Question

A solution of carbon tetrachloride and acetic acid CH3COOH is boiling at 107.6°C. A sample of the vapor above the solution is cooled until it condenses. This condensed sample is analyzed, and turns out to be 50.% carbon tetrachloride by mass.

Calculate the percent by mass of carbon tetrachloride in the boiling solution. Here's some data you may need:

	normal boiling point	density	vapor pressure at 107.6°C
carbon tetrachloride	77.°C	1.6gmL	1733.torr
acetic acid	118.°C	1.1gmL	623.torr

Be sure your answer has the correct number of significant digits.

Note for advanced students: you may assume the solution and vapor above it are ideal.

Answer 1

= 50 nicht x 100 mccent m chacool - I ----- mean = mchacool - ncla * necer x 153.82 = nchacoon 460 nccea = 0.39 nchacoon - ncat YL = - = 0.28 Mccent chacoon It-39 80 Y2 = 0.72 Y Y → mole fraction → mole fraction of of cele in CH₂ coon condensed one in condensed one .

8. x, ev - о-20 4 6 3 3 o-139 0.22 233 → x xo mole mole fraction of cale in Boiling soln fraction of Chacook in Boiling Soon са - о: 139 тисоси Гm . ) 5383) - о. 139 ( си свои во )

Micle = О• 139 x 153-93 телесоод - о• 3563 во ./ mass of cele in Boiling solution = mecet x 100 Mcces & Mckgcoon 1 X (оо 1 + 1 " . 3563 26 • 26 :(.