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90401.

A compound of Xe and F is found to have 53.3%Xe in this compound is :

Answer»

`-4`
ZERO
`+4`
`+6`

ANSWER :D
Discussion
90402.

A compound of Zinc which is white in cold state and yellow in hot state, is

Answer»

`ZnS`
`ZnO`
`ZnCl_(2)`
`ZnZO_(4)`

Solution :ZnO when cold it is pure and when hot it BECOMES impure and this impurity CHANGES ZnO white to yellow. This impurity is known as METAL DEFICIENCY DEFECT.
Discussion
90403.

A compound of Xe and F is found to have 53.3% Xe (atomic weight=133). Oxidation number of Xe in this compound is

Answer»

`+2`
0
`+4`
`+6`

Solution :`XE :F=( 53.3)/(133):(46 .7 )/(19)`
`= 0.4 : 2.45= 1 :6`
thusE.Fofxenonfluorideis `XeF_6`
` thereforeO.Nof XeinXeF_6` is + 6
Discussion
90404.

A compound of X and Y has equal mass of tham. If their atomic weights are 30 and 20 respectively. The molecular formula of compound is :-

Answer»

`X_(2)Y_(2)`
`X_(3)Y_(3)`
`X_(2)Y_(3)`
`X_(3)Y_(2)`

ANSWER :C
Discussion
90405.

A compound of vanadium has a magnetic moment of 1.73 BM, Work out the electronic configuration of the vanadium ion in the compound. (V = 23)

Answer»

SOLUTION :`mu = SQRT(N(n+2)) BM, ` for `mu = 1.83 BM, n=1, V^(4+) : 1s^22s^22p^63s^23p^63d^1`
Discussion
90406.

A compound of two elements P and Q crystallises in cubic structure. If P occupies corners and occupies face centres, what is the composition of the compound ? If atoms of Q along with one direction are removed, what is the composition ?

Answer»

SOLUTION :`PQ_(3):PQ_(2)`
Discussion
90407.

A compound of sodium which when heated gives CO_2 is

Answer»

`Na_2CO_3. 10 H_2O`
`NaHCO_3`
`Na_2CO_3.7H_2O`
`Na_2CO_3.H_2O`

SOLUTION :`2NaHCO_3 OVERSET(HEAT)(to) Na_2CO_3 + H_2O + CO_2`
Discussion
90408.

A compound of non-sugar and glucose which yields glucose on hydrolysis found in plants, is called:

Answer»

Alkoxide
Glucoside
Glycoside
None of these

Answer :B
Discussion
90409.

A compoundof molecular formulaC_(8)H_(8)O_(2) reacts with acetophenone is form a single cross-aldol product in the presence of base. The same compound on reaction with conc. NaOH forms benzyl alcohol as one of the products. The structure of the compounds is :

Answer»




ANSWER :A
Discussion
90410.

A compound of mol. Wt 180 is acetyle to give a compound of mol. Wt 390. the number of amino groups in the initial compound is

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2
4
5
6

Solution :Difference in MASS of COMPOUND `=390-180=210`
WT. of `CH_(3)CO-` GROUP is =43
THEREFORE, no. of `-NH_(2)` group `=210/43 =4.88 =5`
Discussion
90411.

Acompound of iron and chlorine (Fe) soluble in water . An excess of silver nitrates was added to precipitate the chloride ion as silver chloride . If a 134.8 mg of the compound gave 304.8mg of AgCl. What is the value of x ?

Answer»


SOLUTION :N/A
Discussion
90412.

A compoundof mol. Wt.180 is acetylated to give a compound of mol. Wt .390.The number of amino groups in the initial compound is

Answer»

2
4
5
6

Solution :DIFFERENCE in mass of compounds= 390 - 180 = 210
Weightof ` CH_(3)CO - " group"= 43 `
Replacement of - Hby ` - COCH_(3)`group will cause in - CREASE of 42 in mass. Therefore, no.Of `-NH_(2)`groupswith ACCETONE,it gives diacetone amine.
Discussion
90413.

A compound of mol wt. 180 is acetylated to give a compound of molecular wt. 390. What is the number of amine in the initial compounds?

Answer»

Solution :Difference in MASS of compound `=390-180=210`
Wt. of `CH_(3)CO`- group = 43
Therefore, NUMBER of `-NH_(2)` group `(210)/(43)`.
Discussion
90414.

A compound of carbon, hydrogen and nitrogen contains three elements in the respective weight ratio of (:1:3. Calculate its empirical formula. If its molecular weight is 108, what is its molecular formula?

