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50451.

An organic compound on analysis was found to contain 0.032% of sulphur. The molecular mass of the compound. If its molecule contains two sulphur atoms, is

Answer»

2
3
4
5

Solution :4
Discussion
50452.

A compound contains carbon and hydrogen in the mass ratio 3:1. The formula of the compound is

Answer»

`CH_(2)`
`CH_(3)`
`CH_(4)`
`C_(2)H_(6)`

Answer :C
Discussion
50453.

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

Answer»

`XY_(1)Z_(2)`
`Y_2(XZ_3)_2`
`X_(3)(YZ_4)_2`
`X_3(Y_4Z)_2`

SOLUTION :`X_3(YZ_4)_(2)=3 (+2)+2(+5)+8(-2)=0`
Discussion
50454.

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

Answer»

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

Answer :A
Discussion
50455.

A compound contains 90% C and 10% H. The empirical formula of the compound is

Answer»

`C_(8)H_(10)`
`C_(15)H_(30)`
`C_(3)H_(4)`
`C_(15)H_(40)`

ANSWER :C
Discussion
50456.

A compound contains 69.5% oxygen and 30.5% nitrogen and its molecualr weight is 92. The molecualr formula of that compound is:

Answer»

`N_(2)O`
`NO_(2)`
`N_(2)O_(4)`
`N_(2)O_(5)`

Answer :C
Discussion
50457.

A compound contains 50% of X (atomic mass 10) and 50% Y (atomic mass 20). Give its empirical formula.

Answer»

Solution :
`therefore`The EMPIRICAL FORMULA is `X_(2)Y`
Empirical Formula mass=20+20=40
`n=1,therefore` MOLECULAR formula=`("Empirical Formula")_(n)=(X_(2)Y)_(1)=X_(2)Y`.
Discussion
50458.

A compound contains 50%of X (atomic mass 10)and 50% Y (atomic mass 20) . Give its molecular formula .

Answer»

SOLUTION :
`:.` The Empirical Formula is `X_(2)Y`
Empirical Formula MASS = 20 + 20 = 40
Molecular mass = Sum of atomic mass = 40
N = 1, `:.` Molecular formula = `("Empirical Formula")_(n)` = `(X_(2)Y)_(1) X_(2)Y .`
Discussion
50459.

A compound contains 5 gm sulphur and 5 gm oxygen atom. The empirical formula of compound is:

Answer»

SO
`SO_(2)`
`S_(2)O`
`SO_(3)`

ANSWER :B
Discussion
50460.

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96g. What are its empirical and molecular formulas ?

Answer»



Solution :Step-1 : Conversion of mass per cent to grams :
Since we are having mass percent. It is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen is present, 24.27 g carbon is present and 71.65 g chlorine is present.
Step-2 : Convert into number moles of each ELEMENT :
Divide the MASSES obtained above by respective atomic masses of various elements .
Moles of Hydrogen `=(4.07g)/(1.008g) = 4.04`
Moles of Carbon `=(24.27g)/(12.01g) = 2.021`
Moles of Chlorine `=(71.65 g)/(35.453g)=2.021`
Step-3 : Divide the mole value obtained above by the smallest number :
Since 2.021 is the smallest value, division by it gives a ratio of `2:1:1` for `H:C:Cl`
In case the ratios are not whole number, then, they may be converted into whole number by multiplying by the suitable coessicient.
Step-4 : WRITE empirical FORMULA by mentioning the numbers after writing the symbols of respective elements :
`CH_(2)Cl` is the empirical formula of the above compound.
Step-5 : Writing molecular formula :
(a) Determine empirical formula mass
Add the atomic masses of various ATOMS present in the empirical formula .
For, `CH_(2)Cl` empirical formula mass is
`12.01 + 2 xx 1.008 + 35.453=49.48g`
(b) Divide molar mass by empirical formula mass : `("Molar mass")/("Empirical formula mass")=(98.96g)/(49.48g) =2 =(n)`
(c) Multiply empirical formula by n obtained above to get the molecular formula.
Empirical formula `=CH_(2)Cl, n=2`
Hence molecular formula is `=C_(2)H_(4)Cl_(2)`
Discussion
50461.

