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88501.

A mixture of methane, ethylene, ethyne gases is passed through a Woulfe's bottle containing ammonical AgNO_3. The gas not coming out from bottle is

Answer»

METHANE
ETHYNE
Ethylene
all of the above

Answer :B
Discussion
88502.

A mixture of methane, ethylene and acetylene gases is passed through aWolf bottle containing ammoniacal cuprous chloride. The gas coming out is

Answer»

METHANE
Acetylene
A mixture of methane and ethylene
The original mixture

Solution :Acetylene reacts with AMMONICAL cuprous chloride to form red ppt. of copper acetylide while methane and ethylene do not REACT (since they do not have acidic hydrogen ) . They come out from the BOTTLE
`CH-=CH+2[Cu(NH_3)_2]OHto Cu-C-=C-Cu+4NH_3+2H_2O`
`CH_4`+Ammonical `CuCl_2to` No reaction
`C_2H_4 `+Ammonical `Cu_2Cl_2 to ` No reaction
Discussion
88503.

A mixture of Li_(2)CO_(3) and Na_(2)CO_(3) is heated strongly in an open vessel .If the loss in the mass of mixture is (220)/(9) % .The molar ratio of Li_(2)CO_(3) and Na_(2)CO_(3) in the initial mixture is (Li = 7,Na = 23)

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`1:1`
`1:2`
`2:1`
`11:4`

ANSWER :A
Discussion
88504.

A mixture of lime paste is sand, water and

Answer»

Gypsum
Slacked LIME
QUICK lime
Lime STONE

ANSWER :C
Discussion
88505.

A mixture of KMnO_(4) and K_(2)Cr_(2)O_(7) weighing 0.24 g on being treated with KI in acid solution liberates just sufficient I_(2) to react with 60 mL of 0.1 N Ha_(2)S_(2)O_(3) solution . Calculate percentage of Cr and Mn in the mixture .

Answer»


Solution :`underset(+7) (KMnO_(4)) to underset(+2) (Mn^(2+))and underset(+12)(K_(2)Cr_(2)O_(7))to underset(+6)(2Cr^(3+)]`
Discussion
88506.

A mixture of KBr and NaBr weighing 0.560 gm was treated with aqueous Ah^(+) and all the bromide ion was recovered as 0.970 gm of pure AgBr. The fraction by weight of KBr in the sample is

Answer»

`0.25`
`0.2378`
`0.36`
`0.285`

SOLUTION :N/A
Discussion
88507.

A mixture of KBr and NaBr weighing 0.560 g wastreated with aqueousAg^(+) and all the bromide ion was recoveredas 0.970 g ofpure AgBr . What was the fraction by weight of KBrin the sample ? (K =n 39, Br = 80 , Ag = 108 , Na = 23 )

Answer»

SOLUTION :`{:("KBr+NaBr+" AG^(+)to" " AGBr),(" xg(0.56-x)0.97 g"):}`
SINCE Br atoms are conserved, APPLYING POAC for Br atoms,
Moles of Br in KBr + moles of Br in NaBr = molesof Br in AgBr
or ` 1 xx ` moles of KBr +` 1 xx ` moles of NaBr = `1 xx ` moles of AGBr
x = 0.1332 g
Fraction of KBr in the SAMPLE`= (01332)/(0.560) = 0.2378`
Discussion
88508.

A mixture of inorganic substances on treatment with dilute H_2SO_4 gives out a gas which turns lead acetate paper black while potassium dichromate paper is turned green. The mixture contains

Answer»

Sulphite
sulphide
nitrate
nitrite

Answer :B
Discussion
88509.

A mixture of hydrogen and oxygen is evolved when a dilute solution of NaOH is electrolysed. How many moles of each gas would be liberated by a current which deposited 20.942 of Ag ?

