Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

For a first order reaction half life period t(1/2)is independent of initial concentration of its reacting species. What is meant by half life period of a reaction?

Answer»

SOLUTION :The TIME REQUIRED to reducethe concentration of its initial VALUE is called half life period`t_(1/2)`.
2.

For a first order reaction, half life does not depend on.............

Answer»

SOLUTION :INITIAL CONCENTRATION
3.

For a first order reaction, derive expression for the degree of dissociation of the reactant in the exponential form.

Answer»

SOLUTION :`(X)/(a)=1-e^(-KT)`
4.

For a first order reaction -(d[A])/(dt)=K[A]_(0), the reaction is carried out by taking 100mol//L then concentration of A decayed after time (1)/(K) is :

Answer»

`53.21 mol//L`
`36.79 mol//L`
`61.21 mol//L`
`26.79 mol//L`

SOLUTION :`[A]=[A]_(0)E^(-Kt)=100xxe^(-k//k)`
`=(100)/(e )=(100)/(2.718)=36.79mol//L`
`:.[A]` decayed `=16.21`
5.

For a first order reaction AtoB the reaction rate at reactant concentration of 0.01 M is found to be 2.0xx10^(5)Msec^(-1). The half life period of the reaction is [Give your answere divide by 247]

Answer»


SOLUTION :NA
6.

For a first order reaction Ato products 93.75% of A initilly taken (2M), reacts in 80 min, then

Answer»

Rate constant is `3.465xx10^(-2)"min"^(-1)`
units of rate constant are `"min"^(-1)`
INITIAL rate is `6.93xx10^(-2)` mole `"LIT"^(-1)"min"^(-1)`
`t_(1//2)` is independent of initial concentration

SOLUTION :`Kxxt=2.303log((C_(0))/(C_(t))),K=2.303/80xxlog(100/6.25)=2.303/80xxlgo(10/2.5)=2.303/80xxlog(2)`
`=2.303/80xx4xx0.3010=0.03466=3.466xx10^(-2)"min"^(-1),rK.[A]=3.416xx10^(-2)xx2=6.93xx10^(-2)` m/min
7.

For a first order reaction (A)to products the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is

Answer»

`3.47xx10^(-4)`M/in
`3.47xx10^(-5)`M/min
`1.73xx10^(-4)`M/min
`1.73xx10^(-5)`M/min

Solution :`0.1M OVERSET(20)(to)0.050Moverset(20)(to)0.025M,r=K.[A]^(1)=0.693/20xx0.01=0.00346=3.46xx10^(-4)`M/min
8.

For a first order reaction at 27^(@)C the rato of time required for 75% completion of 25% completion of reaction is

Answer»

3
2.303
4.8
0.477

Answer :C
9.

For a first order reaction A(g) to 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t=0) and at time t and P_(0) and P_(t) respectively, Initially, only A is present with concentration [A]_(0) and t_(1//3) is the required for the partial pressure of A to reach 1//3rd of its initial value. The correct option(s) is are (Assume that all these gases behave as ideal gases)

Answer»




Solution :(a,d) `A(g) underset("ORDER")OVERSET("First") to2B(g) + C(g)`
time =0 `P_(0)`

For first order reaction,
`t=1/k In P_(0)/(P_(0)-x)`
`t=1/k In (P_(0))/(P_(0) - (P_(t)-P_(0))/2)`
`=1/k In (2P_(0))/(2P_(0) - P_(t) + P_(0))`
`kt = In (2P_(0))/(3P_(0)-P_(t))`
`= In 2P_(0) - In(3P_(0)-P_(t))`
or `In (3P_(0)-P_(t)) = In 2P_(0)-kt`
Graph between in `(3P_(0)-P_(t)) = In 2P_(0)-kt`
Graph between in `(3P_(0)-P_(t))` Vs t is a straight line with NEGATIVE slope
`therefore` Graph(a) is a correct option.
Since rate constant (k) is independent of initial concentration
`therefore` graph (d) is also a correct option.
10.

For a first order reaction ArarrB the rate constant is x "min"^(-1) . If the initial concentration of A is 0.01 M, the concentration of A after one hour is given by the expression.

