Question

A mixture of two immiscible liquids nitrobenzene and water boiling at ${{99}^{\circ }}C$ has a partial pressure of water 733 mm and of nitrobenzene 27 mm. The ratio of masses of water and nitrobenzene in mixture is:

Answer 1

Hint: The partial pressure of a gas is studied under Henry’s law. It is an important law that depicts volatilization as a function of contaminant concentration and its partial pressure. It relates the partial pressure of gases with their mole fractions and we can easily find the ratio of masses when we know the ratio of mole fraction.

Complete step by step answer:
-Henry’s law states that: at a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
-The constant of proportionality of the law is called the Henry’s law constant and it is a constant at a given temperature only.

-Henry’s law can be written in mathematical form as
   \[pa=kH[c]\]
Where c=molar concentration of the gas
           \[pa\]= partial pressure of the gas
          \[{{k}_{H}}\] =Henry’s law constant
-Ideal gas law is written as PV= nRT where R is ideal gas constant and n is the amount of substance. It is the same for all the gases. There is one more form for the law which is written mathematically as $pV=n{{k}_{B}}{{N}_{A}}T$ where
${{k}_{B}}$ is Boltzmann constant and ${{N}_{A}}$ is Avogadro constant.

-Coming to the question, we can see that the ratio of masses will depend on the mole of the gases. From Henry’s law, we can say that the ratio of moles is equal to the ratio of the partial pressures of the two gases.
\[\Rightarrow \dfrac{{{n}_{1}}}{{{n}_{2}}}=\dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{733}{27}\]

-Now we know that the mole is given by the ratio of the weight to the molecular weight of a compound. The molecular weight of water is 18 and that of nitrobenzene is 123. So we can say that $\begin{align}
 & \frac{733}{27}=\dfrac{{{w}_{W}}}{18}/\dfrac{{{w}_{N}}}{123} \\
 & \Rightarrow \dfrac{{{w}_{W}}}{{{w}_{N}}}=\dfrac{733 X 18}{27 X 123} \\
\end{align}$
Where $\begin{align}
 & {{w}_{W}}=\text{wt}\text{. of water} \\
 & {{w}_{N}}=wt.\text{ }of\text{ nitrobenzene} \\
\end{align}$
Thus the ratio of water to nitrobenzene mass is approximately equal to 4.

Note: Henry’s law is applicable only at moderate temperature and pressure. At higher temperature and pressure, the law fails.
Also, if a gas reacts with the liquid, then Henry’s law is not responsible for the solubility of the gas in liquid. Henry’s law does not hold in those circumstances.
Eg, HCl and $N{{H}_{3}}$ react with water. So Henry’s law does not determine the solubility of these gases in water.