Question

Benzene and toluene from two ideal solution A and B at 313 K. Solution A (total pressure PA​) contains equal mole of toluene and benzene. Solution B contains equal masses of both A (total pressure PB​). The vapour pressure of benzene and toluene are 160 and 60 mm Hg respectively at  313 K. Calculate the value of PA/PB​

Answer 1

Answer:

PA/PB=0.964

Explanation:

Given Benzene and Toluene form two ideal solution A and B at 313 k. Solution  A ( total pressure PA) contains equal mole of Toluene and benzene and solution B contains equal masses of both A( total pressure PB). And it is also given that vapor pressure of benzene and toluene are 160 and 60 mm Hg at 313 k.

To Calculate--- the ratio of PA and PB i.e. PA/PB

As per the Raoult's law the vapor pressure of a component is equal to the mole fraction of that component times the vapor pressure of that component in a pure state at a given temperature.

As we can see that the no. of moles of Toluene and benzene are same so the mole fraction will also same i.e 0.5, we can calculate it by the following method-

Let the mole fraction of toluene and benzene is Xt And Xb respectively

So, Xt = nt/nt +nb ,( here nt and nb are the moles of toluene and benzene)

Xt = 1/1+1 => 1/2                                             Xb= nb/nt+nb

Xt= 0.5                                                             Xb= 1/1+1 => 1/2 =>0.5      

The partial pressure of the Toluene is :

     Pt = Ptnod *Xt    ==> 60 mmHg * 0.5 ==>30 mm Hg ( Ptnod is the vapor pressure of pure toluene

The partial pressure of the Benzene is :

  Pb = Pbnod * Xb ==> 160 mmHg * 0.5 ==> 80 mmHg ( Pbnod is the vapor pressure of pure benzene)

So the total pressure in the case of solution A is ;

     PA= Pt + Pb  ==> 30 mmHg + 80 mmHg

  PA = 110 mmHg

Now for the solution B , the masses of toluene and benzene are same where as the molar masses of the Toluene and Benzene are 92.14 g/mol and 78.11 g/mole respectively .

Let us suppose that 78.11 g of Benzene And also & 78.11 g of Toluene are present in the solution mixture . And if we calculate it we will get 1 mole of benzene and 0.847 mole of toluene. we can calculate as follows:

As n = w/m so nb = 78.11/78.11 ==> 1 mole of benzene

     And nt = 78.11/92.14 ==>  0.847 mole of toluene

Now we have to calculate the mole fraction in this case as like earlier

  And if we do the calculation we get Xb= 0.541 & Xt= 0.458

The Partial pressure of benzene in this case :

Pb= Pbnod * Xb ==> 160 mmHg * 0.541

Pb= 86.56 mmHg

And The Partial pressure of Toluene in this case :

Pt= Ptnod * Xt ==> 60 mmHg *0.458

Pt=27.48 mmHg

Hence the Total Partial Pressure in this case becomes

PB = Pt+Pb ==> 27.48 mmHg + 86.56 mmHg

PB = 114.04 mmHg

Thus , the value of the ratio

PA/PB= 110 mmHg/ 114.04 mmHg

==> PA/PB=0.964