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Answer»

SOLUTION : Empirical FORMULA is `CH_2Cl` Molar mass = 98.96 G (given), Empirical formula mass of`CH_2Cl=49.4"Molar mass"divEFmass =98.96div49.48=2` Molecular formula `=(EF)xxn=(CH_2Cl)xx2=C_2H_4Cl_2`
Discussion
50462.

A compound contains 4.07% hydrogen, 24.27% Carbon and 71.65%, Chlorine. Its molar mass is 98.96. Calculate is Empiricla and Molecular formulae.

Answer»

Solution : Empirical FORMULA `=CH_(2)CL`
MOLECULAR formula `=nxx` EMPIRICALFORMULA
`n=("molar mass")/("Empirical formula mass")=89.96/48.5=2.04=2`
Molecular formula `=2xxCH_(2)Cl=C_(2)H_(4)Cl_(2)`
Discussion
50463.

A compound contains 38.8% C, 16.0% H and 45.2% N. The empirical formular of the compound would be:

Answer»

`CH_(3)NH_(2)`
`CH_(3)CN`
`C_(2)H_(5)CN`
`CH_(2)(NH)_(2)`

Answer :A
Discussion
50464.

A compound contains 40%, C, 6.66% H and 53.33%O. An examination reveals that 9.0g of the compound dissolved in 500g of water raises the boiling point of water by 0.051^(@)C. What is the molecular formula of the compound (K_(f)=0.51 K mol^(-1)kg)

Answer»


ANSWER :`C_(6)H_(12)O_(6)`
Discussion
50465.

A compound contains 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75. Calculate the empirical and molecular formula.

Answer»

SOLUTION :`C_2H_3O_3, C_4H_6O_6`
Discussion
50466.

A compound contains 28% of nitrogen and 72% of a metal by weight. 3 atoms of the metal combine with 2 atoms of nitrogen. Find the atomic weight of the metal.

Answer»


Solution :SUPPOSE, the symbol of the GIVEN metal is M. SINCE, 3 atoms of the metal COMBINE with 2 atoms of nitrogen, the formula of the compound should be `M_3N_2`. Suppose, the atomic mass of M is x.
The MOLECULAR mass of `M_(3)N_(2) = (3 xx x) + (2 xx 14.01) = 3x + 28.02`
The percentage of nitrogen in `M_(3)N_(2) = (28.02)/(3x + 28.02)xx 100`
Since, the given compound contains 28% of nitrogen, we have
`=(28.02 xx 100)/(3x + 28.02) = 28`
Which gives `x=24.02`
Hence, the atomic mass (weight) of the metal is 24.02 amu.
Discussion
50467.

A compound contains 28% of nitrogen and 72% of a metal by weight. 3 atoms of the metal combine with 2 atoms of nitrogen. Find the atomic weight of the metal. (N = 14)

Answer»

Solution :The formula of the compound is `M_(2)N_(2)` (M representing the metal) `2 xx `moles of `M= 3 xx` moles of N.
Now if the WEIGHT of the compound is 1 gram
then weight of M = 0.72 G and weight of N=0-28 g.
`:. 2XX(0.72)/("atomic weight of M") =3xx(0.28)/(14)` .
Atomic weight of the metal =24
Discussion
50468.

A compound contains 26.57% potassium, 35.36% chromium and the remaining oxygen. What is its empirical formula? (At.wt. of K=39.1, Cr=52, 0=16)

Answer»

SOLUTION :RATIO of K : Cr : O = `(26.6)/(39.1) : (35.4)/(52) : (38.1)/(16) = 0.68 : 0.68 : 2.38 = 2 : 2 : 7`
The empirical formula is ` K_(2)Cr_(2)O_(7)`.
Discussion
50469.

A compound contains 20% sulphur. The molecular weight of the compound could be

Answer»

80
240
400
640

Answer :D
Discussion
50470.

A compound containing two -OH groups attached with same carbon is unstable, but which one of the following is stable ?