Answer»

Solution :1 mole (faraday) ELECTRIC yields 1 equivalent of matter.
`therefore` equivalent of Ag `= (20.942)/(108)`
`therefore` mole of electric CHARGE `= (20942)/(108)` faraday.
SINCE the same amount of electricity is passed through diliute NaOH solution,
equivalent of hydrogen liberated `= (20.942)(108)`
and equivalent of OXYGEN liberated `= (20.942)/(108)`
`{:(therefore " mole of " H_(2)" liberated " = (20.942)/(108) xx (1)/(2) = 0.97),("mole of " O_(2) " liberated " = (20.942)/(108) xx (1)/(4) = 0.0485):}}`
Discussion
88510.

A mixture of H_(2) and O_(2) in 2:1 volume is allowed to diffuse through a porous partition what is the composition of gas coming out initially

Answer»

Solution :We have,
`(r_(H_(2)))/(r_(O_(2))) = sqrt((M_(O_(2)))/(M_(H_(2)))) = sqrt((32)/(2)) = 4`.
Now since volum RATIO of `H_(2)` and `O_(2)` in the INITIAL stage is 2 :1 (given) and the ratio of there ratio is 4:1, the overall volume ratio of the GASES diffusing through the porous PARTITION at the initial stage will be 8:1.
Discussion
88511.

A mixture of hydrogen and oxygen in 3:1 volume ratio is allowed to diffuse through a porous partition. What should be the composition of the initial gas diffusing out of the vessel ?

Answer»


ANSWER :`12:1`
Discussion
88512.

A mixture of hydrogen and oxygen contains 20% by mass of hydrogen. What is the total number of molecules present per gram of the mixture?

Answer»


SOLUTION :In 1 gram of the mixture 0.2g of HYDROGEN and 0.6g of OXYGEN are PRESENT. Moles of `H_(2)=(0.2)/(2)=0.1`. Moles of oxygen `=(0.8)/(32)=0.025`. Calculate the number of molecules of hydrogen and oxygen and then add.
Discussion
88513.

A mixture of hexane (b.p. 342K) and toluene (b.p. 384K) can be seprated by

Answer»

SIMPLE distillation
vacuum distillation
Sublimation
Diiferential extraction

Answer :D
Discussion
88514.

A mixture of hydrazine and 40 to 60 percent of H_2O_2 solution is :

Answer»

Antiseptic
Rocket fuel
Germicide
Insecticide

Answer :B
Discussion
88515.

A mixture of helium and argon contains 3 mole of He for every 2 mole of Ar. The partial pressure of argon is :

Answer»

2/3 the TOTAL pressure
1/3 the total pressure
2/5 the total pressure
1/5 the total pressure

Answer :C
Discussion
88516.

A mixture of He(4) and Ne(20) in a 5-litre flask at 300 K and 1 atm weighs 4 g. The mole % of He is

Answer»

2
0.02
20
4

Answer :A
Discussion
88517.

A mixture of He and O_(2) has density 1.3 gm//litre at NTP. Them mole fraction of He is

Answer»

`0.1`
`0.9`
`0.4`
`0.6`

Solution :`d = 1.3 GM//"litre" = (PM)/(RT)`
`1.3 = (1 XX M)/(0.08 xx 273)`
`M = 1.3 xx 22.4`
`M = 29.12 gm//mol`
`M_(AVG) = M_(1)X_(1) xx M_(2) X_(2)`
`29.12 = 4a + 32 - 32a`
`29.12 = 32 - 28a`
`28a = 32 - 29.12`
`a = (2.88)/(28)`
`a = 0.1` Hence, `X_(O_(2)) = 0.9`
`X_(He) = 0.1`
Discussion
88518.

A mixture of He and Ne has a density of 1.36 xx 10^(-3) gm //ml at 0^(@)C and 2.24 atm. Then mole fraction of neon in this mixture is "________"

Answer»

0.4
0.6
0.3
0.7

Solution :`d = 1.36 XX 10^(-3) gm//ml = 1.36 gm// "litre"`
`X_(He) = a`
`X_(Ne) = 1 - a`
`M_(AVG) = x_(1)m_(1) + x_(2) m_(2)`
`M_(avg) = 4a + 20 (1 - a)`
`d = (PM_(avg))/(RT)`
`M_(avg) = (dRT)/(P) = (1.36 xx 0.0821 xx 273)/(2.24)`
`= 13.6 gm//mol`
`M_(avg) = 13.6 = 4a + 20 (1 - a)`
`a = 0.4`
`x_(He) = 0.4`
`x_(Ne) = 0.6`
Discussion
88519.

