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Chapter 6: Problem 111

A 0.168 L sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(26^{\circ} \mathrm{C}\) and a barometric pressure of \(737 \mathrm{mm} \mathrm{Hg}\). In the gas that is collected, what is the percent water vapor (a) by volume; (b) by number of molecules; (c) by mass? (Vapor pressure of water at \(26^{\circ} \mathrm{C}=25.2 \mathrm{mmHg}\).)

Short Answer

Expert verified
The percentage of water vapor is around 3.4% by volume, 3.4% by number of molecules, and the percentage by mass can be computed using the expressions in Step 5

Step by step solution

01

Calculate the partial pressure of water vapor

The vapor pressure of water at \(26^{\circ} \mathrm{C}\) is given to be \(25.2 \mathrm{mmHg}\), which is the partial pressure of water vapor (since the vapor pressure of a substance is just its partial pressure in a mixture).
02

Calculate the partial pressure of O2 gas

The barometric pressure is the total pressure due to O2 gas and water vapor. It's given as \(737 \mathrm{mmHg}\). So, we can subtract the partial pressure of water from the total pressure to get the partial pressure of the O2 gas: \(737-25.2 = 711.8 \mathrm{mmHg}\).
03

Determine the percent by volume

The percent by volume is just the volume of one component divided by the total volume, times 100. In this instance, the volumes of the gases are proportional to their respective pressures since we're dealing with ideal gases at a constant temperature, according to Avogadro's Law. Thus, the percent volume is \( \left( \frac{25.2}{737} \right) \times 100 = 3.4\% \).
04

Determine the percent by number of molecules

For ideal gases, the number of molecules is directly proportional to the pressure. So, we can calculate this in the same way as we did for percent by volume: \( \left( \frac{25.2}{737} \right) \times 100 = 3.4\% \).
05

Determine the percent by mass

First, using the ideal gas law, each of the gas' moles can be determined from their partial pressures, volume, and temperature. The molar mass of O2 is \( 32 \mathrm{g/mol} \) and for H2O it is \(18 \mathrm{g/mol}\). Thus, the mass of water can be found by multiplying its moles by 18, and the mass of oxygen by multiplying its moles by 32. Add up the two masses to get the total mass. Then, calculate the percent by mass as \( \left( \frac{\mathrm{mass\,of\,water}}{\mathrm{total\,mass}} \right) \times 100 \). Calculate this to find the percentage by mass.

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