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88551.

A mixture of chloroform and acetone forms a solution with negative deviation. Why?

Answer»

SOLUTION :A HYDROGEN bond is FORMED between ACETONE and CHLOROFORM as shown below :
Discussion
88552.

A mixture of chlorides of copper, cadmium, chromium, iron and aluminium was dissolved in water acidified with HCl and hydrogen sulphide was was passed for sufficient time. It was filtered, boiled and a few drops of nitric acid were added while boiling. To this solution ammonium chloride and sodium hydroxide were added in excess and filtrate shall give test for-

Answer»

SODIUM ion
sodium, chromium and ALUMINIUM ion
aluminium and IRON ion
sodium, iron, cadmium and aluminium ion

Answer :B
Discussion
88553.

A mixture of CH_4steam on passing over nickel suspension on alumina at 800^@Cgives :

Answer»

CO only
`H_2 only`
CO and `H_2`
NONE of these

Answer :C
Discussion
88554.

A mixture of CH_(4) & O_(2) is used as an optimal fuel if O_(2) is present in thrice the amount required theoretically for combustion of CH_(4). Calcualte the number of effusions steps required to convert a mixture containing1 part of CH_(4) in 193 parts mixture ( parts by volume) . If calorific value (heat evolved when 1 moleis burnt) of CH_(4) is 100 cal// mole & if after each effusion 90% of CH_(4) is collected,find out what initial mole of each gas in initial mixture required for producing 1000 cal of energy after processing.[Given ( 0.9)^(5) = 0.6 ]

Answer»


ANSWER :10 STEPS, 27.78 mol `CH_(4)`, 5333.3 mol `O_(2)`
Discussion
88555.

A mixture of CH_(4) and C_(2)H_(4) was completely burnt in excess of oxygen, yielding equal volumes of CO_(2) and steam. Calculate the percentages of the compounds in the original mixture:

Answer»


ANSWER :d
Discussion
88556.

A mixture of CH_(4) and C_(2)H_(2) occupies a certain volume at a total pressure of 70.5 mm Hg. The sample is burnt, formed CO_(2) and H_(2)O. The H_(2)O is removed and the remaining CO_(2) is found to have pressure of 96.4 mmHg at the same volume and temperature as the original mixture. What mole fraction of the gas was C_(2)H_(2) ?

Answer»

Solution :Let the number of MOLES of `CH_(4)` and `C_(2)H_(2)` be X and y respectively.
`{:(CH_(4),+,C_(2)H_(2),+,O_(2),rarr,CO_(2) ,+,H_(2)O),("x moles",,"y moles",,,,,,):}`
Applying POAC for C atoms,
`1 xx` moles of `CH_(2) + 2 xx` moles of `C_(2)H_(2) = 1 xx` moles of `CO_(2)`
`x + 2y = "moles of " CO_(2)`
As no. of moles `alpha` pressure at const. TEMPERATURE and volume.
`("no. of moles of " CH_(4) " and "C_(2)H_(2))/("no. of moles of " CO_(2)) = (70.5)/(96.4)`
or `(x+y)/(x+2y) = (70.5)/(96.4)`
`therefore (y)/(x+y)` = mole fraction of `C_(2)H_(2) =0.368`.
Discussion
88557.

A mixture of carboxylic acids (X) and (Y) cannot be separated by normal methods however when this mixture is treated with optically active quinine give optically active salts (P) and (Q). These two salis can be separated by fractional crystallization, separated and acidified with HCl to form amines and carboxylic acids. What is the diastereoisorner of (+) -tartaric acid ?

Answer»

(-) - Tartarie ACID
Maleic acid
MESO-TARTARIC acid
All of these

Solution :Meso tartaric Acid
Discussion
88558.

