What will be the order of solubility of NaCl in \(H_2O,CH_3OH,and C_2H_5OH \)
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I can guess that it will be :
\[S(H_2O) > S(CH_3OH) > S(C_2H_5OH)\]
But how can we calculate it, by using the basic of "Force of attraction between particles?"
If I am "correct" then the expression for the force of attraction between particles is :
F = z+z-e^2 / Dr^2
NaCl dissolves in water because the polar water molecules are able to solvate the Na+ and CL- ions. So the more polar the solvent is, the more it will be able to solvate the sodium and chloride ions. As you predicted CH3OH is more polar than ethanol. Hence the order you have predicted is correct.
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z+ and z- are the charge of the ions..
D is dielectric constant.
e is the charge of \(1\) electron.
r is inter ionic distance.
Oh! Thanks @Preetha ma'am . I thought something different. Here is what I did.
(the first part is done above)
As we know that when D increases, F decreases and lesser the F easier will be the break up of the lattice and hence more will be the solubility.
^ that is correct ma'am ?