Question

Calculate the vapor pressure of water over each of the following ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) solutions at \(22^{\circ} \mathrm{C}\) (vp pure water \(=19.83 \mathrm{~mm} \mathrm{Hg})\). Ethylene glycol can be assumed to be nonvolatile. (a) \(X_{\text {ethylene glycol }}=0.288\) (b) percent ethylene glycol by mass \(=39.0 \%\) (c) \(2.42 m\) ethylene glycol

Short answer

Answer: (a) 14.12 mm Hg, (b) 19.07 mm Hg, (c) 19.01 mm Hg.

Step-by-step solution

(a) Finding the mole fraction of water in the solution

Given the mole fraction of ethylene glycol, we can find the mole fraction of water by subtracting it from 1, as there are only two components in the solution: \(x_{\text{water}} = 1 - x_{\text{ethylene glycol}} = 1 - 0.288 = 0.712\)

(a) Calculating the vapor pressure of water

Using Raoult's Law, we can calculate the vapor pressure of water over the solution, given the mole fraction of water and the vapor pressure of pure water at \(22^\circ \mathrm{C}\): \(P_{\text{solution}} = x_{\text{water}} \times P^\circ_{\text{water}} = 0.712 \times 19.83\,\mathrm{mm\, Hg} = 14.12\,\mathrm{mm\,Hg}\) (rounded to 2 decimal places)

(b) Finding the mass fractions of water and ethylene glycol

Given the percent mass of ethylene glycol by mass, we can find the mass fraction of ethylene glycol and water: Mass fraction of ethylene glycol \(= 0.390\) Mass fraction of water \(= 1 - 0.390 = 0.610\)

(b) Calculating mole fraction of water

Next, we convert mass fractions to mole fractions. We need the molar masses of water and ethylene glycol. Molar mass of water (H\(_2\)O) \(= 18.02\,\mathrm{g/mol}\) Molar mass of ethylene glycol (C\(_2\)H\(_6\)O\(_2\)) \(= 62.08\,\mathrm{g/mol}\) Now we can calculate mole fractions: \(x_{\text{water}}=\frac{0.610/18.02}{(0.610/18.02) + (0.390/62.08)} = 0.961\)

(b) Calculating the vapor pressure of water

Using Raoult's Law, we can now calculate the vapor pressure of water over the solution: \(P_{\text{solution}} = x_{\text{water}} \times P^\circ_{\text{water}} = 0.961 \times 19.83\,\mathrm{mm\, Hg} = 19.07\,\mathrm{mm\, Hg}\) (rounded to 2 decimal places)

(c) Finding moles of ethylene glycol and water

Given the molality of ethylene glycol (\(2.42\,\mathrm{m}\)), we can find the moles of ethylene glycol and water in \(1\,\mathrm{kg}\) of water: Moles of ethylene glycol \(= 2.42\,\mathrm{mol}\) Moles of water \(= \frac{1000\,\mathrm{g}}{18.02\,\mathrm{g/mol}} = 55.49\,\mathrm{mol}\)

(c) Calculating mole fraction of water

Next, we can calculate the mole fraction of water: \(x_{\text{water}}=\frac{55.49}{55.49+2.42} = 0.958\)

(c) Calculating the vapor pressure of water

Using Raoult's Law, we can now calculate the vapor pressure of water over the solution: \(P_{\text{solution}} = x_{\text{water}} \times P^\circ_{\text{water}} = 0.958 \times 19.83\,\mathrm{mm\, Hg} = 19.01\,\mathrm{mm\, Hg}\) (rounded to 2 decimal places) In conclusion, the vapor pressures of water over the three ethylene glycol solutions are: (a) \(P_{\text{solution}}=14.12\,\mathrm{mm\,Hg}\) (b) \(P_{\text{solution}}=19.07\,\mathrm{mm\,Hg}\) (c) \(P_{\text{solution}}=19.01\,\mathrm{mm\,Hg}\)