Part 2 21. The density of a 50% water and 50% ethanol mixture, by mass is 0913 g

Question

Part 2 21. The density of a 50% water and 50% ethanol mixture, by mass is 0913 g/mL Note the molar masses of water and ethanol are 18.0 g and 46.0 g; the universal gas constant R is 0.0821 L-at m/mol-K a. Calculate the molality of this solution assuming that water is the solvent in this case b. Calculate the molarity of this solution assuming that water is the solvent in this case. c. Calculate the mole fractions of water and ethanol in this solution. d. Calculate the total vapor pressure of this solution if the vapor pressure of pure water at 2S is 0.0313 atm and the vapor pressure of ethanol is 0.0724 atm. Assuming that water is the solvent, calculate the osmotic pressure of this solution at 25

Explanation / Answer

Given data,

Mass of 1 mL solution = 0.913 g.

In that,

Mass of ethanol = 0.913 * 50 / 100 = 0.4565 g.

Mass of water = 0.4565 g.

Moles of water = mass / molar mass = 0.4565 / 18.0 = 0.02536 mol

Moles of ethanol = 0.4565 / 46.0 = 0.009924 mol

Mole fraction of water = moles of water / total number of moles of solution

= 0.02536 / (0.02536+0.009924) = 0.719

Moles of ethanol = 0.009924 / (0.02536 + 0.009924) = 0.281

(a)

Molality = moles of solute / mass of solvent in kg = 0.009924 / 0.0004565 = 21.7 m

(b)

Molarity = moles of solute / volume of solution in L = 0.009924 / 0.001 = 9.92 M

(c)

Mole fraction of water = 0.719

Mole fraction of ethanol = 0.281

(d)

Vapour pressure in the mixture = vapour pressure in pure state * mole fraction

vapour pressure of water = 0.0313 * 0.719 = 0.0225 atm

vapour pressure of ethanol = 0.0724 * 0.281 = 0.00203 atm

Total vapour pressure of solution = 0.0225 + 0.00203 = 0.0245 atm

(e)

Osmatic pressure = Molarity * R * T = 9.92 * 0.0821 * 298.15 = 242.8 atm

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