Question

Determine the mass percent composition of a solution made by dissolving 3.9 moles CH₃CH₂OH in 471 g of H₂O.

Answer 1

Molar mass of CH3CH2OH,

MM = 2*MM(C) + 6*MM(H) + 1*MM(O)

= 2*12.01 + 6*1.008 + 1*16.0

= 46.068 g/mol

use:

mass of CH3CH2OH,

m = number of mol * molar mass

= 3.9 mol * 46.07 g/mol

= 179.7 g

Use:

Mass % of CH3CH2OH = mass of CH3CH2OH * 100 / mass of solution

= 179.7 * 100 / (179.7 + 471)

= 27.62 %

Answer: 28 %