The boiling point elevation constant for toilet is 3.3 to 12 in Kj per mole the normal boiling point of tall in 110 degree Celsius .enthalpy of vaporization of touline would nearly be

To determine the enthalpy of vaporization of toluene using the boiling point elevation constant, we can apply the principles of colligative properties. The boiling point elevation constant (Kb) gives us insight into how much the boiling point of a solvent increases when a solute is added. In this case, we are looking at toluene, which has a normal boiling point of 110 degrees Celsius. Let's break down the steps to find the enthalpy of vaporization.

Understanding the Relationship

The boiling point elevation can be expressed with the formula:

ΔT_b = K_b * m

Where:

• ΔT_b is the change in boiling point.
• K_b is the boiling point elevation constant.
• m is the molality of the solution.

However, to find the enthalpy of vaporization (ΔH_vap), we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and enthalpy of vaporization:

ln(P2/P1) = -ΔH_vap/R * (1/T2 - 1/T1)

Where:

• P1 and P2 are the vapor pressures at temperatures T1 and T2.
• R is the universal gas constant (8.314 J/mol·K).

Calculating Enthalpy of Vaporization

To find the enthalpy of vaporization, we need to know the vapor pressures at two different temperatures. However, if we don't have specific vapor pressure values, we can estimate ΔH_vap using the boiling point elevation constant.

Given that the boiling point elevation constant for toluene is between 3.3 to 12 kJ/mol, we can use an average value for our calculations. Let's take an average Kb of about 7.65 kJ/mol for our example.

Using the Average Kb Value

We can relate the boiling point elevation to the enthalpy of vaporization using the following approximation:

ΔH_vap ≈ K_b * T_b

Where T_b is the boiling point in Kelvin. First, we need to convert the boiling point of toluene from Celsius to Kelvin:

T_b = 110 + 273.15 = 383.15 K

Now, substituting the values into the equation:

ΔH_vap ≈ 7.65 kJ/mol * 383.15 K

This gives us:

ΔH_vap ≈ 2935.5 kJ/mol

Final Thoughts

Thus, the estimated enthalpy of vaporization for toluene, using the average boiling point elevation constant, is approximately 2935.5 kJ/mol. This value provides a good estimate, but keep in mind that actual experimental values may vary slightly based on specific conditions and purity of the substance.