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QUESTION

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?

"140 g"

So, the idea with steam distillation is that the number of moles of each component of the mixture will be proportional to the 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

P_1 - the vapor pressure of pure water at 91""^@"C"; P_2 - the vapor pressure of pure chlorobenzene at 91""^@"C".

This means that if you take n_1 to be the number of moles of water and n_2 to be the number of moles of chlorobenzene in the vapor, you can say that

n_1/n_2 = P_1^0/P_2^0

In your case, you want to know what mass of distillate will contain "100 g" of chlorobenzene. This means that you will have to use the molar masses of water and chlorobenzene to write their masses as a function of the number of moles.

You know that

M_M = m/n implies m = n * M_M

If you miltiply both sides of the equation by M_"M 1"/M_"M 2", you will have

(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 m_2 = "100 g", M_"M 1" = "18.02 g/mol", and M_"M 2" = "112.56 g/mol".

Moreover, you can find the vapor pressure of pure water at 91""^@"C" and "100 kPa" by using

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

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 .