Answer»

`0.01e^(-x)`
`1xx10^(-2)(1-E^(-60x))`
`(1xx10^(-2))e^(-60x)`
NONE of these

Solution :`k=((2.303)/(t))log(([A_0])/([A]))`
`k=(1/t)ln(([A_0])/([A]))`
`e^(kt)=(([A_0])/([A]))`
`[A]=[A_0]e^(kt)` In this case
`k = x "min"^(-1)and [A_0]=0.01M`
`=1xx10^(-2)M`
t = 1 HOUR = 60 min
`[A]=1xx10^(-2)(e^(-60x))`
11.

For a first order reaction at 27^@C, the ratio of time required for 75% completion to 25% completion of reaction is

Answer»

`3.0`
`2.303`
`4.8`
`0.477`

ANSWER :C
12.

For a first-order reaction, Aoverset(k)toB, the degree of dissociation is equal to

Answer»

`e^(-kt)`
`1-e^(-kt)`
`e^(kt)`
`(1+e^(-kt))`

SOLUTION :For `underset((1-X))overset(1M)(A)tounderset(x)(B)` INTEGRATE `dx/dt=k(1-x)`
13.

For a first order reaction A(g) to 2 B (g) + C(g) at constant volume and 300 K , the total pressure at the beginning (t = 0 ) and at time t are P_(0) and P_(t), respectively , Initially , only A is present with concentration [A]_(0) and t_(1//3) is the time required for the partial pressure of A to reach 1//3^(rd) of its initial value , the correct option (s) is (are) (Assume that all these gases behave as ideal gases)

Answer»




Solution :`A to 2 B + C`
t = 0 `"" P_(0)`
t = t `"" P_(0) - P "" 2P "" P`
`P_(0) + 2P = P_(t)`
`K = (1)/(t)` LN `(P_(0))/(P_(0) - P) = (1)/(t)` ln `(P_(0))/(P_(0) - ((P_(t) - P_(0))/(2))`
`K = (1)/(t)` ln `(2P_(0))/(3P_(0) - P_(t)) implies - Kt` + ln `2 P_(0)`= ln `(3P_(0) - P_(t))`
and `t_(1//3) = (1)/(K)` ln `(P_(0))/(P_(0)//3) = (1)/(K)` ln 3 = constant
RATE constant does not depends on concentration
14.

For a first order reaction A to P , the temperature (T) dependent rate constant (k) was found to follow the equation log k = -(2000) (1)/(T) + 6.0 . The pre-exponential factor A and the activation energy E_(a) , respectively , are

Answer»

`1.0 xx 10^(6) s^(-1)` and `9.2 K J mol^(-1)`
`6.0 s^(-1)` and `16.6 kJ mol^(-1)`
`1.0 xx 10^(6) s^(-1)` and `16.6 kJ mol^(-1)`
`1.0 xx 10^(6) s^(-1)` and `38.3 kJ mol^(-1)`

Solution :From Arrhenius EQUATION k = `Ae^(-E_(a)//RT)`
LN k = ln A `- (E_(a))/(RT)`
`2.303` log k = `2.303` log `A - (E_(a))/(RT)`
log k `= (-E_(a))/(2.303R) xx (1)/(T) + ` log A `"" … (i)`
log k = `-(2000) (1)/(T) + 6 "" … (ii)`
On comparing equation (i) and (ii)
`(-E_(a))/(2.303R) = -2000`
`E_(a) = 2.303 xx 8.314 xx 2000 = 38.29 kJ` M
and log `A = 6 implies A = 10^(6)`
15.

For a first order reaction : A(g) rarr B(g) + C(g)+ D(g) Occurring at 1 bar pressure and 300 K if initial volume of the container containing only A is V_(0) and after 10 minutes it is V_(10) then average life of A will be (in minutes) :

Answer»

`0.1"LN"(2V_(0))/(3V_(0)-V_(10))`
`(10)/("ln"(2V_(0))/(3V_(0)-V_(10)))`
`"0.1 ln"(V_(0))/(V_(0)-V_(10))`
`(10)/("ln"(V_(0))/(V_(0)-V_(10)))`

Answer :B
16.

For a first order reaction A to P, the temperature (T) dependent rate constsant (k) was found to follow the equation logk=-(2000)(1)/(T)+6.0 The per-exponential factor A and the activation energy E_(a), respectively, are

Answer»

`1.0xx10^(6)s^(-1)" and "9.2" kJ mol"^(-1)`
`6.0" s"^(-1)" and "16.6" kJ mol"^(-1)`
`1.0xx10^(-1)s^(-1)" and "16.6" kJ mol"^(-1)`
`1.0xx10^(6)s^(-1)" and "38.3" kJ mol"^(-1)`

Solution :GIVEN log `K=6-2000/T`
COMPARING with log `k=log A - E_a/(2*303RT)`
`log A=6, i.e., A = 10^(6)s^(-1)" and "E_a/(2*303R)=2000`
or `E_a=2000xx2*303xx8*314" J mol"^-1`
`= 38294" J mol"^-1 = 38*3" kJ mol"^-1`
17.