Answer»




None

Answer :C
Discussion
50471.

A compound contains 11.99% N, 13.70% O,9.25% B and 65.06% F. Find its empirical formula .

Answer»

SOLUTION :
`THEREFORE` EMPIRICAL FORMULA of the COMPOUND in `NOBF_4`.
Discussion
50472.

A compound consisting of the monvalent ions,A^(+) , B^(-)crystallizes in the body -centred cubic lattice. (i) What is the formula of the compound ?(ii)If one ofA^(+)ions from the corner is replaced by a monovalent ion C^(+). What would be the simplest formula of the resulting compound ?

Answer»

SOLUTION :(i) Contribution of ` A^(+)`ions present at the corners to WARDS unit cell=` 8 XX 1/8 =1`
Contribution of `B^(-)`ION present at the body - centre =1
Ration of ` A^(+) : B^(-)`= 1:1 . Hence , the formula is AB.
(ii)` A^(+)` ions at the corners = 7. Hence, their contribution towards unit cell = ` 7/8`
` C^(+)` ion at one corner contributes = `1/8`
` B^(-)` ion at the body centre has contribution =1 .
Hence, ratio of A:B :C =` 7/8 : 1/8 :1 = 7: 1 : 8`
Formula will be ` A_(7)BC_(8)`
Discussion
50473.

A compound consisting of the monovalent ions A^+ ,B^- crystallizes in the body-centred cubic lattice. (i)What is the formula of the compound ? (ii)If one of A^+ ions from the corner is replaced by a monovalent ions C^+, what would be the simplest formula of the resulting compound ?

Answer»

Solution :(i)Contribution of `A^+` ions present at the CORNERS towards unit CELL =`8xx1/8=1`
Contribution of `B^-` ion present at the body-centre=1
`therefore` Ratio of `A^+ :B^-` =1:1 .Hence , the formula is AB.
(ii)`A^+` ions at the corners =7. Hence, their contribution towards unit cell=`7/8`
`C^+` ion at ONE CORNER contributes =`1/8`
`B^-` ion at the body centre has contribution =1
Hence, ratio of A:B:C=`7/8:1/8:1=7:1:8 therefore ` Formula will be `A_7BC_8`
Discussion
50474.

A compound C is produced on an industrial scale bt oxidation of 2-ethylanthraquinol by air. Compound C decolourises an acidic solutio of KMnO_(4) with the evolution of O_(2) Compound C produces a brown precipitate when it reacts with MnSO_(4) in alkaline solution. In idustrial preparation of compound C,2-ethyl anthraquinone is also produced which can be converted back 2-tehyl anthraquinol by

Answer»

Addintion of strong acid
Addition of strong base
Catalytic OXIDATION
Catalytic reduction suninh `H_(2)//Pd`

Answer :D
Discussion
50475.

A compound C is produced on an industrial scale bt oxidation of 2-ethylanthraquinol by air. Compound C decolourises an acidic solutio of KMnO_(4) with the evolution of O_(2) Compound C produces a brown precipitate when it reacts with MnSO_(4) in alkaline solution. Compound C reacts with K_(2)Cr_(2)O_(7) in acidic solution in presence of an organic sovent impart__colour to the orgaanic phase

Answer»

orange
YELLOW
blue
GREEN

ANSWER :C
Discussion
50476.

A compound AB_2 possesses the CaF_2 type crystal structure. Write the coordination numbers of A^(2+) and B^- ions in its crystals

Answer»

SOLUTION :COORDINATION no. of A =8, Coordination no. of B =4
Discussion
50477.

A compound AB_(2) l pssessses the CaF_(2)type crystal struture. Write the coordination numbers of A^(2+)and B^(-) ions in its crystals.

Answer»

SOLUTION :Corrdination no. of A = 8 COORDINATION no .of B = 4.
Discussion
50478.

