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89451.

(a) Give plausible explanation for each of the following : (i) The presence of a base is needed in the ammonolysis of alkyl halides. (ii) Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis. (b) Write the IUPAC name of ""CH_(3)-underset(overset(|)(C_(2))H_(5)overset(||)(O))(N-C)-CH_(3)

Answer»

SOLUTION :(i) An alkyl halide reacts with ammonia to FORM primary amine. Primary amine formed reacts further with alkyl halide to form secondary and TERTIARY amines. Presence of a base (B) in this reaction increases the rate of reaction.
`""B +:NH_(3) rarr B-H+NH_(2)^(-)`
`""R-X+NH_(2)^(-) rarr R-NH_(2)+X^(-)`.
(b) `""CH_(3)-UNDERSET(overset(|)(C_(2))H_(5)overset(||)(O))(N-C)-CH_(3)`
Discussion
89452.

(a) Give one example of a paramagnetic substance. (b) Which type of binding force existing in ice?

Answer»

SOLUTION :(a) OXYGEN molecule
(b) HYDROGEN BONDING
Discussion
89453.

(a) Give one example each of the following: (i) Acidic flux(ii) Basic flux (b) What happens when (i) Cu_2 O undergoes self reduction in a silica line converter (ii) Haematite oxidises carbon to carbon monoxide.

Answer»

Solution :(a) (i) ACIDIC flux : `SiO_2 "" (ii) ` Basic flux : `CaO`
(b) (i) `Cu_2 O ` undergoes SELF REDUCTION to form BLISTER copper as :
`2Cu_2 O + Cu_2 S to underset("Blister copper")(6Cu) + SO_2`
(ii) Haematite oxidises carbon to carbon monoxide forming IRON .
`Fe_2 O_3 + 3C to 3CO + 2Fe`
Discussion
89454.

(a) Give one chemical test to distinguish between the following pairs of compounds : (i) Methylamine and dimethylamine (ii) Aniline and benzylamine. (b) Write the structures of different isomers corresponding to the molecular formula C_(3)H_(9)N, which will liberate nitrogen gas on treatment with nitrous acid.

Answer»

Solution :(a) (i) Methylamine and dimethylamine : Methylamine reacts with nitrous acid to form ALCOHOL with the liberation of `N_(2)`. Dimethylamine does not give this test.
`""CH_(3)NH_(2)+HNO_(2) overset(NaNO_(2)+HCl)(rarr) underset("Methanol")(CH_(3)OH)+N_(2)+HCl`
(ii) Aniline and benzylamine : Aniline reacts with nitrous acid at 273-278 K to form benzene DIAZONIUM chloride which reacts further with phenol for form an ORANGE dye. Benzylamine does not give this test.

(b) Structures of isomers :
`""underset("n-Propyl amine")(CH_(3)-CH_(2)CH_(2)NH_(2))""underset("ISOPROPYL amine")(CH_(3)-underset(underset(CH_(3))(|))(CH)-NH_(2))`
Discussion
89455.

(a) Give names of some oxoacids of halogens. (b) Arrange the following in the decreasing order of acidic strength : (i) HClO , HClO_2 , HClO_3 , HClO_4 (ii)HCIO, HBrO, HIO.

Answer»

Solution : (a) Oxoacids of halogens. Fluorine forms only one oxoacid hypofluorous ACID, HFO. The outer halogens form several oxoacids though some of them can be isolated in the form of their salts. The important oxoacids of halogens are GIVEN in Table.

(b) Acidic strength
`(i) HClO_4 gt HClO_3 gt HClO_2 gt HClO(ii) HClO gt HBRO gt HIO`
Discussion
89456.

a. Give mechanism of preparation of ethoy ethane from ethanol. b. How is totuene obtained from phenol?