For a first order reaction (A) toproducts the concentration of A changes from 0.1M to 0.025M in 40 minutes. The rate of reaction of when the concentration of A is 0.01M is:

Answer»

`1.73times10^(-5)M*mi n^(-1)`
`3.47times10^(-4)M*mi n^(-1)`
`3.47times10^(-5)M*mi n^(-1)`
`1.73times10^(-4)M*mi n^(-1)`

SOLUTION :For a FIRST order REACTION `k=(2.303)/(t)"log"(a)/(a-x)=(2.303)/(40)"log"(0.1)/(0.025)=(2.303)/(40)"log"4=(2.303times0.6020)/(40)=3.47times10^(-2) mi n^(-1)`
Rate`=k[A]=3.47times10^(-2)times0.1=3.47times10^(-4)`M/min
18.

For a first order reaction A toproduct the rate of reaction at [A] = 0.2 mol l^(-1) is 1.0 xx 10^(-2) "min"^(-1) . The half life period for the reaction is

Answer»

832 s
440 s
416 s
13.86 s

Solution :r = K [REACTANT] `"" therefore k = (1.0 XX 10^(-2))/(0.2) = 0.05`
`t_(1//2) = (0.693)/(0.05)` = 13.86 MINT = `13.86 xx 60 = 831.6` sec .
19.

For a first order reaction : A to nB (with n-possibly fractional), the concentration of the product varies with time as (A_(0) is the concentration of A at time t=0).

Answer»

`[B]=nA_(0)(1-E^(-2kt))`
`[B]=nA_(0)t`
`[B]=nA_(0)(1-e^(-kt))`
`[B]=nA_(0)(e^(-kt))`

Answer :C
20.

For a first order reaction A to Bthe reaction rate at reactant concentration of 0.01 M is found to be 2.0 xx 10^(-5) mol L^(-1) s^(-1) . The half life period of the reaction is

Answer»

220 s
30 s
300 s
347 s

SOLUTION :R = k[A]
`2 xx 10^(-5) = k xx 10^(-2)`
`k= 2 xx 10^(-3) sec^(-1)`
`t_(1//2) = (0.693)/(k) = (0.693)/(2 xx 10^(-3)) = (693)/(2) = 347` sec .
21.

For a first order reaction A toB the reaction rate at reactant concentration of 0.01 M is found to be2.0xx10^(-5) M sec^(-1) . The half life period of the reaction is [Give your answer divide by 347]:-

Answer»


ANSWER :A
22.

For a first order reaction AtoB the reaction rate at reactant concetration of 0.01 M is found to be 2.0xx10^(-5)mol L^(-1)s^(-1). The half life period of the reactionis

Answer»

`220`s
`30` s
`374` s
`347` s

Answer :D
23.

For a first order reaction A rarr Products, the rate of reaction at [A] = 0.2 M is 10^-2 mol litre^-1 min^-1. The half life period for the reaction is:

Answer»

832 SEC
440 sec
416 sec
14 sec

Answer :A
24.

For a first order reaction A rarr B, A is optically active and B is optically inactive, and following experimental data were observed : |{:("Time","0","60 min",oo),("Optical Rotation",82^(@),22^(@),2^(@)):}| If some impurity, find optical rotation after 2 hours :

Answer»

`7^(@)`
`12^(@)`
`22^(@)`
`42^(@)`

ANSWER :A
25.

For a first order reaction (A) rarr products the concentration of A changes from 0.1 M to 0.025 M in 40 minutes . The rate of reaction when the concentration of A is 0.01 M is :

Answer»

`1.73xx10^(-4) M//MIN`
`1.73xx10^(-5)` M/min
`3.47xx10^(-4)` M/min
`3.47 xx10^(-5)` M/min

Solution :( C) `k= (2.303)/(t) "log" ([A]_(0))/([A])`
`k = (2.303)/(40) "log"(0.1)/(0.025) = 0.03466`
Rate =`k[A] = 0.03466xx0.01`
`=3.466xx10^(-4) M "min"^(-1)`
26.

