When chlorobenzene is steam distilled at 100kPa,the boiling point of the mixture is 91°C. At this temperature,the vapor pressure of chlorobenzene is 29kPa.What is the mass of the distillate that contains 100g of chlorobenzene?
1 Answer
Explanation:
So, the idea with steam distillation is that the number of moles of each component of the mixture will be proportional to the partial pressure of each component in the total pressure.
The total pressure of the mixture can be calculated by using the vapor pressures of the pure components at the temperature at which the distillation takes place
P_"mixture" = P_1^0 + P_2^0" " , where
This means that if you take
n_1/n_2 = P_1^0/P_2^0
In your case, you want to know what mass of distillate will contain
You know that
M_M = m/n implies m = n * M_M
If you miltiply both sides of the equation by
(n_1 * M_("M 1"))/(n_2 * M_"M 2") = P_1^0/P_2^0 * M_"M 1"/M_"M 2"
which is equivalent to
m_1/m_2 = P_1^0/P_2^0 * M_"M 1"/M_"M 2"
You know that
Moreover, you can find the vapor pressure of pure water at
P_1 = P_"total" - P_2
P_1 = "100 kPa" - "29 kPa" = "71 kPa"
Plug in your values into the equation and solve for
m_1/"100 g" = (71color(red)(cancel(color(black)("kPa"))))/(29color(red)(cancel(color(black)("kPa")))) * (18.02color(red)(cancel(color(black)("g/mol"))))/(112.56color(red)(cancel(color(black)("g/mol"))))
m_1 = 71/29 * 18.02/112.56 * "100 g" = "39.2 g"
This means that the total mass of the mixture will be
m_"total" = m_1 + m_2 = "39.2 g" + "100 g" = color(green)("140 g")
I'll leave the answer rounded to two sig figs.