An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for water.) Determine the freezing point and boiling point of the solution. (Assume a density of 1.00 g/ mL for water.)

The freezing point and boiling point of the solution are and respectively.

Explanation:

Depression in freezing point:

where,

= freezing point of solution = ?

= freezing point of water =

= freezing point constant of water =

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

= mass of solute (ethylene glycol) = 21.4 g

= mass of solvent (water) =

= molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

Therefore,the freezing point of the solution is

Elevation in boiling point :

where,

= boiling point of solution = ?

= boiling point of water =

= boiling point constant of water =

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

= mass of solute (ethylene glycol) = 21.4 g

= mass of solvent (water) =

= molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

Thus the boiling point of the solution is

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