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