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93351.

1. 6 gm of pyrolusite was treated with 50 ml of 0.5 M oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 250 ml in a flask . 25 ml of this solution when treated with 0: 02 M KMnO_(4) required 32 ml of the Solution. Find the % of MnO_(2) in the sample and also the percentage of available oxyen .

Answer»

Solution :Redox changes are
`C_(2)O_(4)^(-2) to 2CO_(2) ` (n-factpr = 2)
`MnO_(4)^(-) to Mn^(2+) ` (n-factor = 5)
`MnO_(2) to Mn^(2+)` (n-factor = 2)
Meq. Of `MnO_(2) ` = Meq. Of oxalic acid taken - Meq. Of oxalic acid left
` = 50 xx 0.5 xx 2 - 32 xx 0.02 xx 5 xx 10 ` (in250ml) = 18
`W_(MnO_(2))/M_(MnO_(2)) xx 2 xx 1000 = 18 rArrW_(MnO_(2))/87 xx 2 xx 1000 = 18 `,
`:. W_(MnO_(2)) = 0.7821 ` gm
` :.% " of " MnO_(2) = (0.7821)/ (1.6) xx 100 = 48.88 %`
Meq. of `MnO_(2)` = Meq. of `O_(2) `
`W_(O_(2)) /16 xx 2 xx 1000 = 18, :. w_(o_(2)) = 0.144` g
% of AVAILABLE `O_(2) = (0.144)/(1.6) xx 100 = 9 `
Discussion
93352.

How will you convert (i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline Cl-(CH_(2))_(4)- Cl into hexane- 1,6- diamine ?

Answer»

Solution :`underset("1, 4-Dichlorobutane")(Cl-(CH_(2))_(4)-Cl)+underset(("alc."))("2KCN")underset("-2KCl")rarrNC-(CH_(2))_(4)-CNunderset("(II) "H_(3)O^(+))OVERSET("(i)"LiAlH_(4)//"ether")rarrunderset("Hexane-1, 6-diamine")(H_(2)N-(CH_(2))_(6)-NH_(2)`
Discussion
93353.

1, 3-Dibromopropane reacts with metallic zinc to form :

Answer»

Propene
Propane
Hexane
Cyclopropane

SOLUTION :
Discussion
93354.

1, 3-Butadiene and styrene on polymerisation give

Answer»

BAKELITE
TERYLENE
Buna-S
Teflon

Answer :C
Discussion
93355.

M is a metal which forms an oxide, M_(2)O and (1)/(2) M_(2)O(s) rarr M(s) + (1)/(4) O_(2)(g), Delta H = 90 kJ mol^(-1) When a sample of the metal M reacts with 1 mole of O_(2)(g) to form M_(2)O, DeltaH for the reaction is :

Answer»

`+ 180 KJ `
`-180 kJ`
`-360 kJ`
`360 kJ`
Discussion
93356.

1–bromo–3–chlorocyclobutane when treated with two equivalents of Na, in the presence of ether which of the following will be formed ?

Answer»




ANSWER :D
Discussion
93357.

1–bromo–3–chlorocyclobutane when treated with two equivalents of Na, in the presence of ether which of the following will be formed?

Answer»




ANSWER :D
Discussion
93358.

1, 2-Dichloroethane is a gem-dihalide.

Answer»
Discussion
93359.

1//2 H_(2(g)) to H^(+)+ ?(oxidation)

Answer»

`E^(-)`
`2E^(-)`
`3e^(-)`
no electron

Solution :`(1)/(2)H_(2(g)) to H^(+)+e^(-)`
Discussion
93360.

1, 2, and 3 by appropriately matching the information given in the three columns of the following table. The only INCORRECT combination is:

Answer»

<P>[A] (IV) (III) (Q)
[B] (IV) (iii) (P)
[C] (II) (iii) (P)
[D] (I) (i) (S)

Answer :C
Discussion
93361.

