vanthofffactor

# vanthofffactor - 3 CH 2 CH 2 CH 3 This is a nonelectrolyte...

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Colligative Properties: van’t Hoff Factor The equation for freezing point depression and boiling point elevation contains the letter i. i stands for the van’t Hoff Factor. ∆T = imK f Since freezing point depression and boiling point elevation depend only on the number of particles ( it does not matter what the particles are), we need only determine the total m of the particles. If a solution is 0.2 m in NaCl the i would be about 2. In water, NaCl ionizes as follows: NaCl Na + + Cl - As can be seen, a 0.2 m solution of NaCl becomes 0.4 m in 0.2 m 0.2m 0.2 m particles. In other words the molality doubles. Making i two will take care of this. i is approximately equal to the number of ions into which a species breaks. The true van’t Hoff factor is not exactly 2, but is close enough to call it 2. So, if NaCl is used in a problem, then ∆T = 2 mK f . i is replaced by a 2. Another example, 0.2 m CH

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Unformatted text preview: 3 CH 2 CH 2 CH 3 This is a nonelectrolyte so it does not ionize in water. Therefore, i would be one and ∆T = 1mK f Another example, CaCl 2 Ca 2+ + 2 Cl-In this case i would be 3, and ∆T = 3mK f . The greater the total m of the solution, the (a) higher the boiling point (b) lower the freezing point (c) lower the vapor pressure (d) higher the heat of vaporization To determine if a substance ionized in water or not, you can use the chart below. All species in this chart ionize 100% in water. A. Strong Acids: HCl, HBr, HI, H 2 SO 4 , HNO 3 , HClO 4 , HClO 3 B. Strong Bases: Hydroxides of group IA and II A, Except Be and Mg C. Soluble Salts ( ionic compounds: metal/nonmetal ionize 100%) Always Soluble and Ionize 100% Except with NO 3-, Group IA, NH 4 + , CH 3 COO-, ClO 4-, ClO 3-No Exceptions Cl-Br-, I-Pb, Ag, Hg 2 2+ SO 4 2-Ag, Pb, Hg 2+ Ca, Sr, Ba...
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vanthofffactor - 3 CH 2 CH 2 CH 3 This is a nonelectrolyte...

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