Practice Quiz

SAT Chemistry

Category - Multiple Choice

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What is the freezing point of 1.5 liters of water after .250 mol CaCl2 is added (Kf = -1.86 degrees C·kg/mol for water)?
  1. -.93 degrees C
  2. -.20 degrees C
  3. .93 degrees C
  4. -.23 degrees C
  5. .45 degrees C
Explanation
Answer: -.93 degrees C

Ionic solids lower the freezing points of liquids according to the equation (change in freezing point) = i * Kf * m. In this case, i is three because CaCl2 forms three ions upon dissociation. Lowercase m, or molal concentration, is equal to the moles of solute divided by the kilograms of solvent, or (.250 mol)/(1.5 liters). Entering the variables into the equation gives the final solution, -.93 degrees C.
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