1^(@), 2^(@) and 3^(@) nitroalkanes can be distinguished by 1. acid hydrolysis 2. halogenation 3. reaction with nitrous acid

Answer»

1,3
1,2
only 3
1,2,3

Answer :D
Discussion
93362.

1, 2-Dibromo propane on treatment with X moles of NaNH_(2) followed by treatment with ethyl bromide gives a pentyne. The value of X is :

Answer»

1
2
3
4

Answer :C
Discussion
93363.

1, 2 - dichloro ethane is which type of halide ?

Answer»

GEMINAL HALIDE
Vicinal halide
ALKYLIDENE halide
Allylic halide

Answer :B
Discussion
93364.

1, 2, and 3 by appropriately matching the information given in the three columns of the following table. The only CORRECT combination in which the reaction proceeds through radical mechanism is:

Answer»

`[A] (I) (III) (R)`
[B] (II) (iii) (Q)
[C] (III) (ii) (P)
`[D] (IV) (i) (R)`

Answer :D
Discussion
93365.

1, 2 and 3 by appropriately matching the information given in the three columns of the following table. The CORRECT combination is:

Answer»

(I) (III) (S)
(III) (II) (Q)
(IV) (i) (S)
`(III) (iii) (R)`

Answer :D
Discussion
93366.

1, 2, and 3 by appropriately matching the information given in the three columns of the following table. For the synthesis of berizoic aicd, the only CORRECT combination is

Answer»

<P>[A] (III) (IV) S
(B) (IV) (ii) (P)
`(C)(I)(iv) (Q)`
(D) (I) (i) (Q)

Answer :D
Discussion
93367.

1, 2 and 3 by appropriately matching the information given in the three columns of the following table. The CORRECT combination is:

Answer»

<P>`(IV) (iv)(R)`
(I) (II) (P)
(II) (ii) (P)
(III) (iv) (S)

Answer :C
Discussion
93368.

1, 2 and 3 by appropriately matching the information given in the three columns of the following table. The CORRECT combination is:

Answer»

<P>(I) (i) (Q)
(II) (ii) (P)
`(III) (iv) (R)`
(III) (iii) (S)

ANSWER :D
Discussion
93369.

1^@, 2^@ and 3^@ alcohols are identified by

Answer»

LUCAS TEST
OXIDATION test
HALOFORM test
all of these

ANSWER :D
Discussion
93370.

1, 2, 3-Tribromobenzene can be prepared in pure state from

Answer»

benzene by bromination with `Br_(2) (aq)`<BR>benzene by bromination with `Br_(2)` in `CS_(2)`
benzene by bromination with `Br_(2)` in the PRESENCE of `FeBr_(3)`
None of these.

Solution :Pure 1, 2, 3-tribromobenzene can only be prepared from p-nitroaniline by FOLLOWING sequence of REACTIONS.
Discussion
93371.

1^@ , 2^@ , 3^@amines Can be best distinguished by:

Answer»

`HNO_2` TREATMENT
EXHAUSTIVE alkylation
Mustard OIL reaction
Carbylamine reaction

Answer :A
Discussion
93372.

1/12 of the gram atom of carbon

Answer»

contains 1 ATOM of CARBON
contains the same number of carbon atoms as are present in 2.5 GRAMS of glucose `(C_6H_12O_6)`
contains Avogardo number of carbon atoms
CORRESPONDS to 1 a.m.u.

Solution : contains the same number of carbon atoms as are present in 2.5 grams of glucose `(C_6H_12O_6)`
Discussion
93373.

1, 1-diphenyl methanol is reacted with, HI give

Answer»

1, 1-DIPHENYL IODOMETHANE
1, 1-diphenyl IODOETHANE
diphenyl
NONE of these

Answer :A
Discussion
93374.

