Question

van't Hoff Factor

Chem 202 Freezing Point of Aqueous Solutions Results Experimental freezing point values determined from graphs (sce directions on p. 3-5, steps 2,3): Sample Ty measured /°C AT measured / °C Distilled Water (solvent) -0.1°C Follow your instructor's Solution D (Naci) -1.8°C 1.7°C directions for submitting these values Solution G (lalla) -1.0°C 0.9°C before you leave lab! van't Hoff factor values (see directions on p. 3-6, steps 4-5): Solution D Solute Compound Naci Molality 0.505 mol/kg Ideal van't Hoff factor; lideal = Observed van't Hoff factor; iobsd = calculations: SolutionG Solute Compound Calla Molality o.201 mol/kg Ideal van't Hoff factor; lideal = Observed van't Hoff factor; iobsd = calculations:

Answer 1

Answer

Solution D:

The solute compound is NaCl. When it is in water it releases Na⁺ and Cl^- ions . Thus one mole releases two moles of ions.

Therefore the ideal vant hoff factor assuming 100 dissociation of NaCl = 2

The observed vant hoff factor = observed colligative property/ normal colligative property.

Observed colligative property $\Delta$ T_f observed = 1.7

normal colligative property = $\Delta$ T_f normal = m * K_f K_f for water = 1.86 deg Kg / mol    

               = 0.505 x 1.86

   = 0.9393 deg

Therefore observed vanthoff factor = $\Delta$ T_f observed / $\Delta$ T_f normal

= 1.7/0.9393 = 1.81

Solution G

The solute component is CaCl₂ . On dissolution in water it releases Ca²⁺ ion and 2 Cl^- ions. Therefore one mole of CaCl₂ will release 3 moles of ions assuming 100 % dissociation of CaCl₂

Hence the ideal vant hoff factor for CaCl₂ for 100% dissociation = 3

The observed vant hoff factor = observed colligative property/ normal colligative property.

Observed colligative property $\Delta$ T_f observed = 0.9

normal colligative property = $\Delta$ T_f normal = m * K_f K_f for water = 1.86 deg Kg / mol    

               = 0.201 x 1.86

   = 0.374 deg

Therefore observed vanthoff factor = $\Delta$ T_f observed / $\Delta$ T_f normal

= 0.9/0.374 = 2.41