How many grams of the nonelectrolyte ethylene glycol, (CH2OH)2, must be dissolved in 318 g of water to give a solution that boils at 101.0°C?

Some physical properties of water are shown below.

freezing point 0.0°C boiling point 100.0°C Kf 1.86°C/m Kb 0.512°C/m

How many grams of the nonelectrolyte ethylene glycol, (CH2OH)2, must be dissolved in 318 g of water to give a solution that boils at 101.0°C? I can't seem to get the right answer no matter how much I try.

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You know the physical property data that a 1 molal solution containing ethylene glycol will give a 0.512 degree increase in boiling point. Since your boiling point increase is > 0.512 degrees the concentration has to be > a 1 molal solution. Divide the 1 degree increase by the Kf and you get a 1.95 m solution. That means you need 1.95 moles of ethylene glycol in 1 kg of water.

The molar mass of ethylene glycol is 46 g/mol times 1.95 moles = 89.84 g in 1 kg.

Now you can set up a ratio:

89.84/1000 g = x/318g and solve for x which is 28.57 g

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