Question

At a certain temperature the vapor pressure of pure heptane

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