Question

At a certain temperature the vapor pressure of pure heptane

C7H16

is measured to be

454.mmHg

. Suppose a solution is prepared by mixing

102.g

of heptane and

135.g

of chloroform

CHCl3

.

Calculate the partial pressure of heptane vapor above this solution. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

215 mm Hg

Explanation:

This is a problem we can solve by utilizing Raoult´s law which states that the partial pressure of component A, P (A) in a solution is given by:

P (A) = X (A ) Pº(A)

where A is the mol fraction of component A, and Pº (A) is the pressure of pure component A.

Now here we have the masses so we can calculate the number of moles  and the mol fraction of component A

n = number of moles = mass / MW

n( C₇H₁₆ ) =  102 g / 100.21 g/mol  = 1.02 mol

n( CHCl₃ ) = 135 g / 119.38 g/mol ) = 1.13 mol

The mol fraction of C₇H₁₆ is given by:

X (C₇H₁₆) =  n( C₇H₁₆ )/ [n( C₇H₁₆ ) + n( CHCl₃ )]

                = 1.02 / ( 1.02 + 1.13 ) = 0.474

Then

P(C₇H₁₆) = X (C₇H₁₆) x (C₇H₁₆)

             = 0.474 x 454 mm Hg = 215 mm Hg