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Between a 2% (w/v) aqueous solutions of $\ce{NaCl}$ and $\ce{RbCl}$, which will have a higher boiling point. Here there are two competing factors. First there's the fact that $\ce{NaCl}$ has a higher concentration. So it's solution should have a higher boiling point. Secondly according to Fajan's rule, $\ce{NaCl}$ is more polarised. So it should have a lower boiling point. So which factor is dominant?

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    $\begingroup$ Fajans is irrelevant. Both compounds are going to be fully dissociated anyway. $\endgroup$ – Ivan Neretin Feb 14 '19 at 12:56
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    $\begingroup$ Which solution has the higher molar concentration? $\endgroup$ – matt_black Feb 14 '19 at 16:36
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Molar mass of $\ce{RbCl}$ is $\pu{120.92 g/mol}$, Molar mass of $\ce{NaCl}$ is $\pu{58.44 g/mol}$. Both compounds are highly ionic and dissociate in water solutions. Thus, 2% solution of $\ce{NaCl}$ has approximately double amount of ions than that in 2% solution of $\ce{RbCl}$ (2% is fairly dilute as well, considering their solubility in water). Thus, theoretically, the $\ce{NaCl}$ solution should have higher boiling point (higher solute particles).

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Boiling points depend on the strength of the intermolecular forces of the substance.So NaCl has a higher boiling point than RbCl.

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