Volume of wet hydrogen collected at room temperature: $\pu{40.48 mL}$, ambient pressure: $\pu{29.7 inHg}$ ($\pu{754 torr}$), ambient temperature: $\pu{295 K}$ ($\pu{22^\circ C}$), vapor pressure of water at $\pu{22^\circ C}$: $\pu{19.83 torr}$. Partial pressure of dry hydrogen: $\pu{9.87 inHg}$.
Find the volume of hydrogen under standard conditions.
The equation of combined gas law is $p_1V_1/T_1 = p_2V_2/T_2$.
Say if the volume of wet hydrogen gas is $\pu{40.48 mL}$, does this equation still work if part of that volume is water vapor? Or, in other words, does the same volume of water vapor and hydrogen gas exert the same pressure?
One of the correct way to do this problem is find the moles of dry hydrogen gas using the room temperature and volume of wet hydrogen gas, gas constant and room temp. Right here, I am kinda confused why the volume of wet hydrogen if we are trying to find the moles of dry hydrogen gas. Then rest of this solution is plugging the number of moles into the equation along with STP to find the volume of dry hydrogen gas.
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