Of 100% of the NaCL molecules dissociated, the Van\'t Hoff factor i, would be eq

Question

Of 100% of the NaCL molecules dissociated, the Van't Hoff factor i, would be equal to 2 for NaCl. We assumed it equaled two when we figured out Kb of the NaCl solution.

a. Now we want it find the percent dissociation of the NaCl molecules by finding i using the equation below (use Kb = 0.512/m and 3.456m molarity of NaCl ) T = Tsolution - Tpure = 98.6 - 95.825 = 2.775 show all work i = ( Tb) / Kb x m) i experimental = 1.8 [ {1.8/2} * 100% = 90% dissociation] i theoretical = 2 (100% dissociation)

b. Do you feel comfortable in making the assumption that the NaCl in your experiment was completely dissociated? Why or why not?

c. How does making this assumption (as we did) affect the boiling point elevation of Kb?

Explanation / Answer

a)

Given that,

i = ( Tb) / Kb x m

= 2.775 /(0.512 * 3.456)

i = 1.56

b) yes, we can take the assumption of complete dissociation of NaCl as it is an ionic compound and dissociates completely.

c) In the presence of an electrolyte which dissociates completely in the water the boiling point of the solvent increases.

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