compound formed by elements A and B crystallizes in the cubic arrangement in which atoms A are at the corners of the cube and atoms B at the face centres. What is the formula of the compound?

Answer»


SOLUTION :Number of atoms of A: (8 cornes, `1/8` FRACTION), i.e., `8 x 1/8` = 1, number of atoms of B (6 faces, `1/2` fraction), i.e., `6 xx 1/2` =3
`:.` Number of cations and anions in ONE unit cell = 1 + 3 = 4. So simplest formula is `AB_3`.
Discussion
90486.

A compound does not react with 2,4 di-nitrophenyl hydrazine and Na, compound is

Answer»

Acetone
ACETALDEHYDE
`CH_(3)OH`
`CH_(2)=CHOCH_(3)`

SOLUTION :It is not acetaldehyde or acetone as does not react with hydrazine. It is not `CH_(3)OH` as does not react with NA.
Discussion
90487.

A compound formed by elements A and B crystallizes in cubic structure where A atoms are at the corners of a cube and B atoms are at the face centre. The formula of the compound is :

Answer»

`AB_3`
`AB_2`
`AB_4`
NONE of these

Answer :A
Discussion
90488.

A compound D(C_(6) H_(10)O) upon treatment with alkaline solution of iodine gives a yellow precipitate. The filtrate on acidification gives a white solid E(C_(7)H_(6)O_(2)). Write the structures of D and E and explain the formation of E.

Answer»

SOLUTION :Since compound `D(C_(8)H_(10)O)` upon treatment with ALKALINE solution of `I_(2)` gives YELLOW ppt, THEREFORE, D may be either methyl ketone `(-underset(O) underset(||)(C)-CH_(3))` or methyl carbinol `(-underset(OH) underset(|)(CH)-CH_(3))`
The molecular formula of suggests it to be 1-phenyl ethanol, `C_(6)H_(5) underset(OH) underset(|)(CH)CH_(3)` The reaction may be EXPLAINED as:
Discussion
90489.

A compound D onhydrolysisformedacetamide whichfurtherhydrolysedto giveaceticacid andNH_(3).ThecompoundD is

Answer»

`CH_(3) C-= N`
`C_(2) H_(5)Ona`
`C_(2) H_(5) OH`
`C_(2) H_(5)I`

Solution :`CH_(3)C-=underset( D ) ( N ) +2N_(2) O toCH_(3)CONH_(2)`
`CH_(2) CONH_(2)+ H_(2) Ooverset( H^(+)) (to)CH_(3) COOH+ NH_(3)`
Discussion
90490.

A compound 'D' (C_(8)H_(16)O) has following observation (a) It reacts with Lucas reagent with in 5 min. (b) On heating in presence of acid given a compound X, which can decolorize KMnO_(4) as well as Br_(2)//CCl_(4). (c ) compound Xoverset("ozonolysis")rarr Yoverset(I_(2)//NaOH)rarr2 moles of iodoform + sodium salt of dibasic acid overset("Sodalime")rarr n - butane Then D, X, Y are

Answer»




ANSWER :A
Discussion
90491.

A compound (D) (C_(8)H_(10)O) upon treatement with alkaline solution of iodine gives a yellow precipitate. The filterate on acidification gives a white solid (E) (C_(7)H_(6)O_(2)). Write the structures of (D) and (E), and explain the formation of (E).

Answer»

Solution :`DU` in `(D) =((2n_(C )+2))/2=((2xx8+2)-10)/2=4^(@) (C:H~~1:1)`
So, it must contain benzene ring, and `(MeCO-)` GROUP, since it gives IODOFORM TEST. The structure of COMPOUND `(D)` may be `(Ph-CO-Me)`.
Discussion
90492.

A compound crystallises as follows : ions ''A'' are at corners of a cubic unit cell and ''B'' ions at face centres of a cubic unit cell and ''C'' ions in 1//4^(th) of the total tetrahedral void. Assuming if this is dissolved, only the ions in the tetrahedral voids are dissociated completely in water, which one of the following statements is true. (Assuming all are univalent ions) and also A and C are cations and B is an anions.

Answer»

Boiling point of same concentration of `AlCl_(3)` solution (100% dissociation) will be GREATER than that of this solution.
Boiling point of same concentration of `Ca_(3)(PO_(4))_(2)` (100% dissociation) will be greater than that of this solution.
Boiling point of same concentration of sucrose will be greater than this solution
DATA insufficient to predict.

Answer :B
Discussion
90493.

A compound contains two dissimilar asymmetric carbon atoms. The number of stereoisomers is:

Answer»

2
3
4
1

Answer :C
Discussion
90494.

A compound CuCl has face centred cubic structure. Its density is 3.4g cm^-3.The length of unit cell is :

Answer»

`5.783overset@A`
`6.783overset@A`
`7.783overset@A`
`8.783overset@A`

ANSWER :A
Discussion
90495.

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

Answer»

SOLUTION :`X = 8XX(1)/(8)=1`
Y=1
SIMPLEST formula = XY.
Discussion
90496.

A compound contains elements X and Y. Y atoms form CCP lattice and atoms of X occupy l/3rd of tetrahedral voids. What is the molecular formula of the compound ?

Answer»

xy
`x_(2)y_(3)`
`x_(3)y_(3)`
`xy_(2)`

ANSWER :B
Discussion
90497.

A compound contains three elements A, B and C, if the oxidation number of a+2, B=+5 and C=-2, the possible formula of the compound is

Answer»

`A_(3)(B_(4)C)_(2)`
`A_(3)(BC_(4))_(2)`
`A_(2)(BC_(3))_(2)`
`ABC_(2)`

SOLUTION :Sum of oxidation numbers of all atoms in a neutral molecule is ZERO.
`to A_(3)(B_(4)C):2xx3+2xx[4xx5-2]=42ne0`
`to A_(3)(BC_(4))_(2):+2xx3+2xx[+5-8]=0`
`to A_(2)(BC_(3))_(2):+2xx2+2xx[+5-6]=+2 NE0`
`to ABC_(2):+2+5+(-2)xx2=+3 ne0`
Discussion
90498.

A compound contains three elements A,B and C, if the oxidation number of A=+2B=+5 and C=-2 then possible formula of the compound is

Answer»

`A_(3)(B_(4)C)_(2)`
`A_(3)(BC_(4))_(2)`
`A_(2)(BC_(3))_(2)`
`ABC_(2)`

Solution :`A^(2)+BC_(4)^(3) implies A_(3)(BC_(4))_(2)`
Discussion
90499.

A compound contains atoms X, Y, Z. The oxidation number of X is +2 , Y is +5, and Z is -2. The possible formula of the compound is :

Answer»

`XY_1Z_2`
`Y_2(XZ_3)_2`
`X_3(YZ_4)_2`
`X_3(Y_4Z)_2`

ANSWER :C
Discussion
90500.

A compound contains atoms of three elements in A, B C. If the oxidation number of A is +2, B is +5 and that of C is -2, the possible formula of the compound is

Answer»

`A_(3)(BC_(4))_(2)`
`A_(3) (B_(4) C)_(2)`
`ABC_(2)`
`A_(2) (BC_(3))_(2)`

SOLUTION :`A_(3)(BC_(4))_(2)=3xx2+[5+(-2xx4)]_(2)=0`
Discussion