Answer»

SOLUTION :
Discussion
89457.

a. Give chemical tests to distinguish between the following: (i) Benzoic acid and ethyl benzoate, (ii) Benzaldehyde and acetophenone b. Complete each synthesis by giving missing reagents or products in the following:

Answer»

Solution :a. (i) Benzoic acid and ethyl benzoate:
Sodium bicarbonate test: Warm each compound with `NaHCO_(3)` Benzoic acid GIVES brisk EFFERVESENCE of `CO_(2)` gas whereas ethyl benzoate does not RESPOND to this test.
(Other relevant test can be accepted)
(ii) Benzaldehyde and Acetaphenone:
Iodoform test: Warm each organic compound with `I_(2)` and NaOH solution with acetophenone `(C_(6)H_(5)COCH_(3))` yellow precipitate of iodoform is formed which responds to idoform test, whereas benzaldehyde does not respond to this test.

(ii) `C_(6)H_(5)CH="NN"HCONH_(2)`
(iii) a. `B_(2)H_(6).H_(2)O_(2)//OH` b. PCC
Discussion
89458.

(a) Give chemical test to distinguish between phenol and ethanol in seemingly similar conditions. (b) Write the reaction for what happens when tertiary butyl alcohol is heated with reduced copper at about 573 K.

Answer»

Solution :Add `FeCl_(3)` solution. Phenol will GIVE violet colour WHEREAS ethanol will not.
`underset("tert-Butyl alcohol")(CH_(3)-underset(OH)underset("|")overset(CH_(3))overset("|")"C "-CH_(3))overset(Cu)underset(573K)rarr underset("2-Methylpropene")(CH_(3)-overset(CH_(3))overset("|")"C "=CH_(2)+H_(2)O)`
Discussion
89459.

(a) Give chemical tests to distinguish between the following compounds (One test in each case) : (i) Aniline and ethyl amine. ""(ii) Methyl amine and dimethyl amine. (b) How will you convert aniline to sulphanilic acid ?

Answer»

Solution :(a) (i) Distinction between aniline and ETHYL amine :
Ethyl amine reacts with nitrous ACID to form ALIPHATIC diazonium salt which being unstable liberates nitrogen gas and alcohol. Aniline reacts with nitrous acid at low temperature (273-278 K) to form diazonium salts.
(ii) Distinction between METHYL amine and dimethyl amine :
`underset(underset("(PRIMARY amine)")("Methyl amine"))(CH_(3)NH_(2))""underset(underset("(Secondary amine)")("Dimethyl amine"))(CH_(3)-NH-CH_(3))`
`CH_(3)NH_(2)` reacts with benzene sulphonyl chloride to form N-methylbenzene sulphonamide which is soluble in alkali.
`CH_(3)NHCH_(3)` reacts with benzene sulphonyl chloride to form N, N-Dimethyl sulphonamide which is insoluble in alkali.
This is how the two compounds can be distinguished.
(b) Aniline to sulphanilic acid :
Discussion
89460.

(a) Give any two differences between lyophilic and lyophobic colloids. (b) Write the two steps involved in the mechanism of enzyme catalysed reaction. (c) What is the entropy change (Delta s) for adsorption ?

Answer»

Solution :(a)
(b) Step 1 : Binding of enzyme to substrate to FORM an activated complex
`E + S to ES^(**)`
Step 2 : DECOMPOSITION of the activated complex to form product
`Es^(**) to E + P`
(C) Decreases OR `Delta S = -ve OR DeltaS < O OR Delta S` is negative.
Discussion
89461.

(a) Give any three differences between Physisorption and chemisorption (b) (i) Mention the role of alum in the purification of drinking water. (ii) Give an example of an oil dispersed in water emulsion.

Answer»

Solution :(b)(i) It COAGULATES SUSPENDED impurities or It ACTS as a coagulant.
(II) Milk or VANISHING cream.
Discussion
89462.

Write any two differences between physisorption and chemisorption.

Answer»

SOLUTION :
Discussion
89463.

(a) Give any three differences between physical adsorption and chemical adsorption. (b) What is (i) Tyndall effect (ii) Peptisation ?

Answer»

SOLUTION :
(b) (i) Scattering of LIGHT by COLLOIDAL PARTICLES.
(ii) Conversion of freshly prepared precipitate into a colloid by adding a suitable electrolyte.
Discussion
89464.

