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50351.

A cubic solid is made up of two elements X and Y. Atoms Y are present at the corners of the cube and atoms X at the body centre. What is the formula of the compound ? What are the coordination numbers of X and Y ?

Answer»

Solution :As ATOMS Y are present at the 8 CORNERS of the cube, THEREFORE , number of atoms of Y in the UNIT cell `=1/8xx8=1`
As atoms X are present at the body centre, therefore, number of atoms of X in the unit cell=1
`therefore` Ratio of atoms X :Y =1:1
Hence, the formula of the COMPOUND is XY
Coordination number of each of X and Y =8
Discussion
50352.

A cubic solid is made up of two elements P and Q. Atoms Q are present at the corners of the cube and atoms P at the body centre. What is the formula of the compound ?What are the coordination numbers of P and Q ?

Answer»

<P>

SOLUTION :( Usesymbols P and Q in PLACE of X and Y)
Discussion
50353.

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is 8 g cm^(-3), then the number of atoms present in 256 g of the crystal is Nxx10^24 . The value ofN is

Answer»


Solution :`rho=(ZxxM)/(a^3xxN_0xx10^(-30))`where a is in pm and M in g `MOL^(-1)`
`8=(4xxM)/((400)^3xx6xx10^23xx10^(-30))` or M=76.8
MOLES in 256 g =`256/76.8`=3.33
`THEREFORE` No. of atoms =`3.33xx6xx10^23=2xx10^24`
Discussion
50354.

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is 8g cm^(-3) , then the number of atoms present in 256 g of the crystal isN xx 10^(24). The value of N is

Answer»


Solution :`p ( Z XX M)/( a^(3) xx N_(0) xx 10^(-30)`
Where a is in pm and M in g ` Mol^(-1)`
` 8 = ( 4 xx M)/((400)^(3) xx 6 xx 10^(23) xx 10^(-30))) ORM = 76.8`
Moles in 256 g = ` 256/76.8 = 3.33`
No, of ATOMS= ` 3.33 xx 6 xx 10^(23)`
`= 2XX 10^(24)`
Discussion
50355.

A crystalline solid has a cubic structure in which tungsten (W) atoms are located at the cube corners of the unit cell,oxygen atoms atthe cube edgesand sodium atom at the cubecentre. The molecular formula of the compund is

Answer»

`Na_(2)WO_(3)`
`NaWO_(4)`
`NaWO_(3)`
`Na_(2)WO_(4)`

Solution :No. of W ATOMS=`8times1/8=1`
No. of O atoms=`12times1/4=3`
No. of NA atoms=`1times1=1` ltbnrgt FORMULA of COMPOUND: `NaWO_(3)`.
Discussion
50356.

A crystalline solid have

Answer»

DISORDERED arrangement
Long range order
Short range order
None of these

SOLUTION :A CRYSTALLINE solid have long range order.
Discussion
50357.

A crystalline salt on being rendered anhydrous loses 45.6 % of its weight. The percentage composition of anhydrous salt is : Al = 10.5 % , K = 15.1 % , S = 24.96 %, O = 49.92 % Find the simplest formula of the anhydrous and crystalline salt.

Answer»


Solution :Step I. Formula of anhydrous SALT
`{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Al",10.5,27,(10.50)/(27)=0.39,(0.39)/(0.38)=1.03,1),("K",15.1,39,(15.1)/(39)=0.38,(0.38)/(0.38)=1.00,1),("S",24.96,32,(24.96)/(32)=0.78,(0.78)/(0.38)=2.05,2),("O",49.92,16,(49.92)/(16)=3.12,(3.12)/(0.380)=8.21,8):}`
Formula of anhydrous salt `= AlKS_(2)O_(8)`
Empirical formula mass `= 27 + 39 + 2 xx 32 + 8 xx 16 = 258 u`
Step II. Formula of crystalline salt
LET the WEIGHT of hydrated salt = 100 u , Weight of water lost = 45.6 u
Weight of anhydrous salt `= 100 - 45.6 u = 54.4 u`
If the weight the anhydrous salt is 54.4 u, that of water = 45.6 u
If the weight of anhydrous salt is 258 u, that of water `= ((45.6u))/((54.4u))xx(258u)=216.3u`
No. of `H_(2)O` molecules `= ("Weight of water")/("Molecular mass of water")=((216.3u))/((18u))=12` (approx)
`:.` The formula of crystalline salt `= KAlS_(2)O_(8).12H_(2)O`.
Discussion
50358.