1 . 0 g of an alloy of Aland Mg when treated withexcess of dil. HCI forms MgCl_(2), AlCl_(3) and hydrogen. Theevolved hydrogen,collected over Hg at 0^(@) C has a volume of 1.2 litters at 0.92 atm pressure. Calculate the composition ofthe alloy. (al = 27 and Mg = 24)

Answer»

Solution :The equations,`{:(AI+3h^(+)" "to Ai^(3+) + (3)/(2) H_(2)),(Mg + 2H^(+) to Mg^(2+) + H_(2) ):}`
show that 1 MOLE of Al produces `(3)/(2)` moles of hydrogenand
1 moleof Mg produces 1 mole of hydrogen.
Thusthe moleequation is ,
`(3)/(2)` moles of Al + 1 mole of Mg = 1 mole of `H_(2)`
Let the weight of Al be x g
`:. ` wt. of Mg = (1 - x) g
`(3)/(2) xx (x)/( 27) + (1 - x)/( 24) = ("vol. of " H_(2) " at NTP")/( 22. 4)`
Volumeof `H_(2)` at NTP ` = (1 . 2 xx 0 . 92)/( 273) xx (273)/( 1) = 1 . 1 .104` liters
(using `(p_(1)V_(1))/(T_(1)) = (p_(2)V_(2))/(T_(2)))`
`:. (3)/(2) xx (x)/(2) + (1 - x)/( 24) = (1 . 104)/( 22.4) , x = 0 . 55`
Wt. of Al = 0 . 55 g, wt. of Mg0 . 45
Thus`%` of Al `= ( 0 . 55)/( 1) xx 100 = 55 % `
and `%" of Mg " = (0.45)/( 1) xx 100 = 45%`
Discussion
93375.

0.96 gm a metal oxide MO is dissolved in excess dilute sulphuric acid, KMnO_(4) of strength M/10 requires to react with it and the volume needed to reach the equivalent point is 20 ml assuming that MO is converting into M_(2)O_(3) then what is the atomic mass of M?

Answer»

80
96
70
50

Solution :N/A
Discussion
93376.

0.9546 g of a Rochelle salt , NaKC_(4)H_(4)O_(6).4H_(2)O on ignition ,gave NaKCO_(3) , which was treated with 41.72 mL of 0.1307 N H_(2)SO_(4) . The unreacted H_(2)SO_(4) was then neutralised by 1.91 mL of 0.1297 N NaOH. Find the percentage purity of the Rochelle salt in the sample .

Answer»

SOLUTION :`76.87` %
Discussion
93377.

0.90g of a non-electrolyte was dissolved in 90 g of benzene. This raised the boiling point of benzeneby 0.25^(@)C. If the molecular mass of the non-electrolyte is 100.0 g mol^(-1), calculate the molar elevation constant for benzene.

Answer»

SOLUTION :The given VALUES are:
`W_("SOLUTE")= 0.90 g`
`W_("solvent")= 90.00 g`
`DeltaT_(b) = 0.25 ^(@)C`
`Mw_("solute")= 100.0 g mol^(-1)`
`K_(b)` = ?
Using the formula
`K_(b) = (DeltaT_(b) xx W_("solvent") xx Mw_("solvent")) / (1000 xx Mw_("solute")) = 2.5 K m^(-1)`
`:. K_(b) = (0.25 xx 100 xx 90.0)/(1000 xx 0.90) = 2.5 K m^(-1)`
Thus, `K_(b)` is `2.5 K m^(-1)`.
Discussion
93378.

0.90 g of non-electrolyte was dissolved in 87.90 g of benzene. This raised the boiling point of benzene by 0.25^(@)C. If the molar mass of non-electrolyte is 103.0g mol^(-1), calculate the molat elevation constant for benzene.

Answer»


Solution :`M_(2)=(1000M_(b)w_(2))/(w_(1)DeltaT_(b))"or"K_(b)=(M_(2)w_(1)DeltaT_(b))/(1000w_(2))=("103 G MOL"^(-1)xx87.90g xx0.25K)/("1000 g kg"^(-1)xx0.90g)="2.515 K kg mol"^(-1).`
Discussion
93379.