(a) Give an example for non-narcotic analgesic. (b) Why the use of Aspartame is limited to cold foods and soft drinks ?

Answer»

SOLUTION :(a) Paracetamol or aspirin
(B) It is unstable at cooking temperature.
(a) Non-narcotic analgesis are medications used to CONTROL pain and inflammation. These are not habit FORMING. The EXAMPLES are :
(i) Analgin :

(ii) Diclofenac sodium :

(b) Aspartame is about 150 times sweeten than sucrose but it is unstable at cooking temperature, therefore it is limited to use in cold foods and soft-drinks.
Discussion
89465.

(A) Give a blue solution in H_(2)O. On passing H_(2)S, a black precipitate (B) is formed which is soluble in HNO_(3). On addition of NaOH, the solution gives glue precipitate (C ) which becomes black on boiling in NaOH. On passing ammonia into solution of (A) in water, a deep blue precipitate is ofrmed, which dissolves in excess of NH_(3) giving deep blue colouration (D). Treatment of KCN with aqueous solution of (A) gives a yellow ppt. (E) which dissloves in excess of KCN giving a colourless solution.

Answer»

SOLUTION :
Discussion
89466.

A : Generally, n-hexane and onwards can be sulphonated but isobutane and isopentane can also be sulphonated. R : Isobutane and isopentane can produce tertiary free radical.

Answer»

If both ASSERTION & REASON are true and the reason is the correct explanation of the assertion, then MARK (1).
If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2)
if Assertion is true STATEMENT but Reason is false, then mark (3)
If both Assertion and Reason are false statements, then mark (4)

Answer :1
Discussion
89467.

Find out the correct statement from the following: (a) General formula of silicones R_(2)CO Si (b) Silicones are high - temperature polymers ( c ) Silicones are used for water proofing clothes (d) They are good thermal and electrical conductors.

Answer»


ANSWER :a & d
Discussion
89468.

A gene is a segment ofmolecule of ……….. .

Answer»

DNA
m-RNA
t-RNA
protein

ANSWER :C
Discussion
89469.

A gene is a segment of a molecule of

Answer»

DNA
m-RNA
t-RNA
Protein

Answer :A
Discussion
89470.

A gelatin sol at pH less then the isoelectric value is subjected to an electric field. The sol particles migrate toward

Answer»

ANODE
Cathode
Both anode and cathode
Neither anode nor cathode

Solution :At low pH the basic GROUP will be ionized ( protonated) so willhave positive CHARGE and hence sol particles will MOVE towards cathode.
Discussion
89471.

A Geiger Muller counter is used to study the radioactive process. In the absence of radioactive substances A, it counts 3 disintegration per second (dps). At the start in the presence of A, it records 23 dps, and after 10 m in 13 dps, i. What does it count after 20 m in ?

Answer»

8 dps, 10 min
5 dps, 10 min
5 dps, 20 min
5 dps, 5 min

Solution :In the absence of A, 3 dps is ZERO error, hence
INITIAL count=23- 3 = 20 dps
After 10 min=13-3=10 dps
After 20 min = 5 dps (RECORDED =5+3 = 8 dps)
(50% fall in 10 min, `T_(50)` = 10 min)
Discussion
89472.

(A) : Gases between themselves cannot form a colloidal solution. (R): Gases give homogenous mixture

Answer»

Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A)
Both (A)and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is false
Both (A) and (R ) are false

Answer :A
Discussion
89473.

A gaseous system during a thermodynamic process does not undertake any volume change. It is called

Answer»

ISOCHORIC PROCESS
ISOBARIC process
ISOTHERMAL process
Isoentropic process

ANSWER :A
Discussion
89474.

A gaseous system change form state A(P_1, V_1, T_1) to B(P_2, V_2, T_2), B to C(P_3, V_3, T_3) and finally from C to A. The whole procoess may be called:

Answer»

REVERSIBLE process
Cyclic process
Isobaric process
Spontaneous process

Answer :B
Discussion
89475.