A compound AB crystallises in bcc lattice with the unit cell edge length of 380 pm. Calculate (i) the distance between oppositely charged ions in the lattice ,(ii) radius of B^- if the radius of A^+ is 190 pm

Answer»


Solution :(i)For BCC, distance between `A^+` and `B^- =sqrt3/2a` , i.e., it is half of BODY diagonal
(ii)As the CATIONS and the ANIONS touch each other ,
`r_(A^+)+r_(B^-)`329 pm `therefore r_(B^-)`=329-190=139 pm
Discussion
50479.

A compound AB crystallises in bcc lattice with the unit cell edge length of 380 pm. Calculate (i)The distance between oppositely charged ions in the lattice. (ii) radius ofB^(-) if he radius ofA^(+) is 190 pm.

Answer»


Solution :(i) For BCC, DISTANCE ` A^(+) an B^(-) = sqrt3/2 ` a, i.e, it is half of BODY diagonal
Discussion
50480.

A compound A with formula C_(4)H_(10)O_(3) on acylation with acetic anhydride gives another compound with molecular mass 190. The number of hydroxyl groups in the compound A are

Answer»

2
1
3
Unprdictable

Solution :`-OH underset(-H^(+))overset(+CH_(3)COO^(-))RARR - OCOCH_(3)`
One `-OH` group on acylation BRINGS about a change in MOLECULAR mass = 42
Molecular mass of `C_(4)H_(10)O_(3)`
`= 4 xx 12 + 10 xx 1 + 3 xx 16 = 106`
Actual increase in molecular mass `= 190 - 106 = 84`
`:.` Number of `-OH` groups `= (84)/(42) = 2`
Discussion
50481.

A compound A of molecular formula C_(9)H_(7)O_(2)Cl exists in keto form and predominantly in enolic form 'B' On oxidation with KMnO_(4)'A' gives m-Chlorobenzoic acid. Identify 'A' and 'B'.

Answer»


ANSWER :`(##RES_CHM_GOC_II_E03_003_A01##)`
Discussion
50482.

A compound A on heating gives a colourless gas and a residue that is dissolved in water to obtained B . Excess of CO_(2) is bubbled through aqueous solution of B , C is formed which is recovered in the solid form . Solid C on gentle heating gives back A . The compound is

Answer»

`CaCO_(3)`
`Na_(2)CO_(3)`
`K_(2)CO_(3)`
`CaSO_(4) . 2 H_(2)O`

Solution :`underset((A))(CaC)O_(3)(s) overset(Delta)(to) CO_(2) (g) + CaO(s)`
`CaO (s) + H_(2)O (l) to underset((B))(Ca(OH)_(2)) (aq)`
`underset((B)) (Ca(OH)_(2)) (aq) + underset(("excess"))(2 CO_(2)) (g) to Caunderset((C))((HCO_(3))_(2)) (aq) `
`Caunderset((C))((HCO_(3))_(2)) (aq) overset(Delta)(to) Ca underset((A))(CO_(3)) (s) + CO_(2) (g) + H_(2)O(l)`
Discussion
50483.

A compound (A) of boron reacts with NMe_(3) togive an adduct(B) whichon hydrolysis gives a compound (C ) and hydrogengas. Compound (C )is a acid. Identify the compounds A, B and C. Givethe reactionsinvolved.

Answer»

Solution :Since compound(A) of BORON reacts with `Nme_(3)` to forman ADDUCT (B), THEREFORE, (A) must be a Lewisacid.
Since adduct (B)on hydrolysis gives an acid (C )and hydrogengas, therefore (A) must be `B_(2)H_(6)` and (C ) must be boric acid.
`underset("Diborane (A)")(B_(2)H_(6)) + underset("Adduct (B)")(2NMe)rarr 2BH_(3).NMe_3, underset((B))(BH_(3).NMe_(3))+3H_(2)O rarr underset("Boric acid (C) ")(H_(3)BO_(3)) + 6H_(2) +NMe_(3)`
Discussion
50484.

A compound (A) of boron reacts with NMe_3 to give an adduct (B) which on hydrolysis gives a compound (C) and hydrogen gas. Compound (C) is an acid. Identify the compounds A, B and C. Give the reactions involved.

