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I reason that as mole fraction is a measure of colligative properties, increase in the Vant Hoff factor ( and hence increase in the value of colligative properties) should also leave a mark on the mole fraction of the solute components. However, my instructor tells differently.

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  • $\begingroup$ Equal doesn't always means causation. Vant Hoff factor does not change for very dilute solutions. It does change when mole fraction is high enough to cause non ideal behaviour, but it can decrease, for instance when x is high enough that the solute does not fully dissociate. $\endgroup$ – Alchimista Mar 18 at 15:04
  • $\begingroup$ That is not what I asked for. My instructor told me that if a solute component dissociates in solution to yield ions, the mole fraction of the solute increases, and this is so because the no. of solute particles increases,but the total no. of moles remain constant (why?) If the no of moles of solute increases, shouldn't the total no of moles also increase? I know that will still lead to a greater mole fraction, but doesn't the total no of moles (solute + solvent) also increase? $\endgroup$ – Aabesh Ghosh Mar 18 at 17:30
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    $\begingroup$ It depends on how you define the mole fraction, in NaCl:water or sodium ion : water chloride ion . I don't see much sense on the latter here because is account for dissociation and solvent molecules are in excess. What do you call the solute? Is NaCl, and you account for dissociation. I am afraid you mean in case you add practically more solute. Then of course you change the overall number of moles, not only x. $\endgroup$ – Alchimista Mar 18 at 18:06
  • $\begingroup$ I will explain with an example. Suppose 'x' moles of glucose are dissolved in 'y' moles of a solvent (assume water, also assume x<y). The mole fraction of glucose (solute) is x/(x+y). Now if in a similar situation, 'x' moles of NaCl are dissolved in 'y' moles of water and 100% dissociation is assumed, then is the mole fraction of solute components 2x/(2x+y) or 2x/y ?(In either case, the mole fraction increases) $\endgroup$ – Aabesh Ghosh Mar 20 at 3:31
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    $\begingroup$ In your example, for each ion it will be x/(2x+y). So the latter. Consider that in dilute solutions x + y is about y, too. $\endgroup$ – Alchimista Mar 20 at 7:29

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