A mixture of He and CO_2has a volume of 63.5 mL at 1.0 bar and 28^@C . The systemcontaining the mixture is cooled in liquid nitrogen, and the remaining gas is evacuated. The system is restored to 1.0 bar and 28^@C , and the volume is 40.5 mL. Find the composition of the original mixture.

Answer»

SOLUTION :0.638, 0.362
Discussion
88520.

A mixture of (H_(2)O+C_(6)H_(5)NO_(2)) boils at 90^(@)C. In the vapours of mixture partial vapour pressures of H_(2)O and C_(6)H_(5)NO_(2) are 733 mm Hg and 27 mm Hg respectively. The W_(H_(2)O)//W_(C_(6)H_(5)NO_(2)) is

Answer»


Solution :`(P'_(H_(2)O))/(P'_(C_(6)H_(5)NO_(2)))=(W_(H_(2)O))/(m_(H_(2)O))XX(M_(C_(6)H_(5)NO_(2)))/(W_(C_(6)H_(5)NO_(2)))`
`P'_(H_(2)O)=733mm HG, P'_(C_(6)H_(5)NO_(2))=27MM Hg`
and `m_(H_(2)O)=18, m_(C_(6)H_(5)NO_(2))=123`
`(W_(H_(2)O))/(W_(C_(6)H_(5)NO_(2)))=(733)/(27)xx(18)/(123)=3.972~~4`
Discussion
88521.

A mixture of H_(2)O vapour, CO_(2) and N_(2) was trapped in a glass apparatus with a volume of 0.731 mL. The pressure of the total mixture was 1.74 mm Hg at 27^(@)C. The sample was transferred to a bulb in contact with dry ice (-75^(@)C) so that H_(2)O vapour was frozen out. When the sample was returned to the measured volume, the pressure was 1.32 mm Hg. The sample was then transferred to a bulb in contact with liquid nitrogen (-195^(@)C) to freeze out the CO_(2). In the measured volume, the pressure was 0.53 mm Hg. How many moles of each constituents are in the mixture ?

Answer»

Solution :`p_(H_(2)O) + p_(CO_(2)) + p_(N_(2)) = 1.74 MM`
`p_(CO_(2)) + p_(N_(2)) = 1.32 mm`
`p_(N_(2)) = 0.53 mm`
`therefore p_(CO_(2)) = 1.32 - 0.53 - 0.79 mm`
and `p_(H_(2)O) = 1.74 - 1.32 = 0.42 mm`.
Number of moles of each constituent is calculated USING the EQUATION
`pV = nRT`.
For `H_(2)O : pH_(H_(2)O) = (0.42)/(760)` atm, `V = (0.731)/(1000)` lit.
`T = 27 + 273 = 300 K, R = 0.082` lit. atm/K/mol
`(0.42)/(760) xx (0.731)/(1000) = n_(H_(2)O) xx 0.082 xx 300`
`n_(H_(2)O) = 1.64 xx 10^(-8)`
Similarly,
for `CO_(2) : (0.79)/(760) xx (0.731)/(1000) = n_(CO_(2)) xx 0.082 xx 300`
`n_(CO_(2)) = 3.08 xx 10^(-8)`
and for `N_(2) : (0.53)/(760) xx (0.731)/(1000) = n_(N_(2)) xx 0.082 xx 300`
`n_(N_(2)) = 2.07 xx 10^(-8)`
Discussion
88522.

A mixture of H_(2(g)), N_(2(g)) and O_(2(g)) occuping 10 ml underwent reaction so as to from H_(2)O_(2(l)) and N_(2)H_(2(g)) as the only products causing the volume to contacts by 6ml. The remaining mixture was passed through Pyragallol causing a contraction of 1ml. to the remaining mixture excess H_(2) was added and the above reaction was repeated, causing a reduction in volume of 1ml. assuming no other products to be forming calculate the volume of H_(2) (in ml) in the initial mixture.