A gaseous substance (AB_(3)) decomposes according to the overall equation :AB_(3)rarr (1)/(2) A_(2)+(3)/(2) B_(2) The variation of the partial pressure of AB_(3) with time (starting with pure AB_(3)) is given below at 200^(@)C : {:(Time//h,0,5.0,15.0,35.0),(P_(AB_(3))//mmHg,660,330,165,82.5):} The order of the reqaction is

Answer»

zero
three
one
two

Solution :From the given data, we can observe that when we start our EXPERIMENT `(t=0),P_(AB)` , is `660mmHg` . At the end of `5` hours, it is reduced to HALF `(330mmHg)` . Therefore, half life of reaction is `5` hours. Now, when we start next with `P_(AB)` , equal to `330mmHg` , it BECOMES again half `(165mmHg)` but at the end of `15` hours. Thus, half life is `10HRS` .
We know
`t_(1//2)prop(1)/(a^(n-1))`
Thus `((t_(1//2))_(t))/((t_(1//2))_(2)`
It is a second order reaction.
Alternatively, note that when we start our experiment with `660mmHg` , the half life id=s `5` hours bvut when we start our experiment with `330mmHg` , the half life is `10` hours, i.e., when pressure is halved, the half life is doubled. THis IMPLIES that half life is inversely proportional to initial pressure (or concentration). This is the characteriic of a second order reaction only.
Discussion
89476.

A gaseous substance ABhas a radioactive element A with a half life of 2 days. The gaseous substance decomposes to A_((g)) and B_((g)) in a closed container and an equilibrium gets established rapidly withina fraction of second . The substance to which A decays into is non gaseous and does not effect volume of the container. If equilibrium constant is 9 atm then calculate partial pressure of AB after 4 days if initial pressure of AB taken was 4 atm.

Answer»

`(1)/(4)` atm
`(1)/(2)` atm
1atm
2 atm

Solution :`{:(AB(g),hArr,A(g),+,B(g),K_(P)=9 "atm",),(4 "atm",,0,,0,,),(4 "atm"-x,,x,,x,,),(,,K_(P)=(x xxx)/(4-x)=9,,,,):}`
x=3, -12 x can't be -12
Partial pressure of ABat equilibrium=1 atm
Now after 4 day (TWO half LIFE)the pressure of AB willbecame =`(1)/(4) atm`]
Discussion
89477.

A gaseous substance ABhas a radioactive element A with a half life of 2 days. The gaseous substance decomposes to A_((g)) and B_((g)) in a closed container and an equilibrium gets established rapidly withina fraction of second . The substance to which A decays into is non gaseous and does not effect volume of the container. If at equilibrium total pressure is 4 atm and molar ratio of AB : Bis 2 :1then equilibrium constant will be

Answer»

`(1)/(2) ` PA
`(1)/(2) xx 101325""`Pa
`4 "atm"^(2)`
4 atm

Solution :`{:(AB(g),HARR,A(g),+,B(g),"fractopm of second",),("a mole",,,,,,),(a-X,,x,,x,,),("2 moles",,1 "mole",1 "mole","Total pressure 4 atm",,),(,"mole RATIO"=AB : B,,,,,),(,2:1,,,,,),("Parital",2 "atm",1"atm",1 "atm",,,),(,k_(P)=(1)/(2)"atm",,,,,),(,k_(P)=(1)/(2)xx1.01325 xx 10^(5) "pascal",,,,,):}`
Discussion
89478.

A gaseous sample is generally allowed to do only expansion//compression type work against its surroundings The work done in case of an irreversible expansion ( in the intermediate stages of expansion//compression the states of gases are not defined). The work done can be calculated using dw= -P_(ext)dV while in case of reversible process the work done can be calculated using dw= -PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process. Since P=(nRT)/(V), so w=intdW= - underset(v_(i))overset(v_(f))int(nRT)/(V).dV= -nRT ln(V_(f)/(V_(i))) Since dw= PdV so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. In the above problem

Answer»

work done by the gas in `I^(ST)` SAMPLE `gt` work done by gas in `II^(nd)` sample
work done by gas in `II^(nd)` sample `gt` work done by gas in `I^(st)` sample
work done by gas in `I^(st)` sample`=` work done by gas in `II^(nd)` sample
None of these

Solution :Work done in isothermal PROCESS is more than in ADIABATIC process as shown in the DIAGRAM above.
Discussion
89479.

