Question

A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to that of pure 2-propanol when its mole fraction is 0.41 at its boiling point (82.3°C).

Answer 1

Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

$\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)$

Where:

μ (l): is the chemical potential of 2-propanol in solution    

μ° (l): is the chemical potential of pure 2-propanol  

R: is the gas constant = 8.314 J K⁻¹ mol⁻¹

T: is the temperature = 82.3 °C = 355.3 K

x: is the mole fraction of 2-propanol = 0.41

$\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)$

$\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}$

$\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}$  

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

I hope it helps you!