Answer»


Solution :`{:(H_(2(g)),,N_(2)(g),,O_(2(g))),(X+y,,x+1,,1+y),(H_(2(g)),+,O_(2(g)),to, H_(2)O_(2(L))),(x,,x,,),(H_(2(g)),+,N_(2(g)),to,N_(2)H_(2(g))),(y,,y,,y):}`
2x+2y=6
2x+2y+2=10
x+y=4
`UL(2x+y=6)`
x=2
y=2
Discussion
88523.

A mixture of H_(2)C_(2)O_(4) and NaHC_(2)O_(4) weighing 2.02 g was dissolved in water and the solution made uptp one litre. 10 mL of this solution required 3.0 mL of 0.1 N NaOH solution for complete neutralization. In another experiment 10 mL of same solution in hot dilute H_(2)SO_(4) medium required 4 mL of 0.1N KMnO_(4) KMnO_(4) for compltete neutralization. Calculate the amount of H_(2)C_(2)O_(4) and NaHC_(2)O_(4) in mixture.

Answer»

SOLUTION :Let the wt.Of `H_(2)C_(2)O_(4)`in 10 mL of the solution be x g . The weight of `NaHC_(2)O_(4)` in 10 mLwill be `(0.0202 - x) g `.The weightof `NaHC_(2)O_(4)` iin 10 mL will be `(0.0202 -x)` g
In the first experiment, `H_(2)C_(2)O_(4) and NaHC_(2)O_(4)` are neutralised by NaOH changing into `Na_(2)C_(2)O_(4)` . Theeq.wtof `H_(2)C_(2)O_(4) and NaHC_(2)O_(4)` will therefore be 90/2 and 112 respectively .
THUS ,
m.e of `H_(2)C_(2)O_(4) + " m.e of " NaH_(2)O_(4) =" m.e of " NaOH `
`x/(90//2) xx 1000 + ((0.0202-x))/112 xx 100 = 0.1 xx 3 ""..(1)`
In the second experiment , both `H_(2)C_(2)O_(4) and NaHC_(2)O_(4)` are oxidised t `CO_(2)` by `KMnO_(4)` . the equivalent weight of `H_(2)C_(2)O_(4) and NaHC_(2)O_(4)` will , therefore be 90/2 and 112/2 respectively `({:(C_(2)O_(4)^(2-) to, 2CO_(2)),(+6,+8):})`
Thus ,
m.eof `H_(2)C_(2)O_(4) +" m.eof " NaHC_(2)O_(4) = " m.e of " KMnO_(4)`
`x/(90//2) xx 1000 + ((0.202-x))/(112//2) xx 1000 = 0.1 xx 4 "" ...(2)`
Subtracting (1) from (2) we get ,
`(0.202-x)/112 = (0.1)/1000`
` :. "" x = 0.009` g/10 mL of solution
The 1000 mL of solution contains
`H_(2)C_(2)O_(4) = 0.9 g `
and `NaHC_(2)O_(4) = 2.02 - 0.9 = 1.12 g ` .
Discussion
88524.

A mixture of H_(2)andO_(2) taken in a bulb in 2 : 1 mole ratio is allowed to diffuse through a fine hole. The composition of the gases (mole ratio) coming out initially is 8 : 1. The composition of the gases after some time may be

Answer»

`8:1`
`9:1`
`7:1`
`4:1`

ANSWER :C::D
Discussion
88525.

A mixture of gases O_2H_2and OCare taken in a closed vessel containing charcoal. The graph that represents the correct behavior

Answer»





SOLUTION :THEORY BASED
Discussion
88526.

A mixture of gases contains H_(2) and O_(2) gases in the ratio of 1:4 (w//w). What is the molar ratio of the two gases in the mixture?