A crystalline compound when heated become anhydrous by losing 51.2 % of the mass. On analysis, the anhydrous compound gave the following percentage composition : Mg = 20.0 % , S = 26.66 % and O = 53.33 %. Calculate the molecular formula of the anhydrous compound and crystalline compound. The molecular mass of anhydride compound is 120 u.

Answer»

Solution :Step I. DETERMINATION of empirical formula of the anhydrous compound
`{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("MG",20.0,24,(24.0)/(24)=0.83,(0.83)/(0.83)=1,1),("S",26.66,32,(26.66)/(32)=0.83,(0.83)/(0.83)=1,1),("O",53.33,16,(53.33)/(16)=3.33,(3.33)/(0.83)=4.01,4):}`
`:.` The empirical formula of anhydrous compound is `MgSO_(4)`
Step II. Determination of the molecular formula of the anhydrous compound
Empirical formula mass `= 24+32 +4 XX 16 = 24 + 32 + 64 = 120` u
Molecular mass = 120 u (Given) , `n=("Molecular mass")/("Empirical formula mass")=(120)/(120)=1`
`:.` Molecular formula `=nxx` Empirical formula `=1 xx MgSO_(4)=MgSO_(4)`
`:.` Molecular formula of anhydrous compound `= MgSO_(4)`
Step III. Determination of molecular formula of the crystalline compound
Let us first calculate the number of molecules of water of crystallisation in the compound. For that,
Let the weight of hydrated compound = 100 u
Weight of water lost = 51.2 u
`:.` Weight of anhydrous compound `= 100-51.2 = 48.9` u
Now, if the weight of anhydrous compound is 48.8 u, then that of water 51.2 u
If the weight of anhydrous compound is 120 u, then that of water `= (51.2)/(48.8)xx120 = 125.9` u
Molecular mass of water `(H_(2)O)=2xx1+16 = 18` u
`:.` No. of `H_(2)O` molecules `= ("Weight of water")/("Molecular mass of water")=(125.9)/(18)=7` (approx)
Thus, the formula of crystalline compound `= MgSO_(4).7H_(2)O`.
Discussion
50359.

A crystal of Lead (II) suphide has NaCl structure. In this crystal the shortest distance between thePb^(2+) ion ands^(2-) ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide ?Also calculate the unit cell volume.

Answer»


Solution :EDGE 'a' of unit cell = 2 ` xx` DISTANCE between` Pb^(2+) and S^(2-)` IONS .
Discussion
50360.

A crystal of Lead (II) sulphide has NaCl structure. In this crystal the shortest distance between the Pb^(2+) ion and S^(2-) ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide ? Also calculate the unit cell volume.

Answer»


SOLUTION :Edge 'a' of UNIT CELL = 2 x Distance between `Pb^(2+)` and `S^(2-)`ions .
Discussion
50361.

A crystal of KCl containing a small amount of CaCl_(2) will have

Answer»

vacant `CL^(-)` sites
vacant `K^(+)` sites and a higher DENSITY as compared to pure KCl
vacant `K^(+)` sites and a LOWER density as compared to pure KCl
`K^(+)` ions in the interstitial sites

Answer :C
Discussion
50362.

A crystal may have a number of planes or axes of symmetry but it possesses only one ----- of symmetry.

Answer»

SOLUTION :CENTRE
Discussion
50363.

A crucially important species formed by oxygen in stratosphere is ______.

Answer»
Discussion
50364.