A mixture of carboxylic acids (X) and (Y) cannot be separated by normal methods however when this mixture is treated with optically active quinine give optically active salts (P) and (Q). These two salis can be separated by fractional crystallization, separated and acidified with HCl to form amines and carboxylic acids. The compound (X) and (Y) may be

Answer»

`CH_(3)COOH, CH_(3)CH_(2)COOH`

(+) and (-) - Tartaric ACID
`CH_(3)CH_(2)CH_(2)COOH, CH_(3)CH_(2)CH_(2)CH_(2)COOH`

SOLUTION :dl - Tartaric Acid
Discussion
88559.

A mixture of carboxylic acids (X) and (Y) cannot be separated by normal methods however when this mixture is treated with optically active quinine give optically active salts (P) and (Q). These two salis can be separated by fractional crystallization, separated and acidified with HCl to form amines and carboxylic acids. (P) and (Q) are separated by fractional crystallization due to

Answer»

DIFFERENCE in BOILING point
Difference in SOLUBILITY
Difference in MELTING point
Difference in THERMAL stability

Solution :Solubility difference
Discussion
88560.

A mixture of carboxylic acid (A) and alcohol (B) on heating give and ester (C ) having molecular mass 74. What is (C ) ?

Answer»

`HCOOC_(2)H_(5)`
`CH_(3)COOCH_(3)`
Both a and b
NONE of these

Solution :`(CH_(3) CO)_(2)O + CH_(3)OH overset(DELTA) to CH_(3)COOH + underset("mol. mass 74") (CH_(3)COOCH_(3))`
`(HCO)_(2)O+C_(2)H_(5)OH overset(Delta) to HCOOH + underset("mol. mass 74")(HCOOC_(2)H_(5)) `
Discussion
88561.

A mixture of carbon monoxide and carbon dioxide is found to have a density of 1.7g/lit at S.T.P. The mole fraction of carbon monoxide is

Answer»

0.37
`0.40`
`0.30`
`0.50`

SOLUTION :`M_("MIX")=(DRT)/P_("mix")implies M_("mix")=(1.7xx0.0821xx273)/1=1.7xx22.4` gms
`M_("mix")=X_(CO)xxM_(CO)+(1-X_(CO).M_(CO_2)=X_(CO)xx18+(1-X_(CO))xx44 implies X_(CO)=0.37`
Discussion
88562.

A mixture of camphor and KCI can be separated by :

Answer»

Sulblimation
Evaporation
Filtration
Decantation

Answer :A
Discussion
88563.

A mixture of camphor and calcium sulphate can be separated by :

Answer»

Crystallisation
Steam distillation
Solvent extraction
Sublimation

Answer :D
Discussion
88564.

A mixture of camphar and benozic acid can be separated by

Answer»

Chemical method
Sublimation
Fractional distillation
Extraction with a solvent

SOLUTION :Chemical method USING `NA(HCO_(3))` solution.
Discussion
88565.

A mixture of calcium acetate and calcium formate on heating gives :

Answer»

`CH_3COCH_3`
`CH_3CHO`
`HCHO`
All of the above

Answer :D
Discussion
88566.

A mixture of CaCl_2and NaCl weighing 4.44gis treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56g of CaO. The percentage of NaCl in the mixture is [Atomic mass of Ca = 40]

Answer»

31.5
75
25
40.2

Solution :GIVEN that all the calcium PRESENT in the miture will convert into its carbonates and again on heating it gives its OXIDE CaO of weight 0.56 g.

Moles of CA in `CaCl_(2)` =moles of Ca in `CaCO_(3)`
Moles of Ca in `CaCO_(3)`=moles of Ca in CaO
`implies(X)/(111)=(0.56)/(56)impliesx(0.56)/(56)xx111=1.11g`
[m.wt. of `CaCl_(2)`=111,m.wt. of CaO=56]
We know that `NaCl+CaCl_(2)`=4.44
y+1.11=4.44`implies`y=3.33g
Percentage of NaCl=`3.33xx(100)/(4.44)`=75%
Discussion
88567.

