Calculate the freezing point of a bottle of wine which contains 12.5% ethanol (C

Question

Calculate the freezing point of a bottle of wine which contains 12.5% ethanol (C2H5OH) by volume. The wine has a density of 0.978 g/mL (at 20 degrees Celsius, assume density is contestant as temperature changes). Assume the wine is composed of only ethanol and water. Calculate the freezing point of a bottle of wine which contains 12.5% ethanol (C2H5OH) by volume. The wine has a density of 0.978 g/mL (at 20 degrees Celsius, assume density is contestant as temperature changes). Assume the wine is composed of only ethanol and water. Calculate the freezing point of a bottle of wine which contains 12.5% ethanol (C2H5OH) by volume. The wine has a density of 0.978 g/mL (at 20 degrees Celsius, assume density is contestant as temperature changes). Assume the wine is composed of only ethanol and water.

Explanation / Answer

Let the volume of wine bottle be 100 mL

So the volume of ethanol = 12.5 mL

Mass of ethanol = density x volume

= 0.978 x 12.5

= 12.225 g

And that of water = 100 - 12.5 = 87.5 mL

Since the density of water is 1 g/mL , so the mass of water = 87.5 g

Number of moles of ethanol = 12.225 / molar mass

=12.225 / 46.07

= 0.265 mol

molality = no. of moles of ethanol / mass of water in kg

= 0.265 / 0.0875

= 3.03 m

Freezing point depression = kf. m

kf = freezing point constant = 1.86 oC/m

So putting all the values we get :

= 1.86 x 3.03

= 5.64oC

So the freezing point od wine = 0 - 5.64oC

= -5.64oC

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