A cricket ball of mass 0.5 kg is moving with a velocity of 100 ms^(-1), the wavelength associated with its motion is

Answer»

`13.25xx10^(-26) m`
`13.25 xx10^(-34)` m
`13.25xx10^(-36) m`
`6.6xx10^(-34) m`

Solution :`LAMBDA = (H)/(mv) = (6.6xx10^(-34))/(0.5xx100) = 13.25 xx 10^(-36) m `
Discussion
50365.

A covalent molecule AB_(3) has trigonal pyramidal structure. The number of lone pair and bonding pair of electrons in the molecule are ...... .

Answer»

2 and 2
3 and 1
1 and 3
0 and 4

Solution :1 and 3
`AB_(3) and AB_(3)` E are the same type ATOM. There are `sigma` type three A - B BOND in it and one nonbonding electron pair on A.

Due to this non bonding electron pair and as per the principal of VSEPR three A - B bonds come closer and FORM the pyramidal triangle structure.
Discussion
50366.

A covalent bond is likely to be fonny between two elements which

Answer»

have HIGH electronegativities
have low ionization energies
have low MELTING POINTS
FORM IONS with a small charge

Answer :A
Discussion
50367.

A correct IUPAC name for the following compound is:

Answer»

3,6,7-trimethyl-4-bromo-1-octene
4-bromo-3-methyl-6-isopropyl-1-heptene
4-bromo-3,6,7-triethyl-1-octene
4-bromo-6-isopropyl-3-methyl-1-heptene

Answer :C
Discussion
50368.

A correct IUPAC name for the following compound is:

Answer»

2,5-Dimethyl-3-propyl HEPTANE
3,6-Dimethyl-5-propyl heptane
6-Methyl-4-(1-methylethyl)OCTANE
3-Methyl-5-(1-methylethyl)octane

ANSWER :D
Discussion
50369.

A correct IUPAC name for the following compound is :

Answer»

4-propyl-5-chloro-3-heptanol
4-(1-chloropropyl)-3-heptanol
4-propyl-3-chloro-5-heptanol
5-chloro-4-propyl-3-heptanol

ANSWER :D
Discussion
50370.

A copper wire is dipped in silver nitrate solution in beaker A and a silver wire is dipped in a solution of copper sulphate kept in beaker B. If the standard electrode potential for Cu^(2+) +2e^(-) rarr Cu " is " +0.34 and for Ag^(+) +e^(-) rarr Agis 0.80 V . Given E^@(Ni^(2+)|Ni)=-0.25V and E^@(Cu^(2+) |Cu) = 0.34 V Predict in which beacker the ions present will get reduced ?

Answer»


ANSWER :In BEAKER A , `AG^(+)` will be REDUCED.
Discussion
50371.

A copper plate of 20 cm x 10 cm is to be plated with silver of 1 mm thickness on both the sides. Number of moles of silver required for plating is (density of silver =10.8 g m /cc)____________

Answer»


SOLUTION :VOLUME = `20xx10xx0.1=20C C`
`1C C-10.8g`
`20C C-20xx10.8=2169`
On both sides `=2xx216=432g`
`therefore` No. of moles of `Ag=(432)/(108)=4`
Discussion
50372.

A container of just 10cc has 2.69 xx 10^20 gas molecules at 0^@C. What is the pressure exerted?

Answer»

Solution :Number of MOLES of the gas = `(2.69 xx 10^20)/(6.022 xx 10^(23)) = 4.47 xx 10^(-4)`
Pressure (P) exerted by gas MOLECULES :
`P = (nRT)/V = (4.47 xx 10^(-4) xx 82.1 xx273)/10 = 1 atm`.
Discussion
50373.

A container is separated into two 1 litre compartments by a piston of negligible mass as shown below. In lower compartment 0.1 bar pressure and the other compartment is empty. Now the stoppers are removed so that the gas expands to 2 litre. Heat is supplied to the gas so that finally pressure of gas equals to 0.1 bar. Calculate DeltaHof the process in joule. [1 "bar" liter = 100 J]

Answer»


ANSWER :25
Discussion
50374.

