Question

How many molecules of ethanol (C2H5OH) (the alcohol in alcoholic beverages) are present in 135 mL of ethanol? The density of ethanol is 0.789 g/cm3.

Answer 1

density = 0.789 g/cm^3

density = 0.789 g/mL

volume = 135 mL

use:

density = mass / volume

0.789 g/mL = mass / 135 mL

mass = 106.5 g

Molar mass of C2H5OH,

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

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

= 46.068 g/mol

mass(C2H5OH)= 106.5 g

number of mol of C2H5OH,

n = mass of C2H5OH/molar mass of C2H5OH

=(106.5 g)/(46.068 g/mol)

= 2.312 mol

number of molecules = number of mol * Avogadro’s number

number of molecules = 2.312 * 6.022*10^23 molecules

number of molecules = 1.39*10^24 molecules

Answer: 1.39*10^24 molecules