For a first order reaction A rarr product with initial concentration x " mol L'^(-1) m has a half life period of 2.5 hours. For the same reaction with initial concentration . (x/2) " mol L"^(-1) the half life is ......

Answer»

`(2.5xx2)` hours
`((2.5)/2)` hours
2.5 hours
Without knowing the rate constant , `t_(1//2)` cannot be determined from the given data

SOLUTION :For a FIRST order reaction
`t_(1//2) = (0.693)/K`
`t_(1//2)` does not depend on the initial concentration and it remains constant (WHATEVER may be the initial concentration )
27.

For a first-order reaction

Answer»

the degree of dissociation = `(1-e^(-KT))`
a plot of RECIPROCAL of concentration of the reactant vs time GIVES a straight line
the time taken for the completion of 75% reaction is THRICE the `t_(1//2)` of the reaction
the pre-exponential factor in the Arrhenius's equation has the dimension of time, `T^(-1)`

Answer :A::D
28.

For a first order reaction

Answer»

the degree of DISSOCIATION is equal to `(1-e^(-kt))`
a plot of reciprocal concentration of the reactant vs. TIME GIVES a STRAIGHT line
the time taken for the concentration of `75%` reaction is thrice to `t_(1//2)` of the reaction
the pre-expoential factor in the Arrhenius equation has the dimensions of `"time"^(-1)`

Answer :A::D
29.

For a first order reaction ,

Answer»

The DEGREE of dissociation is equal to `(1 - e^(-kt))`
A PLOT of reciprocal concentration of the reactant v/s time gives a straight line
The time TAKEN for the completion of 75% reaction is thrice the `t_(1//2)` of the reaction
The pre-exponential factor in the Arrhenius EQUATION has the dimension of time , `T^(-1)`.

Solution :In first orderreaction , if `alpha`is the degree of dissociation therefore
Kt = `log_(e)(1)/((1-alpha)) = -log_(e) (1- alpha) ` or `e^(-kt) = 1 - alpha `
`therefore a = 1 - e^(-kt)`
The Arrhenius equation , is `k = AE^(-E_(a)//RT)`
Plot of reciprocal concentration of the reactant v/s time is linear . Dimensions of pre-exponential factor 'A' are equivalent to dimensions of K ,which is `T^(-1)` for a first order reaction .
30.

For a first order process, (-Ea)/(RT)value is -23.03, then value of K/A is

Answer»

`10^(2.303)`
`10^(-10)`
`10^(-23.03)`
`10^(10)`

ANSWER :B
31.

For a first order polymerisation reaction: nA(g) to A_(n) (g) occuring at constant volume and temperature the half-life of polymerisation of 'A' is 20 min. If the total pressure of system is 2atm at t=0 "and" 1.2 "atm at" t= 20 "min". then the value of 'n' is :

Answer»


ANSWER :5
32.

For a first order process A toB, rate constant k_1 = 0.693 "min"^(-1) & another first order process Cto D , k_2 = x "min"^(-1). If 99.9% of C to D requires time same as 50% of reaction A toB , value of x?

Answer»

0.0693
6.93
23.03
13.86

Answer :B
33.

For a first order homogeneous gaseous reaction Ato2B+C, if the total pressure after time t was P_(t) and after a long time (to to oo) was P_(oo) then k in terms of P_(t),P_(oo) and t is

Answer»

`k=2.303/tlog((P_(oo))/(P_(oo)-P_(t)))`
`k=2.303/tlog((2P_(oo))/(P_(oo)-P_(t)))`
`2.303/tlog((2P_(oo))/(3(P_(oo)-P_(YT))))`
None of these

Solution :`Ato2BC+C`
`P_(0)""0""0`
`(P_(0)-x)""2X""x`
`0""2P_(0)""P_(0)`
So `x=((3Pt-P_(alpha))/6),P_(0)=(P_(alpha))/3,K=2.303/txxlog((P_(0))/(P_(0)-x))=2.303/txxlog[(P_(alpha))/3xx6/(3(P_(alpha)-P_(t)))]`
`=2.303/txxlog[(2P_(alpha))/(3(P_(alpha)-P_(t)))]`
34.

For a first order decomposition of N_(2)O_(5)(g) to give NO_(2)(g)and O_(2)(g), what will be the rate constant if at initial instant, after 10 minutes and after a very long time, tolal pressure is 200 mm of Hg, 325 mm of Hg and 450 mm of Hg?

