\(109.6 \mathrm{~cm}^3\)
given:
\begin{array}{l}
V_1=120 \mathrm{~cm}^3 \\
T_1=20^{\circ} \mathrm{C}+273=293 \mathrm{~K} \\
P_1=745 \mathrm{mmHg}
\end{array}
at s.t.p.
\(\mathrm{P}_2=760 \mathrm{mmHg}\)
\(T_2=273 \mathrm{~K}\)
using the general gas equation
\begin{array}{l}
\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2} \\
\Rightarrow \mathrm{V}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1 \mathrm{~T}_2}{\mathrm{~T}_1 \mathrm{P}_2}
\end{array}
\begin{array}{l}
V_2=\frac{745 \times 120 \times 273}{293 \times 760}=\frac{24406200}{22680} \\
V_2=109.60 \mathrm{~cm}^3
\end{array}