The hydrogen gas is collected by displacement of water at 35.0 °C at a total pressure of 748 torr. The vapor pressure of water at 35.0 °C is 42 torr. What is the partial pressure of H_2 if 307.8 mL of gas is collected over water from this method?

Dec 10, 2015

$\text{706 torr}$

Explanation:

The trick here is to not get distracted by the given volume of the hydrogen gas - water vapor mixture.

All you need to focus on is the total pressure of the gaseous mixture and the vapor pressure of water at that temperature.

The partial pressure of hydrogen can be calculated using Dalton's Law of Partial Pressures, which tells you that the total pressure of a gaseous mixture is equal to the sum of the partial pressure of each gas that's a part of that mixture. Simply put, each gas will contribute to the total pressure of the mixture proportionally to the number of molecules it has in the mixture.

Mathematically, you can write this as

$\textcolor{b l u e}{{P}_{\text{total}} = {\sum}_{i} {P}_{i}}$

In this case, the total pressure of the mixture will be equal to the partial pressure of water and the partial pressure of hydrogen gas

${P}_{\text{total" = P_"water}} + {P}_{{H}_{2}}$

This of course means that the partial pressure of hydrogen gas will be

${P}_{{H}_{2}} = {P}_{\text{total" - P_"water}}$

P_(H_2) = "748 torr" - "42 torr" = color(green)("706 torr")