Molality and dissociation
The short answer is because $\ce{HF}$ doesn't dissociate as much as $\ce{HCl}$.  Let's examine why this is by introducing some terms like $\ce{pH}$ and $\ce{pK_a}$.  


$\ce{pH}$ does not measure the strength of a given acid, but rather the acidity of a given solution.
$\ce{pK_a}$ does measure the relative strength of an acid, look at the following equilibrium


$$\ce{H2O + HX&lt;=&gt;H3O^+ + X^-}$$
$$\ce{so~~~K_{eq}=\frac{[H3O^+][X^{-}]}{[H2O][HX]}}$$ but in dilute solutions the concentration of water stays constant so we can rewrite this equation as
$$\ce{K_{a}=\frac{[H3O^+][X^{-}]}{[HX]}}$$ and finally, $$\ce{pK_{a}~=~-Log_{10}K_{a}}$$ Now we can see that the more highly dissociated our acid $\ce{HX}$, the higher the concentrations of $\ce{H3O^+}$ and $\ce{X^{-}}$, and in turn the larger the value for $\ce{K_{a}}$, and the more negative the value of $\ce{pK_{a}}$.

Bottom line: highly dissociated acids have large negative $\ce{pK_{a}}$ values

Now, here is a Table from SparkNotes.  It compares the $\ce{pK_{a}s}$ (degree of ionization or strength) of the various haloacids.  You can see that as we move down the Table from $\ce{HF}$ to $\ce{HI}$ the $\ce{pK_{a}}$ becomes much more negative; our acids become much more dissociated in water.  So we see that $\ce{HF}$ is not very dissociated ($\ce{K_{a}=10^{-3.1}}$, equilibrium lies to the left). On the other hand, 1 mole of $\ce{HCl}$ will produce about 1 mole of $\ce{H^+}$ ($\ce{K_{a}=10^{+7}}$, equilibrium lies far to the right).  Note too that the $\ce{pK_{a}s}$ of these acids correlate with their bond strength which makes sense.  As the $\ce{H-X}$ bond gets weaker, it becomes easier for the acid to ionize (dissociate).