A container holds 3 liter of N_(2(g)) and H_(2)O_((l)) at 29^(@)C The prssure is found to be 1 atm The water in container is instantaneously electrolysed to give H_(2) and O_(2) following the reaction H_(2)O_(l) rarrH_(2(g)) +(1)/(2)O_(2_(g)) At the end of electrolysis the pressure was found to be 1.86atm calculate the amount of water present in the container If the aqueous tension of water at 29^(@)C is 0.04 atm.

Answer»


ANSWER :`1.24 G`
Discussion
50375.

A conjugated alkediene having molecular formula C_13H_22 on ozonolysis yielded ethyl methyl ketone and cyclohexanecarbaldehyde. Identify the diene, write its structural formula and give its IUPAC name.

Answer»

Solution :The structures of the products of ozonolysis are :

Since, the total number of carbon atoms of the TWO products is 11 (4+7) while the MF of the CONJUGATED diene is `C_13H_22`, therefore , the ozonolysis must have also produced another two carbon product. Further since the GIVEN COMPOUND `(C_13H_22)` is an alkadiene, therefore , this two carbon product. Further since the given compound `(C_13H_22)` is an alkadiene, therefore , this two carbon product must be glyoxal, O=CH-CH=O. Replace the oxygen atoms from these three products by double bonds, the structure of the alkadiene is
Discussion
50376.

A conjugated alkadiene having molecular formula C_(13)H_(22), on ozonolysis, yielded ethyl methylketone and cyclohexanecarbaldehyde. The IUPAC name of the diene is

Answer»

1-cyclohexyl-2-methylhexa-1,3-diene
1-cyclohexyl-4-methylhexa-1,3-diene
1-cyclohexyl-1-methylhexa-1,4-diene
1-cyclohexyl-3-methylhexa-1,4-diene

Solution :The STRUCTURES of the products of ozonolysis are

Since the number of carbon atoms of the two products is 11(4+7), therefore, the ozonolysis must have ALSO produced another two carbon ATOM product. Further, since the given compound `(C_(13) H_(22))` cis an alkadiene, this two carbon compound must be glyoxal. OHCCHO. Replace the oxygen atoms from these THREE products by double bonds, the structure of alkadiene is
Discussion
50377.

(A): Concentration of soluble F^(-) ions in drinking water about 1 ppm is essential. (R) Enamel on teeth is made much harder by converting hydroxyapatile 3Ca_(3)(PO_4)_2Ca(OH)_2 into 3Ca_(3)(PO_4)_2 CaF_2

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

SOLUTION :Both (A) and (R) are true and (R) is not the correct explanation of (A)
Discussion
50378.

The compound 'A' on heating gives a colourless gas and a residue. The residue dissolves in water to obtain (B). When excess of CO_(2) is bubbled through aqueous solution of B gentle heating gives back A. Compound 'C' is

Answer»

`CaCO_(3)`
`CA(HCO_(3))_(2)`
`NaCO_(3)`
`NaHCO_(3)`

Solution :`CaCO_(3)(X)overset(Delta)rarrCaO+CO_(2)uarroverset(H_(2)O)rarrCa(OH)_(2)(Y)overset(CO_(2))rarrCa(HCO_(3))_(2)(Z)`
Discussion
50379.

A compound X of boron, reacts with NH_(3)on heatingto giveanother compound Y which is called inorganicbenzene. The compound X canbe prepared by treatingBF_(3) withLithium aluminiumhydride. The compounds X and Y are representedby the formules.

Answer»

`B_(2)H_(6), B_(3)N_(3)H_(6)`
`B_(2)O_(3), B_(3)N_(3)H_(6)`
`BF_(3),B_(3)N_(3)H_(6)`
`B_(3)N_(3)H_(6), B_(2)H_(6)`

Solution :`underset("Diborane (X)")(B_(2)H_(6)) + 6NH_(3) overset(473 K)RARR underset("Inorganic benzene (Y)")(2B_(3)N_(3)H_(6)) + 12H_(2)`
`4BF_(3)+3LiAlH_(4)rarr 2B_(2)H_(6) + 3LiF+3AlF_(3)`
Discussion
50380.