A gaseous substance A undergoes first order dissociation to give B ,Cand D as shown . A(g) to 2B (g) + C(g) + D(g) If molar masses of A,B,C and D are 450 , 100 , 50 and 200 respectively and rate constant of disappearance of A "is" 0.693 xx 10 ^(-3) sec^(-1), then calculate ratio of rate of effusion initially and after 2000 seconds. [Multiply the answer by 26sqrt(52) and fill the value in the OMR.]

Answer»


ANSWER :32
Discussion
89480.

A gaseous reaction was carried out first keeping the volume constant and then keeping the pressure constant. In the second experiment, there was an increase in volume. The heats of reaction were different because

Answer»

In the first CASE ENERGY was SPENT to keep the volume constant
In the second case energy was spent to expand the gases
Specific heats of COMPRESSED gases is more
Specific heats of rarefied gases is more

Answer :B
Discussion
89481.

A gaseous reaction ( hypothetical ) takes place with the following kinetics : AB( g) + B_(2) (g)rarr AB_(3) ( g) AB_(3)(g) + AB(g) overset("slow")(rarr) 2AB_(2)(g) During the kinetic studies it was found that {:(P_(AB)^(@),P_(a_(2))^(@),"Rate"),("(atm)","(atom)","atm" mi n^(-1)),(1,0.5,4.167 xx 10^(-2)),(0.5, 0.5,x):}

Answer»

x is 1.04117
x is `4.167 xx 10^(-2)`
Rate `= K [ AB_(3) ] [AB]` as PER rate LAW
Rate `= k [AB]^(2) [ B_(2) ]` as per rate law

Answer :A,D
Discussion
89482.

A gaseous reaction A(g) rarr 2B (g) +C(g) is found to be of first order . If the reaction is started with p_(A) =90 mm Hg the total pressure after 10 min is found to be 180 min Hg. The rate constant of the reaction is :

Answer»

`4.60xx10^(-3) s^(-1)`
`6.9xx10^(-3) s^(-1)`
`1.15xx10^(-3) s^(-1)`
`4.90 xx10^(-3) s^(-1)`

SOLUTION :( C) ` A(g)rarr 2B (g) + C(g) `
After 10 MIN , `p_(0)-p 2p p `
Pressure after 10 min `P_(0)-p+2p+p=p_(0) +2p`
180 mm HG =90 mm Hg +2P
or 2p= 90 mm mg
`:. P` = 45 mm Hg
Now `K = (2.303)/(t) "log" (p_(0))/(P)`
`=(2.303)/(10 "min") "log" (90)/(45)`
`= (2.303)/(10xx60s)xx0.303`
`= 1.155xx10^(-3) s^(-1)`
Discussion
89483.

A gaseous phase reaction A_2 toB +0.5Cshows an increase in pressure from 100 mm to 120 mm in5 min. Now,- (Delta [A_2])/(Delta t) should be

Answer»

`8mm - "MIN"^(-1)`
`4MM - "min"^(-1)`
`16mm - "min"^(-1)`
`2MM - "min"^(-1)`

Answer :A
Discussion
89484.

A gaseous mixture was prepared by taking equal mole of CO and N_(2) . If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen ( N_(2)) in the mixture is :

Answer»

0.5 atm
0.8 atm
0.9 atm
1 atm

Solution :`N(CO)= n ( N_(2))`
`p_(CO) + p_(N_(2)) = 1 atm `
MOLE fraction of `N_(2l) . x_(N_(2)) = ( 1)/( 1+1) = 0.5 `
`p(N_(2)) = p xx ""^(x)N_(2)= 1 xx 0.5 = 0.5 atm`
Discussion
89485.