Answer»

SOLUTION :`C_(3)H_(4)N, C_(6)H_(8)N_(2))`
Discussion
90415.

A compound of carbon, hydrogen and nitrogen contins these elements in the ratio 9:1:3.5. calcualte the empirical formula. If its molecular mass is 108, what is the molecular formula?

Answer»

Solution :
The EMPIRICAL FORMULA `=C_(3)H_(4)N`
Empirical formula mass `=(3xx12)+(4xx1)+14=54`
`n=("Mol.mass")/("Emp.mass")=(108)/(54)=2`
Thus, molecular formula of the compound
`=2xx` Empirical formula
`=2xxC_(3)H_(4)N`
`=C_(6)H_(8)N_(2)`.
Discussion
90416.

A compound of formula NH_2CH_2COOH may behave:

Answer»

Only as an ACID
Only as a base
Both acid and base
Neither acid nor base

Answer :B
Discussion
90417.

A compound of boron, A on heating swells up which on further heating forms glassy transparent mass B. The chemical constituent B is/are

Answer»

`Na_2 B_4 O_7` only
`NaBO_2` only
`NaBO_2 + B_2 O_3`
`Na_2 B_4 O_7 + B_2 O_3`

ANSWER :C
Discussion
90418.

A compound of aluminium and chlorine is composed of 9.0g Al for every 35.5 g of chlorine. The empirical formula of the compound is:

Answer»

AICl
`AICl_(3)`
`AICl_(2)`
`AICl_(4)`

ANSWER :B
Discussion
90419.

A compound not isomeric with diethyl ether is:

Answer»

METHYL PROPYL ether
Butan-1-ol
2-methylpropan-2-ol
Butanone

Answer :D
Discussion
90420.

A compound MX_(2)has observed and normal molar masses 65.6 and 164respectively .Calculated the apparent degree of ionization of MX_(2):

Answer»

0.75
0.85
0.65
0.25

Solution :`i=("Nromal MOLAR MASS")/("Osberved molar mass")`
`(164)/(65.6)=2.5`
`ALPHA=(i-1)/(n-1)=(2.5-1)/(3-1)`
`=(1.5)/2=0.75%`
Discussion
90421.

A compound made up of two elementsA and B is found to contain 25% A (at. mass = 12.5) and 75% B (at. mass = 37.5). The simplest formula of the compound is :

Answer»

`AB`
`AB_2`
`AB_3`
`A_3B_2`

SOLUTION :
SIMPLE FORMULA AB
Discussion
90422.

A compound made up of two elements A and B has A=70%, B=30%. Their relative number of moles in the compound are 1.25 and 1.88. Calculate atomic masses of the elements A and B

Answer»

SOLUTION :`"Relative no. of moles of an ELEMENT"=(%" of the element")/("Atomic MASS") or "Atomic mass"=(%" of the element")/("Relative no. of moles")`
`THEREFORE"Atomic mass of A"=(70)/(1.25)=56,"Atomic mass of B"=(30)/(1.88)=16`
Discussion
90423.

A compound made of particles A and B. A forms fc c packing and B occupies all the OV_(s). If all the particles along the plane as shown in the figure below are removeed, then the simplest formula of the compound is

Answer»

`A_(5)B_(7)`
`A_(7)B_(5)`
`AB`
`AB_(3.75)`

Solution :c. Since the lattice is `fc c (Z_(eff) = 4)`, therefore, numbered of `A` ions `= 4 (1` corner `+ 3` face centre)
`(OV_(s)` are FORMED at body centre andedge centre). ltbr. Lons are removed from the plane `(PQRS)` as shown in the figure. From the figure, it is clear, `4`corner ions and two edge centre ions on `PQ` and `RS` are alos removed.
Also two face centre ionslying on `PR` and `QS` along with ONE ions on present in the body centre are removed as shown in the figure below.
b

`:.` Number of `A` ions removed
`= 4 xx (1)/(8)` (corner SHARE) `+ 2 xx (1)/(2)` (face centre share)
`= (1)/(2) + 1 = 1(1)/(2)`
Number of `B` ions removed
`2 xx (1)/(4)` (edge centre share) `+ (1)/(1)` (body centre share)
`= (1)/(2) + 1 = 1(1)/(2)`
Number of `A` ions left `= 4 -1(1)/(2) = 2.5`

Number of `B` ions left `= 4 -1(1)/(2) = 2.5`
Thus, formula `= A_(2.5) B_(2.5) = 2.5 AB`
Simplest formula `= AB`
Discussion
90424.

