in between a 2% (w/v) aqueous services of $ceNaCl$ and $ceRbCl$, which will have a greater boiling point.Here there room two contending factors. Very first there"s the fact that $ceNaCl$ has actually a greater concentration. Therefore it"s equipment should have actually a higher boiling point. Secondly follow to Fajan"s rule, $ceNaCl$ is much more polarised. So it should have actually a lower boiling point. Therefore which factor is dominant?



Molar massive of $ceRbCl$ is $pu120.92 g/mol$, Molar mass of $ceNaCl$ is $pu58.44 g/mol$. Both link are highly ionic and dissociate in water solutions. Thus, 2% solution of $ceNaCl$ has actually approximately dual amount that ions 보다 that in 2% systems of $ceRbCl$ (2% is fairly dilute together well, considering your solubility in water). Thus, theoretically, the $ceNaCl$ solution must have greater boiling suggest (higher solute particles).

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Boiling points rely on the toughness of the intermolecular pressures of the substance.So NaCl has actually a higher boiling point than RbCl.


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