$f : [0, 1] \rightarrow \text{ cardinals }$ Collection of all cardinality can not be a set.
Can we find $f : [0, 1] \rightarrow \text{ cardinals }$ s.t. $f$ is 1-1 ?
 A: This is a great question!
First, note that any map $f$ from $[0, 1]$ into $Card$ would yield a well-ordering of $[0,1]$ (put $r\preccurlyeq s$ if $f(r)\le f(s)$); so if there is such a map, then $[0, 1]$ can be well-ordered.
Conversely, any set of ordinals is in bijection with a set of cardinals: think about the (class) function sending $\alpha$ to $\aleph_\alpha$. So if $[0, 1]$ can be well-ordered, then - since every well-ordering is isomorphic to an ordinal, and hence any well-orderable set is in bijection with a set of ordinals - it admits an injection into $[0, 1]$.
So we have:

The answer is yes iff $[0, 1]$ can be well-ordered.

In particular the axiom of choice implies that the answer is yes.

Note that there's nothing special about $[0, 1]$ here: in general, a set  can be put in bijection with a set of cardinals iff it can be well-ordered.

EDIT: There's an important bit of clarification, here. In the above, by "cardinal" I mean "initial ordinal" or $\aleph$-number. In ZF alone, though, we can have non-well-orderable sets - these sets obviously aren't equinumerous with any $\aleph_\alpha$. So in the ZF context, we can also speak of cardinalities more general than just cardinals.
In this context - as Andres observes below - your question can be rephrased as:

Given a set $X$, is there a function $f$ taking each element $x$ of $X$ to some set $f(x)$, where $f(x)\equiv f(y)$ iff $x=y$?

I don't know anything, off the top of my head, about this more complicated question and its various versions.
