Calculate the freezing point of a 1.0 m solution of Ca(NO₃)₂ in water? Remember that Mg(NO₃)₂ is an electrolyte.The normal freezing point of pure water is 0.00 $\degree$C.K_f(water) = -1.86 $\degree$C/m A)1.56 $\degree$C B)5.58 $\degree$C C)1.86 $\degree$C D)-5.58 $\degree$C E)-1.86 $\degree$C

Question

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Ca(NO₃)₂ dissociates into 3 ions (1 Ca²⁺ and 2 NO₃⁻). The van't Hoff factor (i) is 3. The freezing point depression is ΔTf = i * m * Kf = 3 * 1.0 m * (-1.86 °C/m) = -5.58 °C. The new freezing point is 0.00 °C - 5.58 °C = -5.58 °C.

Calculate the Freezing Point of a 1 °\degree° CKf (water) = -1

Calculate the Freezing Point of a 1 $\degree$ CK_f(water) = -1