A solution is prepared by mixing 0.3355 moles of NaNO₃ with 235.0 g of water. Its density is 1.0733 g mL^-1. If K_f = 1.86 °C m^-1 and the solute is completely ionized, what is the expected freezing point of the solution?Hint: be sure to classify the solute as an electrolyte or non-electrolyte.

Question

Quizplus · Accepted Answer

NaNO₃ is an electrolyte and dissociates into 2 ions (Na⁺ and NO₃⁻). The molality of the solution is calculated using the moles of solute and mass of solvent. The freezing point depression is calculated using ΔTf = i * Kf * m, where i is the van't Hoff factor (2 for NaNO₃). Subtract ΔTf from the normal freezing point of water (0°C) to find the expected freezing point.

A Solution Is Prepared by Mixing 0