How is the mixture composed? Find expressions for mass, volume and amount of substance.
\begin{align}
\tag1\label1
V_\text{total} &= V_{\ce{O2}} + V_{\ce{Ar}}\\
\tag3\label2
n_\text{total} &= n_{\ce{O2}} + n_{\ce{Ar}}\\
\tag3\label3
m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}}
\end{align}
Transform $V_\text{total}$ with the ideal gas and solve for $n_{\ce{Ar}}$
\begin{align}
\tag4\label4
V_\text{total} &= \frac{RT}{p_\text{total}}(n_{\ce{O2}} + n_{\ce{Ar}})\\
\tag5\label5
n_{\ce{Ar}} &= \frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}
\end{align}
Use total mass \eqref{3} and transform it with the with molar mass to amount of substance equation ($m = nM$), then plug in \eqref{5}:
\begin{align}
\tag3
m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}}\\
\tag6\label6
m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + n_{\ce{Ar}}M_{\ce{Ar}}\\
\tag7\label7
m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + \left(\frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}\right)M_{\ce{Ar}}
\end{align}
Now there is only one variable left, which you do not know:
\begin{align}
\tag8\label8
m_\text{total} &= n_{\ce{O2}}\left(M_{\ce{O2}} - M_{\ce{Ar}}\right) + \frac{p_\text{total}V_\text{total}}{RT}\cdot M_{\ce{Ar}}\\
\tag9\label9
n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}
\end{align}
We need the following data (It would be by far better to only use SI [derived] units):
\begin{align}
M_\ce{Ar} &= \pu{39.9 g//mol}\\
M_\ce{O2} &= \pu{32.0 g//mol}\\
R &= \pu{8.314 L kPa //K mol}\\
\pu{1 mmHg} &= \pu{133.322 Pa}\\
p_\text{total} &\approx \pu{90 kPa}\\
T &= \pu{316.5 K}\\
m_\text{total} &= \pu{19.3 g}\\
V_\text{total} &= \pu{16.2 L}
\end{align}
Plugging this into \eqref{9}:
\begin{align}
\tag9
n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}\\
n_{\ce{O2}} &= \left(\pu{19.3 g} - \frac{\pu{90 kPa}\cdot\pu{16.2 L}}{\pu{8.314 L kPa //K mol}\cdot\pu{316.5 K}}\pu{39.9 g//mol}\right)\cdot \left(\pu{32.0 g//mol} - \pu{39.9 g//mol}\right)^{-1}\\
n_{\ce{O2}} &= \pu{0.36 mol}
\end{align}
Using the ideal gas with the data provided yields:
\begin{align}
n_\text{total} &=\frac{pV}{RT}\\
n_\text{total} &=\frac{\pu{90 kPa}\times\pu{16.2 L}}{\pu{8.314 L kPa K-1 mol-1}\times\pu{316.5 K}}\\
n_\text{total} &=\pu{0.55 mol}
\end{align}
Therefore the mole fraction is
$$x_\ce{O2} = \frac{n_\ce{O2}}{n_\text{total}} \approx 0.64.$$
And the partial pressure is therefore $\pu{57.7 kPa}$ (or $\pu{433 mmHg}$ which is close enough to the solution, given that I rounded the medieval witchcraft units).