0.85 g aqueous solution of NaNO_(3) is approximately 90% dessociated at 27^(@)C. Calculate the osmotic pressure (R=0.0821 L atm K^(-1) mol^(-1))

Answer»


Solution :Calculation of Van't HOFF factor
`NaNO_(3)hArrNa^(+)("aq")+NO_(3)^(-)("aq")`
`alpha=(i-1)/(n-1)or0.9=(i-1)/(2-1)or i=0.9+1=1.9`
STEP II. Calculation of osmotic PRESSURE
`"Initial no. of moles of "NaNO_(3)(n_(B))=((0.85g))/((85" g mol"^(-1)))=0.01mol`
`"Osmotic pressure "(pi)=iCRT=(i n_(B)RT)/V`
`=(1.9xx(0.01"mol")xx(0.0821" L atm"^(-1)mol^(-1))xx300K)/((0.1K))=4.68"atm".`
Discussion
93380.

0.85% aqueous solution of NaNO_(3) is apparently 90% dissociated . The osmotic pressure of solution at 300 K is

Answer»

4.674 atm
46.74 atm
2.46 atm
4.674 mm Hg

Solution :Molality, `C=(0.85xx1000)/(100 xx 85)=0.1 "mol L"^(-1)`
`{:(NaNO_(3),rarr,NA^(+),+,NO_(3^(-))),(1-alpha,,alpha,,alpha):}`
`i=(1-alpha+alpha+alpha)/(1)=1+alpha=1+0.90=1.90`
`pi=i"CRT" = 1.90xx0.1xx0.082xx300`
`=4.674` atm
Discussion
93381.

0.85% aqueous solution of NaNO_(3) is 90% dissociated at 27^(@) C. Calculate its osmotic pressure. (R= 0.082 L atm K^(-1) "mol"^(-1))[d = 1 g cm^(-3)]

Answer»


ANSWER :`-5` ATM
Discussion
93382.

0.85% aqueous solution of NaNO_(3) is apparently 90% dissociated at 27^(@)C. Calculate its osmotic pressure. (R = 0.0821 atm K^(-1) mol^(-1))

Answer»

Solution :Volume of `NaNO_(3)` solution
`=((100g))/((1"G cm"^(-3)))=100cm^(3)=0.1 L`
`NaON_(3)` dissociates as :
`NaON_(3)toNa^(+)+NO_(3)^(-)`
`alpha=((i-1))/((n-1))i.e.0.9(i-1)/(2-1)`
i-1+0.9=1.9
`M_(B)=23+14+48=85" g mol"^(-1)`
`pi=W_(B)/M_(B)XX(RxxTxxi)/V`
`=((0.85))/((85" g mol"^(-1)))`
`xx((0.082" L ATM K"^(-1)"mol"^(-1))xx(300L)xx(1.0))/((0.1 L))=4.67~~5`
Discussion
93383.

0.80 g of chloroplatinate of a monoacid organic base on ignition gave 0.25 g of Pt. The molecular weight of the base is (Pt = 195)

Answer»

624
214
107
312

Answer :C
Discussion
93384.

0.8 g of silver salt of a dibasic organic acid on ignition yielded 0.54 g of metallic silver. Molecular weight of the acid is (Ag = 108)

Answer»

106
108
320
85

Answer :A
Discussion
93385.

0.788g of a substance after digestion with H_(2)SO_(4) was distilled with an excess of NaOH. The liberatd NH_(3) was absorbed in 100mL of N H_(2)SO_(4) solution. The remaining acid required 73.7 mL of N NaOH solution for neutralization. Find the percentage of nitrogen in the compound

Answer»


ANSWER :`46.7%`
Discussion
93386.

0.76gm of the silver salt of a diabasic acid was ignited. It gave 0.54gm of pure silver. The molecular mass of acid is ........ x10.

Answer»


Solution :`E = (("MASS of silver SALT")/("mass of silver") xx 108-107)`
`=((0.76)/(0.54) xx 108-107)` , E = 45
Molecular mass = `45 xx 2` = 90
Discussion
93387.

