How to calculate what mass of halothane is inspired in one minute given the partial pressure and temperature? If the anesthetic mixture is inspired at the rate of $\pu{100 mL/min}$, what mass of halothane, $\ce{CHCIBrCF3}$, molecular weight $M = \pu{197.4 g/mol}$, is inspired in one minute if the partial pressure of halothane is $\pu{7.6 torr}$ and the temperature is $\pu{21 ^\circ C}$.
  
  
  $\pu{0.08g}$
  $\pu{0.80g}$
  $\pu{1.80g}$
  $\pu{3.36g}$
  


My working:
\begin{align}
P &amp;= \pu{0.01 atm} &amp;&amp;(\pu{1 atm} = \pu{760 torr})\\
T &amp;= \pu{294 K}    &amp;&amp;(273 + 21) \\
V &amp;= \pu{1 L}      &amp;&amp;(\pu{1000 mL} = \pu{1 L})\\
R &amp;= 0.08          &amp;&amp;(\text{L-atm})\\
\end{align}
Equation: 
\begin{align}
  0.01 \times 1 &amp;= n \times 0.08 \times 294\\
  0.01 &amp;= n \times 23.52\\
  n &amp;= 4.3
\end{align}

Which suggest $\pu{4.3 mol}$ of $\ce{CHCIBrCF3}$ which would give the mass $\approx\pu{849g}$.

That is way off any of the answers. Can someone see what I have done wrong?
Volumetric Flow Rate (VFR) is given to you. 
By applying ideal gas equation to the gas, in terms of volumetric flow rate, 
$$\frac{PV}{t} = \frac{nRT}{t}$$

And in an ideal gas mixture (the assumption), 
the partial pressure of a gas is what is exerts alone in same volume in same temperature as of the mixture.
$$\frac{P_\mathrm{gas}\cdot V_\mathrm{tot}}{t} = 
\frac{n_\mathrm{gas}\cdot R\cdot T_\mathrm{total}}{t}$$

amount of substance = mass of the compound/molar mass (or molecular weight of the compound)
$$n = \frac{m}{M}$$

$$\frac{P_\mathrm{gas}\cdot V_\mathrm{total}}{t} =
 \frac{(\frac{m_\mathrm{gas}}{M})\cdot R\cdot T_\mathrm{total}}{t}$$

Apply the values, mind the units:
\begin{align}
V &amp;=\pu{0.1 L}\\
T &amp;= (273+21)\pu{ K}\\
P_\mathrm{gas} &amp;= \pu{0.01 atm}\\
R &amp;=\pu{0.082057 L atm mol^-1K^-1}\\
M_\mathrm{gas} &amp;=\pu{197.4 g mol^-1}\\
t &amp;= \pu{1 min}\\
\end{align}

$$\frac{\pu{0.01 atm}\cdot\pu{0.1 L}}{\pu{1 min}} = 
\frac{
  \frac{m_\mathrm{gas}}{\pu{197.4 g mol^-1}}
  \cdot\pu{0.082057 L atm mol^-1 K^-1}\cdot\pu{294 K}}{\pu{1 min}}$$

For $m_\mathrm{gas}$ I can see the value in your multiple choice.