Calculate the vapor pressure of water over each of the following ethylene glycol C2H6O2 solutions at 80 degree C (vp=pure water=657.6 mm Hg). Ethylene glycol can be assumed to be nonvolatile.
(a) X (ethylene glycol)= 0.288
(b) % ethylene glycol by mass =39%
(c) 2.42 m ethylene glycol
calculate a, b, c.
This is an exercise in Raoult's Law.
If you have not heard of it, review:
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You will need the Mole Fraction X of each mixture. In (2), they have told you that X = 0.288, so the answer is
0.288 * 657.6 mm Hg = _____
My answer was wrong for (a)
The mole fraction of water is Xh2o = 1 - 0.288 = 0.712. Use that value in the formula
Could you provide a little bit more clues i am still confused.
Vapour pressure find formula
To calculate the vapor pressure of water over each of the given ethylene glycol solutions, we will use Raoult's law, which states that the vapor pressure of a component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction.
(a) First, let's calculate the mole fraction (X) of water and ethylene glycol in solution (a).
Mole fraction (X) of water = 1 - X (ethylene glycol)
Mole fraction (X) of water = 1 - 0.288 = 0.712
The vapor pressure of water over solution (a) can be calculated using Raoult's law:
Vapor pressure (vp) of water over solution (a) = X (water) * vp(pure water)
Vapor pressure (vp) of water over solution (a) = 0.712 * 657.6 mm Hg
Vapor pressure (vp) of water over solution (a) ≈ 468.07 mm Hg
(b) To calculate the mole fraction of water and ethylene glycol in solution (b), we need to convert the given mass percentage of ethylene glycol to mole fraction.
Mass percentage of ethylene glycol = 39%
Mass percentage of water = 100% - 39% = 61%
Mole fraction (X) of water = mass percentage of water / molar mass of water
Mole fraction (X) of ethylene glycol = mass percentage of ethylene glycol / molar mass of ethylene glycol
Using the molar masses:
Molar mass of water (H2O) = 18 g/mol
Molar mass of ethylene glycol (C2H6O2) = 62 g/mol
Mole fraction (X) of water = 61 / 18 ≈ 3.39
Mole fraction (X) of ethylene glycol = 39 / 62 ≈ 0.629
Using Raoult's law:
Vapor pressure (vp) of water over solution (b) = X (water) * vp(pure water)
Vapor pressure (vp) of water over solution (b) = 3.39 * 657.6 mm Hg
Vapor pressure (vp) of water over solution (b) ≈ 2229.64 mm Hg
(c) To calculate the vapor pressure of solution (c), we need to convert the given molarity of ethylene glycol to mole fraction.
Molarity of ethylene glycol = 2.42 mol/L
Mole fraction (X) of water = 1 / (1 + molarity of ethylene glycol)
Mole fraction (X) of water = 1 / (1 + 2.42) ≈ 0.292
Using Raoult's law:
Vapor pressure (vp) of water over solution (c) = X (water) * vp(pure water)
Vapor pressure (vp) of water over solution (c) = 0.292 * 657.6 mm Hg
Vapor pressure (vp) of water over solution (c) ≈ 191.82 mm Hg
So, the calculated vapor pressures are:
(a) Vapor pressure (vp) of water over solution (a) ≈ 468.07 mm Hg
(b) Vapor pressure (vp) of water over solution (b) ≈ 2229.64 mm Hg
(c) Vapor pressure (vp) of water over solution (c) ≈ 191.82 mm Hg