A mixture of CaCl_(2) and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all the Ca^(2+) ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g of CaO. The percentage of NaCl in the mixture (atomic mass of Ca = 40) is

Answer»

75
30.6
25
69.4

Solution :Suppose `CaCl_(2)` in the mixture = x G
`underset(111g)(CaCl_(2))+Na_(2)CO_(3)RARR underset(100g)(CaCO_(3))+2NaCl`
111 g `CaCl_(2)` PRODUCE `CaCO_(3)=100g`
`therefore"x g "CaCl_(2)" will produce "CaCO_(3)=(100)/(111)xg`
`underset(100g)(CaCO_(3))rarrunderset(56g)(CaO)+CO_(2)`
100 g `CaCO_(3)` produce `CaO=56g`
`therefore (100x)/(111)g" will produce CaO"=(56)/(100)xx(100x)/(111)`
`=(56x)/(111)g`
Thus, `(56x)/(111)=0.56"or"x=1.11 g`
`therefore %" of NACL in the mixture "=(3.33)/(4.44)xx100=755`
Alternatively,
`underset(111g)(CaCl_(2))rarrCaCO_(3)rarrunderset(56g)(CaO)`
x g `CaCl_(2)` will GIVE `CaO=(56x)/(111)g`
`therefore""(56x)/(111)=0.56"(given)or"x=1.11g`
Discussion
88568.

A mixture of CaCl_(2) and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all the Ca^(+2)ions as calcium carbonate The calcium carbonate so obtained is heated strongly to get 0.56 g of CaO . The percentage of NaCl in the mixture (atomic mass of Ca=40) is

Answer»

A) 75
B) `30.6`
C) 25
D) `69.4`

Solution :Concerned reaction is
`underset("100 G")(CaCO_(3))overset(Delta)(to)underset("56 g")(CaO)+CO_(2)`
56 g of `CaO` is obtained from 100 g of `CaCO_(3)" "0.56` g `CaO` is obtained by `(100)/(56)xx0.56=1"g of "CaCO_(3)`.
we know that,
`underset("111 g")(CaCl_(2))+Na_(2)CO_(3)tounderset("100 g")(CaCO_(3))+2NaCl`
100 g of `CaCO_(3)` is obtained by 111 g of `CaCl_(2)`
1 g of `CaCO_(3)` is obtained by `(111)/(100)=1.11` g of `CaCl_(2)`
Weight of `NaCl=4.44-1.11=3.33g`
`%` age `NaCl=(3.33)/(4.44)xx100=75%`
Discussion
88569.

A mixture of C_(2)H_(6) and C_(4)H_(4) occupies 40 litres at 1 atm and at 400 K. The mixture reacts completely with 130 g of O_(2) to produce CO_(2) and H_(2)O. Assuming ideal gas behaviour, calculate the mole fractions of C_(2)H_(4) and C_(2)H_(6) in the mixture.

Answer»

Solution :Let the moles of `C_(2)H_(6)` and `C_(2)H_(4)` be `n_(1)` and `n_(2)` respectively. Applying ideal gas equation, `pV = n RT`,
`1 xx 40= (n_(1) + n_(2)) xx 0.0821 xx 400`
or, `(n_(1) + n_(2)) = (40)/(0.0821 xx 400)` ……(1)
`C_(2)H_(6) + C_(2)H_(4) + O_(2) rarr CO_(2) + H_(2)O`
Applying POAC for C,
`2n_(1) + 2n_(2)` = moles of `CO_(2)` ......(2)
Applying POAC for H,
`6 n_(1) + 4n_(2) = 2 xx "moles of " H_(2)O` ........(3) Applying POAC for o,
`2 xx (130)/(32) = 2 xx " moles of " CO_(2) + " moles of " H_(2)O` ........(4)
From equations(2), (3) and (4) we get,
`7 n_(1) + 6n_(2) = (260)/(32)` .......(5)
Solving equations (1) and (5) we get,
`n_(1) = 0.8168`
`n_(2) = 0.4012`
`therefore` MOLE FRACTION of `C_(2)H_(6) = (0.8168)/(0.8168 + 0.4012) = 0.67`
Mole fraction of `C_(2)H_(4) = 1-0.67 = 0.33`.
Discussion
88570.