A gaseous mixture of propane and butane of volume 3L on complete combustion produces 10L of CO_(2) under standard conditions of temperature and pressure. The ratio of volumes of propane to butane is:

Answer»


ANSWER :B
Discussion
89486.

A gaseous mixture of propane, acetylene and CO_(2) is burnt in excess of air. Total 4800 KJ heat is evolved . The volume of CO_(2) formed during combustion is 224 liters at STP. The total evolved heat is used to perform two separate process: (i) Vapourising 87.5% of water (liquid) obtained in the process of burning the original mixture. (ii) Forming 3808 litres ethylenemeasured at STP from its elements. DeltaH_(H-H) = 435 KJ//mol "" DeltaH_(C-H) = 416 KJ//mol "" DeltaH_(C-C) = 347 KJ//mol DeltaH_(C-C) = 615 KJ//mol, "" DeltaH_(C-C) = 812 KJ//mol"" DeltaH_("sublimation") "of " (C,s) = 718 KJ//mol DeltaH_(f)^(@)(C_(2)g) = -394 KJ//mol""DeltaH_(f)^(@)(H_(2)O, l) = -286 KJ//mol. "" DeltaH_(f)^(@)(H_(2)O,g) = -247 KJ//mol. Sum of enthalpies of combustion of C_(3)H_(8)(g) and C_(2)H_(2)g) is :

Answer»

`-2198` KJ/mol
`-3499` KJ/mol
`-2798` KJ/mol
`-3099` KJ/mol

Solution :N//A
Discussion
89487.

A gaseous mixture of propane, acetylene and CO_(2) is burnt in excess of air. Total 4800 KJ heat is evolved . The volume of CO_(2) formed during combustion is 224 liters at STP. The total evolved heat is used to perform two separate process: (i) Vapourising 87.5% of water (liquid) obtained in the process of burning the original mixture. (ii) Forming 3808 litres ethylenemeasured at STP from its elements. DeltaH_(H-H) = 435 KJ//mol "" DeltaH_(C-H) = 416 KJ//mol "" DeltaH_(C-C) = 347 KJ//mol DeltaH_(C-C) = 615 KJ//mol, "" DeltaH_(C-C) = 812 KJ//mol"" DeltaH_("sublimation") "of " (C,s) = 718 KJ//mol DeltaH_(f)^(@)(C_(2)g) = -394 KJ//mol""DeltaH_(f)^(@)(H_(2)O, l) = -286 KJ//mol. "" DeltaH_(f)^(@)(H_(2)O,g) = -247 KJ//mol.Total moles of hydrocarbon gases taken in the initial mixture

Answer»

