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The van't Hoff factor is the ratio between the amount of particles produced when a substance is dissolved and the amount of substance being dissolved. If ionization is 0 percent then it would mean that the substance do not dissociate into particles so the van't Hoff factor would be 1. However, when the percent ionization is 100%, then it fully dissociates into ions. For H2SO4 at 100% ionization, the vant hoff factor should be three, two hydrogen ions and one sulfate ion.

Answer:

Van´t Hoff factor assuming 0 % ionization is 1, while when ionization is 100 % the factor is 3.

Explanation:

The study of colligative properties is different for ionic substances than for non-ionic substances. This is related with the fact that when  ionic compounds are dissolved, they separates in their constitutive ions. In contrast, when a non-ionic compound is dissolved, there are only one type o molecule in the solution. This fact affect the estimation of colligative properties and for this reason Van´t Hoff factor was introduced.

Van´t Hoff factor = number of particles after disociation/ number of particles before disociation.

The disociation of H2SO4 is represented as:

H2SO4 ⇔ 2 H+  + SO4-

- Assuming 0 % inization:

Van´t Hoff factor = 1/ 1 ⇒Van´t Hoff factor = 1.

-Assuming 100 % Ionization

Van´t Hoff factor = 3/ 1 ⇒Van´t Hoff factor = 3.

Summarizing, Van´t Hoff factor assuming 0 % ionization is 1, while when ionization is 100 % the factor is 3.

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