A mixture of benzoic acid and naphthalene can be separated by crystallization from :

Answer»

HOT water
cold water
ether
benzene

Answer :A
Discussion
88571.

A mixture of benzene and toluene forms :

Answer»

An IDEAL SOLUTION
Non-ideal solution
Suspension
Emulsion

Answer :A
Discussion
88572.

A mixture of benzaldehyde and formaldehyde when treated with 50% NaOH yields.

Answer»

Sodium benzoate and sodium formate
Sodium formate and BENZYL alcohol
Sodium benzoate and methyl alcohol
Benzyl alcohol and methyl alcohol

Solution :More REACTIVE aldehyde undergoes oxidation to form sodium formate and benzyl alcohol,
`H-CHO +C_(6)H_(5)-CHO + NaOH to H-COONa + C_(6)H_(5) -CH_(2)OH`
Formaldehyde is more reactive than benzaldehyde.
Discussion
88573.

A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives

Answer»

benzyl ALCOHOL and SODIUM formate.
sodium BENZOATE and methyl alcohol.
sodium benzoate and sodium formate.
benzyl alcohol and methyl alcohol.

Answer :A
Discussion
88574.

Amixtureof benzaldehyde and formaldehyde when treated with 50% NaOH yields

Answer»

sodium benzoate and sodium formate
sodium formate and benzyl alcohol
sodiumbenzoate and methyl alcohol
benzyl alcohol and methyl alcohol

Solution :When a MIXTUREOF benzaldehyde and FORMALDEHYDE treated with 50% NaOH it forms sodium formate and benzy alcohol this reactionis called cross CANNIZZARO reaction this reaction LOOKS like
Discussion
88575.

A mixture of benzaldehyde and formaldehyde on heating with aqueous NaO solution gives

Answer»

BENZYL ALCOHOL and sodium formate
sodium benzoate and methyl alcohol
sodium benzoate and sodium formate
None of the above

Solution :`C_(6)H_(5)CHO+CH_(2)O OVERSET("NAOH")to C_(6)H_(5)CH_(2)OH+HCOONa`
In cross Cannizzaro reaction of formaldehyde, formaldehyde is always oxidized.
Hence, (A) is the correct answer.
Discussion
88576.

A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives :

Answer»

SODIUM BENZOATE and methyl ALCOHOL
sodium benzoate and sodium formate
BENZYL alcohol and sodium formate
benzyl alcohol and methyl alcohol

Answer :C
Discussion
88577.

A mixture of benzaldehyde and formaldehyde on heatig with aqueous NaOH solution gives

Answer»

Benzyl alcohol and SODIUM formate
sodium benzoate and methyl alcohol
sodium benzoate and sodium formate
benzyl alcohol and methyl alcohol.

Solution :In cross CANNIZZARO reaction, HCHO gets OXIDIZED and `C_(6)H_(5)CHO` gets reduced.
Discussion
88578.

A mixture of an organic liquid A and water distilled , under one atmospheric pressure at 90.2^(@)C . How many grams of steam will be condensed to obtain 1.0g of liquid A in thedistillate ? (Vapour pressure of water at 99.2^(@)C is 739mm Hg . Molecular weight of A=123)

Answer»
Discussion
88579.

A mixture of benzaldehyde and acetophenone on heating with dilute NaOH solution gives

Answer»

Benzolphenone
Benzalacetophenone
Phenyl BENZOATE
BENZYL ALCOHOL and SODIUM benzoate

Answer :B
Discussion
88580.

