Question

Two-component solid–liquid system of naphthalene–benzene forms a simple eutectic mixture. Assuming an ideal solution, the mole fraction of naphthalene in benzene at \(300~\text{K}\) and \(1~\text{bar}\) is _____ (rounded off to two decimal places).
(Given: freezing point \(T_{\mathrm{fp}}\) and enthalpy of fusion \(\Delta H_{\mathrm{fus}}\) of naphthalene are \(353~\text{K}\) and \(19.28~\text{kJ mol}^{-1}\), respectively; \(R=8.31~\text{J K}^{-1}\text{mol}^{-1}\))

Hint:

Ideal solubility of a solid: \(\ln x = -\dfrac{\Delta H_{\mathrm{fus}}}{R}\left(\dfrac{1}{T}-\dfrac{1}{T_{\mathrm{m}}}\right)\).
Works well for simple eutectic systems far from solid–solid interactions and at pressures near 1 bar.

Answer 1

Correct Answer: 0.3

Step 1: Ideal solubility relation for a solid in a liquid.
For equilibrium between solid naphthalene and its ideal solution at temperature T 

Step 2: Insert data. 
\[ \frac{1}{T}-\frac{1}{T_{\mathrm{m}}}=\frac{1}{300}-\frac{1}{353}\approx 4.99\times10^{-4}\ \text{K}^{-1}, \] \[ \frac{\Delta H_{\mathrm{fus}}}{R}=\frac{19.28\times10^{3}}{8.31}\approx 2.32\times10^{3}. \] Hence, \[ \ln x_{\mathrm{Naph}}\approx - (2.32\times10^{3})(4.99\times10^{-4})\approx -1.16. \] 

Step 3: Solve for \(x_{\mathrm{Naph}}\). 
\[ x_{\mathrm{Naph}}=\exp(-1.16)\approx 0.313 \ \Rightarrow \ \boxed{0.31}\ \text{(to two decimals)}. \]