Solution:

The total flux due to a charge \(Q\) is given by Gauss's law:

\[
\Phi_{\text{total}} = \frac{Q}{\varepsilon_0}.
\]

Since the charge \(Q\) is placed at the mouth of the conical flask, the flux through the flask corresponds to half the total flux (because the charge is symmetrically distributed above and below the mouth):

\[
\Phi_{\text{flask}} = \frac{1}{2} \cdot \Phi_{\text{total}} = \frac{1}{2} \cdot \frac{Q}{\varepsilon_0}.
\]

Thus, the flux through the conical flask is:

\[
\Phi = \frac{Q}{2\varepsilon_0}.
\]