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

the earth would become to overheated and everything alive on it will eventually die from overheating. plants will eventually die as well, water will evaporate, and it will be nothing but a hot planet without anything living on it.

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8