`3`
`4`
`2`
`5`

Solution :Let moles of `C_(3)H_(8) = X , "" "moles of" C_(2)H_(2) = y`
& moles of `CO_(2) =Z.`
Calculation of `DeltaH_(f)^(@) "of" C_(3)H_(8) (g)`
`""3C(s) + 4 H_(2) (g) rarr C_(3)H_(8)(g)`
`DeltaH_(f)^(@) "of" C_(3)H_(8)(g) =[3(718)+ 4(435)] - [2(347) + 8(416)]`
`= 3894- 4022 =-128KJ//mol`
Calculation of `DeltaH_(f)^(@) "of" C_(2)H_(2) (g)`
`""2C(s) + H_(2) (g) rarr C_(2)H_(2)(g)`
` DeltaH_(f)^(@) C_(2)H_(2)(g)=[2(718) + (435)] - [(812) + 2(416)]`
`=(1436 + 435) - [1644]`
`=227 KJ//mol`.
Calculation of `DeltaH_(f)^(@) "of" C_(2)H_(4) (g)`
`2C(s) + 2H_(2) (g) rarr C_(2)H_(4)(g)`
` DeltaH_(f)^(@)"of" C_(2)H_(4)(g) = [2(718) + 2(435)] - [615 + 4(416)]`
`= 2306- 2279 = 27 KJ//mol`.
Calculation of `DeltaH_("Comb")^(@) "of" C_(3)H_(8)(g)`
`""C_(3)H_(8)(g) + O_(2)(g) rarr 3CO_(2)(g) + 4H_(2)O(l)`
`"" DeltaH_("Comb")^(@) "of" C_(3)H_(8) = [3 DeltaH_(f^(@))(CO_(2)) + 4Delta_(F^(@)) (H_(2)O,l)] - DeltaH_(F^(@)) (C_(3)H_(8),g)`
`"" = [3(-394)+ 4(-286)] - (-128) =-2198 KJ//mol`.
Calculation of `DeltaH_("Comb")^(@) "of" 2C_(2)H_(2)(g)`
`C_(2)H_(2)(g) + 2.50_(2)(g) rarr 2CO_(2)(g) + H_(2)O(l)`
`DeltaH_("Comb")^(@) " of " C_(2)H_(2) = [2DeltaH_(f^(@)) (CO_(2)) + DeltaH_(F^(@))(H_(2)O, l)] - Delta_(F^(@))(C_(2)H_(2))`
`=[2(-394)+ (-286)] - 227`
`=-1301 KJ//mol.`
Now total heat released during combustion
`"" 2198 x + 1301 y = 4800""(i)`
`""` Combustion `=3x + 2y + z =10 ""(ii)`
Total moles of `H_(2)O(l) "FORMED" = 4x + y.`
moles of `C_(2)H_(4)(g)` to be prepared `=(3808)/(22.4) = 170.`
Total heat absorbed during evaporation of water formation of `170` moles `C_(2)H_(4)= 4800.`
`"" [(4x + y ) xx 0.875 xx 40] + (170 xx 27) = 4800`
`"" 4x + y =6"(iii)`
on solvent `(i), (ii)` and `(iii)`
we get `x=1, y=2 "and" z=3`
Discussion
89488.

A gaseous mixture of propane, acetylene and CO_(2) is burnt in excess of air. Total 4800 KJ heat is evolved . The volume of CO_(2) formed during combustion is 224 liters at STP. The total evolved heat is used to perform two separate process: (i) Vapourising 87.5% of water (liquid) obtained in the process of burning the original mixture. (ii) Forming 3808 litres ethylenemeasured at STP from its elements. DeltaH_(H-H) = 435 KJ//mol "" DeltaH_(C-H) = 416 KJ//mol "" DeltaH_(C-C) = 347 KJ//mol DeltaH_(C-C) = 615 KJ//mol, "" DeltaH_(C-C) = 812 KJ//mol"" DeltaH_("sublimation") "of " (C,s) = 718 KJ//mol DeltaH_(f)^(@)(C_(2)g) = -394 KJ//mol""DeltaH_(f)^(@)(H_(2)O, l) = -286 KJ//mol. "" DeltaH_(f)^(@)(H_(2)O,g) = -247 KJ//mol. DeltaH_(rxxn)^(@) "for" C_(2)H_(2)(g) + H_(2)g rarr C_(2)H_(4)(g)

Answer»

`-220 kJ//mol`
`-180 KJ//mol`
`-250 KJ//mol`
`-200 KJ//mol`

SOLUTION :N//A
Discussion
89489.

A gaseous mixture of methane and heptane is 8.25 ratio (by weight) respectively is allowed to effuse through a pin ole in the flask. How many methane molecules would come out by the time when the first molecule of heptane is out?

Answer»


ANSWER :5
Discussion
89490.

A gaseous mixture of O_(2) and X containing 20% (mole %) of X, diffused through a small hole in 234 seconds while pure O_(2) takes 224 seconds to diffuse through the same hole. Find the molecular weight of X.

Answer»

SOLUTION :We have ,
`(t_("mix"))/(t_(O_(2))) = sqrt((M_("mix"))/(M_(O_(2))))`
`(234)/(224) = sqrt((M_("mix"))/(32))`,
`therefore M_("mix") = 34.921`
As the mixture contains 20% (mole %) of X, the molar ration of `O_(2)` and X may be represented as 0.8 N : 0.2n, n being the total no. of moles.
`therefore M_("mix") = (32 XX 0.8 n + M_(x) xx 0.2 n)/(n) = 34.921`
`therefore M_(x)` (mol. wt. of X) = 46.6.
Discussion
89491.

