Question

A solution of carbon tetrachloride and acetic acid CH3COOH is boiling at 87°C. A sample of the vapor above the solution is cooled until it condenses. This condensed sample is analyzed, and turns out to be 90.% 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 87°C
carbon tetrachloride	77.°C	1.6gmL	1011.torr
acetic acid	118.°C	1.1gmL	406.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

Given that In vapour phase Mass % of CCl₄ = 90%

CCl₄ = A and acetic acid = B
let the mass of CCl₄ be m_A and that of acetic acid be m_B in vapour phase
Given: m_A = 0.90 * (m_A+m_B)
m_A = 9*m_B
molar mass of CCl₄ = 154 g/mol
molar mass of acetic acid = 60 g/mol
In vapour phase, as number of moles of CCl₄, n_A = m_A/molar mass of CCl₄ = m_A/154
similarly, number of moles of acetic acid, n_B = m_B/molar mass of acetic acid= m_B/60
or n_B = m_A/540
thus n_A+n_B = m_A (1/154 + 1/540)
mole fraction of CCl₄ in vapour phase, y_A = n_A/(n_A+n_B)
upon substitution we get: y_A = (m_A/154)/(m_A/154 + m_A/540)
y_A = 0.78
Total pressure in vapour phase = pressure exerted by each fraction
P_T = P_A + P_B
From Raoult's Law P_A = x_A. P_A⁰ and P_B = x_B. P_B⁰ where x_A and x_B are the mole fractions of the components in solution
where x_A+x_B = 1
upon substitution we get: P_T = P_B⁰ + (P_A⁰-P_B⁰)x_A
using P_A⁰ = 1011 torr and P_B⁰ = 406 torr
P_T = 406 + (406-1011)x_A = 406 - 605 x_A
and from Daltons's Law P_A = y_A.P_T
thus x_A. P_A⁰ = y_A.P_T
x_A . 1011 = 0.78 *(406 - 605 x_A)
x_A = 0.21 and x_B = 1-x_A = 0.79
using x_A/x_B = m_AM_B/m_BM_A
0.21/0.79 = 60*m_A/154*m_B
m_B/m_A = 1.466
add 1 both sides and then inverse
m_A+m_B/m_A = 2.466
or m_A/(m_A+m_B) *100 = 40.55%
thus in solution mass percent of CCl₄ is 40.5 %