A student dissolves 1.0 mol of glucose in 1.0 kg of water. What will be the value of the van't Hoff factor if glucose does not ionise?
Answer and explanation
Correct answer: 1
Glucose behaves as a non-electrolyte in this idealised situation and remains as intact molecules rather than splitting into ions. Therefore one dissolved molecule gives one effective solute particle, so the ratio of actual to calculated particles is i = 1. The amount of water and glucose affects molality, but not this factor when there is no association or dissociation.
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What is the correct answer to this question?
1
Why is this the correct answer?
Glucose behaves as a non-electrolyte in this idealised situation and remains as intact molecules rather than splitting into ions. Therefore one dissolved molecule gives one effective solute particle, so the ratio of actual to calculated particles is i = 1. The amount of water and glucose affects molality, but not this factor when there is no association or dissociation.
Which subject and chapter does this question cover?
This is a Class 12 Chemistry question. Chapter: Chapter 01: Solutions. Topic: 5: Colligative Properties.
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