To solve this problem, we need to use Raoult's Law, which states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

Step 1: Determine the mole fractions of Benzene and Toluene. Since the solution is equimolar, the mole fractions of Benzene and Toluene are both 0.5.

Step 2: Calculate the partial pressures of Benzene and Toluene. According to Raoult's Law, the partial pressure is the product of the mole fraction and the vapor pressure. So, for Benzene, it's 0.5 * 135.4 kPa = 67.7 kPa. For Toluene, it's 0.5 * 54 kPa = 27 kPa.

Step 3: The total pressure at the beginning of the vaporization process is the sum of the partial pressures of Benzene and Toluene. So, it's 67.7 kPa + 27 kPa = 94.7 kPa.

Step 4: At the end of the vaporization process, all the liquid has been converted into vapor. So, the pressures of Benzene and Toluene are their vapor pressures, which are 135.4 kPa and 54 kPa respectively.

So, the pressure at the beginning of the vaporization process is 94.7 kPa, and the pressures at the end of the vaporization process are 135.4 kPa for Benzene and 54 kPa for Toluene.

Question

To solve this problem, we need to use Raoult's Law, which states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

Step 1: Determine the mole fractions of Benzene and Toluene. Since the solution is equimolar, the mole fractions of Benzene and Toluene are both 0.5.

Step 2: Calculate the partial pressures of Benzene and Toluene. According to Raoult's Law, the partial pressure is the product of the mole fraction and the vapor pressure. So, for Benzene, it's 0.5 * 135.4 kPa = 67.7 kPa. For Toluene, it's 0.5 * 54 kPa = 27 kPa.

Step 3: The total pressure at the beginning of the vaporization process is the sum of the partial pressures of Benzene and Toluene. So, it's 67.7 kPa + 27 kPa = 94.7 kPa.

Step 4: At the end of the vaporization process, all the liquid has been converted into vapor. So, the pressures of Benzene and Toluene are their vapor pressures, which are 135.4 kPa and 54 kPa respectively.

So, the pressure at the beginning of the vaporization process is 94.7 kPa, and the pressures at the end of the vaporization process are 135.4 kPa for Benzene and 54 kPa for Toluene.

Knowee AI · Accepted Answer

To solve this problem, we need to use Raoult's Law, which states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

Step 1: Determine the mole fractions of Benzene and Toluene. Since the solution is equimolar, the mole fractions of Benzene and Toluene are both 0.5.

Step 2: Calculate the partial pressures of Benzene and Toluene. According to Raoult's Law, the partial pressure is the product of the mole fraction and the vapor pressure. So, for Benzene, it's 0.5 * 135.4 kPa = 67.7 kPa. For Toluene, it's 0.5 * 54 kPa = 27 kPa.

Step 3: The total pressure at the beginning of the vaporization process is the sum of the partial pressures of Benzene and Toluene. So, it's 67.7 kPa + 27 kPa = 94.7 kPa.

Step 4: At the end of the vaporization process, all the liquid has been converted into vapor. So, the pressures of Benzene and Toluene are their vapor pressures, which are 135.4 kPa and 54 kPa respectively.

So, the pressure at the beginning of the vaporization process is 94.7 kPa, and the pressures at the end of the vaporization process are 135.4 kPa for Benzene and 54 kPa for Toluene.

An equimolar solution of Benzene and Toluene is totally evaporated at a constant temperature, the vapor pressure of benzene and toluene are 135.4 and 54 kPa respectively. What are the pressures at the beginning and the end of the vaporization process?

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