A compound liberates CO_(2) with NaHCO_(3) and also gives colour with neutral FeCl_(3) solution. Compound can be

Answer»




ANSWER :B
Discussion
90425.

A compound is yellow when hot and white when cold. The compound is :

Answer»

`Al_2O_3`
PbO
CaO
ZnO

Answer :D
Discussion
90426.

A compound is soluble in waterif ammonia is added to aqeous solution of the compound, a brown precipitate appears which is solution in dil HCI .The compound has

Answer»

Aluminium
Zine
Iron
Cadmium

Solution :`FE(OH)_(3)` is BROWN
Discussion
90427.

A compound is soluble in water. If ammonia is added, a red precipitate appears which is soluble in dilute HCl. The compound has

Answer»

Aluminium
Zinc
Iron
Cadmium

Solution :`FeCl_(3)+3NH_(4)OH to underset(("red"))(Fe(OH)_(3))+3NH_(4)Cl`
`2FE(OH)_(3)+6HCL to underset("soluble in HCL")(FeCl_(3))+6H_(2)O`
Discussion
90428.

A compound is precipitated when its:

Answer»

ionic product exceeds the solubility product
ionic product is LESS than its solubility product
ionic product is equal to the solubility product
NONE of the above

Answer :A
Discussion
90429.

A compound is made up of atoms 'A' and 'B' atoms A are present at the corners of the cube as well as at the centre of two alternate faces while atoms B are present in all the tetrahedral voids. Find out the formula of the compound.

Answer»

AB
`AB_(4)`
`A_(2+)B`
`A_(7)B_(8)`

ANSWER :B
Discussion
90430.

A compound is made by mixing cobalt (III) nitrite and potassium nitrite solution in the ratio of 1 : 3. The aqueous solution of the compound showed 4 particles per molecules whereas molar conductivity reveals the presence of six electrial charges. The formula of the compound is :

Answer»

`Co(NO_(2))_(3).2KNO_(2)`
`Co(NO_(2))_(3).3KNO_(2)`
`K_(3)[CO(NO_(2))_(6)]`
`NI[(CO)_(4)]`

Solution :`K_(3)[Co(NO_(2))_(6)]to3K^(+)+[Co(NO_(2))_(6)]^(3-)`
Discussion
90431.

A compound is made of three elements A, B and C. Atoms A form face centred cubic cell. Atoms B occupies all octahedral voids and atoms C occupies all tetrahedral voids. If all atoms across one body diagonal are removed, calculate the formula of a compound.

Answer»


ANSWER :`A_5B_4C_8`
Discussion
90432.

A compound is made by mixing cobalt (III) nitrite and potassium nitrite solution in the ratio of 1 : 3. The aqueous solution of the compound showed 4 particles per molecule whereas molar conductivity reaveals the presence of six electrical charges. The formula of the compound is

Answer»

`Co(NO_(2))_(3). 2KNO_(2)`
`Co(NO_(2))_(3). 3KNO_(2)`
`K_(3)[Co(NO_(2))_(6)]`
`K[Co(NO_(2))_(4)]`

Solution :`Co(NO_(2))_(2)3KNO_(2)`
`K_(3)[Co(NO_(2))_(6)] to 3K^(+) +[Co(NO_(2))_(6)]^(3-)`
Discussion
90433.

A compound is in its high oxidation state. Then its will be

Answer»

HIGHLY ACIDIC
Highly basic
Highest OXIDISING property
HALF acidic, half basic

Answer :C
Discussion
90434.

A compound is heated with zinc dust and amonium cholride and the raction mixture is filterted into Tollen's ragment .Formation of silver mirror indicates the presence of following group:

Answer»

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

SOLUTION :N//A
Discussion
90435.

A compound is heated with zinc dust and ammonium chloride followed by addition of the Tollen's reagent. Formation of silver mirror indicates the presence of following group

Answer»

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

SOLUTION :N//A
Discussion
90436.

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

Answer»

SOLUTION :
Discussion
90437.

A compound is formed of two element "A" and "B". The atoms of element "A" forms face centred cubic close packing and atoms of "B" occupies all the tetrahedral voids. The formula of compound is ...

Answer»

`AB_2`
`AB_8`
`A_4B`
`A_2B`

Solution :NUMBER of A atoms = 4
`therefore` Number of TETRAHEDRAL VOIDS = 8
`therefore` TOTAL number of B atoms = 8
`therefore A: B = 4:8 = 1:2 implies AB_2`.
Discussion
90438.

