solutions

benzene and toluene form nearly ideal solutions.at 300K p°toluene=32.06mm and p°benzene=103.01mm.

a liquid mixture is composed of 3 mole of toluene and 2 mole of benzene.if the presurre over mixture is reduced,at what pressure does last trace of liquid disappear.

21 Answers

1
JOHNCENA IS BACK ·

try this one also!!!!!!

1
skygirl ·

YES! I GOT IT NOW!

the concept is dat only which i had explained above...

we need to fing that pressure for which the mole fractions of A and B in vapour becomes 3/5 and 2/5 ...

why--->> that is the mole fraction in liq state... so, if whole liq becomes vapour, then the mole fractions of A and B in the vapour becomes same as that in liq initially..

now, if P is the vapour pressure at that instant ;
mole fraction of A =xA
...................... B =xB

then, 1/P = xA/P0A + xB/P0B ... [the same formula used by mrnobody1]

=> 1/P = 3/5/32.06 + 2/5 /103.01

=> P= 44.183 mm ≈ 44.2 mm

1
skygirl ·

chcek this..

11
Mani Pal Singh ·

YA=p1Xa/p1Xa + p2Xb
and the expression 4 YB is similar

so

after solving we get
YA= 0.308

and

YB= 0.692

1
JOHNCENA IS BACK ·

plz write clearly,what is Ya, Yb,Xa,Xb.send clear replies.

11
Mani Pal Singh ·

YA= mole fraction in the vapour phase
and p1= the po of toluene
and similarly the other 2

1
JOHNCENA IS BACK ·

completely incorrect!!!!!!!!!!!!!!

1
skygirl ·

is the ans 60.44mm ??

1
JOHNCENA IS BACK ·

no,didi.ans. is given as 44.25mm

1
JOHNCENA IS BACK ·

it is correct!!!!!

1
skygirl ·

am missing soomething... this is a real good question...

26
eureka123 ·

i am getting P=41 atm
use Po-Psoln/Po=nsolute/nsolute+nsolvent

26
eureka123 ·

i maybe commiting blunder

1
Sameer Yadav ·

Answer is 44.4mm(approx.)

Here's the solution:

For last drop the mole fraction is 3/5 and 2/5 approximately...but these are the mole fraction in Vapour Phase and not in liquid phase.

A is toluene...and B is benzene

according to formula:

\frac{1}{Ps}=\frac{Ya}{Pa}+\frac{Yb}{Pb}

Put values:

Ya = 3/5

Pa = 32.06 mm

Yb = 2/5

Pb = 103.01 mm

Value of Ps is 44.4(approx.)

Thank You

I Hope This will Help You

1
Kalyan Pilla ·

Partial pressure of solution is

P=paxa+pbxb

wherein pais partial pressure of toulene
& xa is mole fraction of toulene.

pb is partial pressure of benzene
& xb is mole fraction of benzene

This gives vapour pressure as 60.44 mm

It completely turns as liquid when vapour pressure = atmospheric pressure

But in order to find this out we either need the volume of container or the information as to pressure is measured in mm of which substance (eg water, mercury etc.)[2][2]

1
skygirl ·

this is not the way... there is a very nice concept in this question...
which i am able to 'feel' but cudn workout..

see, let the vap pr of soln initially was Po.

now, if we reduce the pressure by ΔP to P' ... then to maintain the initial vap pr some A and some B vaporizes to become vapours such dat the mole fractions of A and B in the vapour is retained.

now, mole fraction of A and B in vapour is obviously their mole fractions in the solution.

1
skygirl ·

@kalyan, to turn the whole liquid out, the vap pressure need not be equal to atm pressure [boiling point]

coz... the whole story is going on in some closed container i think .. orelse .. all these wont occur...

.......................... processing############################

1
skygirl ·

@kalyan, dhakkan lagake pani ubalna... saare pani becomes air or wat ?! .. waise... dekhke ati hun :P

kalyan i saw it jus now :P :P :P
water doesnt disappear[3]

1
JOHNCENA IS BACK ·

everybody nice going....well try.i hope soon we will get the concept as well as correct answer behind the question.

1
Samarth Kashyap ·

derive the expression for total pressuer using mole fraction in the vapour phase(it can be done using daltons law... everybody please try it)

(1/p)=Xa/pa0 + Xb/pb0

substituting for mole fractions 2/5 and 3/5 and p0 values u get
P=44.26mm

1
JOHNCENA IS BACK ·

but mrnobody why to substitue mole fractions 2/5 and 3/5.according to what concept u substituted it.please explain.

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