As Parsa explained in his comment, we **always** have $\textrm{Pic}(C)=\mathbb{Z}$ by Grothendieck.-Lefschetz. However, when $C$ is not smooth this **does not** mean that $C$ is factorial, that is that every Weil divisor is Cartier.

So we must understand when this happens.

I do not know whether there are satisfactory results in every dimension and for any type of singularities.

Let me give an answer for $N=4$, under the condition that $C$ has only isolated ordinary double points ("nodes").

Then there is the following result:

**Theorem.** Let $C \subset \mathbb{P}^4$ be a hypersurface of degree $d$ with at most ordinary double points as singularity. Let $\Sigma:=\textrm{Sing}(C)$. Then the following are equivalent:

- every divisor on the threefold $C$ is Cartier;
- every surface $S \subset C$ is cut out on $C$ by an hypersurface in $\mathbb{P}^4;$
- the set $\Sigma$ imposes independent linear conditions on linear forms of degree $2d-5$.

In other words, $C$ is factorial if and only if

$$H^1(\mathcal{O}_{\mathbb{P}^4}(2d-5) \otimes \mathcal{I}_{\Sigma})=0. \quad (\star)$$

If you have an explicit equation for $C$, you can easily check whether condition $(\star)$ holds by using Macauley2.

Cheltsov showed that that if $|\Sigma| <(d-1)^2$ then $C$ is factorial. For instance, a nodal cubic with at most $8$ nodes is factorial.

This result does not hold if $|\Sigma|=(d-1)^2$: in fact, any hypersurface of the form

$$x_0F+x_1G=0,$$

with $F$ and $G$ general linear forms of degree $d-1$, is not factorial since it contains the $2$-plane $x_o=x_1=0$: notice that there are $(d-1)^2$ nodes on this plane.

For more details on these topics see [I. Cheltsov, Factorial Threefold hypersurfaces, J. Algebraic geometry
19 (2010), no. 4, 781–791] and the references given there.