# Napthalene is dissolved in a liquid with which it forms an ideal solution at 25C. The normal boiling point of napthalene is 80.2 C and ^fusH for napthalene is 19.0 KJ/mol. Determine the solublity of napthalene in this liquid at 25 C.

## To determine the solubility of naphthalene in the liquid at 25 °C, we can use the Raoult's law equation for ideal solutions:

P₁ = X₁ * P₀₁

Where:

P₁ = Vapor pressure of the liquid solution at a given temperature

X₁ = Mole fraction of naphthalene in the liquid solution

P₀₁ = Vapor pressure of pure naphthalene at the same temperature

Since we know the normal boiling point of naphthalene and its standard enthalpy of fusion, we can use them to calculate the vapor pressure of pure naphthalene at 25 °C.

First, let's calculate ΔH_sub, the enthalpy of sublimation of naphthalene, using the enthalpy of fusion:

ΔH_sub = ΔH_fus / n

Where:

ΔH_sub = Enthalpy of sublimation

ΔH_fus = Enthalpy of fusion

n = Number of moles

Assuming the enthalpy of fusion and the normal boiling point are given per mole of naphthalene, we can directly use them.

ΔH_sub = 19.0 kJ/mol

Next, we can calculate the vapor pressure of pure naphthalene at 25 °C using the Clausius-Clapeyron equation:

ln(P₀₁ / P₁) = ΔH_sub / R * (1 / T₁ - 1 / T₀₁)

Where:

P₀₁ = Vapor pressure of pure naphthalene at the normal boiling point

P₁ = Vapor pressure of pure naphthalene at 25 °C

ΔH_sub = Enthalpy of sublimation

R = Gas constant (8.314 J/mol·K)

T₁ = Temperature at which we want to find the vapor pressure (25 °C)

T₀₁ = Normal boiling point of naphthalene

Let's convert the temperatures:

T₁ = 25 + 273.15 = 298.15 K

T₀₁ = 80.2 + 273.15 = 353.35 K

Now, substituting the values:

ln(P₀₁ / P₁) = (19.0 × 10³) / (8.314) * (1 / 298.15 - 1 / 353.35)

To find P₀₁ / P₁, you can rearrange the equation:

P₀₁ / P₁ = e^[(19.0 × 10³) / (8.314) * (1 / 298.15 - 1 / 353.35)]

Finally, let's solve for X₁, the mole fraction of naphthalene:

X₁ = P₁ / P_total

Where P_total = Vapor pressure of the liquid solution at 25 °C.

Now that you have the equation and the values, you can plug them in to calculate the mole fraction of naphthalene (X₁), which gives you the solubility of naphthalene in the liquid at 25 °C.