0.759 g of silver salt of a dibasic organic acid on ignition left 0.463 g metallic silver. The equivalent weight of acid is :

Answer»

70
108
60
50

Answer :A
Discussion
93388.

0.75 mol of ethylene bromide were mixed with 0.25 mol ofpropylene bromide at 358 K to form nearly ideal solution. Vapour pressures of pure ethylene bromide and propylene bromide at 358 K are 2.77xx10^(4)"N m"^(-2) and 1.73xx10^(4)"N m"^(-2) respectively. Calculate the vapour pressure of the solution.

Answer»

SOLUTION :`2.51xx10^(4)"N m"^(-2)`
Discussion
93389.

0.75 g platinic chloride, a monoacid base on ignition gave 0.245 g playinum. The molecular weight of the base is

Answer»

`75.0`
`93.5`
`100`
`80.0`

Solution :`("Wt. of "B_(2)H_(2)PtCl_(6))/(2B+410)=("Wt. of Pt")/195`
`:. 0.75/(2B+410)=0.245/195 implies B=93.5`
EQ. wt. of base `=93.5`, since it is monoacidic.
`:.` Mol. Wt. of base `=93.5xx1=93.5`
Discussion
93390.

0.6mL of glacial acetic acid with density 1.06gmL^(-1) is dissolved in 1kg water and the solution froze at -0.0205^(@)C. Calculated van't Hoff factor. (K_(f) for water is 1.86J jg mol^(-1))

Answer»


SOLUTION :Volume of glacial acetic acid=V=0.6mL
Density of glacial acetic acid `=p=1.06gmL^(-1)`
`W_(1)=1kg=1000g`
Freezing POINT of solution `=t_(f)=-0.0205^(@)C`
`=T_(f)=273+(-0.0205)`
`=272.9795K`
`K_(f)=1.86"K kg MOL"^(-1)`
Molar mass of acetic acid=`M_2=60"g mol"^(-1)`
van't HOFF factor =i=?
Depriciation in freezing point= `=DeltaT_((ob))=273-272.9795`
`=0.0205K`
`"Density"=("Mass")/("Volume")`
`therefore "Mass"="Density"xx"Volume"`
`therefore W_(2)=1.06xx0.6=0636g`
`DeltaT_((th))=K_(f)xx(Wxx1000)/(W_(1)xxM_(2))=(1.86xx0.636xx1000)/(1000xx60)`
=0.01972K
`i=(DeltaT_(f_((obs))))/(DeltaT_(f_((th))))=(0.0205)/(0.01972)=1.0396`
Discussion
93391.

0.66g of an organic compound containing C, H and O gave on combustion 0.968g of CO_(2) and 0.792 g of H_(2)O. Calculate the percentage of O in the compound. (C= 12, H= 1, O= 16)

Answer»

Solution :Moles of C in `CO_(2)=1 XX` moles of `CO_(2)`
`=1 xx (0.968)/(44)= 0.022`
Wt. of `C= 0.022 xx 12= 0.264g`
Moles of H in `H_(2)O=2 xx` moles of `H_(2)O`
`=2 xx (0.792)/(18)= 0.088`
Weight of H= `0.088 xx1= 0.088g`
Total weight of C and H in the COMPOUND `=(0.264 + 0.088)G = 0.352g`
`therefore` weight of O in the compound `= (0.66- 0.352)g= 0.308g`
`therefore` % of O in the compound `= (0.308)/(0.66) xx 100= 46.67%`
Discussion
93392.

0.7 g of Na_(2)CO_(3).xH_(2)O is dissolved in 100 mL of water , 20 mL of which required 19.8 mL of 0.1 N HCl. The value of x is

Answer»

4
3
2
1

Answer :C
Discussion
93393.