Answer»

`16:1`
`2:1`
`1:4`
`4:1`

SOLUTION :Suppose `H_(2)=W`, then `O_(2)=4wg`
MOLES of `H_(2)=(w)/(2)", Moles of O"_(2)=(4w)/(32)=(w)/(8)`
`"MOLAR ratio of "H_(2):O_(2)=(w)/(2):(w)/(8)=(1)/(1):(1)/(4)=4:1`
Discussion
88527.

A mixture of formic acid and oxalic acid is heated with conc. H_(2)SO_(4) The case produced is collected and on its treatment with KOH solution, the volume of the gas decreased by 1/6 th. Calculated the molar ratio of the two acids in the original mixture.

Answer»

Solution :SUPPOSE no. of moles of formic acid is x and that of oxalic acid is y. Then
`underset(x)("mole")overset(Conc.)underset(H_(2)SO_(4))rarrH_(2)O_(4)(l)+underset("x mole")(CO(g))`
`underset("y mole"){:(" "COOH),("|"),(" "COOH):}rarr underset("y mole")(CO(g))+underset("y mole")(CO_(2)(g))+H_(2)O(l)`
Total no. of moles of gaseous product `=x+2y`
As only `CO_(2)` is ABSORBED by KOH, therefore, fraction of `CO_(2)=(y)/(x+2y)=(1)/(6)" (GIVEN)"`
`"or"6y=x+2y or 4y=x or (x)/(y)=(4)/(1)`
Hence, molar ratio of `HCOOH` to `(COOH)_(2)` is `4:1`
Discussion
88528.

A mixture of formaldehyde and benzaldehyde 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.

Solution :Benzyl alcohol and sodium formate will be formed by cross cannizzaro's REACTION.
`C_(6)H_(5)CHO + HCHO overset(NaOH)RARR C_(6)H_(5)CH_(2)OH + HCOONa`
Discussion
88529.

A mixture of ferrie alum, chrome alum and potash alum is dissolved in water and treated with an excess of an NH_(3) solution. The precipitate is filtered and the residue is warmed with a mixture of NaOH and H202 and filtered. We will get

Answer»

a green residue and a YELLOW filtrate
a BROWN residue and a yellow filtrate
a brown residue and a green filtrate
a BLUE residue and a green filtrate

Solution :Brown PRECIPITATE and yellow filtrate
Discussion
88530.

A mixture of ferric oxide (Fe_(2)O_(3)) and Al is used as a solid rocket fuel which reacts to give Al_(2)O_(3) and Fe . No other reactants and products are involved . On complete reaction of 1 mole of Fe_(2)O_(3), 200 units of energy is released. (a) Write a balance reaction representing the above change. (b) What should be the ratio of masses of Fe_(2)O_(3) and Al taken so that maximum energy per unit of fuel is released. (c) What would be energy released if 16 kg of Fe_(2)O_(3) reacts with 2.7 kg of Al.

Answer»


ANSWER :`(i)Fe_(2)O_(3)+2Al toAl_(2)O_(3)+2Fe;(ii)80:27;(III)10,000` units
Discussion
88531.

A mixtureof FeO and Fe_(3)O_(4)whenheated in air to a constantweight,gains 5% in weight. Findthe compositionof the initialmixture.

Answer»

Solution :When FeO and `Fe_(3) O_(4)` are heated, both change to ` Fe_(2) O_(3)` . Let the weights of FeO and `Fe_(3) O_(4)` be X g and y respectively .
`:.` when FeO and `Fe_(3) O_(4)` change completelyto `Fe_(2) O_(3)`
the wt. of `Fe_(2)O_(3) = (105)/( 100) XX ( x + y) = 1. 05 ( x + y) g`
Now,
`{:(FeO + Fe3 O_(4) to Fe_(2) O_(3)),("xgyg1 . 05 (x + y) g "):}`
Applying POAC for Fe atoms,
molesof Fe in FeO + moles of Fe in `Fe_(3) O_(4)` = moles of Fe in `Fe_(2) O_(3)`
`1 xx` moles of FeO + `3 xx ` moles of `Fe_(3) O_(4) = 2 xx ` moles of `Fe_(2) O_(3)`
`(x)/(72) + (3Y)/( 232) = (2 xx 1.05 (x + y))/( 160) [{:(" "FeO = 72),(Fe_(3)O_(4)=232),(Fe_(2)O_(3)=160):}]`
Dividingby y, we get
`(1)/( 72) xx (x)/( y) + (3)/( 232) = ( 2 xx 1.05) xx (x)/(y) + ( 2 + 1.05)/( 160)`
`(x)/(y) ((1)/( 27) - (2 . 1)/( 160)) = (2 . 1)/( 160) - (3)/( 232)`
`(x)/( y) = ( 81)/( 319)`
`:. % " of FeO" = (81)/( ( 81 + 310)) xx 100 = 20.02 %`
and `% of Fe_(3) O_(4) = (319)/(( 81 + 319)) xx 100 = 70 . 98 %`
Discussion
88532.

