A variety is $\mathbb{Q}$-factorial if every global Weil divisor is $\mathbb{Q}$-Cartier. How bad singularities are allowed so that the algebraic variety is still $\mathbb{Q}$-factorial? Is a singular curve $\mathbb{Q}$-factorial? For example is a nodal-cuspidal plane curve $\mathbb{Q}$-factorial?



When is an algebraic variety $\mathbb{Q}$-factorial?
I saw this a while ago, and I assumed that someone would give you an answer, but it doesn't
look like anyone will. So I'll make an attempt. 
As Ulrich suggests, it is better to stick to
normal varieties, so your nodal/cuspidal curves are immediately eliminated. So we should move to dimension at least two. 


A very simple class of examples are cones. Let $V\subset \mathbb{P}^N$ be a nontrivial smooth variety, 
and let $R$ be the local ring of the vertex of the cone over $V$. This is normal and $\dim R\ge 2$. $R$ (or $Spec R$) is $\mathbb{Q}$-factorial exactly 
when $Cl(R)\otimes\mathbb{Q}=0$ where $Cl(R)$ is the (Weil) divisor class group. Now
according to Lipman "Unique factorization in complete local rings", Alg. Geom, Arcata (1974)
$$Cl(R)= Cl(V)/\langle\text{hyperplane}\rangle$$



> 
> Thus $R$ is $\mathbb{Q}$-factorial if and only if the 
>  $Cl(\mathbb{P}^N)\otimes \mathbb{Q}\to Cl(V)\otimes \mathbb{Q}$ is surjective. 
> 
> 
> 


This is generally *not* true. For example, if $V$ is a curve, $R$ is $\mathbb{Q}$-factorial if and only if $V$ has genus $0$. 


**Added Notes**: (1) In the last sentence, I meant that a cone over a higher genus
curve $V$ *cannot* be $\mathbb{Q}$-factorial because $\dim Cl(V)\otimes \mathbb{Q}=\infty$
in this case. (2) As Ulrich & Francesco point out $V$ should be projectively normal. For
example, it would suffice to assume the $V$ is a hypersurface. (3) I forgot to say, that
I'm working over an algebraically closed ground field.


