A MIXTURE OF ETHYL ALCOHOL AND PROPYL ALCOHOL HAS A VAPOUR PRESSURE OF 290mm OF HG AT 300K.THE VAPOUR PRESSURE OF PROPYL ALCOHOL IS 200mm.IF THE MOLE FRACTION OF ETHYL ALCOHOL IS 0.6 THEN, ITS VAPOUR PRESSURE IN mm AT THE SAME TEMPERTURE IS______

A)360

B)350

C)300

D)700

To determine the vapor pressure of ethyl alcohol in a mixture with propyl alcohol, we can use Raoult's Law. This law states that the vapor 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. Let's break down the problem step by step.

Given Information

• Vapor pressure of propyl alcohol (P_propyl) = 200 mmHg
• Vapor pressure of the mixture (P_mixture) = 290 mmHg
• Mole fraction of ethyl alcohol (X_ethyl) = 0.6
• Mole fraction of propyl alcohol (X_propyl) = 1 - X_ethyl = 0.4

Applying Raoult's Law

According to Raoult's Law, the total vapor pressure of the mixture can be expressed as:

P_mixture = P_ethyl * X_ethyl + P_propyl * X_propyl

Finding the Vapor Pressure of Ethyl Alcohol

We need to find the vapor pressure of ethyl alcohol (P_ethyl). Rearranging the equation gives us:

P_ethyl = (P_mixture - P_propyl * X_propyl) / X_ethyl

Substituting the Values

Now, substituting the known values into the equation:

• P_mixture = 290 mmHg
• P_propyl = 200 mmHg
• X_propyl = 0.4
• X_ethyl = 0.6

Plugging these values in:

P_ethyl = (290 mmHg - 200 mmHg * 0.4) / 0.6

P_ethyl = (290 mmHg - 80 mmHg) / 0.6

P_ethyl = 210 mmHg / 0.6

P_ethyl = 350 mmHg

Final Answer

Thus, the vapor pressure of ethyl alcohol in the mixture at 300 K is 350 mmHg. Therefore, the correct answer is B) 350.