A mixture of FeO and Fe_(3)O_(4) was heated in air to a constant mass. It was found to gian 10% in its mass. Calculate the percentage composition of the origincal mixture.

Answer»

Solution :On heating in air, both FeO and `Fe_(3)O_(4)` are oxidized to `Fe_(2)O_(3)` as follows :
`underset(=144g)underset(2(56+16))(2FE)+(1)/(2)O_(2) rarrunderset(=160g)underset(2xx56+3xx16)(Fe_(2)O_(3)),""underset(=464 g)underset(2(3xx56+4xx16))(2Fe_(3)O_(4))+(1)/(2)O_(2) rarr underset(=480g)underset(3xx160)(3Fe_(2)O_(3))`
Suppose `FeO` in the mixture `=X%`
Then `Fe_(3)O_(4)` in the mixture `=(100-x)%`
`Fe_(2)O_(3)` produced from x g of `FeO=(160)/(144)xx xg`
Total `Fe_(2)O_(3)` produced from 100 g of the mixtuer `=100+10=110g`
`therefore""(160x)/(144)+(480(100-x))/(464)=110`
`"or"(10x)/(9)+(30(100-x))/(29)=110`
`"or290x+270(100-x)=110xx9xx29 or 20x=1710 or x=85.5`
`therefore FeO` present in the mixture `=85.5% and Fe_(3)O_(4)=100-85.5=12.4%`.
Discussion
88533.

A mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction.The hydrocarbon that will not be formed is

Answer»

n-butane
n-propane
n-pentane
n-hexane

Answer :B
Discussion
88534.

A mixture of ethyl iodide and n-proplyl iodide is treated with sodim of any one of the followingsubstance is treated with sodium. The correct substace is

Answer»

propene
butane
pentane
hexane

Solution :`3C_(2)H_(5)I+6Na+3CH_(3)CH_(2)CH_(2)L overset("dry ETHER") to C_(6)H_(6)C_(5)H_(12)+C_(4)H_(10)+6Nal`
Discussion
88535.

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 300 mm at 300 K. The vapour pressure of pure propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol in the mixture is 0.375, then the vapour pressure (in mm) of pure ethyl alcohol at the same temperature will be :

Answer»

466
300
700
360

Solution :`300=0.375xxP+0.625xx200, P=466.67 MM`
Discussion
88536.

A mixture of ether and ….. Gives temperature as low as 163 K (1) NaC1 (2) Ice (3) Solid CO_(2) C_(2)H_(5)OH

Answer»

Both A and R are true and R is the CORRECT explanation to A
Both A and R are true and R is not the correct explanation to A
A is true but R is FALSE
A is false but R is true

Answer :C
Discussion
88537.

A mixture of ethene and ethane occupied 35.5 L at 1.0 bar and 45K. The mixture reacted completely with 110.3g of O_(2) to produce CO_(2) and H_(2)O . What was the composition of the original mixture. Assume ideal gas behaviour.

Answer»


ANSWER :`0.57`
Discussion
88538.