0.65 g naphthalene (C_(10)H_(8)) was dissolved in 100 g methyl acetate. Elevation in boiling point of methyl acetate solution was 0.103^(@)C. If boiling point of pure methyl acetate is 57^(@)C, its molar heat of vaporisation will be:

Answer»

8.96 KCAL`MOL^(-1)`
7.946 kcal `mol^(-1)`
6.24 kcal `mol^(-1)`
15.65 kcal `mol^(-1)`

ANSWER :B
Discussion
93394.

0.635 g Cu was dissolved in 5.0 mL hot 60% HNO_(3)(sp. gr. = 1.5). When the reaction came to an end, the volume of the solution was adjusted to 250.0 mL, What is the normality of the solution with respect to HNO_(3) ? (Cu= 63.5)

Answer»

0.256 N
0.126 N
0.324 N
0.425 N

Solution :0.635 g Cu was dissolved in 5.0 mL hot 60% `HNO_(3)` (specific gravity = 1.5)
The following reaction occurs:
`Cu + 4HNO_(3) =Cu(NO_(3))_(2) + 2NO_(2) + 2H_(2)O`
Molecular weight of `HNO_(3) = (1 + 14 + 48) = 63`
In equivlent weight = 63/1 = 63.
63.5 g of Cu reacts with `4 xx 63 = 252 g` of `HNO_(3)`
`therefore 0.635` g of Cu reacts with `252/63.5 xx 0.635 g` of `HNO_(3) = 2.52 g` of `HNO_(3)`
`=1.98/63` g equilvaent of `HNO_(3)`
`therefore 1000 mL` of `HNO_(3)` contains `=1.98/63 xx 1000/250 = 0.126`g equivalent of `HNO_(3)`
The normality of the solution with RESPECT to `HNO_(3)` is 0.126 (N)
Discussion
93395.

0.6 mole of NH_(3) in a reaction vessel of 2 dm^(3) capacity was bougth to equilibrium. The vassel was then found to contain 0.15 mole of H_(2) formed by the reaction 2NH_(3(g))hArrN_(2(g))+3H_(2(g)) Which of the following statement is true

Answer»

`0.15` mole of the original `NH_(3)` had dissociated at equlibrium
`0.55` mole of AMMONIA is LEFT in the vessel
At equilibrium the vessel contained `0.45` mole of `N_(2)`
The CONCENTRATION of `NH_(3)` at equlibrium is `0.25` mole PER `dm^(3)`

Answer :D
Discussion
93396.

0.6 molar MgCl_(2) and 1.2 molar sucrose solution are isotonic then the degree of dissociation of MgCl_(2) is _________

Answer»

`98.3%`
`25%`
`50%`
`76.9%`

ANSWER :C
Discussion
93397.

0.6 mL of acetic acid (CH_(3)COOH), having density 1.06 g mL^(-1), is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205^(@)C. Calculate the van't Hoff factor and the dissociation constant of acid.

Answer»

Solution :Number of moles of acetic acid,
`=(0.6 mL xx 1.06 g mL^(-1))/(60g mol^(-1))=0.0106 mol = n`
Molality `= (0.0106 mol)/(1000 mL xx 1g mL^(-1))`
`=0.0106 mol kg^(-1)`
`therefore Delta T_(F) = 1.86 " K kg mol"^(-1)xx 0.0106 " mol kg"^(-1)`
= 0.0197 K
van.t Hoff factor `(i) =("Observed freezing POINT")/("Calculated freezsing point")`
`=(0.0205 K)/(0.0197 K)=1.041`
Acetic acid is a weak electrolyte and will dissociate into two ions : acetate and hydrogen ions per molecule of acetic acid. If X is the DEGREE of dissociation of acetic acid, then we would have `n(1-X)` moles of undissociated acetic acid, nx moles of `CH_(3)COO^(-)` and nx moles of `H^(+)` ions,
`{:(CH_(3)COOH,hArr,H^(+),+,CH_(3)COO^(-)),("n mol",,0,,0),(n(1-x),,"nx mol",,"nx mol"):}`
Thus TOTAL moles of particles are:
`n(1-x + x + x)=n(1+x)`
`i=(n(1+x))/(n)=1+x=1.041`
Thus degree of dissociation of acetic acid
`=x = 1.041 -1.000=0.041`
Then `[CH_(3)COOH]=n(1-x)`
`=0.0106(1-0.041)`
`[CH_(3)COO^(-)]=nx=0.0106xx0.041`,
`[H^(+)]=nx = 0.0106xx0.041`
`K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])`
`=(0.0106xx0.041xx0.0106xx0.041)/(0.0106(1.00-0.041)=1.86xx10^(-5)`
Discussion
93398.