A gaseous mixture of ethanen, ethane and propane having total volume 200ml is subjected to combustion in excess of oxygen. Percentage of propane in original mixture is 10% then calculate volume of CO_(2)(g) obtained at same temperature and pressure.

Answer»

360 ml
390ml
420ml
can't be determined

Solution :`C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O`
`C_(2)H_(6)+(7)/(2)O_(2)rarr2CO_(2)+3H_(2)O`
`C_(3)H_(8)+5O_(2)rarr3CO_(2)+4H_(2)O`
Total volume of mixture 200ML
Volume of `C_(3)H_(8)=10%` of 200ml=20ml
so `CO_(2)` PRODUCED from propane `20xx3=60ml`
Rest of the gases =180 ml
and both produce double volume of `CO_(2)`
so `CO_(2)` produced `=180xx2=360 ml`
Total `CO_(2)(G)=360+60=420ml`
Discussion
89492.

A gaseous mixture of 3L of propoane and butane on complete combustion at 25^@C produced 10 L CO_2 . Find out composition of the gaseous mixture.

Answer»


ANSWER :2L, 1L
Discussion
89493.

A gaseous mixture of 2 moles of A, 3 moles of B, 5 moles of C and 10 moles of D is contained in a vessel. Assuming that gases are ideal and the partial pressure of C is1.5 atm, total pressure is

Answer»

3 atm
6 atm
9 atm
15 atm

Answer :B
Discussion
89494.

A gaseous mixture of CO and CO_(2) having total volume 150ml with excess of red hot charcoal to causefollowingreaction: CO_(2)(g)+C(s)rarr2CO(g) The volume increases to 250ml. Identify correct statement(s)

Answer»

ORIGINAL mixture contain `(100)/(3)%` of CO.
Original mixture contain 150 ml of `CO_(2)`.
Original mixture contain 100ml of `CO_(2)`.
Original mixture contain 50ml of CO,.

Solution :`CO_(2)+C(s)rarr2CO(g)`
Vml EXCESS
`0ml` remain 2V ml
`DeltaV=2V-V=V ml=250-150=100`
`V_CO_(2))=100ml,V_(CO)=50ml`
`%"of" CO=(V_(CO))/(V_("total"))XX100=(50)/(150)xx100`
`=(100)/(3)%`
Discussion
89495.

A gaseous mixture contains oxygen and nitrogen in the ration of 1:4 by weight therefore, the ratio of their number of molecules is :

Answer»

1:4
1:8
7:32
3:16

Answer :C
Discussion
89496.

A gaseous mixture contains oxygen and nitrogen in the ratio of 1:4 by weight. Therefore, the ratio of their number of molecules is

Answer»

`1:4`
`1:8`
`7:32`
`3:16`

SOLUTION :MOLAR RATIO of `O_2: N_2 = 1/32 : 4/28 = 7 : 32 `
Discussion
89497.

A gaseous mixture contains oxygen and nitrogen in the ratio of 1:4 by mass. Therefore , the ratio of their number of molecules is:

Answer»

`1:4`
`1:8`
`7:32`
`3:16`

ANSWER :C
Discussion
89498.

A gaseous mixture contains H_(2) and N_(2) in the ratio of 1:4 by weight . The ratio of their molecules is :

Answer»

`7:2`
`1:8`
`2:7`
`1:4`

ANSWER :A
Discussion
89499.

A gaseous mixture contains CO_(2) (g) and N_(2)O (g) in a 2 : 5 ratio by mass. The ratio of the number of molecules of CO_(2) (g) and N_(2)O is

Answer»

`5:2`
`2:5`
`1:2`
`5:4`

ANSWER :B
Discussion
89500.

A gaseous mixture contains 56 g of N_(2), 44 g of CO_(2) and 16 g of CH_(4) The total pressure of the mixture is 720 mm Hg. The partial pressure of CH_(4) is:

Answer»

180 mm
360 mm
540 mm
720 mm

Answer :A
Discussion