Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water.The molal freezing point depression constant ( $K_f$ )for water is $1.86 ^ { \circ } \mathrm { Cm } ^ { - 1 }$
A)-0.45 °C
B)+0.45 °C
C)-0.23 °C
D)+0.23 °C
E)1.23 °C

Question

Quizplus · Accepted Answer

The freezing point depression is calculated using the formula: $\Delta T_f = i \cdot K_f \cdot m$, where $i$ is the van't Hoff factor for KCl (which is 2), $K_f$ is the freezing point depression constant, and $m$ is the molality. First, calculate the molality: moles of KCl = 5.0 g / 74.55 g/mol = 0.067 moles; molality = 0.067 moles / 0.550 kg = 0.122 m. Then, $\Delta T_f = 2 \cdot 1.86 \cdot 0.122 = 0.45$. The freezing point is depressed by 0.45°C, so the answer is -0.45°C.

Calculate the Freezing Point of a Solution Containing 5 KfK_fKf​ )For Water Is

Calculate the Freezing Point of a Solution Containing 5 $K_f$ )For Water Is