0.6 mL of acetic acid (CH_(3)COOH)having a density of 1.06 g "mL"^(-1)is dissolved in 1 litre of water . The depression in freezing point observedfor this strength of the acidwas 0.0205^(@)C , Calculate the van't Hoff factor and the dissociation constant of the acidK_(f) for water= 1.86 K kg" mol"^(-1)

Answer»

Solution :Calculation of van't HOFF factor (i)
0.6 mL of acid means solute `(w_(2)) = 0.6 xx 1.6 `
1 Litre of water means solvent `(w_(1)) = 1000 g , (DELTA T_(f))_("observed") = 0.0205 ^(@) C`
` :. `Observed molar mass , `(M_(2)) _("observed") =(1000 xx K_(f) xx w_(2))/(w_(1) xx Delta T _(f))= (1000 " g mol"^(-1) xx 1.86" K kgmol"^(-1) xx 0.636g )/(1000g xx 0.0205 K)`
` = 57.7 g " mol"^(-1)`
Calculatedmolar mass of `CH_(3) COOH = "60 g mol"^(-1)`
` :.`van't Hoff factor (i) = `("Calculated molar mass")/("Observed molar mass ") = (60"g mol"^(-1))/(57.7"g mol"^(-1))= 1.04`
Calculation of degreeof dissociation `(alpha)`from van't Hoff factor (i)
If `alpha` is the degreeof dissociation of ACETIC acid, then
` {:(,CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Intial moles",1,,,,),("Moles at eqm.",1-alpha,,alpha,,alpha",Total" = 1+alpha):}`
`:. i (1+alpha)/1 = 1 + alphaor alpha = 1 - i = 1.04 - 1.0 = 0.04`
Calculation of dissociation constant , If we startwith C mol `L^(-1)`of acetic acid, then
` {:(,CH_(3) COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Intial conc.",C " mol "L^(-1),,,,),("Conc.at eqm.",C(1-alpha),,Calpha,,Calpha):}`
Dissociation constant`(K_(a)) = ([CH_(3)COO^(-1)][H^(+)])/([CH_(3)COOH])=(C alpha. C alpha )/(C(1-alpha)) = (Calpha^(2))/(1-alpha) `
ButC =0.636 g `L^(-1) = (0.636 g L^(-1))/(60 g " mol"^(-1)) = 0.0106 "molL"^(-1) :. K _(a) = ((0.0106)(0.04)^(2))/(1-0.04) = 1.76 xx 10^(-5)`
Discussion
93399.

0.6 ml acetic and having the density 1.06 g mL^(-1) is dissolved in 1 litre water. It's shows the depression in freezing point of 0.020.5^@C. Calculate want haff factor and Ka of an acid.

Answer»


ANSWER :`i=1.041,K_a=1.86xx10^(-5)`
Discussion
93400.

0.6 mL of acetic acid is dissolved in 1 litre of water. The value of van't Hoff factor is 1.04. What will be the degree of dissociation of the acetic acid?

Answer»

0.01
0.02
0.03
0.04

Solution :`{:(CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),("1 MOLE",,0,,0),(1-alpha,,alpha,,alpha),(,,,,"TOTAL "=1+alpha):}`
`i=1+alpha or alpha=i-1=1.04-1=0.04`
Discussion