At a certain temperature a 42.55% (v/v) solution of ethyl alcohol, C2H5OH, in water has a density of 0.9027 g mL-1. The density of ethyl alcohol at this temperature is 0.7782 g mL-1, and that of water is 0.9949 g mL-1.

a. Calculate the molar concentration of ethyl alcohol.
b. Calculate the molal concentration of ethyl alcohol.
c. Calculate the mole fraction of ethyl alcohol in the solution.

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jufsanabriasa jufsanabriasa · Answer 1 · 2021-06-17T02:53:45+02:00

Answer:

a. 7.187M.

b. 7.962m.

c. 0.1846

Explanation:

This solution contains 42.55mL of ethyl alcohol in 100mL of solution.

a. Molar concentration = Moles ethyl alcohol / L solution:

Moles ethyl alcohol -Molar mass: 46.07g/mol-

42.55mL * (0.7782g/mL) * (1mol / 46.07g) = 0.7187 moles ethyl alcohol

Liters solution:

100mL * (1L / 100mL) = 0.100L

Molar concentration: 0.7187mol / 0.100L = 7.187M

b. Molal concentration: Moles ethyl alcohol (0.7187moles) / kg solution

kg solution:

100mL * (0.9027g / mL) * (1kg / 1000g) = 0.09027kg

Molal concentration: 0.7187mol / 0.09027kg = 7.962m

c. Mole fraction: Moles ethyl alcohol / Moles ethyl alcohol + moles water

Moles water: -molar mass: 18.01g/mol-

Volume water = 100mL - 42.55mL = 57.45mL

57.45mL * (0.9949g/mL) * (1mol/18.01g) = 3.1736 moles water

Moles fraction:

0.7187 moles ethyl alcohol / 0.7187 moles ethyl alcohol+3.1736 moles water

= 0.1846