<p>McGraw Hill question on molality
50 gm of AlCl3 (aluminium chloride) in 1.1 kg solvent?
Do we have to multiply 4 in the molality since vant hoff factor for AlCl3 is 4 ?
Another example of the book is of CaCl2 and there 3 has been multiplied in the molality.</p>
<p>You would only multiply by 3 or 4 if the question is asking for the molality of the ions in the solution. However, since it is not clarified, it is okay to assume that this question is only asking for the molality of the COMPOUND AlCl3 in the solvent. Therefore, van’t hoff factor does not influence the answer.</p>
<p>Thanks a lot !!!</p>
<p>so u oculd have done moles over kg despite the vanhoff factor?</p>
<p>mols of solute / kg of sol’n = molal</p>