Final answer:
According to Raoult's law, the vapor pressure of hexane above the solution is 111.825 mmHg and the vapor pressure of pentane is 127.125 mmHg. The total vapor pressure above the solution is 238.95 mmHg.
Step-by-step explanation:
To find the vapor pressures of hexane and pentane above the solution, we can use Raoult's law. Raoult's law states that the vapor pressure of a component in a solution is equal to its mole fraction multiplied by its vapor pressure in the pure state.
For hexane, the mole fraction is given as 0.75, and its vapor pressure in the pure state is 149.1 mmHg.
Therefore, the vapor pressure of hexane above the solution would be:
Vapor pressure of hexane = 0.75 * 149.1 mmHg
= 111.825 mmHg
Similarly, for pentane, the mole fraction is 1 - 0.75 = 0.25,
and its vapor pressure in the pure state is 508.5 mmHg.
Therefore, the vapor pressure of pentane above the solution would be:
Vapor pressure of pentane = 0.25 * 508.5 mmHg
= 127.125 mmHg
The total vapor pressure above the solution is the sum of the vapor pressures of the individual components:
Total vapor pressure = Vapor pressure of hexane + Vapor pressure of pentane
= 111.825 mmHg + 127.125 mmHg
= 238.95 mmHg