Question

The density of liquid methanol, CH3OH (FM=34.05), is 0.791g/mL at 77 degrees F. What is this same temperature in K, and how many methanol molecules will occupy a cubic container, 1mm on each side?

Answer 1

Given that

density of methanol = 0.791 g/mL

side of cube = 1 mm

volume of cube = (side)³ = (1mm) ³ = 1mm³ = 0.001 mL ( 1mm³ = 0.001 mL)

Hence, volume of methanol = volume of cube = 0.001 mL

Mass of methanol:

mass of methanol = density of methanol x volume of methanol

= 0.791 g/mL x 0.001 mL

= 791 x 10^-6 g

Mass of methanol = 791 x 10^-6 g

Moles of methanol:

Mass of methanol = 791 x 10^-6 g

Given that formula mass of methanol = 34.05 g/mol

Moles of methanol = Mass of methanol/ molar mass or formula mass of methanol

= 791 x 10^-6 g / (34.05 g/mol)

= 23.23 x 10^-6 moles

Moles of methanol = 23.23 x 10^-6 moles

Molecules of methanol:

1 mol of any substance contains Avogadro number of molecules i.e. 6.023 x 10²³ molecules

Hence, 23.23 x 10^-6 moles of methanol contains

= 23.23 x 10^{-6 .} 6.023 x 10²³ molecules

= 139.91 x 10¹⁷ molecules

Molecules of methanol = 139.91 x 10¹⁷ molecules

Therefore, 139.91 x 10¹⁷ methanol molecules will occupy a cubic container, 1mm on each side.