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**Gas Laws Fact Sheet**

[tex]
\[
\begin{array}{|l|l|}
\hline
\text{Ideal gas law} & PV = nRT \\
\hline
& R = 8.314 \frac{L \cdot kPa}{mol \cdot K} \\
\text{Ideal gas constant} & \begin{array}{l}
\text{or} \\
R = 0.0821 \frac{L \cdot atm}{mol \cdot K}
\end{array} \\
\hline
\text{Standard atmospheric pressure} & 1 \, atm = 101.3 \, kPa \\
\hline
\text{Celsius to Kelvin conversion} & K = {}^\circ C + 273.15 \\
\hline
\end{array}
\]
[/tex]

**Question:**

When a chemist collects hydrogen gas over water, she ends up with a mixture of hydrogen and water vapor in her collecting bottle. If the pressure in the collecting bottle is 97.1 kilopascals and the vapor pressure of the water is 3.2 kilopascals, what is the partial pressure of the hydrogen?

A. 93.9 kPa
B. 98.1 kPa
C. 100.3 kPa
D. 104.5 kPa

Answer :

To solve this problem, we need to determine the partial pressure of hydrogen gas in the collecting bottle. We're given the following information:

- The total pressure inside the collecting bottle is 97.1 kilopascals (kPa).
- The vapor pressure of the water is 3.2 kPa.

The total pressure in the collecting bottle is the sum of the partial pressures of the gases present, which in this case are hydrogen and water vapor. According to Dalton's Law of Partial Pressures, the total pressure of a gas mixture is the sum of the partial pressures of each individual gas.

To find the partial pressure of the hydrogen gas:
1. Subtract the vapor pressure of the water from the total pressure in the bottle.

[tex]\[
\text{Partial pressure of hydrogen} = \text{Total pressure} - \text{Vapor pressure of water}
\][/tex]

2. Substitute the given values:

[tex]\[
\text{Partial pressure of hydrogen} = 97.1 \, \text{kPa} - 3.2 \, \text{kPa}
\][/tex]

3. Calculate the result:

[tex]\[
\text{Partial pressure of hydrogen} = 93.9 \, \text{kPa}
\][/tex]

Therefore, the partial pressure of the hydrogen gas is 93.9 kPa.

The correct answer is A. [tex]\(93.9 \, \text{kPa}\)[/tex].

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