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 = _____
(a) X (ethylene glycol)= 0.288
(b) % ethylene glycol by mass =39%
(c) 2.42 m ethylene glycol
calculate a, b, c.
If you have not heard of it, review:
(Broken Link Removed)
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 = _____
The mole fraction of water is Xh2o = 1 - 0.288 = 0.712. Use that value in the formula
(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