Answer»

0.693min^(-1)`
`6.93min^(-1)`
`6.93xx10^(-2)MIN^(-1)`
`(6.93)/(2)xx10^(-2)min^(-1)`

ANSWER :C
35.

For a first order chemical reaction : A rarr P. The correct statement(s) is/are :

Answer»

The extent of REACTION completed at any time 't' is dependent on initial concentration of the REACTANT.
The reaction must be an elementary reaction.
The time required for`99%` completion of reaction is `[(2t_(1//2))/("log"_(10)2)]`
Concentration of PRODUCT increases linearly with time.

Answer :C
36.

For a f-orbital the values of m_l are

Answer»

`-1,0,+1`
`0,+1,+2,+3`
`-2,-1,0,+1,+2`
`-3,-2,-1,0,+1,+2,+3`

Solution :For f orbital (L=3) the values of m are `-l` to `+l` i.e., -3,-2,-1,0,+1,+2,+3.
37.

For a diprotic acid, which of the following is true for 1^(st) and 2^(nd) ionization constants (K_(a_(1)) and K_(a_(2)))

Answer»

`K_(a_(1)) = K_(a_(2))`
`K_(a_(1)) gt K_(a_(2))`
`K_(a_(2)) gt K_(a_(1))`
`K_(a_(2)) ge K_(a_(1))`

Solution :`K_(a_(1)) gt K_(a_(2))`
The reason for this is that it is more DIFFICULT to remove a positively charged PROTON from a negative ION due to ELECTROSTATIC forces.
For e.g., `H_(2)X_((AQ)) overset(K_(a_(1)))hArr H_((aq))^(+) + HX_((aq))^(-)`
`HX_((aq))^(-) overset(K_(a_(2)))hArrH_((aq))^(+) + X_((aq))^(2-)`
38.

For a disproportionationreaction the onlycorrect combination is -

Answer»

(I) (II)(R)
(II) (ii) (Q)
(IV) (i) (S)
(III) (ii) (Q)

Answer :D
39.

For a dilute solution , Raoult's law states that:

Answer»

The LOWERING of vapour PRESSURE is equle to the MOLE fraction of solute
The relative lowering of vapour pressure is equal to the mole fraction of solute
The relative lowering of vapour pressure is proportional to the amount of solute in solution
The vapour pressure of the solution is equal to the mole fraction of solvent

Answer :B
40.

For a dilute solution, Raoult's law states that:

Answer»

The relative LOWERING of vapour pressure is equal to the mole fraction of solute.
The relative lowering of vapour pressure is equal to the mole fraction of solvent
The relative lowering of vapour pressure is PROPORTIONAL to the AMOUNT of solute in SOLUTION
The vapour pressure of the solution is equal to the mole fraction of solvent.

Answer :A
41.

For a dilute solution , Raoult.s law states that:

Answer»

The LOWERING of vapour pressure is equle to the MOLE fraction of solute
The RELATIVE lowering of vapour pressure is equal to the mole fraction of solute
The relative lowering of vapour pressure is proportional to the AMOUNT of solute in SOLUTION
The vapour pressure of the solution is equal to the mole fraction of solvent

Answer :B
42.

For a dilute solution having molality m of of a given solute in a solvent of mol.wt.M,b.pt. T_(b) and heat of vaporisation per mole DeltaH:[(aT_(b)/(am)]_(mrarr0) is equal to :

Answer»

Molal elevation constant of SOLVENT
`(RT_(b)^(2)M)/(DeltaH_("vap")), where `M` in `kg DeltaH_("vap")` and `R` in `J` mol`^(-1)`
`(RT_(b)^(2)M)/(DeltaS_("vap")), where `M` in `kg, DeltaS_("vap")` and `R` in `J` mol`^(-1)`
`(RT_(B^(2)M))/(1000DeltaH_("vap"))`, where `M` in `g, R` and `DeltaH_("vap")` expressed in same unit of heat.

SOLUTION :`DeltaT_(b)=mK_(b)i ,K_(b)=(RT^(2))/(1000DeltaH_("vapour"))`
43.

For a dilute solution of a strong electrolyte, which of the folllowing facts is correct?

Answer»

The GRAPH between ` lambda_m` and C is LINEAR
The graph between log ` lambda_m` and C is linear
The graph between` lambda_mand SQRT(C)` is linear
The graph between ` lambda_m and sqrt(C)` has a negative slope.

Solution :
44.

