Question

#5 Ethylene glycol (C₂H₆O₂) is used as an antifreeze. How many grams of ethylene glycol must be added to 1.0 L of water to reduce the freezing point of the glycol-water mixture to

-10.0 degrees C? The freezing point depression constant of water is 1.86 (deg/molal).

Answer 1

Aus Freezing Density point of watu = 0.0°C of water = 1:0g/mL or 1.0kg/L As we know- AT = i kem - AT = Depression in freezing point i= van't hoff factor Kp = depression constant m = molality - _mass of solute moler mass of solute x mass of solveut (kg) Here we have - Volume of water (solvent) = 1.0L " # mass of water = 1.04x 1.0 kg = 11.0kg] # molar mass of Cq HG O₂ (solute ) = 62.07 g/mol mass of Ca HG O2 = ? (let it 'm') 7 kg of water = 1.86°c/m 'i' for Cq H ₂ O = 1 (non-electrolyte) AT = 10°c # plugging values, we get 10 = 1 * 1.86 x m 62.07 X ] 333.71a m Hence, 333.71g of ethylene to 1.0L water to reduce glycol must be added freezing point to -10°c.