General chemistry

Question

Calculate the molality of a solution that is prepared by mixing 259.5 mL of CH3OH 

 (d = 0.792 g/mL) and 1387 mL of CH3CH2CH2OH (d = 0.811 g/mL)

Answer 1

Molality is defined as moles of solute per kg of solvent.

In this problem, CH3OH (methyl alcohol) is the solute and CH3CH2CH2OH (propyl alcohol) is the solvent.

moles of CH3OH = 259.5 ml x 0.792 g/ml x 1 mol/32 g = 6.42 moles

kg of CH3CH2CH2OH = 1387 ml x 0.811 g/ml x 1 kg/1000 g = 1.12 kg

Molality = 6.42 moles/1.12 kg = 5.73 m