Isotonicity

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Prov3:6

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For a compound like Li2Co3 -- what would dissociation factor be ? what would number of particles be ? and what would valence be ?

What a confuddlement !

Thank you for help
 
Lithium carbonate exists only in the anhydrous form... so it doesn't dissociate, meaning the dissociation factor is 1.

Number of particles depends on how much of it you have... the question you are asking makes no sense.

Also the questions you are asking have nothing to do with isotonicity (the title of the thread).

Is English your first language ?
 
All carbonate salts are insoluble except for the group 1 akali metals. While Li is much less soluble than say Na or K....Li2CO3 should dissociate in water. The # of particles would be 3 (2 Li ions + 1 CO3 ion). Someone with more recent/better chemistry background can double check me on that. It would make the solution more hypertonic...as you now have more solutes, 3 ions, versus the 1 molecule before Li2CO3 disassociated.
 
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Thank you Lnsean not just for your help but for choosing to be kind. So if the question on exam uses Li2Co3 asks us to calculate mosm/l i would use 3 as number of diss particles in the formula and if the question asks to calculate meq i would use 3 as valence ( since we have 3 ions ? ) and since it disscociates into 3 particles the dissociation factor would be 2.6 if the question was to calculate E value ?
 
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