A mixture of anhydrous ZnCl_(2)+conc.HCI is known as

Answer»

FEHLING's REAGENT
LUCAS reagent
Tollen's reagent
BENEDICT's reagent

Answer :B
Discussion
88581.

A mixture of an acid anhydride (A) and a monobasic acid (B) on heating produces another monobasic acid (C) of equivalent weight 74 and an anhydride (D). The acids and anhydrides remain in equilibrium. The anhydride (D) contains two identical flouro alkyls groups. The acid (B) contains a trifluoro methyl group and has an equivalent weight of 128. Give the structures of (A) to D) with proper reasoning (Atomic weight of fluorine 19).

Answer»

Solution : Acid anhydride (A) +`underset((CF_(3) "group")) ("Monobasic acid (B)")` `hArr`Monobasic acid(C ) +`underset((Two CF_(3) "group")) ("Anhydride(D)")`
Eq. wt.=128Eq. wt.=74
Probable formula of acid (B)=`CF_(3)(CH_(2))_(n)COOH`
EQUIVALENT weight of (B) =`("MOLECULAR weight")/("BASICITY")`
`128=(57+12+14n+12+32+1)/1`
n=1
`therefore(B)=CF_(3)CH_(2)COOH` and `(D)=(CF_(3)CH_(2)CO)_(2)O`
Similarly, molecular formula of acid (C ) can be calculated to be `CH_(3)CH_(2)COOH`. The reaction, therefore, can be given as,
`underset((A))((CH_(3)CH_(2)CO)_(2)O)+underset((B))(F_(3)"CC"H_(2)COOH)hArrunderset((C ))(CH_(3)CH_(2)COOH)+underset((D))((F_(3)"CC"H_(2)CO)_(2)O)`
Discussion
88582.

A mixture of aluminium and zinc weighning 1.67 g was completely dissolved in acid and evolved 1.69 litres of hydrogen at NTP. What was the weight of

Answer»

Solution :EQ. of METAL= `(0.43)/E ` ( eq. wt of metal `-= `E)
`:. ` m.e of metal ` = (0.43)/E xx 1000 = 430/E "" …(Eqn.3)`
m.e of total `H_(2)SO_(4)`solution` = 1 xx 50 = 50 "" …(Eqn. 1)`
Butm.e of `H_(2)SO_(4)` reacted with metal = m.e of the metal
`= (430)/E `
` :. ` m.e of unreached `H_(2)SO_(4) = ( 50 - 430/E)`
Again m.e of unreacted `H_(2)SO_(4)` = m.e of NaOH . ....(Eqn.2)
` :.E = 12.01 ` .
Discussion
88583.

A mixture of aluminium and zincweighning 1.67g was completely dissolvedin acidand evolved 1.69 litres of hydrogen at NTP . What was the aluminium in the original mixture ? (Al = 27, Zn = 65.4)

Answer»

Solution :Since `H_(2)` is formed by both Al and Zn ,
EQ. of Al + eq of Zn = eq. of `H_(2)` .
Let w be the mass in grams of Al in the MIXTURE.
`:. (w)/("eq. wt . of Al") + ((1.67 - w))/(" eq . wt .of Zn") = (1.69)/( " vol. of 1 eq. of " H_(2)" at NT in lit")`
` OMEGA /(27//3) +(1.67 - omega)/(65.4//2) = (1.69)/(11.2){:{("eq . wt of Al"=27/3),("eq. wt. of Zn"=(65.4)/2 ):}}`
` :."" omega = 1.24 g `
Discussion
88584.

A mixture of alpha-amino acids is obtained when proteins are hydrolysed by

Answer»

ACIDS
Bases
Enzymes
All

Answer :D
Discussion
88585.

A mixture of Al(OH)_3 and Fe (OH)_3 can be separated easily by treating it with:

Answer»

HCL
`NH_4OH`
`HNO_3`
NaOH

Answer :D
Discussion
88586.

