A 2.00 -L sample of \(\mathrm{O}_{2}(g)\) was collected over water at a total pressure of 785 torr and \(25^{\circ} \mathrm{C}\) . When the \(\mathrm{O}_{2}(g)\) was dried (water vapor removed), the gas had a volume of 1.94 \(\mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and 785 torr. Calculate the vapor pressure of water at \(25^{\circ} \mathrm{C}.\)

To calculate the partial pressure of the oxygen gas, we will use Dalton's Law, which states that the total pressure of a gas mixture is the sum of the partial pressures of its components: \[ P_{total} = P_{O_2} + P_{H_2O} \], where \(P_{total}\) is the total pressure, \(P_{O_2}\) is the partial pressure of the oxygen gas, and \(P_{H_2O}\) is the partial pressure of water vapor. We will first determine the partial pressure of the oxygen gas, which can be calculated using the following formula: \[P_{O_{2}} V_{O_{2}} = n_{O_{2}} R T\], where \(P_{O_{2}}\) is the partial pressure of the oxygen gas, \(V_{O_{2}}\) is the volume of the dried oxygen gas, \(n_{O_{2}}\) is the number of moles of oxygen gas, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. We are given: - The volume of the dried oxygen gas, \(V_{O_{2}}\) = 1.94 L - The temperature, \(T\) = 25°C = 298.15 K - The total pressure, \(P_{total}\) = 785 torr - The ideal gas constant, \(R\) = 62.36 L torr / K mol We can now solve for \(P_{O_{2}}\): \[P_{O_{2}} = \frac{n_{O_{2}} R T}{V_{O_{2}}}\]

To calculate the number of moles of oxygen gas, we will use the initial volume of the gas mixture (2.00 L) with the given temperature and total pressure. We will use the following formula: \[n_{O_{2}} = \frac{P_{total} V_{O_{2_{initial}}}}{R T}\], where \(V_{O_{2_{initial}}}\) is the initial volume of the gas mixture. Now we plug in the values: \[n_{O_{2}} = \frac{785 \,\text{torr} \times 2.00 \, \text{L}}{62.36 \, \frac{\text{L torr}}{\text{K mol}} \times 298.15 \, \text{K}}\]

Now we will compute the partial pressure of the oxygen gas using the number of moles of oxygen gas obtained from step 2: \[P_{O_{2}} = \frac{n_{O_{2}} R T}{V_{O_{2}}}\]

Once we have the partial pressure of the oxygen gas, we can use Dalton's Law to find the partial pressure of the water vapor: \[P_{H_2O} = P_{total} - P_{O_2}\]

The partial pressure of the water vapor calculated in step 4 is equal to the vapor pressure of water at 25°C. Therefore, the vapor pressure of water at 25°C is: \[P_{H_2O} = P_{total} - P_{O_2}\]