7. If the irreducible fraction $\frac{p}{q}\left(p, q \in \mathbf{Z}_{+}\right)$ is converted to a decimal as $0.18 \cdots$, then when $q$ is the smallest, the value of $p$ is ( ).
(A) 9
(B) 7
(C) 5
(D) 2
7. D.

From the problem, we have $0.18<\frac{p}{q}<0.19 \Leftrightarrow \frac{100 p}{19}<q<\frac{100 p}{18}$. It is evident that as $p$ increases, the lower bound of $q$ also increases. When $p=1$, there is no integer $q$ that satisfies the condition; when $p=2$, $q=11$ satisfies the condition. Therefore, $q_{\min }=11, p=2$.