Question 9.16: In the northeastern United States during freezing weather, C...
In the northeastern United States during freezing weather, CaCl_{2} is spread on icy highways to melt the ice. Calculate the freezing point of a solution containing 0.50 mole of CaCl_{2} in 1 kg of water.
STEP 1 State the given and needed quantities.
| ANALYZE THE PROBLEM | Given | Need | Connect |
| 0.50 mole of CaCl_{2} 1 kg of water | freezing point of solut ion | CaCl_{2} strong electrolyte, 1 mole/kg lowers freezing point 1.86 °C |
STEP 2 Determine the number of moles of solute particles.
CaCl_{2}(s) → Ca^{2+}(aq) + 2Cl^{-}(aq) = 3 moles of solute particles
\begin{array}{r c}\boxed{\begin{matrix} 1 mole of CaCl_{2} = 3 moles of solute particles \\\frac{ 3 moles solute particles}{1 mole CaCl_{2} } \text{ and } \frac{1 mole CaCl_{2}}{ 3 moles solute particles} \end{matrix}} \end{array}STEP 3 Determine the temperature change using the moles of solute particles and the degrees Celsius change per mole of particles.
\begin{array}{r c}\boxed{\begin{matrix} 1 mole of solute particles = 1.86 °C\\\frac{1.86 °C}{1 moles solute particles } \text{ and } \frac{1 moles solute particles}{1.86 °C} \end{matrix}}\end{array}Temperature change = 0.50 \cancel{mole CaCl_{2}}\times \boxed{\frac{3 \cancel{moles solute particles}}{1 \cancel{mole CaCl_{2}}} } \times \boxed{\frac{1.86 °C}{1 \cancel{ moles solute particles}} }
=2.8 °CSTEP 4 Subtract t he temperature change from the freezing point. The temperature change, ΔT, is subtracted from 0.00 °C to obtain the new freezing point of the CaCl_{2} solution.
T_{solution} = T_{water} – ΔT = 0.00 °C – 2.8 °C = -2.8 °C