If the vapor pressure of a liquid is 0.850 atm at 20°C and 0.897 atm at 25°C, what is the normal boiling point of the liquid?

Use the Clausius-Clapeyron equation but you will need the heat of vaporization.

To find the normal boiling point of a liquid, we need to determine the temperature at which its vapor pressure is equal to atmospheric pressure, which is typically considered to be 1 atm.

Given that the vapor pressure of the liquid is 0.850 atm at 20°C and 0.897 atm at 25°C, we can use the Clausius-Clapeyron equation to calculate the normal boiling point.

The Clausius-Clapeyron equation is given by:

ln(P2/P1) = -ΔH_vap/R * (1/T2 - 1/T1)

Where:
P1 = vapor pressure at T1
P2 = vapor pressure at T2
ΔH_vap = molar heat of vaporization
R = ideal gas constant (8.314 J/(mol·K))
T1 = initial temperature
T2 = final temperature

We can rearrange the equation to solve for the final temperature (T2), which represents the normal boiling point:

T2 = ((ΔH_vap/R) / ((1/T1) - (1/T2))) + T1

Using the given values:
P1 = 0.850 atm
P2 = 1 atm (since we're calculating the normal boiling point)
T1 = 20°C = 293.15 K
T2 = unknown

We can now substitute the values into the equation and solve for T2:

T2 = ((ΔH_vap/R) / ((1/T1) - (1/T2))) + T1

Since we don't have information about ΔH_vap, we cannot calculate the exact boiling point of the liquid. The molar heat of vaporization is needed to find the normal boiling point accurately.

To determine the normal boiling point of a liquid, we need to find the temperature at which its vapor pressure equals the atmospheric pressure, which is typically 1 atm.

Given that the vapor pressure of the liquid is 0.850 atm at 20°C and 0.897 atm at 25°C, we can use these two data points to calculate the boiling point.

First, we need to determine how the vapor pressure changes with temperature. One commonly used equation to describe this relationship is the Clausius-Clapeyron equation:

ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

where:
P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively,
ΔHvap is the enthalpy of vaporization,
R is the ideal gas constant (0.0821 L·atm/(K·mol)),
T1 and T2 are the temperatures in Kelvin.

Since we are interested in the boiling point (normal boiling point is defined at 1 atm), we can rewrite the equation as:

ln(1/0.850) = -(ΔHvap/R) * (1/Tb - 1/293.15)

where Tb is the boiling point in Kelvin (20°C = 293.15 K).

Next, we can rearrange the equation to solve for the boiling point (Tb):

1/Tb = [ln(1/0.850) * R/ΔHvap] + 1/293.15

Now, let's calculate the boiling point using the given data:

1/Tb = [ln(1/0.850) * 0.0821/ΔHvap] + 1/293.15

We need another piece of information, the enthalpy of vaporization (ΔHvap), to evaluate the equation. Unfortunately, it is not provided in the question. The enthalpy of vaporization is a substance-specific value that represents the energy required to convert one mole of the liquid to vapor at its boiling point. Without this data, we cannot calculate the boiling point.

To proceed further, we need to know the enthalpy of vaporization for the liquid in question.