How much ethyl alcohol must be …
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Ajit Sharma 6 years, 6 months ago
Molal depression constant of water (Kf ) = 1.86 K kg mol-1
Density of water = 1 g / cc = 1 kg / L
Given ,Volume of water = 1 L
Therefore mass of water = 1 kg
Freezing point = -20oC = ( 273-20) K= 253 K
Freezing point of water = 273 K
Hence, depression in freezing point = Kf x m (m= molality )
(273-253) K = 1.86 K kg mol-1 x (moles of ethyl alcohol) /mass of solvent in kg
20 = 1.86 x ( moles of ethyl alcohol) / 1 Kg
moles of ethyl alcohol = 20/1.86 = 10.753 mol
Mass of 10. 753 mol of ethyl alcohol = 46 x 10.753 g = 494.638 g.
So , more than 494.638 g will be added to lower the freezing point.
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