# 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

#### 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 **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

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.