What is the molality of a solution prepared by dissolving 75.8 g of ethylene glycol, HOCH2CH2OH, in 1.55 L of water? Assume the density of water is 1.00 g/mL.

0.787 mol. Kg⁻¹ or 0.787 m

Explanation:

Data Given:

Mass = m = 75.8 g

Volume = V = 1.55 L = 1550 mL

M. Mass of Ethylene Glycol = 62.07 g/mol

Density of water = d = 1.0 g/mL

To Find:

Molality = ?

Molality is a unit of concentration and is defined as the number of moles of solute per Kg of a solvent.

Molality = Moles of Solute / Kg of Solvent - - - (1)

First Calculate Moles of Solute as;

Moles = Mass / M. Mass

Moles = 75.8 g / 62.07 g/mol

Moles = 1.22 moles

Secondly calculate mass of water as,

mass of water = density * volume

mass of water = 1.0 g/mL * 1550 mL

mass of water = 1550 g or 1.55 Kg

Now, putting values of moles and mass of solvent in equation 1,

Molality = 1.22 mol / 1.55 Kg

Molality = 0.787 mol. Kg⁻¹ or 0.787 m