5. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1, O$ is the center of the base $A B C D$, and $M, N$ are the midpoints of edges $A_{1} D_{1}$ and $C C_{1}$, respectively, then the volume of the tetrahedron $O-M N B_{1}$ is $\qquad$ .
(5) $\frac{7}{48}$ Hint: Draw a line through $O$ parallel to $AB$, intersecting edges $AD$ and $BC$ at points $E$ and $F$ respectively, and connect $BE$, taking the midpoint of $BF$ as $Q$. It is easy to see that $OQ \parallel BE \parallel B_{1}M$. Therefore, $OQ \parallel$ plane $MNB_{1}$. Hence
$$
V_{O-MNB_{1}}=V_{Q-MNB_{1}}=V_{M-NQB_{1}}.
$$

Also,
$$
\begin{aligned}
S_{\triangle NQB_{1}}= & S_{BCC_{1}B_{1}}-S_{\triangle NC_{1}B_{1}} \\
& -S_{\triangle QCN}-S_{\triangle QBB_{1}} \\
= & 1-\frac{1}{4}-\frac{3}{16}-\frac{1}{8}=\frac{7}{16}.
\end{aligned}
$$

Therefore, $V_{Q-MNB_{1}}=\frac{1}{3} S_{\triangle NQB_{1}} \cdot A_{1}B_{1}=\frac{7}{48}$.