Question

A mixture was prepared that contained 50.0 g

of CCl4 and 50.0 g of CHCl3. At 50 C, the

vapor pressure of pure CCl4 is 317 torr and

that of CHCl3 is 526 torr. What is the vapor

pressure of this mixture at 50 C?

Answer 1

Data given : Mass of CCL₄ = 50 grams

                  Mass of CHCL₃ = 50 grams

                 Vapor pressure of pure CCL₄= 317 torr

                Vapor pressure of pure CHCl₃ = 526 Torr

Molar mass of CCL₄ =152 grams / mole

Molar mass of CHCl₃ =119.5 grams / mole

Moles of CCL₄ = Weight of CCL₄ /Molar mass of CCl₄ = 50/152 = 0.3289 moles

Moles of CHCL₃ = Weight of CHCl₃/ Molar mass of CHCl₃ = 50/119.5 =0.4184 moles

Let us convert mole into mole fraction

Compound	Moles	Mole fraction (xi)
CCl4	0.3289	0.44
CHCl3	0.4184	0.56
Total	0.7473	1

Now we will use raoult's law to obtain the total pressure ( vapour pressure ) of the mixture

Let CCL₄ =a, CHCl₃ =b

Raoult's law is given by,

y_aP = x_a P^o_a                             ..........(1)

Where , y_a = mole fraction of component 'a' in vapour phase

            P =total pressure of the system

           P^o_a= Vapor pressure of component 'a'

SImilarly we can write it for component (b)

y_bP = x_b P^o_b                            ...........(2)

Also for binary mixture we have ,

y_a + y_b =1                            .................(3)

Adding equation (1) and equation (2) and using using equation (3) , we get

P = x_a P^o_a +x_b P^o_b

We know , x_a = 0.44 , x_b= 0.56 , P^o_a=317 torr , P^o_b = 526 torr

P= (0.44*317)+(0.56*526) = 434.04 Torr

Hence total pressure of the mixture at 50 ^oC is 434.04 Torr