For a dilute solution, lowering of vapour pressure prop mole fraction of the solute or, lowering of vapour pressure = K xx mole fraction of the solute, where K is

Answer»

a CONSTANT for the solute
a constant for the solvent
a constant for the solution
vapour PRESSURE of the solvent

Answer :B::D
45.

For a dilute solution having molality m for a given solute in a solvent of mol. Wt. M,b.pt. T_(b) and heat vaporisation per mole DeltaH, [(delT_(b))/(delm)]_(mrarr 0) is equal to

Answer»

Molal elevation constant of solvent
`(RT_(b)^(2)M)/(Delta_(vap)H)`, where M in kg , `Delta_(vap)H` and R is `SI` UNIT
`(RT_(b)^(2)M)/(Delta_(vap)S),` Where M in kg , `Delta_(vap)S` and R in SI unit
`(RT_(b)^(2)M)/(1000Delta_(vap)H),` where M in g, R and `Delta_(vap)H` in `SI` unit

Answer :A::B::D
46.

For a dilute solution containing 2.5 g of a non-volatile, non-electrolytic solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2^(@)C. Assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (take K_(b)="0.76 K kg mol"^(-1))

Answer»

<P>724
740
736
718

Solution :`M_(2)=(1000K_(B)w_(2))/(w_(1)xxDeltaT_(b))=(1000xx0.76xx2.5)/(100xx2)=9.5`
`(p^(@)-p_(s))/(p^(@))=x_(2)=(n_(2))/(n_(1)+n_(2))=(n_(2))/(n_(1))""(" as " n_(2) lt lt n_(1))`
`=(w_(2)//M_(2))/(w_(1)//M_(1))`
`(760-p_(s))/(760)=(2.5//9.5)/(100//18)=0.047`
`or 760-p_(s)=760xx0.047=35.7`
`or""p_(s)=724.3mm`
47.

For a dilute solution containing 2.5 g of non-electroyte solute in 100g of water , the elevation in boiling point at 1 atm pressure is 2^(@)C . Assumingconcentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K_(b)=0.76 "mol"^(-1))

Answer»

724
740
736
718

Answer :A
48.

For a decomposition reaction, the values of rate constant k at two different temmperature are given below: k_(1)=2.15xx10^(-8)"L mol"^(-1)s^(-1) at 650 K k_(2)=2.39xx10^(-7)"L mol"^(-1)s^(-1) at 700 K Calculate the value ofactivation energy for this reaction. [R=8.314JK^(-1)"mol"^(-1)]

Answer»

Solution :Substitute the values in the Arrhenius equation given below:
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`"or log"(2.39xx10^(-7))/(2.15xx10^(-8))=(E_(a))/(2.303xx8.314)[(700-650)/(650xx700)]`
`"ORLOG "11.116=(E_(a))/(19.147)xx(50)/(650xx700) or 1.0461=(E_(a))/(19.147)xx(50)/(455000)`
or`E_(a)=(1.0461xx19.147xx455000)/(50)=182270J or 182.27kJ`
49.

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solution in 100g of water, the elevation in boiling point at 1 atm pressure is 2^(@)C. Assuming concentrationof solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is: (take k_(b) = 0.76 K kg mol^(-1))

Answer»

724
740
736
718

Solution :`DeltaT_(B)=K_(b)xxm`
`2=0.76xxm ""IMPLIES""m=2/0.76`
`m=(w_(2)xx1000)/(M_(2)xxW_(1)("in grams"))=2.5/(M_(2)xx100)xx1000`
`2/0.76=(2.5xx10)/(M_(2))impliesM_(2)=(2.5xx10)/2xx0.76`
`(p_("SOLVENT")^(@)-p_("solutte"))/(p_("solvent")^(@))=x_("solute")`
`p_("solvent")^(@)=1atm=760` MM Hg
`(760-p_("solvent")^(@))/760=(2.5//M_(2))/(100//18)`
`760-p_("solute")=(2.5xx2)/(2.5xx10xx0.76)xx18/100xx760`
`760-p_("solute")=36impliesp_("solute")=724` mm hg
50.

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2^(@)C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K_(b)=0.76"K kg mol"^(-1))

Answer»

<P>724
740
736
718

Solution :`Delta T_(b)=K_(b)xx m`
`2=(0.76xx2.5//M)/(0.1)rArr M=9.5`
Since solute is present in less amount
`(P^(@)-P_(s))/(P^(@))=(n)/(N)`
`(760-P_(s))/(760)=(2.5//9.5)/(100//18)rArr P_(s)=724`