Answer»

Solution :Compound (A) of boron reacts with `NMe_3` and gives an adduct (B) THUS compound (A) is Lewis acid. SINCE (B) on hydrolysis gives an acid (C) and `H_2` gas, THEREFORE (A) is `B_2H_6`, [B] is an adduct `2BH_3 NMe_3` and (C) is boric acid.
Reactions are as follows :
`underset"Diborane (A)"(B_2H_6) + underset"Adduct (B)"(2NMe_3)to underset("Adduct (B)")(2BH_3 NMe_3)`
`underset"(B)"(BH_3NMe_3) + 3H_2O to underset"Boric acid (C)"(H_3BO_3) + NMe_3 + 6H_2`
Discussion
50485.

A compound (A) of boron reacts with Nme_(3) to give an adduct (B) which on hydrolysis gives a compound (C) and hydrogen gas. Compound (C) is an acid. Identify the compounds A,B and C. give the reactions inovolved.

Answer»

SOLUTION :Since, compound (A) of boron reacts with `NMe_(3)` to form an adduct (B) thus, compound (A) is a Lewis ACID. Since, adduct (B) on hydrolysis gives an acid (C) and hydrogen gas, therefore, (A) MUST be `B_(2)H_(6)` and (C) must be boric acid
`underset("DIBORANCE (A)")(B_(2)H_(6))+underset("Adduct (B)")(2MNe_(3))rarrunderset("Adduct (B)")(2BH_(3)NMe_(3))`
`BH_(3).underset((B))(NMe)+3H_(2)Orarrunderset("Boric acid (C)")(H_(3)BO_(3))+NMe_(3)+6H_(2)`.
Discussion
50486.

A compound A has a molecular formula C_(2)Cl_(3)OH. It reduces Fehling solution and on oxidation gives a monocarboxylic acid B. A can be obtained by the action of chlorine on ethyl alcohol. A is

Answer»

METHYL chloride
Monochloroacetic acid
Chloral
Chloroform

Solution :Since the compound gives TOLLEN's REAGENT test, it must cantain `-CHO` group.
`C_(2)H_(5)OH+Cl_(2)underset(-2HCl)rarr H_(3)C-CHO underset(-3HCl)overset(3Cl_(2))rarr underset("Chloral")(Cl_(3)C.CHO)`
Discussion
50487.

A complex of iron contains 45.6% iron by mass. Calculate the number of iron atoms present in 15.0 g of this complex.

Answer»


ANSWER :`7.38 XX 10^(22)`
Discussion
50488.

A completely filled d-orbital (d^10) is

Answer»

SPHERICALLY symmetrical
has octahedral symmetry
Has TETRAHEDRAL symmetry
DEPENDS on the ATOM

Answer :B
Discussion
50489.

A common example of heterogeneous catalysis is the synthesisof NH_(3) N_(2)(g)+3H_(2)(g) overset(Fe) iff 2NH_(3)(g) In this example,

Answer»

adsorption provides the activation energy for the reaction
adsorption increase the CONCENTRATION of REACTANTS on the surface of the catalyst
Catalyst increase the rate of forward reaction more than the rate of backward reaction.
adsorption splits the molecular species into more REACTIVE atomic species.

Solution :A catalyst increase the rate of forward and backward reaction to the same extent.
Discussion
50490.

A commercial sample of hydrogen peroxide is labelled as 10 volume. Its percentage strength is nearly

Answer»

`3%`
`1%`
`90%`
`10%`

ANSWER :D
Discussion
50491.

A commercial sample of hydrogen peroxide marked as 100 volume H_(2)O_(2), it means that.

Answer»

1 ML of `H_2O` will give 100 ml `O_2` at STP
1 L of `H_2O` will give 100 ml `O_2` at STP
1 L of `H_2O` will give 22.4 L `O_2`
1 ml of `H_2O_2` will give 1 MOLE of `O_2` at STP

Answer :A
Discussion
50492.

A commercial sample of hydrogen peroxide is labelled as 10 volume. Its percentage stregth is nearly

Answer»

`3%`
`1%`
`90%`
`10%`.

Solution :SEE COMPREHENSIVE Review.
Discussion
50493.

