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.

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

a:

by the combination of an unlimited number of elements

thats a little if u want the answer just look it up on google thats what i do haha : )