# 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?

Sep 29, 2015

$\text{140 g}$

#### 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

${P}_{\text{mixture" = P_1^0 + P_2^0" }}$, where

${P}_{1}$ - the vapor pressure of pure water at $91 \text{^@"C}$;
${P}_{2}$ - the vapor pressure of pure chlorobenzene at $91 \text{^@"C}$.

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 $\text{100 g}$ 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

${M}_{M} = \frac{m}{n} \implies m = n \cdot {M}_{M}$

If you miltiply both sides of the equation by ${M}_{\text{M 1"/M_"M 2}}$, 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

${m}_{1} / {m}_{2} = {P}_{1}^{0} / {P}_{2}^{0} \cdot {M}_{\text{M 1"/M_"M 2}}$

You know that ${m}_{2} = \text{100 g}$, ${M}_{\text{M 1" = "18.02 g/mol}}$, and ${M}_{\text{M 2" = "112.56 g/mol}}$.

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

${P}_{1} = {P}_{\text{total}} - {P}_{2}$

${P}_{1} = \text{100 kPa" - "29 kPa" = "71 kPa}$

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"))))

${m}_{1} = \frac{71}{29} \cdot \frac{18.02}{112.56} \cdot \text{100 g" = "39.2 g}$

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.