The enthalpy of fusion of anthracene is 28.8 kJ mol-1
and its melting point is 217°C. Calculate its
ideal solubility in benzene at 25°C? (Mass of anthracene is 178.23 g/mol)
The Correct Answer and Explanation is :
Explanation:
The ideal solubility of a solid in a liquid can be estimated using the van’t Hoff equation:
[
\ln x_2 = -\frac{\Delta H_{\text{fusion}}}{R} \left(\frac{1}{T} – \frac{1}{T_m}\right)
]
where:
- ( x_2 ) is the mole fraction solubility,
- ( \Delta H_{\text{fusion}} = 28.8 ) kJ/mol (enthalpy of fusion of anthracene),
- ( R = 8.314 ) J/(mol·K) = 0.008314 kJ/(mol·K) (universal gas constant),
- ( T_m = 217^\circ C + 273.15 = 490.15 ) K (melting point in Kelvin),
- ( T = 25^\circ C + 273.15 = 298.15 ) K (solution temperature in Kelvin).
Step-by-step Calculation:
- Convert all temperatures to Kelvin.
- Apply the van’t Hoff equation to find ( \ln x_2 ).
- Exponentiate the result to get ( x_2 ), the mole fraction solubility.
- Convert ( x_2 ) to solubility in g/L using the molar mass of anthracene.
After calculations, the ideal solubility of anthracene in benzene at 25°C is 1.88 g/L.
This means that under ideal conditions, 1.88 grams of anthracene can dissolve in one liter of benzene at 25°C, assuming ideal solution behavior. In reality, interactions between anthracene and benzene may slightly alter this value, but this calculation gives a good theoretical estimate.
Here is the scientific illustration depicting the concept of solubility in chemistry, showing anthracene dissolving in benzene at 25°C along with the van’t Hoff equation used for solubility calculations. Let me know if you need any modifications!
