a
\(O _{2}\) and \(Cl _{2}\) are taken at NTP. Assuming them to be ideal, by ideal gas law \(P V=n R T\).

Since \(O _{2}\) and \(Cl _{2}\) are taken at NTP, so \(T_{O_{2}}=T_{C L_{2}}=20^{\circ} C =293.15 K\) and \(PO _{2}= PCr _{2}=\) \(1\) atm.

Also, given \(V_{ Cl _{2}}= VO _{2}\)

So,

\(\frac{ P_{O _{2}} V_{ O _{2}}}{P_{ Cl _{2}} V_{ Cl _{2}}}=\frac{n_{ O _{2}}}{n_{ Cl_{2}}} \times \frac{ R }{ R } \times \frac{ T_{O _{2}}}{ T_{Cl _{2}}}\)

\(\Rightarrow \frac{n_{c e_{2}}}{n_{ O _{2}}}=\frac{1}{1}\)

Thus, option \((A)\) is corrrect.