A mixture of ethane (C_(2)H_(6)) and ethene (C_(2)H_(4)) occupies 40 litres at 1.00 atm and at 400K. 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 :Apply PV = nRT and calculate n of the mixture. We get n = 1.22 moles.
Suppose `C_(2)H_(6)="x mole. Then "C_(2)H_(4)=(1.22-x)" mole"`
`2C_(2)H_(6)+7O_(2) rarr 4CO_(2)+6H_(2)O`
`C_(2)H_(4)+3O_(2) rarr 2CO_(2)+2H_(2)O`
`O_(2)" used for x mole of "C_(2)H_(6)=(7)/(2)XX x=3.5x" mole"`
`O_(2)` used for `(1.22-x)` mole of `C_(2)H_(4)=3(1.22-x)` mole
Total `O_(2)` used `=3.5x+3(1.22-x)=(130)/(32)"(GIVEN)"`
Calculate x. We get x = 0.805
`"Mole fraction of "C_(2)H_(6)=(0.805)/(1.22)=0.66`
Discussion
88539.

A mixture of ethane (C_(2)H_(6)) and ethane (C_(2)H_(4)) occupies 40 litres at 1.00 atm and 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 :Calculation of total no. of MOLES in the gaseous mixture
Applying ideal gas equation, PV = nRT
`"1 atm"xx40L= nxx0.0821"L atm K"^(-1)"mol"^(-1)xx400K`
`"or"n=(40)/(0.0821xx400)="1.218 MOLE"`
Calculation of no. of moles of each component
Suppose no. of moles of `C_(2)H_(6)` in the mixture = x
Then no. of moles of `C_(2)H_(4)` in the mixture = `1.218-x`
ALSO, 130 G of `O_(2)=(130)/(32)" moles = 4.0625 moles"`
The reactions for complete combustion of `C_(2)H_(6)` and `C_(2)H_(4)` are
`2C_(2)H_(6)+7O_(2) rarr 4CO_(2)+6H_(2)O"...(i)"`
` C_(2)H_(4)+3O_(2) rarr 2CO_(2)+2H_(2)O"...(ii)"`
From eqn. (i), no. of moles of `O_(2)` required for complete combustion of x moles of `C_(2)H_(6)=(7)/(2)XX x=3`
From eqn. (ii), no. of moles of `O_(2)` required for complete combustion of `(1.218-x)` moles of `C_(2)H_(4)`
`=3(1.218-x)`
`therefore""3.5x+3(1.218-x)=4.0625`
`"or"0.5x=4.0625-3.654=0.4085"or"x=0.8170" mole"`
Calculation of mole fractions
`"Mole fraction of "C_(2)H_(6)=(.^(n)C_(2)H_(6))/(n_("total"))=(0.817)/(1.218)=0.67" and mole fraction of "C_(2)H_(4)=1-0.67=0.33`
Discussion
88540.

A mixture of ethane, ethene and ethyne is passed through ammoniacal AgNO_3 solution. The gases which contain unreacted are

Answer»

Ethane and ETHENE
Ethane and ETHYNE
Ethene and ethyne
Ethane only

Solution :Ethyne reacts with ammonical `AgNO_3` to GIVE WHITE ppt of silver acetylide while ethane and ethene do not react because acidic HYDROGEN is absent.
Discussion
88541.

A mixture of ethane and ethene occupies 41 L at 1 atm and 500 K. The mixture reacts completely with 10/3 mole of oxygen to produce CO_(2) and H_(2)O. The mole fraction of ethane and ethane in the mixture are (R = 0.0821 L atm K^(-1)" mol"^(-1)) respectively

Answer»

0.50, 0.50
0.75, 0.25
0.67, 0.33
0.25, 0.75

Solution :41 L at 1 atm, 500 K = ? L at 1 atm, 273 K (STP)
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`(1xx41)/(500)=(1xxV)/(273) or V=22.4L=1" mole"`
`C_(2)H_(6)+(7)/(2)O_(2)rarr 2CO_(2)+3H_(2)O`
`C_(2)H_(2)+3O_(2)rarr 2CO_(2)+2H_(2)O`
SUPPOSE `C_(2)H_(6)=x` mole, Then `C_(2)H_(4)=1-x `mole `O_(2)` consumed `=(7X)/(2)+3(1-x)`
`=3+(x)/(2)=(10)/(2)"(given)"`
`"or"x=2((10)/(3)-3)=(2)/(3)=0.67`
Discussion
88542.