A compound is formed by two elements x and y. Atoms of the elements y make ccp (cubic close packing) and those of the element x occupy all the octahedral voids. What is the formula of the compound ?

Answer»

Solution :The ccp lattice is FORMED by atoms y.
The number of OCTAHEDRAL VOIDS (x) generated is equal to the number of atoms of the element (y).
THUS, the atoms of the elements x and y are present in 1 : 1 ratio.
THEREFORE the formula of the compound is xy.
Discussion
90439.

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 cations) occupy all the octahedral voids. What is the formula of the compound ?

Answer»

Solution :Let number of Y atoms = n.
Total number of octahedral voids = n
SINCE X occupies all the octahedral voids, the FORMULA of the compound is `X_nY_n` .
Taking `n = 1 implies XY` is the formula of compound.
Discussion
90440.

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

Answer»

Solution :Total no. of N-ATOMS per unit cell = 4
`therefore` Total no. of TETRAHEDRAL VOIDS = 8 
Total number of voids occupied `= 8 xx 1/3`
by atoms M
`therefore` Total no. of M-atom = `8/3`
`therefore `Ratio of M and N atoms per unit cell = `8/3 : 4`
`therefore` Formula of COMPOUND = `M_2N_3`
Discussion
90441.

A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1//3 rd of the 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 atomsM"=(2n)/(3):n=2:3`
HENCE, the formula is `M_(2)N_(3)`
Discussion
90442.

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

Answer»

SOLUTION :Suppose the number of anions = x
Tetrahedral VOIDS = 2X
Number of cation `=(1)/(3)xx2x =(2x)/(3)`
`("Number of cations")/("Number of anions")=(2x// 3)/(x)=(2)/(3)`
HENCE, the FORMULA of compound is `M_(2)N_(3)`.
Discussion
90443.

A compound is formed by two elements A and B. Atoms of the element B (as anions) are in ccp arrangement and those of the element A as cations occupy all the octahedral voids. What is the formula of the compound?

Answer»


Solution :The atoms of B (as anions) form ccp LATTICE and the NUMBER of octahedral VOIDS is equal to the number of atoms of B. Since all the octahedral are occupied, the number of atoms of A = Number of atoms of B or the FORMULA of the compound is AB.
Discussion
90444.

A compound is formed by two element M and N . The element N forms ccp and atoms M occupy frac{1}{3} of the tetral voids. What is the formula of the compound?

Answer»

SOLUTION :Let the number of atoms of N be .n.
`THEREFORE` number of tetrahedral VOIDS = 2n
Number of atoms M =`FRAC{1}{3}xx2n=frac{2n}{3}`
`therefore` Ratio of atoms M and N = frac{2}{3}n:n=frac{2}{3}:1=2:3` Formula is `M_2N_3`
Discussion
90445.

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

Answer»

SOLUTION :
Discussion
90446.

A compound is formed by substitution of two chlorine for two hydrogens in propane. The number of possible isomeric compounds is ……

Answer»

4
3
5
2

Solution :`C_(3)H_(8)overset(-2H)underset(+2Cl)rarr C_(3)H_(6)Cl_(2)` The following FOUR structural isomers are possible.
`CH_(3)CH_(2)CHCl_(2)(I), CH_(3)C Cl_(2)CH_(3)(II)`,
`ClClH_(2)CH_(2)Cl (III), CH_(3)-CHCl-CH_(2)Cl(IV)`.
Since (IV) has a chiral carbon, therefore it has two optical isomers. So in all, five isomers are possible.
Discussion
90447.

A compound is formed by substitution of two chlorine for two hydrogens in propane, the number of possible isomeric compounds is:

Answer»

2
3
4
5

Answer :D
Discussion
90448.

A compound is formed by substitution of two chlorine for two hydrogens in propane. The number of possible isomeric compound is

Answer»

4
3
5
2

Solution :The compound is `C_(3)H_(6)Cl_(2)` and the number of possible ISOMERIC compound are 5.
Discussion
90449.

A compound is formed by substitution of two chlorine for two hydrogens in propane. The number of possible isomeric compounds is

Answer»

4
3
5
2

Answer :C
Discussion
90450.

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 formula of the compound is

Answer»

AB
`A_(2)B`
`AB_(2)`
`AB_(6)`

ANSWER :A
Discussion