A compound X, of boron reacts with NH_3 on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF_3 with lithium aluminum hydride. The compounds X and Y are represented by the formulas…

Answer»

`B_2H_6, B_3N_3H_6`
`B_2O_3 , B_3N_3H_6`
`BF_3, B_3N_3H_6`
`B_3N_3H_6, B_2H_6`

Solution :(i)`B_2H_6` reacts with ammonia and gives `B_2H_6 . 2NH_3`which is formulated as `[BH_2(NH_3)_2]^(+) [BH_4]^(-)`and on HEATING gives `B_3H_3H_6`borazine ALSO called borazole.
`underset"(X)"(3B_2H_6+6NH_3) to 3[BH_2(NH_3)_2]^(+) [BH_4]^(-) oversetDeltato underset"(Y)"(2B_3N_3H_6) + 12H_2`
(ii)`B_2H_6` can be prepared by reduction of `BF_3` with `LiAlH_4 . 4BF_3 + 3LiAlH_4 to underset"(Y)"(2B_2H_6) + 3LiF + 3AlF_3`
Discussion
50381.

A compound X is heated with C_(2)H_(5) OH and H_(2)SO_(4) the fumes produced burn with green flame. The compound X is

Answer»

`H_(3)BO_(3)`
`Na_(2)B_(4)O_(7). 10 H_(2) O`
`K_(3)BO_(3)`
NONE of these

Answer :B
Discussion
50382.

A compound "X" has molecular formula C_5H_(9) Cl. It does not react with bromine in C Cl_4. On treatment with strong base it produces single compound "Y" (C_5H_8) and reacts with Br_(2)(aq). Ozonolysis of "Y" produces a compound C_5H_(8)O_2. The structure of X is

Answer»




SOLUTION :
Discussion
50383.

A compound ''X" has boiling point of 475 K but it starts decomposing at 400 K. Which type of distillation process is suitable for its purification ?

Answer»

SOLUTION :DISTILLATION under REDUCED PRESSURE
Discussion
50384.

A compound X has a molecular formula C_(2)Cl_(3)OH. It reduces Fehling solution and on oxidation produces a monocarboxylic acid Y. X can also be obtained by the action of Cl_(2) on Ethanol. X is

Answer»

Chloral
`CHCl_(3)`
`CH_(3)CL`
Chloroacetic acid

Solution :As it reduces Fehling solution, so it CONTAINS aldehyde group. It is chloral `Cl_(3)C - CHO` since it is obtained by the action of `Cl_(2)` on ethanol
Discussion
50385.

A compound (X ) gives fruity smell. [X] on hydrolysis gives an acid and alcohol . Acid give violet colour with neutral FeCl_(3) while alcohol give yellow precipitate on boiling with I_(2) and NaOH. (X) can be :

Answer»




SOLUTION :`[R-underset(O)underset(||)(C)-OR' OVERSET("Hydrolysis")to R-CO OH +R'-OH]`
Discussion
50386.

A compound X declourises Br_(2) water and reacts slowly with conc H_(2)SO_(4) to give an addition product. X reacts with HBr to form Y. Y reacts with NaOH to form Z. on oxidation Z gives hexan-3-one. X, Y and Z in the reactions are underset("Addition product") underset(darr "conc. "H_(2)SO_(4))(X) overset(Br_(2)" water")to"Decolourisation" X overset(HBr) to Y overset(NaOH) to Z overset([O])to CH_(3)CH_(2)underset("Hexan-3-one")(CH_(2)-underset(O)underset(||)(C)-CH_(2)CH_(3))

Answer»

