At 80 degrees Celsius, the vapor pressure of pure liquids x, y and z are 385, 190 and 66 mm Hg. what pressure must be exerted on the surface of the solution containing one mole of x, three moles of y, and two moles of z for the boiling point to begin at 80 degrees Celsius?
To find the pressure required for the boiling point to begin at 80 degrees Celsius, we need to use Raoult's law, which states that the vapor pressure of a solution is equal to the mole fraction of each component multiplied by its corresponding vapor pressure:
P(solution) = (mole fraction of x * vapor pressure of x) + (mole fraction of y * vapor pressure of y) + (mole fraction of z * vapor pressure of z)
First, we need to calculate the mole fractions of each component in the solution:
Mole fraction of x = moles of x / total moles
= 1 mole / (1 mole + 3 moles + 2 moles)
= 1/6
Mole fraction of y = moles of y / total moles
= 3 moles / (1 mole + 3 moles + 2 moles)
= 3/6 = 1/2
Mole fraction of z = moles of z / total moles
= 2 moles / (1 mole + 3 moles + 2 moles)
= 2/6 = 1/3
Now, we can substitute these values into Raoult's law equation:
P(solution) = (1/6 * 385 mm Hg) + (1/2 * 190 mm Hg) + (1/3 * 66 mm Hg)
Calculating the values:
P(solution) = (64.17 mm Hg) + (95 mm Hg) + (22 mm Hg)
P(solution) = 181.17 mm Hg + 95 mm Hg + 22 mm Hg
P(solution) = 298.17 mm Hg
Therefore, the pressure that must be exerted on the surface of the solution containing one mole of x, three moles of y, and two moles of z for the boiling point to begin at 80 degrees Celsius is approximately 298.17 mm Hg.