How to calculate the partial pressure of oxygen in this mixture of oxygen and argon?
How is the mixture composed? Find expressions for mass, volume and amount of substance.
\begin{align}
\tag1\label1
  V_\text{total} &amp;= V_{\ce{O2}} + V_{\ce{Ar}}\\
\tag3\label2
  n_\text{total} &amp;= n_{\ce{O2}} + n_{\ce{Ar}}\\
\tag3\label3
  m_\text{total} &amp;= 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} &amp;= \frac{RT}{p_\text{total}}(n_{\ce{O2}} + n_{\ce{Ar}})\\
  \tag5\label5
  n_{\ce{Ar}}  &amp;= \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} &amp;= m_{\ce{O2}} + m_{\ce{Ar}}\\
  \tag6\label6
  m_\text{total} &amp;= n_{\ce{O2}}M_{\ce{O2}} + n_{\ce{Ar}}M_{\ce{Ar}}\\
  \tag7\label7
  m_\text{total} &amp;= 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} &amp;= 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}} &amp;= \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} &amp;= \pu{39.9 g//mol}\\
M_\ce{O2} &amp;= \pu{32.0 g//mol}\\
R &amp;= \pu{8.314 L kPa //K mol}\\
\pu{1 mmHg} &amp;= \pu{133.322 Pa}\\
p_\text{total} &amp;\approx \pu{90 kPa}\\
T &amp;= \pu{316.5 K}\\
m_\text{total} &amp;= \pu{19.3 g}\\
V_\text{total} &amp;= \pu{16.2 L}
\end{align}

Plugging this into \eqref{9}:
\begin{align}
  \tag9
  n_{\ce{O2}} &amp;= \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}} &amp;= \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}} &amp;= \pu{0.36 mol}
\end{align}

Using the ideal gas with the data provided yields:
\begin{align}
  n_\text{total} &amp;=\frac{pV}{RT}\\
  n_\text{total} &amp;=\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} &amp;=\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).