Does normalizing a mixed state give a pure state?

An arbitrary one qubit density matrix $rho$:

$$rho = begin{pmatrix} rho_{00} & rho_{01} rho_{10} & rho_{11}end{pmatrix} = frac{I + vec{r} cdot vec{sigma}}{2}$$

where $vec{r} cdot vec{sigma} = r_x sigma_x + r_y sigma_y + r_z sigma_z$, $sigma$s are Pauli matrices, $r$ is the corresponding vector for the density matrix in the Bloch sphere.

The probability of measuring $|0rangle$ state is equal to $rho_{00}$ and the probability of measuring $|1rangle$ state is equal to $rho_{11}$. Thus, the normalization of the probabilities of measuring $|0rangle$ or $|1rangle$ states for the density matrix corresponds to the statement $Tr(rho) = rho_{00} + rho_{11} = 1$. In this sense it is already a normalized state if $Tr(rho) = 1$, but still the vector $vec{r}$ in the Bloch sphere formalizm will be inside the sphere for a mixed state.

I am not sure how the action described in the question can be related to the normalization, because both mixed (inside the sphere) and pure (on the sphere) states should be normalized.

Bloch vector $vec{v}$ of a density matrix $$ rho_C = frac{1}{2}(I + vec{v}cdotvec{sigma}) $$ can be shorter than 1 i.e. $|vec{v}| le 1$, when the the density matrix represents a mixed state.

If you normalize it $vec{n} = vec{v}/|vec{v}|$ then you will get the density matrix of the C' pure state $$ rho_{C'} = frac{1}{2}(I + vec{n}cdotvec{sigma}). $$

When it comes to interpretations, I don't know of any simple relation between C and C' like e.g. measurements of C won't bring you to C.