Benzene and Toluene are mixed to an ideal solution. At temperature of 300𝐾 the vaporpressure of Toluene is 32.06 𝑚𝑚𝐻𝑔 and of Benzene is 103.1 𝑚𝑚𝐻𝑔.a) A liquid mixture contains of 3 moles of Toluene and 2 moles of Benzene. In what pressurethe first sight of vapor is detected if the temperature is kept constant?b) What is the composition of the steam at that point?c) If we continue to lower the pressure at what pressure the last drop of liquid will evaporate?And what is the composition of the last drop of liquid?

This problem can be solved using 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.

a) To find the total pressure at which the first sight of vapor is detected, we need to find the partial pressures of Toluene and Benzene and then add them together.

The mole fraction of Toluene (X_Toluene) is 3/(3+2) = 0.6 and the mole fraction of Benzene (X_Benzene) is 2/(3+2) = 0.4.

Using Raoult's Law, the partial pressure of Toluene (P_Toluene) is X_Toluene * P_Toluene^0 = 0.6 * 32.06 mmHg = 19.236 mmHg and the partial pressure of Benzene (P_Benzene) is X_Benzene * P_Benzene^0 = 0.4 * 103.1 mmHg = 41.24 mmHg.

The total pressure (P_total) is P_Toluene + P_Benzene = 19.236 mmHg + 41.24 mmHg = 60.476 mmHg. So, the first sight of vapor is detected at this pressure.

b) The composition of the steam at this point is the same as the composition of the liquid mixture, which is 60% Toluene and 40% Benzene.

c) The last drop of liquid will evaporate when the total pressure is equal to the vapor pressure of the component with the highest vapor pressure, which is Benzene in this case. So, the last drop of liquid will evaporate at a pressure of 103.1 mmHg. The composition of the last drop of liquid is the same as the composition of the initial liquid mixture, which is 60% Toluene and 40% Benzene.