`X=CH_(3)CH_(2)CH=CHCH_(3),Y=CH_(3)CH_(2)CH(Br)CH(Br)CH_(2)CH_(3),Z=CH_(3)CH_(2)CH_(3)`
`X=CH_(3)CH=CHCH_(3),Y=CH_(3)CH(Br)CH(Br)CH_(3),Z=CH_(3)CH_(2)CH_(2)OH`
`X=CH_(3)CH_(2)CH=CHCH_(2)CH_(3),Y=CH_(3)CH_(2)-UNDERSET(Br)underset(|)(C)H-CH_(2)CH_(2)CH_(3),Z=CH_(3)CH_(2)CH_(2)-underset(OH)underset(|)(C)H-CH_(2)CH_(3)`
`X=CH_(3)CH_(2)CH_(2)CH=CHCH_(3),Y=CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(2)Br, Z=CH_(3)CH_(2)CH_(2)CH_(2)OH`

SOLUTION :
Discussion
50387.

A compound X (C_5H_8) reacts with ammoniacal AgNO_3 to give a white precipitate, and on oxidation with hot alkaline KMnO_4 gives the acid, (CH_3)_2CHCOOH. Therefore , X is

Answer»

`CH_2=CHCH=CHCH_3`
`CH_3(CH_2)_2C-=CH`
`(CH_3)_2CH-C-=CH`
`(CH_3)_2C=C=CH`

Solution :Terminal ALKYNE containing `(CH_3)_2CH` GROUP on oxidation GIVES `(CH_3)_2CHCOOH`.
Discussion
50388.

A compound X (C_(5)H_(10)O) react with 2,4-DNP but does not give silver mirror test and idoform reaction. The possible structure for X is

Answer»


ANSWER :`CH_(3)-CH_(2)-overset(O)overset(||)(C)-CH_(2)-CH_(3)`
Discussion
50389.

A compound with molecular formula C_(8)H_(14), contains 12 secondary and two tertiary H atoms. The compound is :

Answer»




ANSWER :B::C::D
Discussion
50390.

A compound with molecular formula, C_7 H_(16)shows optical isomerism, the compound will be

Answer»

2, 3-dimethylpentane
2,2-dimethylpentane
2-methylhexane
2,3-dimethylhexane

Solution :
2,3 DIMETHYL pentane is optically active due to the PRESENCE of CHIRAL centre
Discussion
50391.

A compound with molecular formula C_(4)H_(10)O_(4) on acylation with acetic anhydride gives a compound with molecular formula C_(12)H_(18)O_(8). How many hydroxyl groups are present in the compound?

Answer»

1
2
3
4

Solution :Mol. Mass of `C_(4)H_(10)O_(4)`
`= 12 xx 4 + 10 xx 1 + 4 xx 16 = 122`
Mol. Mass of acylated product `(C_(12)H_(18)O_(8))`
`= 12 xx 12 + 18 xx 1 + 4 xx 16 = 290`
Increase in mol. Mass `= 290 - 122 = 168`
The reaction involved is `-OH + CH_(3)COCl rarr - O.CO. CH_(3) + HCL`
Increase in mol mass due to one `-OH` group = (Mol. mass of `COCH_(3)`) - At. mass of H `= 2 xx 12 + 16 + 3 - 1 = 42`
`:.` No. of `-OH` groups `= (168)/(42) = 4`
Discussion
50392.

A compound with molecular formula C_(4)H_(10)O_(3) is converted by the action of acetyl chloride to a compound with molecular weight 190. The original compound has :

Answer»

ONE OH group
Two OH group
Three OH group
No OH group

Solution :M. Mass of compound `= 4 xx 12 + 1 xx 10 + 3 xx 16 = 106` and the MOLECULAR mass of its acetate derivative is 190. Hence No. of OH groups `= (190-160)/(42)=2`.
Discussion
50393.

A compound with empirical formula C_2H_5O has a vapour density of 45. The molecular formula of the compound will be

Answer»

`C_(2)H_(5)O`
`C_(2)H_(2)O_(4)`
`C_(5)H_(14)O_(4)`
`C_(4)H_(10)O_(2)`

Solution : empiricalformulamass=45,MOLECULARMASS `=2 xx 45=90, therefore n=2`
Discussion
50394.

