Question

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.

Answer 1

Answer:

                     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