01:160:162 Lecture Notes - Lecture 3: Molar Mass, Stoichiometry, Chief Operating Officer

• School: Rutgers University
• Department: Chemistry
• Course: 01:160:162
• Professor: Tavss

Document Summary

Compare the freezing points of three 0. 10 m aqueous solutions: Tf = kf m = 1. 86 (0. 10) = 0. 186. We would expect all 0. 10 m solutions to have the same freezing point depression of 0. 186: the freezing point depression for the nacl solution is almost double that of the sucrose or urea. Since freezing point depression is a colligative property, depending only on the amount of solute present, the actual amount of nacl doubles when it dissociates. One mol of nacl produces a total of 2 moles of ions. We can take this into account by using the van"t hoff factor i in our equations. Each mol of nacl produces 1 mol of na+ and 1 mol of cl-, or 2 mol of total particles. i = moles particles in solution moles solute dissolved. The i becomes a multiplier for all concentrations of solute.