Question

The mass of ethylene glycol (molar mass 62.1 g / mol), the main component of the antifreeze, must be added to 10.0 L of water to obtain a solidified solution at -23 ° C. Assume a water density of 1 g / mL.

Answer 1

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First we find out the freezing point:

→(-23.0)-(0.0_

→-23.0 degrees C

Calculate target Colligative Molarity:

→ -23=1.86 * Cm

→ 12.37= Cm.

→i* molality

→i= 1

→12.37= 1*m

→12.37= m

molality= moles of solute(C2H6O2)/kg of solvent water

12.37m= moles C2H6O2/ 10.0kg water

123.7= moles of C2H6O2

123.7moles C2H6O2 * 62.26g C2H6O2

= 7701.56 g of C2H6O2

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