A mixture of alcohol and ether is called :

Answer»

Natalite
Power alcohol
Peroxide
None

Answer :A
Discussion
88587.

A mixture of Al_(2)O_(3) and Fe_(2)O_(3) can be separated by using

Answer»

SODIUM hydroxide
Cold water
Ethyl alcohol
Boling wate

Solution : NAOH is the leaching reagent which is used to SEPARATE `Al_(2)O_(3) and Fe_(2)O_(3)`from the mixture. NaOH dissolves `Al_(2)O_(3)` but not `Fe_(2)O_(3)`.
Discussion
88588.

A mixture of Al_(2)O_(2) and Fe_(2) can be separated by using

Answer»

SODIUM hydroxide
Cold water
Ethyl alcohol
Boiling water

Answer :A
Discussion
88589.

A mixture of A(g) and B(g) is formed, where mole fraction of gas A is 0.8. Given : P_(A)^(@) = 400 torr, P_(B)^(@) = 200 torr Mixture of A and B obey Raoult's law.Then :

Answer»

Composition of first drop of condensate is `(X_(A) = 2/3)`
Total PRESSURE when first drop of condensate form is `1000//3` torr
The composition of LAST BUBBLE is `(Y_(A) = 8/9)`
Composition of A in vapour when mole FRACTION of A and B in liquid are equal `(Y_(A) = 2/3)`

ANSWER :A::B::C::D
Discussion
88590.

A mixture of Al_(2) O_(3) and Fe_(2) O_(3) can be separa-ted by using

Answer»

SODIUM hydroxide
Cold WATER
ETHYL alcohol
Boiling water

Answer :A
Discussion
88591.

A mixture of AB_(3)(g) and A_(2)B_(4)(g) is placed in a sealed container. (At. Mass of A = 14 and At. ,mass of B =1 ) The temperature initially maintained is 300K. The total pressure is 0.5 atm. The container is heated to 1200k, when both AB_(3) and A_(2) B_(4) decomposed to A_(2) and B_(2) when the decomposition was complete the pressure observed was 4.5 atm. What is the percentage of AB_(3)(g) in the original mixture ?

Answer»


ANSWER :`0075`
Discussion
88592.

A mixture of acetone and methanol can be separated by :

Answer»

STEAM distillation
vacuum distillation
Fractional distillation
Sublimation

Answer :C
Discussion
88593.

A mixture of acetone and "CCl"_(4) can be separated by

Answer»

Azoeotropic distillation
Fractional distillation
Steam distillation
Vacuum distillation

Solution :(a) is used only to separate azerotrope, i.e., constant boiling MIXTURE.
(B) is correct because `"CCl"_(4)` and acetone DIFFER in their boiling point.
(C ) is used only in CASE of steam volatile substance. It is not correct.
(d) Vacuum distillation is used in liquids which decompose at their boiling point.
`:.` It is not correct.
Discussion
88594.

A mixture of A and B in the molar ratio 1:2 forms a maximum boiling azeotrope. Identify the incorrect statement, if A is more volatile. [ Molar mass of A=100, Molar mass of B=50]

Answer»

A liquid solution of A and B having mass % of A=50 will have VAPOURS having mass % of A=50.
A liquid solution of A and B having mass % of ` Agt 50%` will have vapours having mass % of Agt50%.
A mixture of A and B in the molar mass ratio 1:3 can be seperated into azeotopic mixture and pure A
A mixture of A and B in the molar ratio 2:3 can be seperated into AZEOTROPIC mixture and pure A.

Answer :C
Discussion
88595.

