Volume of dried gas at STP


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> A $54.0\ \mathrm{mL}$ sample of oxygen is collected over water at $20\ \mathrm{^\circ C}$ and $770\ \mathrm{Torr}$ pressure. What is the volume of the dry gas at STP?
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By using the combined gas law, I came up with the answer $\pu{51.2 mL}$.


$770 - 21.1 = 748.9\ \mathrm{Torr}$ pressure for oxygen


$21.1$ = vapor pressure of water at $23\ \mathrm{^\circ C}$


$\frac{P\_1V\_1}{ T\_1 }= \frac{P\_2V\_2 }{ T\_2}$


$\frac{748.9V\_1}{273 }= \frac{770\times54}{296}$


$V\_1 = \pu{51.2 mL}$


Is this how it's supposed to be done?


