I need help with both parts of this quesiton, do not know where to start. What i

Question

I need help with both parts of this quesiton, do not know where to start.

What is the total pressure? What is the mole fraction of toluene in the above solution?

Q12. For an ideal solution of 2.00 mole of benzene and 1.50 mole of toluene at 298. K, what is the total pressure above the solution? What is the mole fraction of toluene above the solution? P*(toluene)=3.572 kPa and P*(benzene)=9.657 kPa at 298 K. Q12b. For a real solution of CS2-acetone with a mole fraction of CS, of 0.4058 at 25°C, the vapor pressure of CS, is found to be 379.6 Torr and that of acetone is found to be 254.5 Torr. Calculate the activity and activity coefficient for CS2. P*cs2=512.3 Torr and P* acetone is 343.8 Torr. What type of deviation is it from Raoult’s law? What does it tell you about interactions between the CS, and Acetone in the solution?

Explanation / Answer

Q12

if ideal

x-benzene = mol/total mol = 2/(2+1.5) = 0.57

x-toluene = 1-0.57 = 0.43

find total P given T = 298K

apply raoult law

x1*P°1 = y1*Ptotal

x2*P°2 = y2*Ptotal

P°vapor benzene = 9.657

P°vapor toluene =3.572

substitute

x1*P°1 = y1*Ptotal

x2*P°2 = y2*Ptotal

0.57*95.1371 = y1*Ptotal

0.43*3.572 = (1-y1)*Ptotal

now, solve for Ptotal

Ptotal = 54.228/y1

1.53596 = (1-y1)*54.228/y1

1.53596 /54.228 = 1/y1 - 1

0.02832+1 = 1/y1

y1 = (0.02832+1 )^-1 =0.9724

y2 = 1-0.9724 = 0.0276

Ptotal = 54.228/y1 = 54.228/0.9724

Ptotal = 55.767 kPa

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