The compound which is not isomeric with diethyl ether is

Answer»

n-Propyl METHYL either
1-Butanol
2-Methyl-2-propanol
Butanone

Answer :D
Discussion
50395.

A compound which does not give a positive test in Lassaigne.s test for nitrogen is

Answer»

UREA
HYDRAZINE
AZOBENZENE
PHENYL hydrazine

SOLUTION :Hydrazine.
Discussion
50396.

A compound which contains one atom of X and two atoms of Y for each three atoms of Z is made by mixing 5.00 g of X, 1.15 xx 10^(23) atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 g of the compound is formed. Calculate the atomic weight of Y if the atomic weights of X and Z are 60 and 80 amu respectively.

Answer»

SOLUTION :Moles of X =`(5)/(60)= 0.08` mole
Moles of Y = `(1.15xx10^(23))/(6.022xx10^(23))=0.19` mole
Moles of Z = 0.03 mole .
Since the formula of the compound is `XY_(2)Z_(3)`
moles of X : moles of Y : moles of Z =1 : 2:3 = 0.01 : 0.02: 0.03 . Comparing these values with the moles of X, Y and Z calculated above, we find that moles of X and Y are in excess and therefore, moles of X and Y associated with 0.03 mole of Z are 0.01 and 0.02 mole respectively. Now,
WT of Z + wt . of Y + wt. of Z = wt. of `XY_(2)Z_(3)`
or `0.01xx60 +0.02 xx ` at. wt. of Y + `0.03xx80 = 4.4 `
`:. ` at .wt. of Y = 70 AMU .
Discussion
50397.

A commercial sample of common salt has 45.5% chlorine. What is the weight percentage of pure sodium chloride in the sample?

Answer»

Solution :Formula weight of SODIUM chloride =23+35.5=58.5
Weight percentage of CHLORINE in pure sodium chloride
`=(35.5)/(58.5)xx 100=60.6`
60.6 PARTS of Cl (100% pure) has 45.5 parts of Cl
Percentage PURITY of sodium chloride `=(45.5)/(60.6) xx 100=75`
Discussion
50398.

A compound (P) on reaction with "Q" in basic medium (KOH) gives a bad smelling compound (CH_(3)CH_(2)NC). Compound Q can be prepare by reaction with calcium hypochlorite [Ca(O Cl)_(2)] P and Qcan :

Answer»

`CH_(3)-CH_(2)-NH_(2) & CHCl_(3)`
`CH_(3)-CH_(2)-NO_(2) & CH_(3)Cl`
`CH_(3)-CH_(2)-NH-CH_(3) & COCl_(2)`
`(CH_(3)-CH_(2))_(3)N & Cl_(2)`

Solution :`CH_(3)-CH_(2)-NH_(2)+CHCl_(3)+KOH to CH_(3)CH_(2)NC`
`CH_(3)-CO-CH_(3)+CA(OCL)_(2) to CHCl_(3) +(CH_(3)CO O)_(2)Ca`
Discussion
50399.

A compound P(C_(5)H_(6)) gives positive Bayer test and on hydrogenation form a hydrocarbon B(C_(5)H_(10)) which gives only one monochloro product. The compound "P" is

Answer»




`CH-=C-CH_(2)-CH=CH_(2)`

SOLUTION :
Discussion
50400.

A compound P(C_(5)H_(10)O) reacts with dilute sulfuric acid to give Q and R as the final products. This reactions is about 10^(15) times faster than of ethylene. Both Q and R give positive iodoform test. Identify the structure of P, Q and R (b) Rationalize the extraordinary reactivity of P.

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Answer :`(##RES_CHM_ROHR_E03_008_A01##)`
(b) The GREATER STABILITY of the OXONIUM intermediate is RESPONSIBLE for the extraordinary reactivity. This is demonstrated for the teo SETS of P, Q, R (case I and ii above) as follows.
`(##RES_CHM_ROHR_E03_008_A02##)`
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