Lens spaces are rational homology sphere. Is all lens spaces are manifold? if not then under what condition lens spaces become a manifold? what is dimension of these spaces. Moreover all these spaces are rational homology spheres in manifold case. 
 A: $\mathbb{Z/pZ}$ acts freely on $S^3$, and resultant quotient spaces are called lens spaces. Since the group is finite and free, so the resultant space is manifold.[you can generalized that into higher dimesional odd spheres as well en.wikipedia.org/wiki/Lens_space#Definition ] 
Observe that Lens spaces are orientable (as Jason DeVito mentioned) since the group action on $S^{2n+1}$ is orientable preserving. So an orientation on sphere descends to an orientation on $M$.
In order to prove that they are rational homology spheres, it is enough to prove that given any $u\in H_i(M,\mathbb Z), 0<i<n$ there exists some fixed integer $k$ s.t $ku=0$. Let $f:S^3\to M$ be the universal covering map. Let $deg(f)=k$ ( since it is a finite cover). Then $f_*(S^3)=k[M]$. Given any $u\in H_i$ there exists an unique $b\in H^{n-i}$ s.t. $[M]\cap b= u$ {Poincare Duality}. Then $ku= k[M]\cap b= f_*([S^3])\cap b= f_*( [S^3]\cap f^*(b)) =0$ (since $f^*(b)=0$). Thus $M$ is a rational homology sphere. 
This same argument works for higher dimension as well.
