A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 235 torr. Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature. What is the mole fraction composition of the mixture?
you would have to use addition. and convert it into a fraction.
To find the mole fraction composition of the mixture, we can use Raoult's Law, which states that the vapor pressure of a component in a mixture is equal to the product of its mole fraction in the mixture and its vapor pressure in its pure state.
Let's denote the mole fraction of pentane as x and the mole fraction of hexane as (1 - x).
According to Raoult's Law, we can write:
Vapor pressure of the mixture = (mole fraction of pentane) * (vapor pressure of pentane) + (mole fraction of hexane) * (vapor pressure of hexane)
Given that the vapor pressure of the mixture is 235 torr, the vapor pressure of pentane is 425 torr, and the vapor pressure of hexane is 151 torr, we have:
235 = x * 425 + (1 - x) * 151
Let's solve this equation for x:
235 = 425x + 151 - 151x
235 - 151 = 425x - 151x
84 = 274x
x = 84/274
x ≈ 0.307
Therefore, the mole fraction of pentane in the mixture is approximately 0.307, and the mole fraction of hexane is (1 - 0.307) ≈ 0.693.