A mixture of ""^(239)Pu and ""^(240)Pu has a specific activity of 6.0xx 10^(9) dis/s. The half-lives of the isotopes are 2.44 xx 10^(4) and 6.58 xx 10^(3) years, respectively. Calculate the isotopic composition of this sample

Answer»

Solution :Specific activity means activity per gram. Sp. Activity of `""^(239)Pu= lamda= N = ((0.6932)/(2.44 xx 10^(4))) ((6.022 xx 10^(23))/(239))`
`=7.15 xx 10^(16)`/YR/g
`=2.27 xx 10^(9)//s//g. (1 yr = 3.15 xx 10^(7)s)`
Sp. activity of `""^(240)Pu= ((0.6932)/(6.58 xx 10^(3))) ((6.022 xx 10^(23))/(240))`
`=2.64 xx 10^(17)//yr//g= 8.37 xx 10^(9)//s//g`
THUS, `(2.27 xx 10^(9)) x + (8.37 xx 10^(9)) (1-x)=6.0 xx 10^(9)` (x=fraction of `""^(239)Pu`) x=0.39 or 39%
Discussion
88596.

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H_2SO_4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

Answer»

2.8
`3.0`
`1.44`
`4.4`

ANSWER :A
Discussion
88597.

A mixture of 20 mL of CO, CH_(4) and N_(2) was burnt in excess of O_(2) resulting in the reduction of 13 mL of volume. The residual gas was then treated with KOH solution to show a contraction of 14 mL in volume. Calculate volume of CO, CH_(4) and N_(2) in the mixture. All measurements are made at constant pressure and temperature.

Answer»

SOLUTION :`"Suppose volume of CO = a mL," CH_(4)="b mL and N"_(2)="c mL. Then a + b + c = 20 mL…(i)"`
The combustion reaction will be
`CO+(1)/(2)O_(2)RARR CO_(2)`
`CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O(l)`
`N_(2)+O_(2) rarr "No reaction"`
Thus, a mL of CO will produce `CO_(2)=amL`
`"b mL of CH"_(4)" will produce CO"_(2)=bmL`
`N_(2)` will remain as such, i.e. = c mL
As `CO_(2)` is absorbed by KOH, decrease in volume on treating with KOH will be `=a+b=14mL` (Given) .....(ii)
The first given decrease is DUE to `O_(2)` consumed.
a mL of CO will consume `O_(2)=(a)/(2)mL`
b mL of `CH_(4)` will consume `O_(2)=2bmL`
`therefore O_(2)` consumed `=(1)/(2)+2b=13mL"(Given )...(iii)"`
From eqns. (i) and (ii), `c = 20 - 14 = 6 mL`
From Eqns. (ii) and (iii),
`(a)/(2)+2(14-a)=13 "or"(3)/(2)a=15"or"a=10mL`
`therefore " From eqn. (ii),"10+b=14"or"b=4mL`
`CO=10mL, CH_(4)=4mL, N_(2)=6mL`
Discussion
88598.

A mixture of 2 moles of CO and 1 mole of O_2 in a closed vessel is ignited to convert CO into CO_2. Then

Answer»

`DeltaH LT DeltaE`
`DeltaH GT DeltaE`
`Delta =DeltaE`
The RELATIONSHIP DEPENDS UPON the capacity of the vessel.

Answer :A
Discussion
88599.

A mixture of 2 moles of carbon monoxide and one mole of oxygen in a closed vessel is ignited to get carbon dioxide. If Delta H is the enthalpy change and Delta U is the change in internal energy, then

Answer»

`DELTA H GT Delta U`
`Delta H lt Delta U`
`Delta H = Delta U`
Not definite

Answer :B
Discussion
88600.

A mixture of 2 mole of carbon monoxide gas and one mole of dioxyen gas is enclosed in a close vessel and is ignited to convert carbon monoxide into carbon dioxide. If the enthalpy change is DeltaH and internal change is DeltaU, then for the above process.

Answer»

`DELTAH = DELTAU`
`DeltaH + DeltaU=1`
`DeltaH - DeltaU GT 0`
`DeltaH LT DeltaU`.

Answer :D
Discussion