A mixture of ethanal and propanal has a vapour pressure of 290 mm at 300 K. The vapour pressure of propanal is 200 mm. If the mole fraction of ethanal is 0.6, its vapour pressure (in mm) at the same temperature will be ……

Answer»

350 MM
300 mm
700 mm
360 mm

ANSWER :A
Discussion
88543.

A mixture of equal parts of any pair of enantiomers is called :

Answer»

diastereomers
mesomers
Racemic mixture
CONFORMATIONAL isomer

ANSWER :C
Discussion
88544.

A mixtureof CS_(2)and H_(2) S onpassingoverheatedcoppergives :

Answer»

methane
ethane
COMPLEX MIXTURE
propane

Answer :A
Discussion
88545.

A mixture of D (+)glucose and D (-) fructose is known as _____.

Answer»

CANE SUGAR
sweetless sugar
INVERT sugar
STARCH sugar

SOLUTION :invert sugar
Discussion
88546.

A mixture of camphor and benzoic acid can be separated by:

Answer»

Sublimation
Chemical methods
Fractional crystallisation
Ectraction with solvent

Answer :B
Discussion
88547.

A mixtureof CO_(2) and CO is passed over red hot graphite when 1 mole of mixture changes to 33.6 L (converted to STP). Hence, mole fraction of CO_(2) in the mixture is

Answer»

0.25
0.33
0.5
0.66

Solution :Suppose `CO_(2)` in the mixture = X MOLE
Then CO in the mixture `=(1-x)` mole
`CO_(2)+Crarr2CO`
Thus, 1 mole of `CO_(2)` gives 2 moles of `CO`
`therefore " x mole of "CO_(2)" will give "CO=2x" mole"`
TOTAL moles of CO after PASSING over RED hot graphite
`=(1-x)+2x=1+x=(1+x)xx22.4L`
Thus, `(1+x)xx22.4=33.6L" (Given)"`
`"or"(1+x)=1.5`
`"or"x=0.5`
Hence, mole fraction of `CO_(2)` in the mixture
`=(0.5)/(1)=0.50`
Discussion
88548.

A mixture of CO and CO_(2) is found to have a density of 1.5 g/L at 30^(@)C and 740 torr. What is the composition of the mixture.

Answer»

`CO = 0.35775, CO_(2) = 0.64225`
`CO = 0.64225, CO_(2) = 0.3575`
`CO = 0.500, CO_(2) = 0.500`
`CO = 0.2500, CO_(2) = 0.7500`

Solution :`because d=(PM)/(RT)implies M=(dRT)/(P)= (1.5 xx 0.0821 xx 303 xx 76)/(74)=38.276`
Let x MOLE of CO and hence mole of `CO_(2) = (1 – x)`,`38.276 = x xx 28 + (1 – x) 44 implies x = 0.35775`. So `1–x = 0.64225`
Discussion
88549.

A mixture of CO and CO_(2) having a volume of 20 mL is mixed with X mL of oxygen and electrically sparked. The volume after explosion is 16 + X mL under the same conditions. What would be the residual volume if 30 mL of the original mixture is treated with aqueous NaOH?

Answer»

12 mL
10 mL
9 mL
8 ml

Solution :`2CO+O_(2) to 2CO_(2)`, VOLUME of `O_(2)` consumed = (20+ X)-(16+ X) = 4 mL
`"Volume of CO in 20 ml"= 2 XX 4 = 8 mL`
`"Volume of CO in 30 mL"= (8)/(20) xx 30 = 12ML` = volume of residual gas (`CO_(2)` is absorbed by NaOH)
Discussion
88550.

A mixture of chloroxylenol and terpinecol acts as

Answer»

ANTISEPTIC
antipyretic
antibiotic
analgesic

Solution :antiseptic
Discussion