A commercial sample of H_(2)O_(2) is labelled 10 volume. Its percentage strength is nearly :

Answer»

0.01
0.03
0.1
0.9

Solution :N//A
Discussion
50494.

A commercial sample of hydrogen peroxide is labelled as 10 Volume, its percentage strength is nearly

Answer»

0.01
0.03
0.1
0.9

Answer :B
Discussion
50495.

A combustible gas is liberated when caustic soda solution is heated with

Answer»

`NH_(4)Cl`
`I_(2)`
Zn
S.

Solution :`Zn + 2NaOH rarrNa_(2)ZnO_(2)+UNDERSET("Inflammable gas")(H_(2))`
Discussion
50496.

A colourless solid substance (A) on heating evolved CO_2 and also gave a white residue, soluble in water. Residue also gave CO_2 when treated with dilute HCI.

Answer»

`Na_(2)C)_(3)`
`NaHCO_(3)`
`CaCO_(3)`
`Ca(HCO_(3))_(2)`

Solution :`2NaHCO_(3)tounderset(downarrow)(2HCI)("Soluble in water")Na_(2)CO_(3)+H_(2)O+CO_(2)uparrow`
`NaCI+H_(2)O+CO_(2)uparrow`
Discussion
50497.

A colourless salt gives violet colour to Bunsen flame and also turns moist litmus paper blue. The salt is

Answer»

`Na_(2)CO_(3)`
`KNO_(3)`
`K_(2)CO_(3)`
`Cu(OH)_(2)`

ANSWER :C
Discussion
50498.

A colourless mixture of two salts (A) and (B) excess is soluble in H_(2)O. Separately (A) turns blue litmus red and (B) turns red litmus blue. (A) given white precipitate with (B), which dissolves in excess of (B) forming (C). (A) when placed in moist air gives fumes and can form dimer. (A) gives white precipitate with NH_(4)CI and NH_(4)OH soluble in (B).(A) also gives white precipitate with AgNO_(3) soluble in NH_(4)OH. White fumes are formed in moist air. It is due to formation of:

Answer»

`AI(OH)_(3)`
`HCI`
`Zn(OH)_(2)`
`NaAIO_(2)`

SOLUTION :`(A) rarr AICI`,
(B) `rarr NaOH, AI^(3+) +H_(2)O hArr AI(OH)^(2+) +H^(+)`
Discussion
50499.

A colourless mixture of two salts (A) and (B) excess is soluble in H_(2)O. Separately (A) turns blue litmus red and (B) turns red litmus blue. (A) given white precipitate with (B), which dissolves in excess of (B) forming (C). (A) when placed in moist air gives fumes and can form dimer. (A) gives white precipitate with NH_(4)CI and NH_(4)OH soluble in (B).(A) also gives white precipitate with AgNO_(3) soluble in NH_(4)OH. (A) remains soluble in H_(2)O in presence of excess of (B). It is due to formation of :

Answer»

`Na_(2)ZN(OH)_(2)`
`NaAI(OH)_(4)`
`Na_(2)Zn(OH)_(2)`
`AI_(2)O_(3)`

Solution :`(A) rarr AICI`,
(B) `rarr NaOH, AI^(3+) +H_(2)O hArr AI(OH)^(2+) +H^(+)`
Discussion
50500.

A colourless mixture of two salts (A) and (B) excess is soluble in H_(2)O. Separately (A) turns blue litmus red and (B) turns red litmus blue. (A) given white precipitate with (B), which dissolves in excess of (B) forming (C). (A) when placed in moist air gives fumes and can form dimer. (A) gives white precipitate with NH_(4)CI and NH_(4)OH soluble in (B).(A) also gives white precipitate with AgNO_(3) soluble in NH_(4)OH. (A) turns blue litmus red. it is due to acidic solution because of:

Answer»

hydrolysis of cation of (A)
hydrolysis of anion of (A)
hydrolysis of both cation and aion
none of the above is correct

Solution :`(A) rarr AICI`,
(B) `rarr NaOH, AI^(3+) +H_(2)O HARR AI(OH)^(2+) +H^(+)`
Discussion